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Transcript 
1
(a)
The power P dissipated in a resistor of resistance R is measured for a range of values of
the potential difference V across it. The results are shown in the table below.
(i)
V/V
V2 / V2
P/W
1.00
1.0
0.21
1.71
2.9
0.58
2.25
1.01
2.67
1.43
3.00
9.0
1.80
3.27
10.7
2.18
3.50
12.3
2.43
Complete the table above.
(1)
(ii)
Complete the graph below by plotting the two remaining points and draw a best fit
straight line.
(2)
(iii)
Determine the gradient of the graph.
gradient = .......................
(3)
(iv)
Use the gradient of the graph to obtain a value for R.
R = .......................
Page 1 of 165
(1)
(b)
The following questions are based on the data in the table above.
(i)
Determine the value of R when V = 3.50 V.
R = ................... Ω
(1)
Page 2 of 165
(ii)
The uncertainty in V is ± 0.01 V. The uncertainty in P is ± 0.05 W.
Calculate the percentage uncertainty in the value of R calculated in part (1).
percentage uncertainty = ................... %
(3)
(iii)
Hence calculate the uncertainty in the value of R.
uncertainty = .......................
(1)
(iv)
State and explain whether the value of R you calculated in part (1) is consistent with
the value of R you determined from the gradient in part (a)(iv).
(2)
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(Total 14 marks)
2
(a)
(i)
Describe how you would make a direct measurement of the emf ɛ of a cell, stating
the type of meter you would use.
...............................................................................................................
...............................................................................................................
(1)
Page 3 of 165
(ii)
Explain why this meter must have a very high resistance.
...............................................................................................................
...............................................................................................................
(1)
(b)
A student is provided with the circuit shown in the diagram below.
The student wishes to determine the efficiency of this circuit.
In this circuit, useful power is dissipated in the external resistor. The total power input is the
power produced by the battery.
Efficiency =
The efficiency can be determined using two readings from a voltmeter.
(i)
Show that the efficiency =
where ɛ is the emf of the cell
and V is the potential difference across the external resistor.
(1)
Page 4 of 165
(ii)
Add a voltmeter to the diagram and explain how you would use this new circuit to
take readings of ɛ and V.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(c)
Describe how you would obtain a set of readings to investigate the relationship between
efficiency and the resistance of the external resistor. State any precautions you would take
to ensure your readings were reliable.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(d)
State and explain how you would expect the efficiency to vary as the value of R is
increased.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 9 marks)
Page 5 of 165
3
(a)
Define the electrical resistance of a component.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
Calculate the total resistance of the arrangement of resistors in the figure below.
total resistance ...............................
(3)
(c)
(i)
Calculate the current in the 3.0 Ω resistor in the figure above when the current in the
9.0 Ω resistor is 2.4 A.
current in the 3.0 Ω resistor ....................................
Page 6 of 165
(ii)
Calculate the total power dissipated by the arrangement of resistors in the figure
above when the current in the 9.0 Ω resistor is 2.4 A.
total power ....................................
(4)
(Total 9 marks)
4
(a)
Define the electromotive force (emf) of an electrical power supply.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
Explain why, when a battery is supplying a current to a circuit, the voltage measured
between its terminals is less than its emf.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 7 of 165
(c)
In the circuit shown in the figure below the voltmeter has a very high resistance and the
resistance of the ammeter is negligible. The motor M is being tested using a battery with an
emf of 9.00 V.
(i)
State the reading on the voltmeter when the switch S is open.
voltmeter reading .................................
(ii)
When S is closed and the motor is allowed to run freely the voltmeter reading is 8.41
V and the ammeter reads 0.82 A. Calculate the internal resistance of the battery.
internal resistance .............................
(iii)
Explain why the ammeter reading is greater than 0.82 A when the motor does work
by lifting a load.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(Total 9 marks)
Page 8 of 165
5
A heating element, as used on the rear window of a car, consists of three strips of a resistive
material, joined, as shown in the diagram, by strips of copper of negligible resistance. The
voltage applied to the unit is 12 V and heat is generated at a rate of 40 W.
(a)
(i)
Calculate the total resistance of the element.
...............................................................................................................
...............................................................................................................
(ii)
Hence show that the resistance of a single strip is about 11 Ω.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(b)
If each strip is 2.6 mm wide and 1.1 mm thick, determine the length of each strip.
resistivity of the resistive material = 4.0 × 10–5 Ω m
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 8 marks)
Page 9 of 165
6
The circuits in Figure 1 and Figure 2 both contain a 6.0 V supply of negligible internal
resistance. Each circuit is designed to operate a 2.5 V, 0.25 A filament lamp L.
The lamp works normally in both circuits.
Figure 1
(a)
Figure 2
Calculate the resistance of the filament lamp when working normally.
resistance .................................................
(2)
(b)
Calculate the resistance of the resistor that should be used for R in Figure 1.
resistance .................................................
(2)
(c)
Calculate the total resistance of the circuit in Figure 2.
total resistance ................................................
(3)
Page 10 of 165
(d)
Explain which circuit dissipates the lower total power.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 10 marks)
7
A battery of e.m.f. 12 V and negligible internal resistance is connected to a resistor network as
shown in the circuit diagram.
(a)
Calculate the total resistance of the circuit.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
Calculate the current through the 50 Ω resistor.
........................................................................................................................
........................................................................................................................
(1)
(Total 4 marks)
Page 11 of 165
8
(a)
Draw, on the axes below, the current/voltage characteristic for a filament lamp.
Do not insert any values for current or voltage.
(3)
(b)
Explain why the characteristic has the shape you have drawn.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The current/voltage characteristic of a filament lamp is to be determined using a datalogger,
the data then being fed into a computer to give a visual display of the characteristic. Draw
the circuit diagram required for such an experiment and state what is varied so as to
produce a range of values.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(5)
(Total 11 marks)
Page 12 of 165
9
A battery of e.m.f. 12 V and internal resistance r is connected in a circuit with three resistors
each having a resistance of 10 Ω as shown. A current of 0.50 A flows through the battery.
Calculate
(a)
the potential difference between the points A and B in the circuit,
........................................................................................................................
........................................................................................................................
........................................................................................................................
(b)
the internal resistance of the battery,
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(c)
the total energy supplied by the battery in 2.0 s,
........................................................................................................................
........................................................................................................................
(d)
the fraction of the energy supplied by the battery that is dissipated within the battery.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(Total 7 marks)
Page 13 of 165
10
In each of the following circuits the battery has negligible internal resistance and the bulbs are
identical.
Figure 1
(a)
Figure 2
For the circuit shown in Figure 1 calculate
(i)
the current flowing through each bulb,
...............................................................................................................
...............................................................................................................
(ii)
the power dissipated in each bulb.
...............................................................................................................
...............................................................................................................
(2)
(b)
In the circuit shown in Figure 2 calculate the current flowing through each bulb.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
Explain how the brightness of the bulbs in Figure 1 compares with the brightness of the
bulbs in Figure 2.
........................................................................................................................
........................................................................................................................
(2)
(Total 7 marks)
Page 14 of 165
11
(a)
For a conductor in the form of a wire of uniform cross-sectional area, give an equation
which relates its resistance to the resistivity of the material of the conductor. Define the
symbols used in the equation.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
(i)
An electrical heating element, made from uniform nichrome wire, is required
to dissipate 500 W when connected to the 230 V mains supply.
The cross-sectional area of the wire is 8.0 × 10–8 m2. Calculate the length of
nichrome wire required.
resistivity of nichrome = 1.1 × 10–6 Ω m
.............................................................................................................
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.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Two heating elements, each rated at 230 V, 500 W are connected to the 230 mains
supply
(A) in series,
(B) in parallel.
Explain why only one of the circuits will provide an output of 1 kW.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
(Total 8 marks)
Page 15 of 165
12
In the circuit shown, the battery has negligible internal resistance.
(a)
(i)
If the emf of the battery = 9.0 V, R1 = 120 Ω and R2 = 60 Ω, calculate the current I
flowing in the circuit.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the voltage reading on the voltmeter.
...............................................................................................................
...............................................................................................................
(4)
(b)
The circuit shown in the diagram acts as a potential divider. The circuit is now modified by
replacing R1 with a temperature sensor, whose resistance decreases as the temperature
increases.
Explain whether the reading on the voltmeter increases or decreases as the temperature
increases from a low value.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 7 marks)
Page 16 of 165
13
In the circuit shown, the battery has an emf of 12 V and an internal resistance of 2.0 Ω. The
resistors A and B each have resistance of 30 Ω.
Calculate
(i)
the total current in the circuit,
.............................................................................................................
.............................................................................................................
(ii)
the voltage between the points P and Q,
......................................................................................................................
......................................................................................................................
(iii)
the power dissipated in resistor A,
......................................................................................................................
......................................................................................................................
......................................................................................................................
(iv)
the energy dissipated by resistor A in 20 s.
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(Total 8 marks)
Page 17 of 165
14
In the circuit shown, the battery has negligible internal resistance.
Calculate the current in the ammeter when
(a)
the terminals X and Y are short-circuited i.e. connected together,
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
the terminals X and Y are connected to a 30 Ω resistor.
........................................................................................................................
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........................................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 6 marks)
15
(a)
A cell of emf ϵ and internal resistance r is connected in series to a resistor of resistance R
as shown. A current I flows in the circuit.
(i)
State an expression which gives ϵ in terms of I, r and R.
...............................................................................................................
Page 18 of 165
(ii)
Hence show how VR, the potential difference across the resistor, is related to ϵ, I and
r.
...............................................................................................................
(2)
(b)
A lamp, rated at 30 W, is connected to a 120 V supply.
(i)
Calculate the current in the lamp.
...............................................................................................................
...............................................................................................................
(ii)
If the resistor in part (a) is replaced by the lamp described in (b), determine how many
cells, each of emf 1.5 V and internal resistance 1.2 Ω, would have to be connected in
series so that the lamp would operate at its proper power.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(Total 7 marks)
16
The circuit diagram shows a light-emitting diode connected in series with a resistor R and a 3.0 V
battery of negligible internal resistance. The potential difference across the terminals of the diode
is 2.0 V and the current through it is 10 mA. The diode emits photons of wavelength 635 nm.
Page 19 of 165
(a)
Calculate the resistance of R.
........................................................................................................................
........................................................................................................................
(b)
Calculate the electrical power supplied to the diode.
........................................................................................................................
........................................................................................................................
(c)
Calculate the energy of a photon of wavelength 635 nm.
........................................................................................................................
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........................................................................................................................
........................................................................................................................
(d)
Estimate the number of photons emitted per second by the diode.
........................................................................................................................
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........................................................................................................................
........................................................................................................................
(e)
State an assumption you made in your estimation in part (d).
........................................................................................................................
........................................................................................................................
(Total 8 marks)
Page 20 of 165
17
A student investigates the variation of electric potential with distance along a strip of conducting
paper of length l and of uniform thickness. The strip tapers uniformly from a width 4w at the broad
end to 2w at the narrow end, as shown in Figure 1. A constant pd is applied across the two ends
of the strip, with the narrow end at positive potential, Vl, and the broad end at zero potential. The
student aims to produce a graph of pd against distance x, measured from the broad end of the
strip.
Figure 1
(a)
Draw a labelled circuit diagram which would be suitable for the investigation.
(2)
Page 21 of 165
(b)
The student obtained some preliminary measurements which are shown below.
pd, V/V
0
2.1
4.5
7.2
Distance, x/m
0
0.100
0.200
0.300
By reference to the physical principles involved, explain why the increase of V with x
is greater than a linear increase.
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(4)
(c)
The potential, V, at a distance x from the broad end is given by
V = k – 1.44Vl ln (2l – x),
where Vl is the potential at the narrow end, and k is a constant.
(i)
The student’s results are given below. Complete the table.
l = 0.400 m
distance x/m
potential V/V
(2l – x)/m
ln (2l – x)
0.100
2.1
0.700
– 0.357
0.200
4.5
0.270
6.4
0.330
8.3
0.360
9.3
0.380
10.1
Page 22 of 165
(ii)
Plot a graph of V against ln (2l – x) and explain whether or not it confirms the
equation.
.............................................................................................................
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.............................................................................................................
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.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Use the graph to calculate Vl.
.............................................................................................................
.............................................................................................................
(10)
(Total 16 marks)
18
(a)
X and Y are two lamps. X is rated at 12 V, 24 W and Y at 6.0 V, 18 W. Calculate the current
through each lamp when it operates at its rated voltage.
X: .................................................................................................................
Y: ..................................................................................................................
(2)
Page 23 of 165
(b)
The two lamps are connected in the circuit shown. The battery has an emf of 27 V and
negligible internal resistance. The resistors R1 and R2 are chosen so that the lamps are
operating at their rated voltage.
(i)
What is the reading on the voltmeter?
.............................................................................................................
(ii)
Calculate the resistance of R2.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Calculate the current through R1.
.............................................................................................................
(iv)
Calculate the voltage across R1.
.............................................................................................................
(v)
Calculate the resistance of R1.
.............................................................................................................
(7)
(Total 9 marks)
Page 24 of 165
19
A battery of emf 24 V and negligible intemal resistance is connected to a resistor network as
shown in the circuit diagram in the diagram below.
(a)
Show that the resistance of the single equivalent resistor that could replace the four
resistors between the points A and B is 50 Ω.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)
(b)
If R1 is 50 Ω, calculate
(i)
the current in R1,
.............................................................................................................
.............................................................................................................
(ii)
the current in the 60 Ω resistor.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 8 marks)
Page 25 of 165
20
The circuit diagram shows a light emitting diode (LED) connected in series with a resistor, R, and
a 3.0 V battery of negligible internal resistance.
(a)
The LED lights normally when the forward voltage across it is 2.2 V and the current in it is
35 mA.
Calculate
(i)
the resistance of R,
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
the number of electrons that pass through the LED each second.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(b)
The LED emits light at a peak wavelength of 635 nm.
(i)
Calculate the energy of a photon of light of this wavelength.
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 26 of 165
(ii)
Estimate the number of photons emitted by the LED each second when the current
through it is 35 mA. Assume all the photons emitted by the LED are of wavelength
635 nm and that all the electrical energy produces light.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(Total 8 marks)
21
(a)
In the circuit in Figure 1, the battery, of emf 15 V and the negligible internal resistance, is
connected in series with two lamps and a resistor. The three components each have a
resistance of 12 Ω.
Figure 1
(i)
What is the voltage across each lamp?
.............................................................................................................
(ii)
Calculate the current through the lamps.
.............................................................................................................
.............................................................................................................
(3)
Page 27 of 165
(b)
The two lamps are now disconnected and reconnected in parallel as shown in Figure 2.
Figure 2
(i)
Show that the current supplied by the battery is 0.83 A.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Hence show that the current in each lamp is the same as the current in the lamps in
the circuit in Figure 1.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(c)
How does the brightness of the lamps in the circuit in Figure 1 compare with the brightness
of the lamps in the circuit in Figure 2?
Explain your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 8 marks)
Page 28 of 165
22
In the circuit shown the battery has emf
(a)
(i)
and internal resistance r.
State what is meant by the emf of a battery.
.............................................................................................................
.............................................................................................................
(ii)
When the switch S is open, the voltmeter, which has infinite resistance, reads 8.0 V.
When the switch is closed, the voltmeter reads 6.0 V.
Determine the current in the circuit when the switch is closed.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Show that r = 0.80 Ω.
.............................................................................................................
.............................................................................................................
(4)
(b)
The switch S remains closed. Calculate
(i)
the power dissipated in the 2.4 Ω resistor,
.............................................................................................................
(ii)
the total power dissipated in the circuit,
.............................................................................................................
.............................................................................................................
Page 29 of 165
(iii)
the energy wasted in the battery in 2 minutes.
.............................................................................................................
(4)
(Total 8 marks)
23
Four resistors, each having resistance of 50 Ω, are connected to form a square. A resistance
meter measured the resistance between different corners of the square. Determine the
resistance the meter records when connected between the following corners.
(a)
Between A and C, as in Figure 1.
Figure 1
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 30 of 165
(b)
Between A and B, as in Figure 2.
Figure 2
......................................................................................................................
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......................................................................................................................
......................................................................................................................
(3)
(Total 5 marks)
Page 31 of 165
24
(a)
Figure 1 shows two possible arrangements of connecting three resistors, each resistor
having a resistance of 40 Ω.
Figure 1
Calculate the equivalent resistance in each case.
(i)
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 32 of 165
(b)
The designer of a heating element for the rear window of a car decides to connect six
separate heating elements together as shown in Figure 2. Each element has a resistance
of 6.0 Ω and the unit is connected to a 12 V dc supply having zero internal resistance.
Figure 2
(i)
Calculate the current in each single element.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
With the aid of a similar calculation give a reason why the heater would not be as
effective if all six were connected in series.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 8 marks)
Page 33 of 165
25
In the circuit shown, a battery of emf and internal resistance r is connected to a variable
resistor R. The current I and the voltage V are read by the ammeter and voltmeter respectively.
(a)
The emf is related to V, I and r by the equation
= V + Ir
Rearrange the equation to give V in terms of
, I and r.
......................................................................................................................
(1)
(b)
In an experiment, the value of R is altered so that a series of values of V and the
corresponding values of I are obtained. Using the axes, sketch the graph you would expect
to obtain as R is changed.
(2)
(c)
State how the values of
and r may be obtained from the graph.
..................................................................................................................
r ....................................................................................................................
(2)
(Total 5 marks)
Page 34 of 165
26
The circuit in Figure 1 has a thermistor connected in series to a 200 Ω resistor and a 12 V
battery of negligible internal resistance. Figure 2 shows how the resistance, Rth, of the thermistor
varies with temperature.
