Download AS Electricity Part I

AS Electricity Part I
342 minutes
332 marks
(a) On the axes below draw the I – V characteristic for a silicon semiconductor diode in
both forward bias and reverse bias. Indicate any relevant voltage values on the axis.
Q1.
(4)
(b)
The figure below shows the I – V characteristic for a filament lamp. Explain the shape of
the characteristic.
You may be awarded additional marks to those shown in brackets for the quality of written
communication in your answer.
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(4)
(Total 8 marks)
Q2.
An oscilloscope is connected to an alternating voltage source of rms value 4.2 V at a
frequency of 2.5 kHz.
(a)
Calculate the peak-to-peak alternating voltage.
peak-to-peak voltage = ....................................
(2)
(b)
Figure 1 represents the screen of the oscilloscope.
Figure 1
Determine
(i)
the voltage sensitivity of the oscilloscope,
voltage sensitivity = ....................................
(ii)
the time base setting of the oscilloscope.
time base setting = .................................
(3)
(c)
The time base of the oscilloscope is switched off and the voltage sensitivity is set to
0.5 V div–1. The oscilloscope is connected across a 1.75 V battery of internal resistance
3.5 Ω which is connected to a 10 Ω resistor as shown in Figure 2. Figure 3 represents
the screen of the oscilloscope which shows the spot when registering zero volts.
Figure 2
Figure 3
(i)
Draw a spot on Figure 3 showing the appearance on the screen when the switch is
open. Label this spot O.
(ii)
When the switch is closed determine the current flowing through the 10 Ω resistor.
current = ....................................
(iii)
Draw a spot on Figure 3 showing the appearance on the screen when the switch is
closed. Label this spot C.
(5)
(Total 10 marks)
Q3.
A heating unit as used on the rear window of a car consists of five strips of resistive material
joined at either end by strips of copper of negligible resistance, shown in the diagram below.
Heat is generated at a rate of 45 W when the unit is connected to a 12 V car battery.
(a)
(i)
Calculate the total resistance of the unit.
total resistance = ..........................
(ii)
Show that the resistance of each strip is about 16 Ω.
(4)
(b)
If each resistive strip is 2.5 mm wide and of length 0.80 m, determine the thickness of
each strip.
Resistivity of the resistive material = 5.0 × 10–5 Ωm.
thickness = ....................................
(3)
(c)
The rear window heater on a car is sometimes set to be switched off automatically after a
period of time to prevent draining the battery.If the charge stored in a fully charged battery
is 1.44 × 105 C, how long, in hours, would it take to fully drain the battery with the heater?
time = ...........................hours
(3)
(Total 10 marks)
Q4.
(a)
(i)
On the axes in Figure 1 draw the I – V characteristic for a silicon semiconductor
diode, giving any relevant voltage values.
Figure 1
(ii)
A very sensitive ammeter can be protected using a silicon semiconductor diode as
shown in Figure 2. Use the characteristic in Figure 1 to explain how the ammeter is
protected when it is connected in a circuit carrying too much current for it to
measure.
Figure 2
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(6)
(b)
Determine the current flowing through the ammeter in Figure 3.
Figure 3
current = ....................................
(5)
(Total 11 marks)
A car battery has an emf of 12 V and an internal resistance of 5.0 × 10–3 Ω.
Q5.
(a)
(i)
Explain what is meant by the emf of the battery.
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(1)
(ii)
Explain what is meant by the internal resistance of the battery.
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(1)
(b)
The battery is used to provide the starting motor of a car with a current of 800 A.
(i)
Calculate the potential difference across the terminals of the battery.
answer = ................................................... V
(2)
(ii)
Calculate the rate of dissipation of energy due to its internal resistance stating an
appropriate unit.
answer = ......................................................
(3)
(c)
State and explain the effect of attempting to use a battery with a much higher internal
resistance to start the car.
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(2)
(Total 9 marks)
Q6.
The diagram below shows an arrangement of resistors.
(a)
Calculate the total resistance between terminals A and B.
answer = ................................................... Ω
(2)
(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)
Q7.
(a)
Some materials exhibit the property of superconductivity under certain conditions.
•
State what is meant by superconductivity.