(a)
(i)
Calculate the current in the circuit when the temperature is 25°C.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the potential difference across the thermistor at 25°C.
...............................................................................................................
...............................................................................................................
(3)
Page 35 of 165
(b)
Without further calculation, explain how you would expect the potential difference across
the thermistor to change as the temperature increases from 25°C.
You may be awarded marks for the quality of written communication in your answer.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The circuit in Figure 1 is modified by removing the 200 Ω resistance to give the circuit in
Figure 3.
The temperature of the thermistor is increased at a steady rate from 25°C to 45°C in 10
minutes.
(i)
Calculate the power dissipated in the thermistor at
25°C ......................................................................................................
...............................................................................................................
45°C ......................................................................................................
...............................................................................................................
(ii)
Use the mean value of the powers determined in part (c)(i) to calculate the energy
supplied by the battery during the period in which the temperature of the thermistor
increases.
...............................................................................................................
...............................................................................................................
Page 36 of 165
(iii)
State why the energy value, determined in part (c)(ii) is not an accurate value.
...............................................................................................................
...............................................................................................................
(6)
(Total 12 marks)
27
The graph shows the current−voltage characteristic of the output from a solar cell when light of
intensity 450 W m−2 is incident on it.
(a)
(i)
Using data from the graph above estimate the maximum power output from the solar
cell.
maximum power ................................................. W
(2)
Page 37 of 165
(ii)
Sketch, on the axes below, a graph to show how the power output varies with voltage
for this solar cell for the same incident light intensity.
(2)
(iii)
When the light intensity is 450 W m−2 the cell has an efficiency of 0.15 at the
maximum power.
Calculate the area of the solar cell.
area ................................................ m2
(3)
(b)
A manufacturer has a supply of solar cells that each have an electromotive force (emf) of
0.70 V and an internal resistance of 0.78 Ω when delivering maximum power.
(i)
Explain what is meant by an emf of 0.70 V.
...............................................................................................................
...............................................................................................................
(1)
Page 38 of 165
(ii)
The manufacturer uses a number of these solar cells in an array to make a power
supply that has an emf of 14 V and an internal resistance of 3.9 Ω when delivering
maximum power.
Describe and explain the arrangement of cells the manufacturer has to use in this
array. Go on to calculate the number of cells the manufacturer needs to make the
power supply.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
number of cells .....................................................
(4)
(c)
Communications satellites use solar cells to generate electrical power.
Discuss why solar cells are appropriate for this task.
Your answer should refer to:
•
any additional features that would be needed to ensure that the satellite’s electrical
systems operate continuously
•
whether solar cell arrays are appropriate for space probes that travel to the edge of
the solar system.
The quality of your written communication will be assessed in your answer.
(6)
(Total 18 marks)
Page 39 of 165
28
The circuit in the diagram below contains four identical new cells, A, B, C and D, each of emf
1.5V and negligible internal resistance.
(a)
The resistance of each resistor is 4.0 Ω.
(i)
Calculate the total resistance of the circuit.
answer = ...................................... Ω
(1)
(ii)
Calculate the total emf of the combination of cells.
answer = ....................................... V
(1)
(iii)
Calculate the current passing through cell A.
answer = ....................................... A
(2)
Page 40 of 165
(iv)
Calculate the charge passing through cell A in five minutes, stating an appropriate
unit.
answer = ......................................
(2)
(b)
Each of the cells can provide the same amount of electrical energy before going flat.
State and explain which two cells in this circuit you would expect to go flat first.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 9 marks)
29
(a)
A set of decorative lights consists of a string of lamps. Each lamp is rated at 5.0 V, 0.40 W
and is connected in series to a 230 V supply.
Calculate
(i)
the number of lamps in the set, so that each lamp operates at the correct rating,
.............................................................................................................
.............................................................................................................
(ii)
the current in the circuit,
.............................................................................................................
.............................................................................................................
(iii)
the resistance of each lamp,
.............................................................................................................
.............................................................................................................
Page 41 of 165
(iv)
the total electrical energy transferred by the set of lights in 2 hours.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(b)
When assembled at the factory, one set of lights inadvertently contains 10 lamps too many.
All are connected in series. Assume that the resistance of each lamp is the same as that
calculated in part (a) (iii).
(i)
Calculate the current in this set of lights when connected to a 230 V supply.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
How would the brightness of each lamp in this set compare with the brightness of
each lamp in the correct set?
.............................................................................................................
(3)
(Total 8 marks)
30
(a)
A student is given three resistors of resistance 3.0 Ω, 4.0 Ω and 6.0 Ω respectively.
(i)
Draw the arrangement, using all three resistors, which will give the largest resistance.
(ii)
Calculate the resistance of the arrangement you have drawn.
.............................................................................................................
.............................................................................................................
Page 42 of 165
(iii)
Draw the arrangement, using all three resistors, which will give the smallest
resistance.
(iv)
Calculate the resistance of the arrangement you have drawn.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(b)
The three resistors are now connected to a battery of emf 12 V and negligible internal
resistance, as shown in Figure 1.
Figure 1
(i)
Calculate the total resistance in the circuit.
.............................................................................................................
.............................................................................................................
(ii)
Calculate the voltage across the 6.0 Ω resistor.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 9 marks)
Page 43 of 165
31
The graph shows how the resistance, RR, of a metal resistor and the resistance, RTh, of a
thermistor change with temperature.
(a)
Give the values of the resistance RR and RTh at a temperature of 200 °C.
RR ..............................................................RTh
....................................................................
(1)
(b)
The resistor and thermistor are connected in series to a 12V battery of negligible internal
resistance, as shown in Figure 1.
Figure 1
(i)
Calculate the voltage across the terminals AB when both the resistor and thermistor
are at 200 °C.
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 44 of 165
(ii)
Assuming that the temperature of the resistor always equals the temperature of
the thermistor, deduce the temperature when the voltage across the resistor equals
the voltage across the thermistor.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(c)
A lamp and a switch are now connected across the terminals AB, as shown in Figure 2.
The temperature of the thermistor does not change from that obtained in part (b)(ii).
Figure 2
(i)
The lamp is rated at 2.0 W at a voltage of 6.0 V. Calculate the resistance of the lamp
at this rating.
.............................................................................................................
.............................................................................................................
(ii)
The switch S is now closed. Explain, without calculation, why the voltage across the
thermistor will fall from the value in part (b)(ii).
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 9 marks)
Page 45 of 165
32
In the circuit shown in Figure 1, the battery, of emf 6.0V, has negligible internal resistance.
Figure 1
(a)
Calculate the current through the ammeter when the switch S is
(i)
open,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
closed.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
The switch S is now replaced with a voltmeter of infinite resistance.
Determine the reading on the voltmeter.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 5 marks)
Page 46 of 165
33
The following figure shows part of the circuit diagram for a car lighting circuit.
The table shows the power rating of the various lamps used in the circuit.
(a)
Lamp
Power/W
Tail light
8.0
Sidelight
5.0
Headlight
60
Explain why all the lamps are connected in parallel.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
The emf of the battery used in the circuit is 12 V and it has negligible internal resistance.
Calculate the current through the battery when the headlights and tail lights are both on.
current .................... A
(3)
Page 47 of 165
(c)
(i)
State which lamp filament has the least resistance.
.............................................................................................................
.............................................................................................................
(ii)
Explain why this resistance is smaller when the lamp is first switched on.
.............................................................................................................
.............................................................................................................
(3)
(d)
The side and tail lamps are accidentally left on for 12 hours when the car is parked.
(i)
Calculate the energy dissipated in the lamps during this time.
energy ................... J
(ii)
The battery used by the car is capable of delivering a current of 1.5 A for 24 hours.
The car’s starter motor needs a current of 100 A which lasts for at least one second
in order to start the engine. State and explain whether the car is likely to start after
the 12 hours.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 13 marks)
Page 48 of 165
34
In the circuit shown in the figure below, the battery, of negligible internal resistance, has an emf
of 30 V. The pd across the lamp is 6.0 V and its resistance is 12 Ω.
(a)
Show that the total resistance of the circuit is 20 Ω.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(b)
Calculate
(i)
the current supplied by the battery,
.............................................................................................................
.............................................................................................................
(ii)
the pd between the points A and B,
.............................................................................................................
.............................................................................................................
(iii)
the current in the lamp.
.............................................................................................................
.............................................................................................................
(4)
Page 49 of 165
(c)
(i)
What is the power of the lamp, in W?
.............................................................................................................
.............................................................................................................
(ii)
What percentage of the power supplied by the battery is dissipated in the lamp?
.............................................................................................................
.............................................................................................................
(3)
(Total 10 marks)
35
(a)
A student wishes to measure the resistivity of the material of a uniform resistance wire.
The available apparatus includes a battery, a switch, a variable resistor, an ammeter and a
voltmeter.
(i)
Draw a circuit diagram which incorporates some or all of this apparatus and which
enables the student to determine the resistivity of the material.
(ii)
State the measurements which must be made to ensure that a reliable value of the
resistivity is obtained.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 50 of 165
(iii)
Explain how a value of the resistivity would be obtained from the measurements.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(10)
(b)
A wire made from tin with cross-sectional area 7.8 × 10–9 m2, has a pd of 2.0 V across it.
Calculate the minimum length of wire needed so that the current through it does not exceed
4.0 A.
resistivity of tin = 1.1 × 10–7 Ω m
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
(Total 12 marks)
36
A student wishes to collect data so he can plot the I-V curve for a semiconductor diode.
(a)
(i)
Draw a suitable diagram of the circuit that would enable the student to collect this
data.
(3)
Page 51 of 165
(ii)
Describe the procedure the student would follow in order to obtain an I-V curve for the
semiconductor diode.
The quality of your written communication will be assessed in this question.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
(b)
The diagram below shows an arrangement of a semiconducting diode and two resistors.
A 12.0 V battery is connected with its positive terminal to A and negative terminal to B.
(i)
Calculate the current in the 8.0 Ω resistor
.............................................................................................................
.............................................................................................................
.............................................................................................................
answer .................................. A
(2)
Page 52 of 165
(ii)
Calculate the current in the 4.0 Ω resistor if the p.d. across the diode, when in forward
bias, is 0.65 V expressing your answer to an appropriate number of significant
figures.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
answer ................................... A
(3)
(Total 14 marks)
Page 53 of 165
37
Figure 1 shows a circuit including a thermistor T in series with a variable resistor R. The battery
has negligible internal resistance.
Figure 1
The resistance–temperature (R−θ) characteristic for T is shown in Figure 2.
Figure 2
(a)
The resistor and thermistor in Figure 1 make up a potential divider.
Explain what is meant by a potential divider.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
Page 54 of 165
(b)
State and explain what happens to the voltmeter reading when the resistance of R is
increased while the temperature is kept constant.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
State and explain what happens to the ammeter reading when the temperature of the
thermistor increases.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(d)
The battery has an emf of 12.0 V. At a temperature of 0 °C the resistance of the thermistor
is 2.5 × 103 Ω.
The voltmeter is replaced by an alarm that sounds when the voltage across it exceeds
3.0 V.
Calculate the resistance of R that would cause the alarm to sound when the temperature of
the thermistor is lowered to 0 °C.
resistance = ............................... Ω
(2)
Page 55 of 165
(e)
State one change that you would make to the circuit so that instead of the alarm coming on
when the temperature falls, it comes on when the temperature rises above a certain value.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(Total 9 marks)
38
The diagram below shows an arrangement of resistors.
(a)
Calculate the total resistance between terminals A and B.
answer = ................................................... Ω
(2)
Page 56 of 165
(b)
A potential difference is applied between the two terminals, A and B, and the power
dissipated in each of the 400 Ω resistors is 1.0 W.
(i)
Calculate the potential difference across the 400 Ω resistors.
answer = ................................................... V
(ii)
Calculate the current through the 25 Ω resistor.
answer = .................................................... A
(iii)
Calculate the potential difference applied to terminals A and B.
answer = ................................................... V
(6)
(Total 8 marks)
Page 57 of 165
39
The diagram below shows the circuit for a small convector heater. Heater elements can be
switched in and out of the circuit using switches X and Y. Each element has a resistance R and
the power supply has an emf V.
(a)
The table shows the possible combinations of open and closed switches.
When a switch is closed, charge can flow through it.
Complete the table. Assume that the internal resistance of the power supply is negligible.
The first row of the table has been done for you.
switch combination
total resistance in circuit
X open, Y closed
R
X closed, Y open
X open, Y open
X closed, Y closed
(3)
(b)
State and explain which switch combination will dissipate least energy.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 58 of 165
(c)
Explain in terms of electron flow how thermal energy is produced in one of the heater
elements when charge flows through it.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(d)
The power supply for the heater is replaced with one that has a higher internal resistance.
Explain how this change will affect the thermal energy output of the heater for a given
switch combination. State which switch combination will be affected most by the change.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(e)
The resistance of each heater element is 68 Ω. Each one is made from 7.2 m of nichrome
wire wound on a rod.
(i)
Calculate the radius of the nichrome wire.
resistivity of nichrome = 1.1 × 10–6 Ω m
radius ................................ m
(2)
(ii)
Suggest two properties that the rod must have to make it suitable in this application.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 14 marks)
Page 59 of 165
40
The Figure below shows a simplified circuit for the main lights on a car.
The battery has an emf of 12 V and no internal resistance.
The table below gives data about the lamps being used in the circuit. The resistances given are
correct when the lamp is operating at its normal operating voltage.
OPERATING VOLTAGE V
RESISTANCE Ω
H, headlight lamp
12
3.8
R, rear lamp
12
5.6
D, dashboard lamp
12
72
LAMP
(a)
(i)
Calculate the power of a single headlight lamp when operating at 12 V.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
power ............................................. W
(2)
Page 60 of 165
(ii)
Calculate the resistance of the combination of lamps when operating at 12 V.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
resistance .............................................. Ω
(3)
(iii)
Calculate the total power of the combination of lamps when operating at 12 V.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
power ............................................ W
(2)
(b)
The battery is replaced with one of a lower emf. State and explain how the resistance of the
lamps would have to change in order to achieve the same brightness.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 9 marks)
Page 61 of 165
41
A battery is connected to a 10 Ω resistor as shown in the diagram below. The emf (electromotive
force) of the battery is 6.0 V.
(a)
(i)
Define the emf of a battery.
.............................................................................................................
.............................................................................................................
(1)
(ii)
When the switch is open the voltmeter reads 6.0 V and when it is closed it reads
5.8 V.
Explain why the readings are different.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
(b)
Calculate the internal resistance of the battery.
answer = ..................................... Ω
(3)
Page 62 of 165
(c)
State and explain why it is important for car batteries to have a very low internal resistance.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 8 marks)
42
A very high resistance voltmeter reads 15.0 V when it is connected across the terminals of a
power supply.
(a)
Explain why the reading on the voltmeter is equal to the emf of the power supply.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
Page 63 of 165
(b)
A resistor of value 470 Ω is connected across the terminals of the power supply in parallel
with the voltmeter, as the figure below shows. The voltmeter reads 14.5 V.
(i)
Calculate the current in the 470 Ω resistor.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
current ...................................... A
(2)
(ii)
Calculate the internal resistance of the power supply.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
internal resistance ....................................... Ω
(3)
(Total 8 marks)
Page 64 of 165
43
The circuit shown below shows a thermistor connected in a circuit with two resistors, an ammeter
and a battery of emf 15V and negligible internal resistance.
(a)
When the thermistor is at a certain temperature the current through the ammeter is
10.0 mA.
(i)
Calculate the pd across the 540 Ω resistor.
answer = ..................................... V
(1)
(ii)
Calculate the pd across the 1200 Ω resistor.
answer = ..................................... V
(1)
Page 65 of 165
(iii)
Calculate the resistance of the parallel combination of the resistor and the thermistor.
answer = ..................................... Ω
(2)
(iv)
Calculate the resistance of the thermistor.
answer = ..................................... Ω
(2)
(b)
The temperature of the thermistor is increased so that its resistance decreases.
State and explain what happens to the pd across the 1200 Ω resistor.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 9 marks)
Page 66 of 165
44
A battery of emf 9.0 V and internal resistance, r, is connected in the circuit shown in the figure
below.
(a)
The current in the battery is 1.0 A.
(i)
Calculate the pd between points A and B in the circuit.
answer = ..................................... V
(2)
(ii)
Calculate the internal resistance, r.
answer = ..................................... Ω
(2)
(iii)
Calculate the total energy transformed by the battery in 5.0 minutes.
answer = ..................................... J
(2)
Page 67 of 165
(iv)
Calculate the percentage of the energy calculated in part (iii) that is dissipated in the
battery in 5.0 minutes.
answer = ..................................... %
(2)
(b)
State and explain one reason why it is an advantage for a rechargeable battery to have a
low internal resistance.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 10 marks)
Page 68 of 165
45
The circuit shown in the figure below shows an arrangement of resistors, W, X, Y, Z, connected
to a battery of negligible internal resistance.
The emf of the battery is 10V and the reading on the ammeter is 2.0 A.
(a)
(i)
Calculate the total resistance of the circuit.
answer = ..................................... Ω
(1)
(ii)
The resistors W, X, Y, and Z all have the same resistance. Show that your answer to
part (a) (i) is consistent with the resistance of each resistor being 3.0 Ω.
answer = ..................................... Ω
(3)
Page 69 of 165
(b)
(i)
Calculate the current through resistor Y.
answer = ...................................... A
(2)
(ii)
Calculate the pd across resistor W.
answer = ...................................... V
(2)
(Total 8 marks)
46
A cell of emf, ε, and internal resistance, r, is connected to a variable resistor R. The current
through the cell and the terminal pd of the cell are measured as R is decreased. The circuit is
shown in the figure below.
Page 70 of 165
The graph below shows the results from the experiment.
(a)
Explain why the terminal pd decreases as the current increases.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
(i)
Use the graph to find the emf, ε, of the cell.
answer = ..................................... V
(1)
Page 71 of 165
(ii)
Use the graph above to find the internal resistance, r, of the cell.
answer = ..................................... Ω
(3)
(c)
Draw a line on the graph above that shows the results obtained from a cell with
(i)
the same emf but double the internal resistance of the first cell labelling your graph A.
(2)
(ii)
the same emf but negligible internal resistance labelling your graph B.
(1)
(d)
In the original circuit shown in part (a), the variable resistor is set at a value such that the
current through the cell is 0.89 A.
(i)
Calculate the charge flowing through the cell in 15 s, stating an appropriate unit.
answer = ......................................
(2)
Page 72 of 165
(ii)
Calculate the energy dissipated in the internal resistance of the cell per second.
answer = ..................................... W
(2)
(Total 13 marks)
47
(a)
The rating of a car headlamp is 12 V, 55 W.
The resistance in this headlamp is due to a thin piece of wire. At its working temperature,
the wire has a length of 5.0 × 10–2 m and a cross-sectional area of 1.9 × 10–8 m2.
Calculate, at the working temperature, the resistivity of the metal used to make the wire.
State an appropriate unit for your answer.
resistivity ............................................unit ............................................
(5)
(b)
(i)
Define the term electromotive force (emf).
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 73 of 165
(ii)
The figure below is a circuit diagram illustrating how two of these headlamps are
connected to a car battery.
The car battery has an emf of 12 V.
When the switch S is closed there is a current of 9.1 A through the battery and a
potential difference of 11.9 V across the headlamps.
Calculate the internal resistance of the car battery.
internal resistance ........................... Ω
(2)
(c)
A fault develops in one of the headlamps in the figure above causing its resistance to
decrease.
State and explain how this fault affects the brightness of the other headlamp.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 12 marks)
Page 74 of 165
48
A battery of negligible internal resistance is connected to lamp P in parallel with lamp Q as
shown in Figure 1. The emf of the battery is 12 V.
Figure 1
(a)
Lamp P is rated at 12 V 36 W and lamp Q is rated at 12 V 6 W.
(i)
Calculate the current in the battery.
answer = ...................................... A
(2)
(ii)
Calculate the resistance of P.
answer = ...................................... Ω
(1)
(iii)
Calculate the resistance of Q.
answer = ...................................... Ω
(1)
Page 75 of 165
(b)
State and explain the effect on the brightness of the lamps in the circuit shown in Figure 1
if the battery has a significant internal resistance.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The lamps are now reconnected to the 12 V battery in series as shown in Figure 2.
Figure 2
(i)
Explain why the lamps will not be at their normal brightness in this circuit.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 76 of 165
(ii)
State and explain which of the lamps will be brighter assuming that the resistance of
the lamps does not change significantly with temperature.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 12 marks)
49
The figure below shows two resistors, R1 and R2, connected in series with a battery of emf 12 V
and negligible internal resistance.
(a)
The reading on the voltmeter is 8.0 V and the resistance of R2 is 60 Ω.
(i)
Calculate the current in the circuit.
answer = ...................................... A
(2)
(ii)
Calculate the resistance of R1.
answer = ..................................... Ω
(1)
Page 77 of 165
(iii)
Calculate the charge passing through the battery in 2.0 minutes. Give an appropriate
unit for your answer.
answer = .............................................. unit = ...............................
(2)
(b)
In the circuit shown in the figure above R2 is replaced with a thermistor. State and explain
what will happen to the reading on the voltmeter as the temperature of the thermistor
increases.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 8 marks)
50
X and Y are two lamps. X is rated at 12 V 36 W and Y at 4.5 V 2.0 W.
(a)
Calculate the current in each lamp when it is operated at its correct working voltage.
X .................................................. A
Y .................................................. A
(2)
Page 78 of 165
(b)
The two lamps are connected in the circuit shown in the figure below. The battery has an
emf of 24 V and negligible internal resistance. The resistors, R1 and R2 are chosen so that
the lamps are operating at their correct working voltage.
(i)
Calculate the pd across R1.
answer ......................................... V
(1)
(ii)
Calculate the current in R1.
answer ......................................... A
(1)
(iii)
Calculate the resistance of R1.
answer ......................................... Ω
(1)
(iv)
Calculate the pd across R2.
answer ......................................... V
(1)
(v)
Calculate the resistance of R2.
answer ......................................... Ω
(1)
Page 79 of 165
(c)
The filament of the lamp in X breaks and the lamp no longer conducts. It is observed that
the voltmeter reading decreases and lamp Y glows more brightly.
(i)
Explain without calculation why the voltmeter reading decreases.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Explain without calculation why the lamp Y glows more brightly.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 11 marks)
51
A copper connecting wire is 0.75 m long and has a cross-sectional area of 1.3 × 10–7 m2.
(a)
Calculate the resistance of the wire.
resistivity of copper = 1.7 × 10–7 Ωm
resistance = ........................................... Ω
(2)
(b)
A 12 V 25 W lamp is connected to a power supply of negligible internal resistance using
two of the connecting wires. The lamp is operating at its rated power.
(i)
Calculate the current flowing in the lamp.
current = ........................................... A
(1)
Page 80 of 165
(ii)
Calculate the pd across each of the wires.
pd = ........................................... V
(1)
(iii)
Calculate the emf (electromotive force) of the power supply.
emf = ........................................... V
(2)
(c)
The lamp used in part (b) is connected by the same two wires to a power supply of the
same emf but whose internal resistance is not negligible.
State and explain what happens to the brightness of the lamp when compared to its
brightness in part (b).
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 8 marks)
Page 81 of 165
52
The circuit diagram below shows a 6.0 V battery of negligible internal resistance connected in
series to a light dependent resistor (LDR), a variable resistor and a fixed resistor, R.
(a)
For a particular light intensity the resistance of the LDR is 50 kΩ. The resistance of
R is 5.0 kΩ and the variable resistor is set to a value of 35 kΩ.
(i)
Calculate the current in the circuit.
current...........................................A
(2)
(ii)
Calculate the reading on the voltmeter.
voltmeter reading ...........................................V
(2)
(b)
State and explain what happens to the reading on the voltmeter if the intensity of the light
incident on the LDR increases.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 82 of 165
(c)
For a certain application at a particular light intensity the pd across R needs to be 0.75 V.
The resistance of the LDR at this intensity is 5.0 kΩ.
Calculate the required resistance of the variable resistor in this situation.
resistance ........................................... Ω
(3)
(Total 9 marks)
53
Three identical cells, each of internal resistance R, are connected in series with an external
resistor of resistance R. The current in the external resistor is I. If one of the cells is reversed in
the circuit, what is the new current in the external resistor?
A
B
C
D
(Total 1 mark)
Page 83 of 165
54
The circuit diagram below shows a 12 V battery of negligible internal resistance connected to a
combination of three resistors and a thermistor.
(a)
When the resistance of the thermistor is 5.0 kΩ
(i)
calculate the total resistance of the circuit,
total resistance = ......................................... kΩ
(3)
(ii)
calculate the current in the battery.
current = ........................................ mA
(1)
Page 84 of 165
(b)
A high-resistance voltmeter is used to measure the potential difference (pd) between points
A-C, D-F and C-D in turn.
Complete the following table indicating the reading of the voltmeter at each of the three
positions.
voltmeter
position
pd / V
A-C
D-F
C-D
(3)
(c)
The thermistor is heated so that its resistance decreases. State and explain the effect this
has on the voltmeter reading in the following positions.
(i)
A–C........................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
D–F........................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 11 marks)
Page 85 of 165
55
(a)
The cell in Figure 1 has an emf of 3.0 V and negligible internal resistance.
Figure 1
Calculate the potential difference across the 8 Ω resistor.
potential difference ............................................... V
(2)
Page 86 of 165
(b)
Figure 2 shows the same circuit with a voltmeter connected across the 8 Ω resistor.
Figure 2
The voltmeter reads 1.8 V. Calculate the resistance of the voltmeter.
resistance ...............................................Ω
(3)
(Total 5 marks)
56
An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5
m s–1. What is the efficiency of the system?
A
5%
B
12%
C
50%
D
100%
(Total 1 mark)
Page 87 of 165
57
The circuit diagram below shows a battery of electromotive force (emf) 12 V and internal
resistance 1.5 Ω connected to a 2.0 Ω resistor in parallel with an unknown resistor, R. The battery
supplies a current of 4.2 A.
(a)
(i)
Show that the potential difference (pd) across the internal resistance is 6.3 V.
(1)
(ii)
Calculate the pd across the 2.0 Ω resistor.
pd ...........................................V
(1)
(iii)
Calculate the current in the 2.0 Ω resistor.
current ...........................................A
(1)
(iv)
Determine the current in R.
current ........................................... A
(1)
Page 88 of 165
(v)
Calculate the resistance of R.
R ........................................... Ω
(1)
(vi)
Calculate the total resistance of the circuit.
circuit resistance ........................................... Ω
(2)
(b)
The battery converts chemical energy into electrical energy that is then dissipated in the
internal resistance and the two external resistors.
(i)
Using appropriate data values that you have calculated, complete the following table
by calculating the rate of energy dissipation in each resistor.
resistor
rate of energy dissipation / W
internal resistance
2.0 Ω
R
(3)
(ii)
Hence show that energy is conserved in the circuit.
...............................................................................................................
...............................................................................................................
(2)
(Total 12 marks)
Page 89 of 165
58
A student investigates how the power dissipated in a variable resistor, Y, varies as the resistance
is altered.
Figure 1 shows the circuit the student uses. Y is connected to a battery of emf ε and internal
resistance r.
Figure 1
Figure 2 shows the results obtained by the student as the resistance of Y is varied from 0.5 Ω to
6.5 Ω.
Figure 2
resistance of Y / Ω
Page 90 of 165
(a)
Describe how the power dissipated in Y varies as its resistance is increased from 0.5 Ω to
6.5 Ω.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
The emf of the battery is 6.0 V and the resistance of Y is set at 0.80 Ω.
(i)
Use data from Figure 2 to calculate the current through the battery.
current .......................................... A
(3)
(ii)
Calculate the voltage across Y.
voltage .......................................... V
(2)
(iii)
Calculate the internal resistance of the battery.
internal resistance .......................................... Ω
(2)
Page 91 of 165
(c)
The student repeats the experiment with a battery of the same emf but negligible internal
resistance. State and explain how you would now expect the power dissipated in Y to vary
as the resistance of Y is increased from 0.5 Ω to 6.5 Ω.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 12 marks)
59
The cells in the circuit shown in the figure below have zero internal resistance. Currents are in
the directions shown by the arrows.
R1 = 0 − 10Ω
R2 = 10Ω
R1 is a variable resistor with a resistance that varies between 0 and 10 Ω.
(a)
Write down the relationship between currents I1, I2 and I3.
........................................................................................................................
(1)
(b)
R1 is adjusted until it has a value of 0 Ω.
State the potential difference across R3.
potential difference = ........................ V
(1)
Page 92 of 165
(c)
Determine the current I2.
current = ........................ J
(2)
(d)
State and explain what happens to the potential difference across R2 as the resistance of
R1 is gradually increased from zero.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 7 marks)
Page 93 of 165
60
In the circuit shown, V is a voltmeter with a very high resistance. The internal resistance of the
cell, r, is equal to the external resistance in the circuit.
external resistance
Which of the following is not equal to the emf of the cell?
A
the reading of the voltmeter when the Switch S is open
B
the chemical energy changed to electrical energy when
unit charge passes through the cell
C
twice the reading of the voltmeter when the switch S is
closed
D
the electrical energy produced when unit current passes
through the cell
(Total 1 mark)
Page 94 of 165
Mark schemes
1
(a)
(i)
5.1 and 7.1 ✓
Exact answers only
1
(ii)
Both plotted points to nearest mm ✓
Best line of fit to points ✓
The line should be a straight line with approximately an equal
number of points on either side of the line
2
(iii)
Large triangle drawn at least 8 cm × 8 cm ✓
Correct values read from graph ✓
Gradient value in range 0.190 to 0.210 to 2 or 3 sf ✓
3
(iv)
) = 5.0 Ω
(R =
Must have unit ✓
Allow ecf from gradient value
No sf penalty
1
(b)
(i)
5.04 (Ω) or 5.0 (Ω) ✓
(Allow also 5.06 Ω or 5.1 Ω, obtained by intermediate rounding up of 3.502)
From R =
1
(ii)
(Uncertainty in V = 0.29% )
Uncertainty in V2 = 0.57%, 0.58% or 0.6% ✓
From uncertainty in V = 0.01 / 3.50 × 100%
Uncertainty in P = 2.1% ✓
From uncertainty in P = 0.05 / 2.43 × 100% = 2.1%
Uncertainty in R =2.6%, 2.7% or 3%
Answer to 1 or 2 sf only ✓
2.1 % + uncty in V2 (0.6%) = 2.7%
Allow ecf from incorrect uncertainty for V2 or P
3
(iii)
(Absolute) uncertainty in R is ( ± ) 0.14 or just 0.1 Ω (using 2.6%)
(or 0.15 or 0.2 Ω using 3%) ✓
Must have unit (Ω)
Must be to 1 or 2 sf and must be consistent with sf used from (ii)
No penalty for omitting ± sign
1
Page 95 of 165
(iv)
Works out possible range of values of R based on uncertainty in
(iii), e.g. R is in range 5.0 to 5.2 Ω using uncertainty of ± 0.1 Ω ✓
No credit for statement to effect that the values are or are not
consistent, without any reference to uncertainty
Allow ecf from (iii)
Value from (a)(iv) is within the calculated range (or not depending on figures,
allowing ecf) ✓
Allow ecf from (a)(iv)
2
[14]
2
(a)
(i)
Voltmeter across terminals with nothing else connected to battery / no additional load.
✓
1
(ii)
This will give zero / virtually no current ✓
1
(b)
(i)
Answer must clearly show power: εI and VI, with I cancelling out to give
formula stated in the question ✓
1
(ii)
Voltmeter connected across cell terminals ✓
Switch open, voltmeter records ε
Switch closed, voltmeter records V
Both statements required for mark ✓
Candidates who put the voltmeter in the wrong place can still
achieve the second mark providing they give a detailed description
which makes it clear that:
To measure emf, the voltmeter should be placed across the cell with
the external resistor disconnected
And
To measure V, the voltmeter should be connected across the
external resistor when a current is being supplied by the cell
2
(c)
Vary external resistor and measure new value of V, for at least 7 different values of
external resistor ✓
Precautions - switch off between readings / take repeat readings (to check that emf or
internal resistance not changed significantly) ✓
2
Page 96 of 165
(d)
Efficiency increases as external resistance increases ✓
Explanation
Efficiency = Power in R / total power generated
I2R / I2(R + r) = R / (R + r)
So as R increases the ratio becomes larger or ratio of power in load to power in
internal resistance increases ✓
Explanation in terms of V and ε is acceptable
2
[9]
3
(a)
R = V/I
M1
with all three variables defined
accept voltage
A1
2
(b)
use of 1/R = 1/R1 + 1/R2
C1
effective resistance of parallel resistors = 2
C1
3
total resistance = 11 Ω
A1
(c)
(i)
ratio 2/3 seen/ V= 4.8 V used
/clear attempt to find pd across parallel resistors
C1
current = 1.6 A
c.a.o.
A1
4
Page 97 of 165
(ii)
use of P= I2R (= 2.42 × 11)
C1
total power = 63.4 W
allow e.c.f. from (b)
A1
[9]
4
(a)
the (total) energy transferred/work done when
one unit/coulomb of charge
B1
is moved around a circuit/provided by the supply
B1
2
(b)
work is done inside the battery/there is resistance
inside the battery
B1
so less energy is available for the external circuit/someoltage
is lost between the terminal/mention of lost volts
B1
2
(c)
(i)
9.00 V
c.a.o.
B1
Page 98 of 165
(ii)
lost voltage = E - V or E = I(R + r)
C1
0.82r = 0.59
C1
5
internal resistance = 0.720 Ω
A1
(iii)
because the battery has to provide more energy/power
B1
[9]
5
(a)
(i)
P=
gives 40
(1)
RT = 3.6 Ω (1)
[or P = VI to give I = 3.3 (A) (1) and R = P / I2 = 3.7 Ω (3.67 Ω) (1)]
(ii)
three resistors in parallel (1)
R = 3.6 × 3 = 10.8 (Ω) (1)
(allow C.E. for RT from (i))
5
(b)
(use of R =
l=
= 0.77 m (1)
= gives) 10.8 =
(1)
(1)
(allow C.E for R from (a)(ii)
3
[8]
Page 99 of 165
6
(a)
R = V/I
C1
10 Ω
A1
2
(b)
total resistance = 6/0.25 = 24 Ω
or correct substitution in potential divider formula
C1
14 Ω
A1
2
(c)
correct substitution in parallel resistance formula
or current through 15 Ω = 0.16 A
C1
or RΩ=
or
resistance of parallel combination = 6.0 Ω
or parallel combination resistance + 8.4
(allow in substitution form)
or total current = 0.41 A
14.4 Ω (cnao)
C1
{
incorrect answer would get C1, C1
(may use incorrect (a) value here)
A1
3
Page 100 of 165
(d)
identifies circuit with lower or higher total resistance correctly
(e.c.f.)