•
Explain the required conditions for the material to become superconducting.
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(3)
(b)
The diagram below shows the cross–section of a cable consisting of parallel filaments that
can be made superconducting, embedded in a cylinder of copper.
(i)
The cross–sectional area of the copper in the cable is 2.28 × 10–7 m2. The resistance
of the copper in a 1.0 m length of the cable is 0.075 Ω. Calculate the resistivity of the
copper, stating an appropriate unit.
answer = ......................................................
(3)
(ii)
State and explain what happens to the resistance of the cable when the embedded
filaments of wire are made superconducting.
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(3)
(Total 9 marks)
Q8.
A car battery has an emf of 12 V and an internal resistance of 9.5 × 10–3 Ω. When the
battery is used to start a car the current through the battery is 420 A.
(a)
Calculate the voltage across the terminals of the battery, when the current through the
battery is 420 A.
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answer ........................................... V
(2)
(b)
The copper cable connecting the starter motor to the battery has a length of 0.75 m and
cross-sectional area of 7.9 × 10–5 m2. The resistance of the cable is 1.6 × 10–3 Ω.
Calculate the resistivity of the copper giving an appropriate unit.
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answer .................................................
(3)
(Total 5 marks)
Q9.
The diagram below shows an ac waveform that is displayed on an oscilloscope screen.
The time base of the oscilloscope is set at 1.5 ms per division and the y-gain at 1.5 V per
division.
(a)
For the ac waveform shown,
(i)
Calculate the frequency
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answer ............................................ Hz
(3)
(ii)
Calculate the peak voltage
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answer ........................................... V
(2)
(iii)
the rms voltage
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answer ............................................ V
(2)
(b)
State and explain the effect on the oscilloscope trace if the time base is switched off.
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(2)
(Total 9 marks)
Q10.
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)
(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.
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(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
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answer .................................. A
(2)
(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.
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answer ................................... A
(3)
(Total 14 marks)
Q11.
An alternating current (ac) source is connected to a resistor to form a complete circuit. The
trace obtained on an oscilloscope connected across the resistor is shown in the diagram below.
The oscilloscope settings are: Y gain 5.0 V per division
base 2.0 ms per division.
(i)
time
Calculate the peak voltage of the ac source.
answer = ....................................... V
(1)
(ii)
Calculate the rms voltage.
answer = ....................................... V
(1)
(iii)
Calculate the time period of the ac signal.
answer = ..................................... ms
(1)
(iv)
Calculate the frequency of the ac signal.
answer = ...................................... Hz
(2)
(Total 5 marks)
Q12.
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)
(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.
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(3)
(Total 9 marks)
Q13.(a)
A sample of conducting putty is rolled into a cylinder which is 6.0 × 10–2 m long and has a
radius of 1.2 × 10–2 m.
resistivity of the putty = 4.0 × 10–3 Ωm.
Calculate the resistance between the ends of the cylinder of conducting putty.Your answer
should be given to an appropriate number of significant figures.
answer = ...................................... Ω
(4)
(b)
Given the original cylinder of the conducting putty described in part (a), describe how you
would use a voltmeter, ammeter and other standard laboratory equipment to determine a
value for the resistivity of the putty.
Your description should include
•
•
•
•
a labelled circuit diagram,
details of the measurements you would make,
an account of how you would use your measurements to determine the result,
details of how to improve the precision of your measurements.
The quality of your written communication will be assessed in this question.
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(8)
(Total 12 marks)
Q14.
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.
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(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.
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(2)
(b)
Calculate the internal resistance of the battery.
answer = ..................................... Ω
(3)
(c)
State and explain why it is important for car batteries to have a very low internal
resistance.
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(2)
(Total 8 marks)
Q15.
(a) A semiconducting diode is an example of a non-ohmic component. State what is
meant by a non-ohmic component.
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(1)
(b)
A filament lamp is also an example of a non-ohmic component.
(i)
Sketch on the axes below the current-voltage characteristic for a filament lamp.
(2)
(ii)
State, with reference to the current-voltage characteristic you have drawn, how the
resistance of the lamp changes as the pd across its terminals changes.
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(1)
(c)
A filament lamp has a power rating of 36 W when there is a pd across its terminals of 12V.