B1
power = V2/R or VI
B1
V is same for both so circuit with higher resistance
dissipates lower power or V is same for both and
circuit with lower current dissipates lower power
or reason by means of calculations
B1
3
[10]
7
(a)
(three parallel resistors) give
R = 10 (Ω) (1)
10 Ω and 50 Ω in series gives 60 Ω (1)
(allow e.c.f. from value of R)
(3)
(b)
(V = IR gives) 12 = I × 60 and I = 0.2 A (1)
(allow e.c.f. from (a))
(1)
[4]
8
(a)
correct curve in positive quadrant (1)
correct curve in negative quadrant (1)
passing through origin (1)
(3)
Page 101 of 165
(b)
the current heats the filament (1)
(temperature rises) resistance increases (1)
pd. and current do not increase proportionally (1)
some reference to mirror image in negative quadrant (1)
The Quality of Written Communication marks were awarded primarily for the quality
of answers to this part
(max 3)
(c)
diagram to show:
battery, variable resistance (or variable supply) and filament (1)
current sensor in series circuit (1)
voltage sensor across filament (1)
the two sensor boxes connected to datalogger (1)
method:
variable resistor or variable supply altered
[or choose recording interval] (1)
thus changing both V and I (1)
The Quality of Written Communication marks were awarded primarily for the quality
of answers to this part
(max 5)
[11]
9
(a)
RAB = 5.0 (Ω) (1)
V (= 5.0 × 0.50) = 2.5 V (1)
(b)
Vr = 12 – 2.5 + 5.0 (1) = 4.5 (V) (1)
= 9.0 Ω (1)
r=
(c)
W(= EIt) = 12 J (1)
(d)
Wr(= VrIt) = 4.5 (J) (1)
= 0.375 (1)
[Max 7]
10
(a)
(i)
I=
(ii)
P = (0.80)2 × 5 = 3.2 W (1) (allow e.c.f. from (a)(i))
= 0.80 A (1)
(2)
Page 102 of 165
(b)
Itot =
(1) = 1.60 (A) (1)
I=
= 0.80 (A) (1) (allow e.c.f. from Itot)
(3)
(c)
same brightness (1)
because same current (1)
[or an answer consistent with their current values]
(2)
[7]
11
(a)
R=
(1)
ρ is resistivity, l is the length of the wire, A is the cross-sectional area (1)
2
(b)
(i)
P=
R=
l=
(1)
= 106(Ω)(1)
=
(105.8 Ω)
= 7.7 m (1)
(7.69 m)
(allow C.E. for incorrect value of R )
(ii)
in series, voltage across each < 230 V or pd shared (1)
power (= V2/R) is less than 500 W in each (1)
in parallel, voltage across each = 230 V (1)
correct rating,
conclusion (1)
[or, in series, high resistance or combined resistance (1)
low current (1)
in parallel, resistance is lower,
higher current (1)
more power, justified (1)]
max 6
[8]
Page 103 of 165
12
(a)
(i)
(use of V = IR gives) V = I(R1 + R2) (1)
= 50 mA ✓
(ii)
Vout (= IR2) = 0.05 × 60 = 3 V (1)
(allow C.E. for value of I from (i))
4
(b)
(temperature increases, resistance decreases), total resistance decreases (1)
current increases (1)
voltage across R2 increases (1)
[or R2 has increased share of (total) resistance (1)
new current is same in both resistors (1) larger share of the 9 V (1)]
[or Vout = Vin
(1) R1 decreases (1) Vout decreases (1)]
3
[7]
13
(i)
(V = IR gives) 12 = (30 + 30 + 2)I (1)
I=
(ii)
= 0.19 A (1)
(0.194 A)
VPQ = 12 – (0.19 × 2) (1)
= 11.6 V (1)
(allow C.E. for incorrect I in (i))
[or VPQ = 0.19 × 60 = 11.6 V]
[or VPQ = 12 ×
(I = 0.194 A gives 11.6 V)
= 11.6 V
Page 104 of 165
(iii)
(PA = I2R gives) PA = (0.19)2 × 30 = 1.08 (1)
[or PA =
W (1)
]
(allow C.E. for incorrect I in (i) or incorrect V in (ii))
(iv)
(E = PAt gives) E = 1.08 × 20 (1)
= 21.6 J (1)
(allow C.E. for incorrect PA in (iii))
[8]
14
(a)
only 30 Ω in the circuit (1)
(use of V = IR gives) 6 = I × 30 and I = 0.20 A (1)
2
(b)
two resistors in parallel gives
and R = 20 (Ω) (1)
total resistance = 20 + 30 = 50 (Ω) (1)
(allow C.E. for value of R)
I=
= 0.12 A (1) (allow C.E for total resistance)
4
[6]
15
(a)
(i)
ϵ = I (R + r) (1)
(ii)
VR = IR gives VR = ϵ - Ir (1)
2
Page 105 of 165
(b)
(i)
P = VI gives 30 = 120 I (1)
I = 0.25 A (1)
(ii)
I through lamp = 0.25 (A) and p.d. across it = 240 V (1)
p.d. due to 1 cell = 1.5 – (0.25 × 1.2) = 1.2 (V) (1)
number of cells =
= 100 (1)
[or RL given by 30 = 0.252 RL and RL = 480 (Ω) (1)
1.5n = 0.25(480 + 1.2n) (1)
1.2n = 120 and n = 100 (1)]
[or ϵ = V + Ir gives 1.5n = 120 + 0.25 × 1.2n (2)
n = 100 (1)]
5
[7]
16
(a)
VR = (3.0 – 2.0) = 1.(0) (V) (1)
R=
= 100 Ω (1)
(b)
(use of P = IV gives) Pdiode (= 10 × 10–3 × 2.0) = 0.02(0) W (1)
(c)
(use of c = fλ gives)
) = (4.7 × 1014Hz) (1)
(use of E = hf gives) E (= 6.63 × 10–34 × 4.7 × 1014) = 3.1 × 10–19 J (1)
(allow C.E. for A.E. in value of f)
(d)
energy supplied in 1 sec = 0.02(0) (J) (1)
(allow C.E. for value of P from (ii))
number of photons emitted in 1 sec =
= 6.5 × 1016 (1)
(allow C.E. for value of E)
(e)
all the energy supplied converted to light energy [or 100% efficient]
[or monochromatic light]
[or all photons (emitted by LED) have the same energy] (1)
[8]
17
(a)
circuit diagram to show:
wide end of conducting strip to – of battery, narrow end to + (1)
voltmeter between wide end and probe (1)
2
Page 106 of 165
(b)
resistance gradient increases as x increases (1)
because strip becomes narrower (as x increases) (1)
current constant throughout strip (1)
voltage gradient = current × resistance gradient, so
voltage gradient increases as x increases (1)
4
(c)
(i)
(2l – x)
(0.700)
0.60(0)
0.53(0)
0.47(0)
0.44(0)
0.42(0)
ln (2l – x)
(–0.357)
–0.511
–0.635
–0.755
–0.821
–0.868
1st column correct to 2 s.f. (1)
2nd column correct to 3.s.f. (1) (1)
(only 4 values correct, (1))
(ii)
suitable scales (1)
axes labelled and units included (1)
5 points correctly plotted (1)
acceptable straight line (1)
straight line confirms equation because equation is of form
y = mx + c with negative gradient (1)
(iii)
gradient = (–)
= (–) 15.4 (V) (1)
1.44 V1 = 15.4 gives V1 = 11 V (1)
(10.7 ± 0.2 V)
[alternative: V = V1 when x = l and ln (2l – x) (= ln 0.4) = 0.92 (1)
at ln (2l – x) = 0.92, graph gives V1 = 11 V (1)
(10.7 ± 0.2 V)]
10
[16]
18
(a)
(i)
for X: (P = VI gives) 24 = 12I and I = 2 A (1)
for Y 18 = 6I and I = 3 A (1)
2
Page 107 of 165
(b)
(i)
12 V (1)
(ii)
voltage across R2 (= 12 – 6) = 6 (V) (1)
I = 3 (A) (1)
(V = IR gives) 6 = 3R2 and R2 = 2Ω (1)
(allow C.E. for I and V from (a) and (b)(i))
[or V = I(Ry + R2) (1) 12 = 3(2 + R2) (1)
R2 = 2Ω (1)]
(iii)
current = 2 (A) + 3 (A) = 5 A (1)
(allow C.E. for values of the currents)
(iv)
27 (V)– 12 (V) = 15 V across R1 (1)
(v)
for R1, 15 = 5 R1 and R1 = 3Ω (1)
(allow C.E. for values of I and V from (iii) and (iv)
7
[9]
19
(a)
first pair in parallel
=
(1)
= gives R’ = 20 ( Ω) (1)
second pair in parallel
gives R” = 30( Ω) (1)
resistance between A and B = 20 + 30 (1) (= 50 Ω)
(allow C.E. for values of R’ and R")
4
(b)
(i)
total resistance = 50 + 50 = 100 Ω (1)
(V = IR gives) 24 = I 100 and I = 0.24 A (1)
(ii)
current in 60 Ω =
I (1)
= 0.080 (A) (1)
[or alternative method]
(allow C.E. for value of I from (b)(i))
4
[8]
Page 108 of 165
20
(a)
(i)
pd across resistor (= 3.0 – 2.2) = 0.8 (V) (1)
(use of V = IR gives) R =
(ii)
= 23 Ω (1) (22.9 Ω)
charge flow in 1 s = 0.035 (C) (1)
no. of electrons (in 1 s)
= 2.2 × 1017 (1)
(2.19 × 1017)
4
(b)
(i)
(use of E = hf =
gives) E =
(1)
= 3.1(3) × 10–19 J (1)
(ii)
(use of P = VI gives) P (= 2.2 × 0.035) = 0.077 (W)
[or use of P = I2R with R
= 63 (Ω)]
maximum no. of photons emitted per sec. =
= 2.5 × 1017 (1) (2.48 × 1017)
(allow C.E. for value of E from (i) and value of P from (ii))
4
[8]
21
(a)
(i)
5 V (1)
(ii)
RT = 36 (Ω)
(use of V = IR gives) 15 = I × 36 and I = 0.42 A (1)
3
(b)
(i)
equivalent resistance of the two lamps
RT = 6 + 12 = 18 (Ω) and 15 = I × 18 (1)
(ii)
(1)
(to give I = 0.83 A)
current divides equally between lamps (to give I = 0.42 A)
(or equivalent statement) (1)
3
(c)
same brightness (1)
(because) same current (1)
2
[8]
Page 109 of 165
22
(a)
(i)
energy changed to electrical energy per unit charge/coulomb
passing through
[or electrical energy produced per coulomb or unit charge]
[or pd when no current passes through/or open circuit] (1)
(ii)
I=
(iii)
(use of
= 2.5 A (1)
= I(R + r) gives)
= V + Ir and 8 = 6 + Ir (1)
substitution gives 8 – 6 = 2.5r (1) (and r = 0.8 Ω)
4
(b)
(i)
(use of P = I2R gives) PR = 2.52 × 2.4 = 15 W
[or P = VI gives P = 6 × 2.5 = 15 W] (1)
(allow C.E. for value of I from (a))
(ii)
PT = 15 + (2.52 × 0.8) (1)
= 20 (W) (1)
(allow C.E. for values of PR and I)
(iii)
E = 5 × 2 × 60 = 600 J (1)
(allow C.E. for value of P from (i) and PT from (ii))
4
[8]
23
(a)
between A and C: (each) series resistance = 100Ω (1)
(parallel resistors give)
+ = gives RAC = 50Ω (1)
2
(allow C.E. for incorrect series resistance)
Page 110 of 165
(b)
between A and B: series resistance = 150Ω (1)
parallel =
(1)
(allow C.E. for series resistance)
RAB = 37.5Ω (1) (38Ω)
3
[5]
24
(a)
(i)
(1)
R=
(ii)
= 13 Ω (1)
(13.3)
two resistors in parallel give 20 (Ω) (1)
R = 20 + 40 = 60 (Ω) (1)
max 3
(b)
(i)
three resistors in parallel give
(= 2 (Ω))
and total resistance = 4 (Ω) (1)
total current =
= 3 (A) (1)
(allow C.E. for value of total resistance)
current in each element 1.0 A (1)
(allow C.E. for value of total current)
[or 6 V across each set
resistance of each set = 2 Ω, gives current through
each set = 3 (A)
current in each element = 1.0 A]
[or 6 V across each set/resistor,
resistance of one resistor = 6 Ω,
gives current in each element = 1.0 A]
Page 111 of 165
(ii)
six resistors in series gives R = 36 (Ω) and I =
= 0.3 (A) (1)
heating effect (I2R) much reduced [or less power] (1)
5
[8]
25
(a)
V = –Ir +
(1)
1
(b)
straight line (within 1st quadrant) (1)
negative gradient (1)
2
(c)
: intercept on voltage axis (1)
r: gradient (1)
2
[5]
26
(a)
(i)
at 25 (°C), total resistance = 300 + 200 = 500 (Ω) (1)
I=
= 24 mA (1)
(allow C.E. for value of total resistance)
(ii)
pd across thermistor = 24 × 10–3 × 300 = 7.2 V (1)
(allow C.E. for value of current from (i) and Rth from graph)
3
Page 112 of 165
(b)
as temperature increases, resistance (of thermistor) decreases (1)
total resistance decreases (1)
current in circuit increases (1)
pd across resistor increases (1)
(since battery remains at 12 V) pd across thermistor decreases (1)
[or Rth decreases (1)
potential divider situation (1)
Vth = 12 ×
(1)
denominator decreases less slowly than numerator (1)
Vth decreases (1)
or for last two marks, thermistor gets smaller share of voltage
explanation of this]
(max 3)
QWC 1
(c)
(i)
(use of P =
gives) at 25°C
P=
= 0.48 W (1)
at 45°C correct reading of R = 30 (Ω) (1)
P=
(ii)
= 4.8 (W) (1)
E = Pt = 2.64 × 10 × 60 (1)
= 1.6 × 103 J (1)
(allow C.E. from part (i)
(iii)
rate of decrease of resistance is not linear
[or resistance not directly proportional to temperature] (1)
6
[12]
Page 113 of 165
27
(a)
(i)
Use of P = VI with pair of valid coordinates from graph
C1
0.52 (W)
Allow 1sf if within 0.49 to 0.52
A1
2
(ii)
Correct general shape
M1
Linear rise between 0.0 ‒ 0.5 V and falls to zero at 0.71 V
A1
2
(iii)
Use of efficiency =
C1
Use of I =
C1
Their (i) / 67.5 (m2)
(7.7 × 10−3 if correct)
A1
3
Page 114 of 165
(b)
(i)
0.7 J of work done (by cell) per 1 C of charge (when moved round circuit)
OR
(Terminal) pd across (solar) cell with no load / current is 0.7 V
Not “per unit charge”
B1
1
(ii)
20 cells in series (to produce 14 V)
B1
Series arrangement has internal resistance of 15.6 Ω
B1
Cells in parallel (needed to reduce total internal resistance of array)
B1
80 cells / 4 parallel sets of 20 cells in series
B1
4
Page 115 of 165
(c)
The marking scheme for this question includes an overall assessment for the quality
of written communication (QWC). There are no discrete marks for the assessment of
QWC but the candidate’s QWC in this answer will be one of the criteria used to
assign a level and award the marks for this question.
Descriptor ‒ an answer will be expected to meet most of the criteria in the level
descriptor.
Level 3 ‒ good
-claims supported by an appropriate range of evidence;
-good use of information or ideas about physics, going beyond those given in the
question;
-argument is well structured with minimal repetition or irrelevant points;
-accurate and clear expression of ideas with only minor errors of grammar,
punctuation and spelling.
Level 2 ‒ modest
-claims partly supported by evidence;
-good use of information or ideas about physics given in the question but limited
beyond this;
-the argument shows some attempt at structure;
-the ideas are expressed with reasonable clarity but with a few errors of grammar,
punctuation and spelling.
Level 1 ‒ limited
-valid points but not clearly linked to an argument structure;
-limited use of information about physics;
-unstructured;
-errors in spelling, punctuation and grammar or lack of fluency.
Level 0
-incorrect, inappropriate or no response.
Some points:
Use on communication satellite:
Continuous supply of energy from Sun
No need for fuel (for power purposes)
Large area of solar cells not needed (but possible)
Low mass
Can be unfolded (after launch)
No environmental hazard
Reliable/no moving parts
Continuous operation:
Arrays need to track sun (to maximise absorption)
Shielding required as can be damaged by meteors or cosmic rays
Need storage system (rechargeable batteries / capacitors) for back
up (if in shadow)
Limit use of energy-intensive operations
Use on space probe:
Light intensity / energy too low at large distance
Intensity falls as inverse-square
Area of array would be too large
Page 116 of 165
Solar cells will have degenerated too much over this time
B6
6
[18]
28
(a)
(i)
6.0 (Ω) (1)
1
(ii)
4.5 (V) (1)
1
(iii)
(use of I = V/R)
I = 4.5/6.0 = 0.75 (A) (1)
current through cell A = 0.75/2 = 0.375 (A) (1)
2
(iv)
charge = 0.375 × 300 = 112 (1) C (1)
2
(b)
cells C and D will go flat first or A and B last longer (1)
current/charge passing through cells C and D (per second) is
double/more than that passing through A or B (1)
energy given to charge passing through cells per second is double
or more than in cells C and D (1) or in terms of power
3
[9]
29
(a)
(i)
no of bulbs =
= 46 (1)
(ii)
(use of P = VI gives) I =
(iii)
resistance of each bulb =
= 0.080 A (1)
= 63 Ω (62.5 Ω)
(allow C.E. for number of bulbs and value of I)
[or R
or
= 62.5 Ω
= 62.5 Ω]
5
Page 117 of 165
(b)
(iv)
energy consumed by the set = 0.4 × 46 × (2 × 60 × 60) (1)
= 132 kJ (1)
(allow C.E. for number of bulbs from (i))
(i)
no of bulbs = 56, gives total resistance = 62.5 × 56 (Ω) (= 3500) (1)
I=
= 0.066 A (1) (0.0657 A)
(use of 63 Ω gives 0.065 A)
(allow C.E. for no. of bulbs in (a) (i) and R in (a) (iii))
(ii)
bulbs would shine less bright (1)
3
[8]
30
(a)
(i)
three resistors in series (1)
(ii)
R = 3.0 + 4.0 + 6.0 = 13 Ω (1)
(iii)
three resistors in parallel (1)
(iv)
(1)
R = 1.3 Ω (1)
5
(b)
(i)
two resistors in parallel give
and R’ = 2.0 (Ω) (1)
total resistance = (2 + 4) = 6.0 Ω (1)
4
(ii)
divide the emf in the ratio of 2 : 4 (1)
to give 4.0 V (1)
[or any suitable method]
[9]
Page 118 of 165
31
(a)
at 200 °C : RR = 130 ± 1 (Ω), RTh = 18 ± 1Ω (1)
1
(b)
(i)
VAB = Vin
= 12 ×
(1)
(1)
= 1.5 V (1)
(1.46 V)
(allow C.E. for values from (a))
(c)
(ii)
Rth = RR occurs at 50 °C (1)
(i)
(use of P =
gives) Rb =
4
= 18 Ω (1)
[or use of P = VI and calculate I]
(ii)
(S open, RTh ≈ 90 Ω)
bulb and thermistor in parallel (1)
gives lower resistance than thermistor on its own (1)
total resistance in circuit decreases (1)
current increases VR > 6 V (1)
(hence Vth < 6 V i.e. decreases)
[or use of potentiometer equation, or ratio of resistances and
share of pd]
max 4
[9]
32
(a)
(i)
(total) resistance = (20 + 60) (Ω) (1)
(V = IR gives) I =
(ii)
= 0 075 A (1)
with S closed, (effective) resistance = 20 (Ω) (1)
I=
=0.3 A (1)
max 3
Page 119 of 165
(b)
use of same current as in part (i) (1)
voltmeter reading = 0.075 × 60 = 4.5 V (1)
[or use potentiometer equation 6 ×
(allow C.E. for value of I from (a)(i)
= 4.5 V]
2
[5]
33
(a)
so that each lamp is connected directly across the battery (1)
if one lamp blows others are still on (1)
2
(b)
use of power = VI (1)
current through each headlight = 60/12 = 5.0 A
or current through each tail light = 8/12 = 0.67 A (1)
total current = 2 × 5.0 + 2 × 0.6667 = 11(.3) A (1)
3
(c)
the lamp with the highest power rating has the least resistance (1)
the resistance is greater because the temperature of
the filament is lower (1)
and resistance increases with temperature (1)
3
(d)
(i)
(use of energy = power × time)
energy dissipated = (8.5) × 2 × 12 × 3600
(any power ´ time) (1)
energy dissipated = 1.1(2) × 106 J (1)
2
Page 120 of 165
(ii)
stored energy in battery =
12 × 1.2 × 24 × 3600 = 1.24 × 106 (1)
energy to start = 12 × 100 × 1 = 1200 J (1)
energy left = (1.24 – 1.12) × 106 = 120 000 J
so hence car will start (1)
(conclusion assuming all working correct)
3
[13]
34
(a)
(for lamp and resistor) 18(Ω) + 12(Ω) = 30(Ω) (1)
(in parallel)
+
=
(1) (gives R = 10(Ω))
(allow C.E.for wrong value in first step)
total resistance =
(= 20 Ω)
+ 10 (1)
3
(b)
(i)
(use of V = IR gives) I =
(ii)
pdAB = 30 – (10 × 1.5) (1)
= 15V (1)
[or alternative method]
= 1.5 A (1)
(allow C.E. for value of I from (i))
(iii)
Ilamp =
= 0.5 A (1)
[or alternative method] (allow C.E. for value of pdAB from (ii))
4
(c)
(i)
(lamp power) (= I2R) = 0.52 × 12 = 3.0 (W) (1)
(allow C.E. for value of Ilamp from (b) (iii))
Page 121 of 165
(ii)
power from battery = 30 × 1.5 = 45 (W) (1)
(allow C.E. for value of I from (b) (i))
(1)
(allow C.E. for power in lamp and/or battery in (i))
3
[10]
35
(a)
(i)
circuit diagram to show:
wire, ammeter, battery, (variable resistor) and switch in series (1)
[or potentiometer with ammeter in correct position]
voltmeter across the wire (1)
(ii)
(method: constant length of wire)
measure length (of wire) (1)
measure diameter (of wire) (1)
measure voltage (across) and current (through wire) (1)
vary resistor to obtain different voltage and current (1)
alternative
[(method: variable length of wire)
measure length (each time) (1)
measure diameter (1)
(for full length of wire) measure voltage and current (1)
voltmeter to shorter lengths, measure voltage (and current) (1)]
(iii)
(use of) ρ =
(to calculate ρ) (1) (for either method)
calculate A from (πr2) (1) (for either method)
(method: constant length of wire)
determine
for (one) length (1)
repeat readings (for same length and) take mean of ρ or R (1)
[or plot graph of V vs I to give mean R (1)
or gradien =
(1)]
alternative
[(method: variable length of wire)
determine
for each length (1)
calculate ρ for each length and take mean (1)
[or graph of R vs l (1) with correct gradient (1)]
10
Page 122 of 165
(b)
(use of
gives)
(1)
l = 0.035 m (1)
2
[12]
36
(a)
(i)
suitable variable input (variable power supply or
variable resistor) (1)
protective resistor and diode forward biased (1)
correct current and pd measuring devices (1)
3
Page 123 of 165
(ii)
the mark scheme for this part of the question includes an
overall assessment for the Quality of Written Communication
QWC
descriptor
mark
range
goodexcellent
Uses accurately appropriate grammar, spelling, punctuation
and legibility.
Uses the most appropriate form and style of writing to give an
explanation or to present an argument in a well structured
piece of extended writing.
[May include bullet points and/or formulae or equations].
Answer refers to at least 5 of the relevant points listed below.
5-6
modestadequate
Only a few errors.
Some structure to answer, style acceptable, arguments or
explanations partially supported by evidence or examples.
Answer refers to at least 3 or the relevant points listed below.
3-4
Several significant errors.
Answer lacking structure, arguments not supported by
evidence and contains limited information.
Answer refers to no more than 2 of the relevant points.
1-2
poorlimited
incorrect,
inappropriate
or no
response
No answer at all or answer refers to unrelated, incorrect or
inappropriate physics.
0
The explanation expected in a competent answer should
include a coherent selection of the following physics ideas.
connect circuit up (1)
measure current (I) and pd/voltage (V) (1)
vary resistance/voltage (1)
obtain a range of results (1)
reverse connections to power supply (and repeat) (1)
plot a graph (of pd against current) (1)
mention of significance of 0.6V or disconnect between readings
or change range on meters when doing reverse bias (1)
Page 124 of 165
(b)
(i)
(use of I = V/R)
I = 12/8 (1)= 1.5A (1)
(ii)
I = (12 – 0.65 (1))/4 = 2.8 A (1) sig figs (1)
5
[14]
37
(a)
A combination of resistors in series connected across a voltage source
(to produce a required pd) ✓
Reference to splitting (not dividing) pd
1
(b)
When R increases, pd across R increases ✓
Pd across R + pd across T = supply pd ✓
So pd across T / voltmeter reading decreases ✓
Alternative:
Use of V=
✓
Vtot and R2 remain constant ✓
So V increases when R1 increases ✓
3
(c)
At higher temp, resistance of T is lower ✓
1
So circuit resistance is lower, so current / ammeter reading increases ✓
1
(d)
Resistance of T = 2500 Ω
Current through T = V / R = 3 / 2500 = 1.2 × 10−3 A ✓
(Allow alternative using V1/R1 = V2/R2)
pd across R = 12 − 3 = 9 V
The first mark is working out the current
1
Resistance of R = V / I = 9 / 1.2 × 10−3 = 7500 Ω✓
The second mark is for the final answer
1
Page 125 of 165
(e)
Connect the alarm across R instead of across T ✓
allow: use a thermistor with a ptc instead of ntc.
1
[9]
38
(a)
(use of 1/Rtotal = 1/R1 + 1/R2)
1/Rtotal = 1/400 + 1/400 = 2/400
Rtotal = 200 Ω (1) (working does not need to be shown)
hence total resistance = 25 + 200 = 225Ω (1)
2
(b)
(i)
(use of P = V2/R)
1 = V2/400 (1)
V2 = 400 (working does not need to be shown)
V = 20V (1)
(ii)
(use of I = V/R)
I = 20/400 = 0.05A (1) (working does not need to be shown)
hence current = 2 × 0.05 = 0.10A (1)
(iii)
(use of V = IR)
pd across 25Ω resistor = 25 × 0.10 = 2.5V (1)
(working does not need to be shown)
hence maximum applied pd = 20 + 2.5 = 22.5V (1)
6
[8]
Page 126 of 165
39
(a)
total resistance in
circuit
(b)
X open, Y closed
R (given)
X closed, Y open
2/3 R
X open, Y open
2R
X closed, Y
closed
R/2
energy dissipation is V2/R
or approach using both I = V/R and P = VI
B1
highest resistance gives least energy
or X open, Y open or their highest tabulated resistance
B1
2
(c)
electrons collide with ions
M1
transferring energy to them/giving them or increasing their
vibrational/kinetic energy
A1
2
Page 127 of 165
(d)
voltage across load lower or load voltage =
B1
or load current reduced or load current =
thermal energy output will decrease (in any stated circuit)
or identifies lowest resistance in table as being most affected
M1
since P =
or since power = I2 R
A1
3
(e)
(i)
resistance =
or substitution or A = 1.16 × 10–7 (m2)
C1
1.93 × 10–4 (m) 0.193mm
A1
(ii)
two properties from
high resistance/resistivity (low electrical conductivity)
B1
high melting point
B1
low thermal capacity/specific heat capacity
B1
4
[14]
Page 128 of 165
40
(a)
(i)
P = V2/R with substitution: 144/any resistance
C1
37.9 (W)
A1
2
(ii)
use of 1/R formula with substitution of some data
even if not all five resistors
C1
correct calculation of 1/R (giving 0.897)
C1
1.11 (Ω)
A1
3
(iii)
144/their aii
C1
129 to 131 (W) ecf
A1
2
(b)
lower resistance needed
B1
(to achieve) higher current (for l 2R to be the same)/
correct use of V2/R
B1
2
[9]
Page 129 of 165
41
(a)
(i)
work (done)/energy (supplied) per unit charge (by battery) (1)
(or pd across terminals when no current passing through
cell or open circuit)
1
(ii)
when switch is closed a current flows (through the battery) (1)
hence a pd/lost volts develops across the internal resistance (1)
2
(b)
(use of ε = V + Ir)
I = 5.8/10 = 0.58 (A) (1)
6.0 = 5.8 + 0.58r (1)
r = 0.2/0.58 = 0.34 (Ω) (1)
3
(c)
need large current/power to start the car (1) (or current too low)
internal resistance limits the current/wastes power(or energy)/reduces
terminal pd/increases lost volts (1)
2
[8]
Page 130 of 165
42
(a)
any three from
voltmeter resistance is high
B1
current in circuit is 0 or low
B1
no (low) energy lost in voltmeter
B1
no lost volts/volts lost overcoming internal resistance
B1
‘load’/voltmeter resistance >> internal resistance
B1
voltage across voltmeter >> voltage across int. resistance
B1
3
(b)
(i)
current = 14.5/470
C1
0.031 or 0.0309 (A) amps penalise 1 sf
A1
2
Page 131 of 165
(ii)
‘lost’ pd in cell = 0.5 V
C1
internal resistance = 0.5 × 470/14.5 or (0.5/0.031 etc)
C1
16 (Ω) (16.2) (16.7)
A1
or 13.9(14) using 0.031 A to calculate total circuit
resistance
15.4(15) using 0.0309
A1
30 using 0.3
3
[8]
43
(a)
(i)
voltage = 0.01 × 540 = 5.4 V (1)
1
(ii)
voltage = 15 – 5.4 = 9.6 V (1)
1
(iii)
(use of resistance = voltage/current)
resistance = 9.6/0.01 (1) = 960 Ω (1)
or RT = 15/0.01 = 1500 Ω (1)
R = 150 – 590 = 960 Ω (1)
or potential divider ratio (1)(1)
2
(iv)
(use of 1/R = 1/R1 + 1/R2)
1/960 = 1/200 + 1/R2 (1)
1/R2 = 1/960 – 1/1200
R2 = 4800 Ω (1)
2
Page 132 of 165
(b)
(voltage of supply constant)
(circuit resistance decreases)
(supply) current increases or potential divider argument (1)
hence pd across 540 Ω resistor increases (1)
hence pd across 1200 Ω decreases (1)
or resistance in parallel combination decreases (1)
pd across parallel resistors decreases (1)
pd across 1200 Ω decreases (1)
3
[9]
44
(a)
(i)
(use of V = IR)
Rtotal = 1 (ohm)
V = 1 × 1 = 1.0 V
2
(ii)
(use of V = IR)
R = 9.0/1.0 = 9.0 Ω
r = 9.0 − 1.0 − 6.0 = 2.0 Ω
or use of (E = I(R + r))
9.0 = 1(7 + r)
r = 9.0 − 7.0 = 2.0 Ω
2
(iii)
(use of W = Vlt)
W = 9.0 × 1.0 × 5 × 60
W = 2700 J
2
(iv)
energy dissipated in internal resistance = 12 × 2.0 × 5 × 60 = 600 (J)
percentage = 100 × 600/2700 = 22%
CE from part aii
2
Page 133 of 165
(b)
internal resistance limits current
hence can provide higher current
or energy wasted in internal resistance/battery
less energy wasted (with lower internal resistance)
or charges quicker
as current higher or less energy wasted
or (lower internal resistance) means higher terminal pd/voltage
as less pd across internal resistance or mention of lost volts
2
[10]
45
(a)
(i)
(use of R = V/l)
R = 10/2.0 = 5.0 Ω
1
(ii)
R = 2 (Ω)
Rtotal = 2 + 3
(= 5 Ω)
3
(b)
(i)
voltage across Y = 10.0 – 2.0 × 3.0 = 4.0 V
current in Y = 4.0/3.0 = 1.3 A
2
(ii)
current through W = 0.67 A
voltage = 0.67 × 3 = 2.0 V
(or 4.0/2 = 2.0 V
)
2
[8]
Page 134 of 165
46
(a)
mention of pd across internal resistance or energy loss
in internal resistance or emf > V
pd across internal resistance/lost volts increases with
current or correct use of equation to demonstrate
2
(b)
(i)
y – intercept 1.52 V (± 0.01 V)
1
(ii)
identifies gradient as r or use of equation
substitution to find gradient or substitution in equation
r = 0.45 ± 0.02 Ω
3
(c)
(i)
same intercept
double gradient (must go through 1.25, 0.40 ± 1.5 squares)
2
(ii)
same intercept horizontal line
1
(d)
(i)
(use of Q = lt)
Q = 0.89 × 15 = 13
C
2
(ii)
use of P = I2r
P = 0.892 × 0.45
P = 0.36 W
2
[13]
Page 135 of 165
47
(a)
correct substitution into P = V 2/R
(condone power of 10 error)
C1
R = 2.62 (Ω) = 144/55 = 122/55
C1
correct substitution into ρ= RA/L
(condone error on R and/or power of 10 errors)
C1
resistivity = 9.9(5) × 10–7 (range 9.9 to 9.95 × 10–7)
A1
unit = Ω m
B1
5
(b)
(i)
joules per coulomb (of charge)/work done per unit charge
(treat reference to force as neutral)
M1
where charge moved (whole way) round circuit
A1
2
(ii)
lost volts = 0.1 (V) or 0.1 seen as voltage
C1
r = 0.011 to 1.09 × 10–2 (Ω)
A1
2
Page 136 of 165
(c)
brightness decreases
B1
increased current (in circuit/battery)
B1
increased lost volts leading to decreased pd across bulb or decreased
terminal pd
B1
3
[12]
48
(a)
(i)
(use of P=VI)
I = 36/12 + 6/12
= 3.5 (A)
2
(ii)
(use of V=IR)
R = 12/3 = 4 (Ω)
1
(iii)
R = 12/0.50 = 24
(Ω)
1
(b)
terminal pd/voltage across lamp is now less OR current is less
due to lost volts across internal resistance OR due to higher resistance
lamps less bright
3
(c)
(i)
current through lamps is reduced as resistance is increased or
pd across lamps is reduced as voltage is shared
hence power is less OR lamps dimmer
2
Page 137 of 165
(ii)
lamp Q is brighter
lamp Q has the higher resistance hence pd/voltage across is greater
current is the same for both
hence power of Q greater
3
[12]
49
(a)
(i)
(use of V = IR)
I = (12-8) / 60
= 0.067 Or 0.066(A)
2
(ii)
(use of V = IR)
R = 8/0.067 = 120 (Ω)
1
(iii)
(use of Q = It)
Q = 0.067 × 120 = 8.0
C
2
(b)
reading will increase
resistance (of thermistor) decreases (as temperature increases)
current in circuit increase (so pd across R1 increases) OR correct potential divider
argument
3
[8]
50
(a)
(use of P = V/l)
l = 36/12 = 3.0 A
l = 2.0/4.5 = 0.44 A
2
(b)
(i)
pd = 24 − 12 = 12 V
1
(ii)
current = 3.0 + 0.44 = 3.44 A
1
(iii)
R1 = 12/3.44 = 3.5 Ω
1
Page 138 of 165
(iv)
pd = 12 − 4.5 − 7.5 V
1
(v)
R2 = 7.5/0.44 = 17 Ω
1
(c)
(i)
(circuit) resistance increases
current is lower (reducing voltmeter reading)
or correct potential divider argument
2
(ii)
pd across Y or current through Y increases
hence power/rate of energy dissipation greater or temperature of lamp
increases
2
[11]
51
(a)
(use of ρ=RA / l)
R = 1.7 × 10−7 × 0.75 / 1.3 × 10−7
R = 0.98 Ω
First mark for sub. and rearranging of equation.
Bald 0.98 gets both marks
Final answer correct to 2 or more sig. figs.
2
(b)
(i)
(use of P=VI) I= 2.08 A
1
(ii)
V=2.08 × 0.98 = 2.04 V
C.E. from (a) and (b)(i)
1
(iii)
emf = 12 + 2
× 2.04 = 16.1 V
C.E. from (b)(ii)
If only use one wire then C.E. for second mark
2
(c)
lamp would be less bright
as energy / power now wasted in internal resistance / battery
OR terminal pd less
OR current lower (due to greater resistance)
No C.E. from first mark
2
[8]
Page 139 of 165
52
(a)
(i)
(use of I = V / R)
first mark for adding resistance values 90 k Ω
I = 6.0 / (50 000 + 35 000 + 5000)
= 6.7 × 10−5A
accept 7 × 10−5 or dotted 6 × 10−5
but not 7.0 × 10 −5 and not 6.6 × 10 −5
2
(ii)
= 0.33 (0.33 − 0.35) V
V = 6.7 × 10−5 × 5000
OR
= 0.33( V)
V = 5 / 90 × 6
CE from (i)
BALD answer full credit
0.3 OK and dotted 0.3
2
(b)
resistance of LDR decreases
need first mark before can qualify for second
reading increase because greater proportion / share of the voltage across R OR higher
current
2
(c)
I = 0.75 / 5000 = 1.5 × 10−4 (A)
(pd across LDR = 0.75 (V))
pd across variable resistor = 6.0 − 0.75 − 0.75 = 4.5 (V)
R = 4.5 / 1.5 × 10−4 = 30 000 Ω
or
I = 0.75 / 5000 = 1.5 × 10−4 (A)
RtotalI = 6.0 / 1.5 × 10−4 = 40 000 Ω
R = 40 000 − 5000 − 5000 = 30 000 Ω
3
[9]
53
54
A
[1]
(a)
(i)
1/R total = 1/(40)
R total = 10.9 kΩ
+1/(10+5)
= 0.09167
3
(ii)
I = 12 / 10.9 k = 1.1 mA
1
Page 140 of 165
(b)
position
pd / V
AC
6.0
DF
4.0
CD
2.0
C.E. for CD
3
(c)
(i)
AC: no change
constant pd across resistors / parallel branches(AE)
no CE from first mark
2
(ii)
DF: decreases
as greater proportion of voltage across fixed / 10 k Ω resistor
no CE from first mark
2
[11]
55
(a)
potential divider formula used or current found to be 0.25 A
C1
A1
allow 1 s.f.
2.0 V
1.0 V (with working) gains 1 mark
2
(b)
main current =1.2 V / 4 Ω = 0.3 (A)
C1
Rtotal = 1.8 V / 0.3 A = 6 Ω or I8 = 0.225 (A)
C1
RV = 24 Ω
A1
3
[5]
56
57
C
[1]
(a)
(i)
(use of V=Ir)
V= 4.2 × 1.5
= 6.3 (V)
1
Page 141 of 165
(ii)
pd = 12 − 6.3 = 5.7 V
NO CE from (i)
1
(iii)
(use of I = V / R)
I = 5.7 / 2.0 = 2.8(5) A
CE from (ii)
(a(ii)/2.0)
accept 2.8 or 2.9
1
(iv)
I = 4.2 – 2.85 = 1.3(5) A
CE from (iii)
(4.2 −(a)(iii))
accept 1.3 or 1.4
1
R= 5.7 / 1.35 =4.2 Ω
(v)
CE from (iv)
(a(ii) / (a)(iv))
Accept range 4.4 to 4.1
1
(vi)
CE from (a)(v)
Rparallel = 1.35 Ω
second mark for adding internal resistance
Rtotal = 1.35 + 1.5
OR
R = 12/4.2
R= 2.85 Ω
= 2.85 Ω
2
(b)
(i)
resistor
Rate of energy dissipation (W)
1.5 Ω internal resistance
4.2 2 × 1.5 = 26.5
2.0 Ω
2.85 2 × 2.0 = 16.2 (15.68 − 16.82)
R
1.352 × 4.2 = 7.7 (7.1 − 8.2)
CE from answers in (a) but not for first value
2.0: a(iii)2×2
R: a(iv)2×a(v)
3
Page 142 of 165
(ii)
energy provided by cell per second = 12 × 4.2 = 50.4 (W)
energy dissipated in resistors per second = 26.5 + 16.2 + 7.7 = 50.4
(hence energy input per second equals energy output)
if not equal can score second mark if an appropriate comment
2
[12]
58
(a)
power increases to a maximum / ( up) to 3.0 (2.8 -3.4) Ω / / (up)to 3.0 W ✓
then decreases ✓
2
(b)
(i)
(use of P = I 2R)
when R = 0.8 Ω power = 1.95 W ✓
1.9 = I 2 × 0.8 ✓
I = √2.375 = 1.5(4) (A) ✓
Range
1.9 - 2.0 W for power (first mark)
Current 1.5 – 1.6 A
3
(ii)
(use of V = IR)
V = 1.54 × 0.8 ✓
V = 1.2 V ✓
CE from part (i)
2
(iii)
(use of ε = V + Ir)
6.0 = 1.2 + 1.54 × r ✓
r = (6.0 – 1.2) / 1.54 = 3.1 (2.9 – 3.2)( Ω) ✓
use of maximum power theorem (quoted) as alternative method can get both
marks i.e. read peak maximum from graph
CE from part (ii)
2
(c)
power would decrease (as R increased) ✓
pd / voltage across R is now constant / equal to emf ✓
and so power proportional to 1 / R / inversely proportional to R OR
can quote P = V2 / R but only if scored second mark ✓
3
[12]
59
(a)
I3 = I1 + I2 ✓
1
(b)
10 V ✓
1
Page 143 of 165
(c)
I2 = (12 – 10) / 10 ✓
Allow ce for 10 V
1
= 0.2 A ✓
The first mark is for the pd
The second is for the final answer
1
(d)
pd across R2 increases
As R1 increases, pd across R1 increases as pd = I1 R1 ✓
First mark is for identifying that pd across R1 increases (from zero).
1
pd across R3 = 10 V – pd across R1
Therefore pd across R3 decreases ✓
Second mark is for identifying that pd across R3 must decrease
1
pd across R2 = 12 – pd across R3
Therefore pd across R2 increases ✓
Third mark is for identifying that this means pd across R2 must
increase
1
[7]
60
D
[1]
Page 144 of 165
Examiner reports
1
(a)
(i)
Correctly answered by almost all students.
(ii)
As usual in this question a small proportion of students failed to accurately plot both
points and an even greater proportion were unable to draw an acceptable line of best
fit.
(iii)
The most common error by weaker students was misreading data from the graph.
Most students were able to calculate a gradient value within the allowed range.
(iv)
A large proportion of students correctly calculated the value of R. A small proportion
of students lost credit by failing to quote the unit.
As anticipated this proved to be a demanding question, with only the very best students
achieving full marks on all four parts.
(b)
2
(a)
(b)
3
(1)
Most students successfully calculated the value of R from the data in the table.
(2)
Only the more able students successfully calculated the percentage uncertainty in R.
The most common mistakes were failure to double the percentage error in V, and
correctly add this to the percentage uncertainty in P. Credit was also lost by the
significant figure penalty on the final answer.
(3)
A relatively easy calculation but made more demanding by the requirements for
correct units and significant figures.
(4)
This was the most demanding part of the question, and required students to use the
uncertainty value calculated to work out the possible range of values of R to decide
whether the two values were compatible.
(i)
Students had to make it clear that the voltmeter ‘alone’ should be connected across
the cell.
(ii)
A correct explanation was given by a large proportion of students.
(i)
Answered well by the more able students.
(ii)
A proportion of students seemed to understand how to use the voltmeter but failed to
show the correct position on the circuit diagram.
(c)
This question discriminated well. Many students failed to give sufficient detail as required
by the mark scheme for the first marking point. The second marking point proved to be
more accessible, with a greater proportion of students able to suggest an appropriate
precaution.
(d)
As anticipated this proved to be very demanding, with only the more able students
successfully stating and explaining why efficiency would increase as external resistance
increases.
(a)
Fewer than half of the candidates were able to fully define electrical resistance.
(b)
This was answered correctly by most. Those who did go wrong usually demonstrated a
weakness in mathematics rather than in physics.
Page 145 of 165
(c)
4
5
This part was generally not answered well.
(i)
Only the more able candidates were able to correctly answer this part.
(ii)
Even with errors carried forward from part (i), only a few more candidates gained full
marks for this question.
A question which discriminated well between the good and the very good candidates, while
many of the less able candidates found it difficult to gain more than one or two marks at best.
(a)
Many candidates were unable to give a good definition of electromotive force. The
examiners had hoped to see statements referring to total energy transfer per unit charge as
it is moved around a circuit. While a high impedance voltmeter can be used to measure the
emf of a supply, on an open circuit this does not define electromotive force.
(b)
A few more candidates were able to give a satisfactory answer here than in part (a).
(c)
(i)
As the emf was given to 3 significant figures in the question, the reading was
expected to also be to 3 significant figures. This proved to be harder than expected.
(ii)
Many candidates simply divided the voltage by the current.
(iii)
Few candidates were able to provide a satisfactory answer to this part. A statement
to the effect that the battery needs to provide more energy to do the extra work was
all that was required, but very many candidates wrote vaguely about decreased
resistance and/or greater voltage.
Answers to this question realised quite high marks. In part (a) the calculation of the total
resistance was invariably correct. Most candidates then realised that the three strips were in
parallel and proceeded accordingly. A significant number of candidates however, proceeded by
assuming that the resistance of the strip was 11 Ω and worked backwards through the
calculation. This method was not acceptable, since it was felt that the candidates had been set a
task to calculate the actual resistance of each strip from the total resistance of the element.
Very few difficulties arose in part (b), apart from several candidates taking the area of cross
section as circular. Many candidates used the approximate value of 11 Ω for the resistance of
each strip instead of 10.8 Ω, which they had obtained in part (a). This was not acceptable and a
penalty was applied.
Page 146 of 165
6
(a)
This was usually completed successfully.
(b)
Again, there were many correct answers. A sizeable proportion of the candidates
calculated the total resistance in the circuit (24Ω) but then did not subtract the filament
resistance.
(c)
This was done poorly. Correct analysis of this simple circuit proved too much for a
significant proportion of the candidates. Although identifying the formula, many could not
identify which components were in parallel and which were in series. Many were
inadequately skilled in the management of the reciprocals.
(d)
Most were able to identify the circuit which had the higher resistance correctly. The next
step was to appreciate that as the voltage is the same in both cases and, as P = V2/R, the
circuit with the higher resistance dissipates lower power. A common error was to choose,
inappropriately, the equation P = 12R and then to assume that the current in both circuits is
the same which leads to the wrong conclusion.
7
Part (a) was, in general, very well done with the mathematical manipulation of three resistors in
parallel posing a problem to only a handful of candidates. There were many attempts however
where the candidates did not seem sure of the expression and had the four resistors in parallel,
or, more frequently, considered the 50 Ω to be in series with 1 / Rt, where Rt was the sum of the
three parallel resistors.
In contrast to part (a), part (b) gave very poor results with some 50% of the candidates failing to
gain the allocated mark. The common error was giving the current as I = 12 / 50 i.e. ignoring the
equivalent resistance of the three parallel resistors. This showed a very poor understanding of
the dc circuit.
8
The characteristic of the filament lamp presented in part (a) was usually acceptable, although
there was a smattering of answers showing the characteristic curving in the opposite way.
Several candidates offered the characteristic of a semiconductor diode and gained no credit.
The explanation for the shape of the characteristic in part (b) was not as complete and logical as
examiners had hoped for. The majority of candidates were aware that the filament was
non-ohmic but could not explain why the characteristic curved the way it did. Most candidates
realised that heat was generated but were vague as to what increased in temperature and what
caused it. Other candidates had the resistance of the filament decreasing with increasing
temperature in order to justify an incorrect curve.
In part (c) credit was usually gained both for the diagram and the account. Some of the diagrams
were very poor and did not show a data logger, but the most significant omission was a variable
resistor (or potentiometer) or variable power supply, in order to vary the voltage across the
filament. The general standard of the diagram and subsequent account showed however that the
majority of candidates were familiar with this type of experiment.
9
Most candidates could successfully analyse the circuit in this question. and the answers were
quite good on the whole. Even though some candidates had problems in calculating the effective
resistance of the parallel combination of resistors they were able to do the rest of the question
effectively.
Page 147 of 165
11
Very few candidates failed to give the correct equation and definition of the symbols in part (a).
Likewise in part (b) (i); calculations on resistivity are well understood and very few incorrect
answers were seen. The usual error occurred in the calculation of the resistance of the wire, but
the incorrect value would be carried forward as a consequential error. However, if the final
answer then turned out to be a ridiculous value, such as 2 0 × 10–6 m, it was not accepted.
Part (b) (ii) gave rise to a few problems when deciding which combination of resistors gave an
output of 1 kW. There were many very good answers, with the candidates showing that the
output from the parallel combination gave the required output. A common error was stating that
in the series combination, since the same current went through each element, then this gave the
required output. Another common misconception was stating that when the resistors were in
series, all, or almost all the voltage would be used up at the first resistor, leaving very little for the
other resistor.
12
Part (a) provided the candidates with a reasonably easy four marks, and very few failed
completely on the calculation. Usual errors such as units and arithmetic errors occurred but, in
general, the candidates knew how to proceed with the calculation.
Part (b) required clear, logical thinking and sadly, the majority of candidates failed to gain the full
three marks. Having been told in the question that the resistance of the sensor decreased with
increasing temperature, many candidates simply wrote that the reading of the voltmeter would
increase. Such a statement, although in itself correct, without any reasoning did not gain marks.
Many candidate realised that the current in the circuit would decrease, but failed to go any
further. The best approach seemed to be using the potential divider equation and candidates who
tackled the question from this angle were usually awarded two or three marks.
13
This question worked well and many candidates gained full marks. The majority of the other
candidates only failed to gain maximum marks because of a unit error or significant figure error.
Disappointingly, many answers were expressed as a fraction. It should be noted that this practice
is not acceptable and the first answer expressed as a fraction was treated as a significant figure
error.
In part (i) the error which occurred most frequently was ignoring the internal resistance of the
battery. The correct answer was 0.19 A, to two significant figures, but many candidates rounded
this down to 0.2 A, which apart from incurring a penalty, also, when carried forward to part (ii),
gave a voltage across the resistors of 12 V. This implied that there was no voltage developed
across the internal resistance of the battery. Although many candidates produced such an
answer no one noted that such a situation was not possible. Many answers to part (ii), when
carrying forward an incorrect value of the current from part (i), gave an answer well in excess of
12 V. Again this did nor seem to worry the candidates.
In part (iii) many candidates made the error of calculating the power dissipated in the total
external resistance instead of in resistor A alone. The unit of power was usually correct as was
the unit of energy in part (iv). Many candidates arrived at the correct answer in part (iv).
Consequential errors were carried forward throughout the whole question. This gave many
candidates the chance to gain some marks even if their initial calculation and subsequent answer
was incorrect.
Page 148 of 165
14
15
Most candidates scored well on this question, although part (a) proved to be the most
troublesome. A considerable number of candidates seemed unfamiliar with the effect of shorting
out, or connecting terminals together and many assumed that doing so would not affect the
effective overall resistance. In part (b) the large majority of candidates realised that two of the
resistors were in parallel and proceeded accordingly to obtain the correct answer. There were
very few errors in calculating the sum of the parallel resistors.
Part (a) was a very straightforward question for 2 marks. The equation in part (i) was usually
correct, which is not surprising since it is given in the data sheet. A surprising number of
candidates failed to rearrange this equation in terms of VR.
Calculating the current in part (b)(i) was also straightforward and the majority of candidates
gained all the marks allocated. The calculation of the number of cells in part (ii) was considered
by the examiners to be reasonably difficult and they were well pleased by the number of
candidates who arrived at the correct answer, through a variety of methods.
16
In part (a), a significant number of candidates obtained a value of 200 Ω or 300 Ω for the resistor
through not using the correct pd across it. In part (b), the same candidates usually proceeded
with an incorrect calculation of power in the diode by using in the expression I2R the resistance
calculated in part (a). Some candidates were not aware of the correct value of the prefix m in mA.
The energy of the photon was usually calculated correctly in part (c), but a small minority wrongly
considered 1 / λ as the frequency or used an incorrect equation. In part (d), most candidates
knew how to proceed and gave a correct calculation. In the final part most candidates gave a
correct assumption made in the previous estimation. Those candidates who were not specific in
stating the assumption were not awarded this mark.
17
The circuit diagrams produced in part (a) were generally disappointing. Few candidates showed
the voltmeter correctly connected to the tapered conducting paper and many had the positive
pole of the battery connected to the broad end of the strip and the negative pole to the narrow
end. The voltmeter was often shown in parallel with the full length of the conducting strip or else
connected between the probe and the narrow end of the conducting strip.
Equally disappointing were the explanations of the non-linearity of voltage with distance along
the strip in part (b). Most candidates realized that the increase of potential was non-linear
because the strip was tapered, but few scored any further marks. Candidates who knew the
correct relationship between resistance, length and cross-sectional area often failed to mention
that the cross-sectional area was the product of the thickness and the width of the strip, although
others did realize that the increase of resistance was due to the decrease of area as well as the
increase of length. Only a few candidates stated that the current was constant or that the pd was
proportional to the resistance.
In the graphical section of part (c) most candidates completed the table satisfactorily and
proceeded to plot an adequate graph. Some candidates failed to score full marks as a result of
careless errors such as omitting the unit of potential from the graph axis or failing to choose
suitable scales. Candidates were usually able to relate the given equation to y = mx + c and
thereby demonstrate that their graph confirmed to the given equation. Both marks were usually
gained in part (iii) although there were some who plotted the potential on the x-axis thereby often
losing the final mark because they failed to realize that the gradient was now equal to 1/(1.44 Vl).
Page 149 of 165
18
This question involved the analysis of a relatively difficult circuit, which included two lamps and
two resistors. The question however, was so structured that the majority of candidates were able
to work through and gain full marks. Others, unfortunately, although making a reasonable
attempt, failed to gain many marks. In part (a), the majority of candidates calculated the correct
value of the currents passing through each lamp.
In part (b), obtaining the correct answers to parts (i) and (ii) depended to a large extent on
realising that the reading on the voltmeter equalled the voltage across lamp X. Many candidates
missed this point, but were still able to gain some marks. In part (ii) the error that was committed
regularly was determining the resistance of lamp Y instead of the resistance of resistor R2. But at
least, more candidates realised that the same current passed through lamp Y and R2. Answers to
parts (iii) and (iv) used the answer to part (a) as a starting point, but many candidates failed to
realise that the current through R1 was the sum of the current through the two lamps.
Considerable guesswork took over at this stage and although most of it was wrong, candidates
could still get a mark for part (v) by using the answers obtained to parts(iii) and (iv).
19
Part (a) was the calculation of the equivalent resistance of a network of resistors consisting of
resistors connected in series and in parallel. The majority of candidates gained full marks on this
section and were not troubled by the calculation. However, it is worth pointing out that since the
final answer of 50 Ω was given in the question, then in order to gain full marks it was necessary
to show that the two equivalent series resistors were being added together.
Part (b) did not prove to be as easy; the problem in (i) was that many candidates gave the total
resistance as 50 Ω rather than 100 Ω. No consequential error for calculating the current was
allowed and frequently no marks were awarded for this section. It was possible in part (b)(ii) to
gain the two marks even if the answer to (i) was incorrect, but very few candidates managed to
gain these marks. The usual error was giving the current in the circuit as 24/20, i.e. ignoring the
second batch of parallel resistors. Again, many candidates, having calculated the total current
correctly, assumed that 2/3 would pass through the 60 Ω resistor, not realising that the greater
the resistor, the lower the current for a given voltage.
20
Part (a) proved to be very accessible and many candidates scored full marks. Most candidates
calculated the resistor pd as 0.8 V and then calculated the resistance, as expected. Other
candidates however, calculated the total circuit resistance, then the diode resistance and
obtained the required resistance by subtraction. In this particular problem some candidates used
an incorrect pd and were not awarded any credit. Many clear and correct answers were seen in
part (ii).
The energy of the photon was calculated correctly in part (b) by many candidates, but some
failed to score because the wavelength was taken as 1 / f or because the energy was taken to be
½QV. The general principle behind the question in part (ii) was understood by most candidates
and many correct answers were seen. A small minority of candidates however, calculated and
used the power supplied to the resistor and not the diode.
Page 150 of 165
21
The question involved straightforward calculations on voltage, resistance and current. In part
(a)(i) it was hoped that candidates would have spotted the correct voltage across each lamp by
inspection. Surprisingly, even those who managed to get the wrong answer in part (i)
nevertheless ignored their answer and proceeded from first principles to obtain the correct
answer to part (ii).
Part (b) involved the same circuit components as in part (a) but connected differently. The
majority of candidates showed that the current from the battery was the value given in the
question. Using this value they then proceeded to argue or calculate the current in each lamp.
Those candidates who merely halved the current value obtained in part (i) without any reasoning
did not gain the mark.
Although the question told the candidates that the current through each lamp was the same in
both circuits it was disappointing to find in part (c) how many candidates tried to argue that the
brightness of the bulbs in the 2nd circuit would be different to that in the first, the main thrust of
their argument being that the voltage across each bulb was different and therefore that the
brightness would be different.
22
Candidates found this question very accessible and many gained full marks. In part (a) the
meaning of emf seems to be reasonably well understood with most candidates opting for the
voltage when no current passed through the circuit. Others defined it correctly in terms of energy
per unit coulomb. There were, unfortunately, many candidates who, apparently, had not
encountered the definition of emf and merely quoted electromagnetic force, or even tried to
define it in terms of a force in the circuit. The calculation of the current in part (ii) was well done
and in part (iii) correct substitution of values into the equation = V + Ir gave r = 0.80 Ω.
Part (b) was involved with calculation of power and energy and although the majority of
candidates obtained the correct answer for the power dissipated in the 2.4 Ω resistor, fewer had
the correct answer for the total power dissipated in the circuit and a disappointing number had
the correct value for the energy wasted the battery. The usual answer to the last part was to give
the energy in the complete circuit. Whether this was due to inaccurate reading of the question or
due to lack of understanding could not be decided.
23
In this example of calculating equivalent resistance, the same resistor network was used twice,
the equivalent resistance being calculated between different terminals. The majority of
candidates had no difficulty with the calculations, but it was worrying to find many answers where
the candidates had attempted a solution, not by calculation, but with phrases such as “electricity
takes the path of least resistance and therefore the effective resistance (in part (b)) is 50 Ω.”
It was surprising to find that a significant number of candidates obtained the correct result in part
(b) but failed on part (a), since part (b) was deemed to be the most difficult of the two.
Considerable arithmetical difficulty was encountered by many candidates with the reciprocal of
the resistance when calculating the resistance of parallel resistors.
Page 151 of 165
24
Candidates are by now well used to questions on resistors in series and in parallel and part (a)
contained no hidden terrors. Invariably, correct answers were gained for both circuits. Examiners
are concerned however at the trend of using an unusual nomenclature for a combination of
resistors in series and in parallel, e.g. the combination of resistors in the second circuit would be
given by candidates in some
centres as RT =
. The fact that they subsequently gave the correct
answer of 60 Ω showed that the candidates had worked out the parallel section first and then
added the series section. Such a system is not to be encouraged since it serves to confuse and if
a wrong answer is given, it does not help the examiner to find out where the error occurred and
then perhaps award a consequential error mark. The other point concerning this section, was
that answers to part (i) were given in many instances as 131/3 Ω or 13.3 Ω (i.e. recurring). These
were treated as significant figure errors. Answers given as fractions are not accepted.
Part (b) was more difficult and many candidates failed to understand the physics of the circuit.
Comparatively few candidates gained full marks. An error which cropped up continually in part (i)
was correctly calculating the resistance of three resistors in parallel (2Ω) but then using 12 V to
calculate the current, not realising that the effect of the other three resistors halved the pd. There
were also many candidates who calculated the correct current from the supply, but split this
equally between the six resistors and not three. In part (ii), although most candidates calculated
the current as 0.33 A and thus correctly concluded that the current was less than that in the part
(i), they failed to capitalise on this and merely said that ‘therefore the heater was less effective’.
In order to gain the final mark it was necessary to mention the heat/power generated by the
current.
25
The response to this question was very disappointing, especially in view of the fact that this topic
has been examined several times previously, including questions on the graphical nature of the
quantities involved. Rearranging the equation in part (a) was intended as a guide to drawing the
graph in part (b). The majority of candidates did rearrange the equation correctly, but some
candidates failed to do this and ended up with a quotient.
Sketching the I-V graph in part (b) was, quite literally, a disaster area. The large majority of
candidates drew a straight line of positive gradient passing through the origin. Obviously this was
the easiest line to draw without applying any thought to the question. No marks were awarded for
such attempts. If a line of positive gradient was drawn, and did not pass through the origin, then 1
mark was awarded. A large number of curved lines, some starting at zero, others at a positive
value of V and decreasing to zero, were also presented. Of the candidates who drew a straight
line with a negative gradient, many lost marks by either extending the line into negative values of
V, or negative values of I. It must be pointed out that when carrying out an experiment to obtain
this graph, it is not possible to obtain zero values of V or I. However, since some textbooks do
show the graph extending to the V axis, this was accepted, but graphs extending to the I axis
were not.
Most candidates gained at least one mark in part (c), but the impression gained was that
candidates had learned the answers parrot fashion with no reference to the graph. The gradient
of a curved graph was often given as the answer.
Page 152 of 165
26
Part (a) provided three reasonably easy marks for the large majority of candidates. The most
common error was forgetting to add the resistance of the thermistor to that of the resistor. Errors
in part (ii) arose from not reading the question properly and many candidates calculated the
potential across the resistor instead of the thermistor.
To obtain full marks in part (b) required a logical answer, starting with a statement that the
resistance of the thermistor decreased with rising temperature. Almost all candidates obtained
credit for this. The next step was realising that this resulted in the total resistance decreasing.
Not many candidates mentioned this, but full marks could still be obtained if they realised that the
current in the circuit increased as the temperature fell. Many candidates got to this point but then
failed to go any further. A common answer was stating, without any justification, that the voltage
across the thermistor decreased because the rate of increase of current was greater than the
rate of decrease of resistance. In order to gain the remaining marks, it was necessary to discuss
the potential across the fixed resistance. As an alternative approach, many candidates
recognised the circuit as a potential divider, gave the potential divider equation and argued
correctly from that point. Part (b) proved to be a good discriminator.
Part (c) also earned high marks. The calculation of power dissipated in the thermistor was
usually correct for both temperatures and very few failed to read the graph correctly. The usual
error was using the same current as in part (a), not realising that the removal of the fixed resistor
increased the current. It was obvious from answers to part (ii) that a significant number of
candidates were not familiar with the term mean value. Several candidates subtracted the two
values before dividing by two, others just added them. Having arrived at the mean power, most
candidates were aware that to change to energy required multiplying by the time, in seconds.
Only the better candidates realised that the answer to part (iii) was that the rate of decrease of
resistance with temperature was not linear, or that the graph was not a straight line.
Page 153 of 165
27
Just over half the candidates scored full marks in part (a)(i). The common error by candidates
who failed to score any marks was to simply take the maximum values of current (0.9 A) and
voltage (0.7 V), rather than read off a pair of corresponding values, for which they would have
gained at least 1 mark.
In part (a)(ii) about one-third of candidates correctly interpreted and sketched an initially linear
portion for the graph, peaking at about 0.6 V and then dropping to zero just past 0.7 V. A further
third of candidates gained 1 mark for a sketch that began at the origin, peaked and returned to 0
W at some voltage.
Nearly 25% of candidates successfully dealt with the 2 calculations required to find the area of
the solar cell in part (a)(iii), but dealing with the efficiency proved to be the biggest difficulty to
many candidates. The common error was to multiply their power output (from (a)(i)) by 0.15.
These candidates would have helped themselves by initially writing the efficiency equation and
then substituting the data.
Most candidates had not learnt the definition of emf sufficiently well to apply it to the context
given in part (b)(i). Very few high-quality answers referring to “moving charge around the circuit”
were seen.
Applying knowledge of the behaviour of emf and resistance in series and parallel circuits in part
(b)(ii) proved challenging for a high majority of the candidates, with only about 10% gaining 3 or
more marks. Many candidates appreciated that a series arrangement of 20 cells would provide
the required emf of 14 V but failed then to consider that the addition of the resistances would be
too great and therefore a parallel arrangement, to lower the internal resistance, was necessary.
Good comprehension skills were rewarded in part (c) which quite clearly asked candidates to
address 3 points regarding the use of solar cells. High marks were obtained by candidates who
did so with clear paragraph structure. Others failed to respond to the question in sufficient detail
and addressed only 1 or 2 of the points, thus limiting their mark. Common, correct answers
pointed out the continuous availability of sun light, the need for an energy storage system, and
the very low intensity of light at the edge of the solar system. Many low-achieving candidates
evaluated the use of solar cells in terms of terrestrial use and addressed limitations of such
resources as HEP and wind power.
Page 154 of 165
28
The majority of candidates seemed to approach this question with confidence and set out their
working well. Many did not appreciate the effect of connecting the two identical cells in parallel
and it was quite common to see them using the combination of parallel resistors formula to
combine the emfs of the two cells. This was something that was not confined to the less able
candidates but was seen across the full ability range. This was not a heavy penalty as
subsequent answers received full credit whatever value candidates had deduced for the total
emf.
Part (a) (iv) assessed the unit for charge and the majority of candidates had no problems with
this.
The deduction required for part (b) proved quite discriminating and only the very best candidates
obtained all three marks. The first mark for identifying cells C and D proved quite straightforward
but the explanation less so. Many candidates appreciated that the greater current in the cells in
series was significant but were unable to take this to the next step and link this with the rate of
energy dissipation.
29
All candidates were able to gain a reasonable number of marks for this question, and many were
awarded full marks. The answer to part (a) (i) was usually correct, but there were some problems
with part (a) (ii). The most common error was obtaining the correct value of the current in the
circuit, as required, but then dividing this value by the number of lamps. The resistance of each
lamp, in part (iii), was usually calculated correctly, although some candidates made heavy
weather of the calculation when using the expression VIR for power. Several recurring errors in
part (iv) resulted in this part not being answered as well as the others. These errors would be
calculating the energy used by one lamp instead of by the set, or omitting the factor of 2 (hours)
when calculating the number of seconds, or omitting the conversion from minutes into seconds.
Incorrect answers in part (a) were allowed to be carried forward into part (b), which resulted in
part (b) performing quite well. The usual error in part (i) was carrying out the calculations for 10
lamps, instead of 56. Answers to part (ii) were usually correct, candidates realising that the
greater the current, the greater the brightness.
30
This is the first time in these series of examinations that candidates have been required to draw
their own arrangement of resistors. The majority of candidates gave the correct answers in part
(a), although some did try an arrangement of resistors similar to that in part (b). There were a few
incorrect calculations in part (a) (ii) even though the three resistors were in series. The usual
error in part (iv) was calculating correctly the value of \IR but then forgetting to invert to obtain R.
In part (b) the calculation for the total resistance was usually correct although there was some
concern amongst the examiners to see the expression RT =
+ 4 occurring quite frequently.
Page 155 of 165
Invariably this resulted in the wrong answer, because candidates would not invert the value for
the parallel resistors. The occurrence of this ‘system’ of calculating resistance was brought to the
attention of teachers in the last report, but it seems to be more common than before. Part (ii) was
not answered well, with candidates just writing numbers down without any reasoning and in the
end confusing themselves. Candidates who just gave an answer of 4 V with no working shown
were not credited, because it was possible to obtain that answer by incorrect physics.
Candidates should be trained to give some explanation of what they are attempting in such
calculations. It was also sad to see candidates obtaining the (correct) answer of 4.0 V across the
parallel resistors, but then shooting themselves in the foot by assuming that the voltage across
the 6.0 Ω was different to that across the 3.0 Ω.
31
Reading the values of the two resistances from the graph in part (a) did not prove to be difficult
although many candidates lost the mark by not looking at the graph carefully and also by drawing
a thick pencil line across, so that it covered the required point. This was especially true for the
resistance of the thermistor.
The calculation in part (b) (i) was very accessible and the majority of candidates obtained the
correct value for the voltage across AB. In part (b) (ii), many candidates gained the mark by
simply writing down 50 °C. Others would fill up the available three lines with voltage calculations
before arriving at the correct answer.
The calculation in part (c) (i) was usually correct, but answers to part (c) (ii) frequently failed to be
awarded full marks. Candidates find this type of question, where the answer requires a logical
sequence of steps, difficult. Usually, candidates would note that the lamp and thermistor were
connected in parallel, thus reducing the resistance across AB. The next step of stating that the
total resistance in the circuit would also decrease was usually ignored. The final steps of relating
the decrease in resistance to a decrease in voltage across the thermistor was rarely done
correctly. Candidates who attempted the answer in terms of the potential divider equation, or in
terms of the ratio of resistances usually fared better.
32
The majority of candidates found this question straightforward and gained the maximum number
of marks. Others, however, were not sure of the effect on the circuit of having the switch open or
closed. A considerable number of candidates reversed the calculations for parts (a) (i) and (ii).
Several candidates, in the situation when the switch was closed, i.e. effectively shorting out the
60Ω, resorted to adding up the two resistances using the expression for parallel resistors.
In part (b) the majority of candidates realized that a voltmeter of infinite resistance had the same
effect on the circuit as an open switch and proceeded accordingly.
Page 156 of 165
34
This question proved to be very accessible and full marks were gained frequently. It is worth,
however, pointing out a few recurring errors. In part (a), because the answer was given, it was
expected that candidates would show full working. Very often the final expression would contain
10Q, being added in series to the product of the parallel
arrangement, the parallel resistance still being in the mathematical form
etc.
Because it was so easy to deduce that the parallel section was equivalent to 10 Ω, examiners did
expect to see the value being worked out from the basic expression, otherwise a mark would be
deducted.
In part (b) the first two calculations were usually correct, but the third part often produced wrong
answers, 1.25 A appearing quite frequently. Part (c) produced many correct answers, with
candidates being quite familiar with calculations for power and being able to calculate the
percentage.
35
Questions requiring a description of an experiment invariably perform well and part (a) was no
exception. Many candidates produced clear, concise and logical answers. This could be partly
due to the fact that measuring the resistivity of a wire is a popular experiment for coursework. In
the circuit diagrams submitted, the only point that attention needs to be drawn to is when
candidates use a potentiometer rather than a variable resistance. It is important that the ammeter
should be placed in the correct position to measure the current through the wire, and not
measure the current through the battery. Many of the circuit diagrams did not include the
resistance wire, the candidates seemingly being confused between the resistance wire and the
variable resistance. Unfortunately, candidates who produced such a circuit invariably penalised
themselves not only in part (i) but also in part (ii).
When listing, in part (ii), the measurements which needed to be taken, a mark was usually lost by
candidates who stated that the area of cross section, rather than the diameter, had to measured.
Several candidates referred to the diameter as the thickness of the wire, a term which was not
acceptable. The other point which was missed quite frequently, was the need to make a series of
measurements of voltage and current by altering the variable resistor. Another point which needs
to be stressed is that many candidates stated that the length of the wire was changed,
contradicting the evidence of their own circuit diagram, which showed a fixed length resistance
wire.
In part (iii), the majority of candidates gave the correct equation relating resistivity to the other
parameters and stated how to calculate A from the diameter and how to calculate the resistance
R of the wire. Most candidates used a graph to obtain R, with those who used a variable length
using a graph of length against R, calculating ρ from the gradient.
Very few candidates failed to obtain the correct value for the length of wire in part (b).
Page 157 of 165
36
The circuit diagrams drawn by candidates in part (a) (i) were generally not done well. Many did
not include a means of varying the potential difference across the diode and the inclusion of a
load resistor was rare. Less able candidates also confused the positioning of the voltmeter. There
were very few occasions where a potential divider was used even though this is best practice for
obtaining the full characteristics for the diode.
The descriptions of experimental procedure required for part (a) (ii) were generally thorough but
some did suffer from a poor structure and this had an impact on the assessment of the Quality of
Written Communication. Many candidates did not mention anything about reverse characteristics
and it was noticeable that a significant minority did not appreciate that it was important to obtain
readings with a potential difference of less than 1.0 V.
The calculation in part (b) (i) was done well and full marks were the norm. Part (b) (ii) proved to
be not so straightforward and it was common to see candidates divide the potential difference
across the diode by the resistance of the resistor. This proved to be one of the most
discriminating questions on the paper.
38
Part (a) was answered well, with many candidates obtaining full marks.
Part (b) caused more problems and the use of the power formula that involves potential
difference and resistance was quite rare. In part (b) (ii) there was some confusion over potential
difference and candidates frequently used their answer from part (b) (i). Part (b) (iii) was
answered much better, with candidates frequently benefiting from consequential error.
39
Many were unable to identify which resistors were in the circuit for different switch settings and
then calculate the total resistance in part (a). The 2R value was most easily identified but only
about 20% of the candidates obtained credit for the other switch settings. Some could correctly
write down the equations for calculating resistors in parallel but then got no further.
Part (b) was not done well. Few appreciated that they need to use V2/R and not I2R because
potential difference was the common factor for each setting and not current.
Part (c) exposed many misconceptions about charge flow through conducting material. Many
thought thermal energy to be produced by electrons bumping into each other or that they excite
atoms which then release the energy as heat. Some thought the nuclei of atoms in the
conducting material to be involved.
Page 158 of 165
In part (d) a good proportion of the candidates appreciated that the thermal energy output would
be decreased and, although there were some very good answers, many went on to give a partial
explanation of why this would happen in terms of the reduced pd across the load or a reduced
load current.
There were many correct answers to part (e) (i), but it was disappointing that a quarter of the
candidates were unable to make any progress with the calculation.
A majority could give at least one sensible property in part (e) (ii), but many candidates had
clearly confused the rod with the resistance wire so gave properties that were ‘opposite’ to those
needed or irrelevant properties. Some gave contradictory properties such as ‘it must have a low
resistivity and be a good conductor’.
40
Almost all of the candidates correctly calculated the power of a headlight lamp. A very few
extracted an incorrect value of resistance from the table. In part (a)(ii), candidates made the
usual errors of using the resistors in series formula instead of resistors in parallel. Others omitted
to invert their answer once they had calculated the sum of the reciprocals of the five resistors. In
part (a)(iii), some candidates revisited their part (a)(i) type calculations to find the powers of all of
the lamps, rather than using the overall resistance value that they had just calculated.
In part (b), most candidates realised that, in this circumstance, lamps of lower resistance ought to
be used. Candidates’ justifications were not always convincing. The best answers referred to the
equation relating power to voltage and resistance. Some argued that lower resistances should be
used so that the current would increase, this was acceptable. What was not acceptable was the
argument that it was necessary to restore the current to its previous value.
41
Candidates’ performance in this question was generally poor and it appears that the effect of
internal resistance on terminal pd is not well understood. While many came up with an
acceptable definition of emf few were able to explain convincingly the effect on the voltmeter if
the switch is closed. A significant proportion of candidates assumed a current was flowing when
the switch was open and it was quite common to see statements such as ‘ when the switch is
closed voltage stops flowing through the voltmeter and so its reading decreases’, which supports
the view that potential difference in circuits is a concept that many candidates struggle with.
Further evidence of this was provided by the explanations given in part (c).
Many candidates did not seem to appreciate the reason why a car battery needs to have a low
internal resistance.
Page 159 of 165
42
Part (a) was very poorly answered by most candidates and highlighted many misconceptions
about the operation of electrical circuits. The relevance of the high resistance of the voltmeter
was often appreciated but relatively few could go on to explain why. Some candidates made
some progress with an approach using either the potential divider formula or E = IR + Ir.
Part (b) (i) was done well but one significant figure answers were penalised. Some from their
working seemed to have made a positive decision to convert 0.309 to 0.3 A as their answer and
considered this to be a two significant figure response.
The correct working was rewarded in part (b) (ii) although using the approach via E = IR + Ir and
a current of 0.3 A yielded an internal resistance of 30 Ω (rather that ≈ 16 Ω) and this would give a
terminal pd that significantly different from the 14.5 V given. The obvious approach was from lost
volts = 0.5 V and then r = 0.5/0.31. Many spoilt their attempt using E = IR + Ir by associating E
with the terminal pd.
43
This question proved to be very discriminating with only the more able candidates able to score
high marks. The calculations involved in part (a) proved too challenging for many candidates.
Part (a) (i) and (ii) generated the most correct responses, but the remainder of the analysis was
only accessible to the more able candidates.
Part (b) required analysis without calculation and the majority of explanations seen were
confused and not self consistent. Many candidates stated that more current goes through the
thermistor and therefore the pd across it falls, resulting in the pd across the parallel 1200 Ω
resistor increasing. Another common misunderstanding was the effect that the decreasing
thermistor resistance had on the current through the battery. Many thought that the current
remained constant and, although this still led them to deduce that the pd fell, their arguments
frequently contained contradictions.
Page 160 of 165
44
Students fared better in the circuit analysis involved in this question than they did in question 6.
Parts (a) (i), (ii) and (iii) were answered well with a significant proportion of students able to
correctly find the total circuit resistance. The calculation of the parallel network was done
correctly by the majority of students, although the working shown by many was sometimes not
set out properly with the reciprocal of total resistance being equated to the total resistance. This
was in part due to the combined resistance being equal to 1 Ω.
Part (a) (iv), in which students had to calculate the energy transformed by the battery in
5.0minutes, was not answered as well. A significant proportion of students did not appreciate that
this was found by multiplying the emf of the battery by the appropriate time. Part (a) (v) caused
students even more problems and only a minority of the more able students were able to
correctly calculate the energy dissipated in the internal resistance of the battery.
The final part of this question was well answered with most students giving sensible suggestions.
However, one out of two marks was quite common due to students mixing up an explanation with
a reason; an example being ‘has a higher terminal pd’ and ‘provides large current’.
45
This question proved to be very discriminating with only the high performing candidates able to
score high marks. The calculations involved in part (a) proved to be straightforward and the
majority of candidates realised that this was 5.0 Ω. Part (a)(ii) caused more problems and there
were many answers in which the calculation of the resistance of the parallel component was
spoilt by poor setting out – equating ½ to 2 Ω was a common occurrence.
Part (b) required candidates to calculate currents in the parallel branches of the circuit. Many
tried to do this by ratio and got the currents the wrong way round, ie quoting a value of 0.67 A
instead of the correct 1.3 A. A more successful approach, used by more able candidates,
involved the calculation the pd across the series resistor and hence the deduction of the pd
across Y. Once this was known the current in Y could be correctly calculated. This approach also
enabled candidates to give the correct pd across W because they realised it was half the value of
the pd they had already calculated for Y.
46
Part (a) of this question generated some of the poorest responses in the paper with over three
quarters of the candidates obtaining no marks. The evidence suggests that candidates find the
concept of internal resistance and its relationship with terminal pd quite challenging and were
unable to convincingly explain what was happening in this circuit as the current increased. A
significant number of candidates assumed that the terminal pd decreased because the internal
resistance was increasing due to an increase in temperature of the cell. There was also a lack of
precision in answers making it hard to determine which resistance was being referred to in many
explanations.
Page 161 of 165
In part (b), the majority were able to find the emf of the cell correctly but the determination of
internal resistance, r, proved to be much more discriminating. The more able candidates
appreciated that the gradient of the graph was equal in magnitude to r and those who did, for the
most part, produced acceptable answers. Alternative solutions using the equation, ε = IR + Ir,
were less successful as there were often careless mistakes made when this approach was used
– an example being the calculation of R using terminal pd and current and then using this value
with a different value of current to find r.
Part (c) required candidates to add lines to the existing graph. Most appreciated that these two
lines had the same intercept as the original line and a significant number realised that the line for
the cell with double the internal resistance would have a gradient double in magnitude. The line
for the cell with zero internal resistance caused more problems and less than half the candidates
drew horizontal lines.
Part (d) produced some mixed responses. The calculation of charge was successfully answered
by the majority and the unit for charge is clearly well known. However, the calculation of energy
dissipated in the internal resistance per second caused far more problems and over half the
candidates did not score any marks in this section – many applied the wrong equation and this
resulted in them either multiplying the terminal pd or the emf of the cell by the given current.
47
The calculation in part (a) was well done by the two–thirds of the students. However, some of
these students lost a mark for either an incorrect unit of resistivity or rounding data in the middle
of the calculation and thereby obtaining an answer outside the range on the mark scheme.
Some students were unable to correctly determine the resistance of the bulb. A common mistake
was the attempted use of V= I R without P =VI
Few students were able to define emf in sufficient detail. Most students had only a vague idea
about the meaning of this term. It was common for students to refer to it as a force or to describe
it as a rate of flow of charge.
Part (b) (ii) was well done by about one quarter of the students. Most students had a very limited
understanding of the situation and at best attempted to use V=IR to find a resistance.
The lack of quality in the descriptive answer in part (c) underlined the students’ lack of
confidence in this area of the specification. Very few students appreciated that the effect was due
to changes in the terminal pd of the battery. Most students adopted an argument based on the
path of least resistance without describing why there was a reduced current in the normal
headlamp. A significant number of students stated that the normal headlamp would be
unaffected by the fault due to it being in parallel with the faulty headlamp.
Page 162 of 165
48
49
50
The quantitative parts of this question were well answered but as is often case, students found
the qualitative aspect the much more challenging. The calculations of current and resistance
caused few problems and the majority of students were able to explain the effect of an
appreciable internal resistance. Part (c) caused far more problems and a significant proportion
could not convincingly explain why the lamps were not at normal brightness when connected in
series. They seemed not to appreciate that the voltage of the 12 V battery was divided between
the lamps or that the circuit resistance is higher when the lamps are in series. They also found it
very difficult to explain which lamp was brighter – many incorrectly assuming that it was lamp P
as it had a higher power rating.
The majority of students were able to analyse the circuit correctly although surprisingly a
significant minority had problems with (a)(i) because they did not appreciate that the pd across
R2 was 4.0 V. This did not affect their subsequent responses however, as the answer they gave
was carried forward to subsequent calculations. The qualitative aspect of the question presented
students with a greater challenge. Many incorrectly stated that the voltmeter reading would
decrease as the thermistor resistance falls seemingly forgetting that the voltmeter was connected
across R1.
With the exception of part (a), students found this question particularly challenging.
The calculations in part (b) were very structured but this did not seem make the analysis of the
circuit straightforward. In part (b) (i), less than half the students were able to calculate the pd
across the resistor correctly with many not appreciating that the pd across the a parallel network
was the same as the pd across lamp X. Part (b) (ii) produced better answers, although a
significant proportion of students did not appreciate that they simply needed to add together the
two currents calculated in part (a).
Part (b) (iii) was answered well, although many students benefited from being allowed to use
incorrect answers from parts (b) (i) and (ii). The remainder of the circuit analysis did cause
problems due to many students not realising that the pd across R2 was simply the difference in
the pd’s across the lamps or that the current through R 2 was the same as the current in lamp Y.
Part (c) required students to consider the effect of lamp X ceasing to conduct. In part (c) (i) they
had to explain the effect on the voltmeter reading. This was not answered well with a significant
proportion of students thinking the voltmeter reading would increase. This was mainly due to the
mistaken assumption that the current in the circuit would increase. Part (c) (ii) generated more
correct responses because many students stated that the current through lamp Y would
increase, although it was clear from their answers many thought that this was due to the current
from lamp X now going through lamp Y. It was not commonly appreciated that although the
overall current in the circuit had decreased, the current through R2 and lamp Y was higher than it
had been when lamp X was conducting.
Page 163 of 165
51
The first part of this question involved the use of the resistivity formula and many were able to do
this successfully. In the vast majority of cases they were also able to calculate the current flowing
in the lamp using the power formula.
Parts (b)(ii) and (iii) were answered less successfully and only about half of the candidates
appreciated that the pd across the wires was found by multiplying their answer in part (a) by the
answer in part (b)(i). In part (b)(iii) candidates were required to calculate the emf of the supply
and this proved to be quite a challenge with only about 23% scripts obtaining full marks. Many
answers gave values of less than 12 V.
Part (c) required a knowledge of the effects of internal resistance and this is a topic that has
caused problems in the past. This time however, fewer confused answers were seen and full
marks were relatively common.
52
This question on a potential divider circuit was a mixture of qualitative and quantitative. As is
often the case with questions involving electric circuits, candidates coped better with quantitative
parts. This was particularly true in part (a) where the calculation involved more than one stage.
Part (b) was not well done and only the strongest candidates manage to relate the changing light
intensity to the voltmeter reading. A significant proportion of candidates were under the
impression that increasing the light intensity increases the ldr resistance.
Part (c) did involve a calculation but this was much more challenging than part (a) because there
were no intermediate stages. Only a third of candidates were able to calculate a correct value for
the resistance of the variable resistor. The majority of those who were successful calculated the
value using a ratio method rather than calculating the current and then using this value with the
correct pd to find the resistance.
54
In this question candidates were required to analyse a bridge network of resistors. The
calculation of the circuit resistance in part (a) proved to be reasonably straightforward with over
two thirds of candidates scoring full marks. The only common error in weaker scripts was the
combining of all the resistors as parallel resistors instead of combining the series branches first.
The calculation of current in part (a) (ii) was done well and with consequential error applied, the
majority of candidates were able to do this successfully.
Part (b) was not well answered and very few candidates were able to give correct answers for
the voltmeter reading in the three positions. The position that proved the most challenging was
the pd between C and D and it is clear that many candidates did not appreciate that this was
found by subtracting the pd across D and F from the pd across C and E.
Part (c) was a qualitative question and previous papers suggest that candidates find these
difficult. Only the very best candidates managed to get full marks in this section and it was the
explanations of the effect on the voltmeter that proved to be the most challenging. For example
over 60% of candidates appreciated that the pd across the thermistor decreased but only about
14% managed to explain why. A common mistake was to try and use current in explanations and
this led them to conclude incorrectly that if current goes up then so does pd or that the increase
in current cancels out the decrease in resistance. Very few used the constant 12 V across the
parallel branches to justify their conclusions.
Page 164 of 165
55
57
The majority of candidates correctly calculated the voltage in (a). Many used the same current as
in (a) to do the calculation in (b); dividing the difference in the voltage across the resistor
[between (a) and (b) = 0.2 V] by the current in (a) to give 0.8 Ω was a very common incorrect
answer. Only a few candidates were able to perform the complete calculation to obtain a
resistance of 24 Ω.
Part (a) was highly structured and led candidates through a full circuit calculation in stages. This
approach appeared to have helped them and more successful solutions were seen than has
been the case in the past with this type of circuit.
The part that caused the most problems was (a) (ii) with a significant proportion of candidates not
appreciating that the pd across the 2.0 Ω resistor was the same as that across resistor R.
Candidates were however, not penalized when they carried their incorrect answer to subsequent
parts and consequently the remaining calculations were often carried out successfully.
Part (b) proved to be much more demanding and only about half the candidates managed to
complete the table for the rate of energy dissipation successfully.
The demonstration of energy conservation in part (b) (ii) provided an even greater challenge and
only about a third of candidates provided a convincing analysis of energy conservation in the
circuit. A fifth of candidates made no attempt at this part of the question.
58
Candidates often find circuit analysis questions challenging if the power supply in the circuit has
an internal resistance. This certainly proved to be the case in this exam.
Most candidates were able to interpret the graph in part (a) but when it came to the calculations
in part (b) (i), only about half of the candidates appreciated that the pd across resistor R was not
6.0 V. This led them to calculate an incorrect value for current. They were allowed consequential
error however, and this meant that higher marks were seen in parts (b) (ii) and (b) (iii).
Part (c) was answered very badly with only about 6% of candidates obtaining full marks and
nearly 70% getting zero. The commonest mistake was the assumption that the new graph would
have the same overall shape as the one shown in figure 2. Very few candidates seemed to
appreciate that with negligible internal resistance, power would be inversely proportional to
resistance.
Page 165 of 165
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