(i)
Calculate the resistance of the filament when the pd across its terminals is 12V.
answer = ..................................... Ω
(2)
(ii)
A student predicts that if the pd across the bulb is reduced to 6.0 V the power rating
of the bulb would be 9.0 W. State and explain how in practice the power rating will
be slightly different from this value.
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(3)
(Total 9 marks)
Q16.
(a) A student wishes to investigate how the resistance of a thermistor changes with
temperature.
(i)
Draw a labelled diagram of a suitable circuit that would enable the student to
measure the resistance of the thermistor.
(2)
(ii)
Describe the procedure the student would follow in order to obtain accurate and
reliable measurements of the resistance of the thermistor at different temperatures.
The quality of your written communication will be assessed in this question.
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(6)
(b)
The diagram below shows a thermistor connected in series with a resistor, R, and battery
of emf 6.0 V and negligible internal resistance.
When the temperature is 50 °C the resistance of the thermistor is 1.2 kΩ. The voltmeter
connected across R reads 1.6V.
(i)
Calculate the pd across the thermistor.
answer = ...................................... V
(1)
(ii)
Calculate the current in the circuit.
answer = ...................................... A
(1)
(iii)
Calculate the resistance of R quoting your answer to an appropriate number of
significant figures.
answer = ..................................... Ω
(2)
(c)
State and explain the effect on the voltmeter reading if the internal resistance of the
battery in the circuit in part (b) was not negligible.
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(2)
(Total 14 marks)
Q17.
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)
(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.
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(3)
(Total 9 marks)
Q18.
When a filament lamp is switched on it takes 0.50 seconds for the filament to reach its
normal operating temperature. The way in which the current changes during the first second
after switching on is shown on the graph below.
(a)
Use the graph to determine the maximum current through the lamp.
answer = ...................................... A
(1)
(b)
Assuming that the lamp is connected to a 12V dc supply of a negligible internal
resistance,
(i)
Calculate the resistance of the lamp when it has reached its normal operating
temperature,
answer = ..................................... Ω
(1)
(ii)
Calculate the power of the lamp when it has reached its normal operating
temperature.
answer = ..................................... W
(1)
(c)
Explain why the current through the lamp decreases between 0.05 s and 0.50 s.
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(2)
(d)
State and explain the change, if any, to the final current through the lamp if it is connected
to the same supply with another similar lamp
(i)
in series,
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(2)
(ii)
in parallel.
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(2)
(e)
State and explain why a filament lamp is most likely to fail as it is switched on.
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(2)
(Total 11 marks)
Q19.
Domestic users in the United Kingdom are supplied with mains electricity at a root mean
square voltage of 230V.
(a)
State what is meant by root mean square voltage.
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(1)
(b)
(i)
Calculate the peak value of the supply voltage.
answer = ...................................... V
(2)
(ii)
Calculate the average power dissipated in a lamp connected to the mains supply
when the rms current is 0.26 A.
answer = ..................................... W
(1)
(c)
The frequency of the voltage supply is 50 Hz. On the axes below draw the waveform of
the supplied voltage labelling the axes with appropriate values.
(4)
(Total 8 marks)
Q20.
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
topart (a) (i) is consistent with the resistance of each resistor being 3.0 Ω.
answer = ..................................... Ω
(3)
(b)
(i)
Calculate the current through resistor Y.
answer = ...................................... A
(2)
(ii)
Calculate the pd across resistor W.
answer = ...................................... V
(2)
(Total 8 marks)
Q21.
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.
The graph below shows the results from the experiment.
(a)
Explain why the terminal pd decreases as the current increases.
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(2)
(b)
(i)
Use the graph to find the emf, ε, of the cell.
answer = ..................................... V
(1)
(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
graphA.
(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)
(ii)
Calculate the energy dissipated in the internal resistance of the cell per second.
answer = ..................................... W
(2)
(Total 13 marks)
Q22.
An oscilloscope is used to investigate various voltage sources. In order to do this a voltage
source is connected to the y-input and the time base is switched off. Figure 1 below shows the
screen of the oscilloscope when the y-input is not connected to a voltage source.
Figure 1
Figure 2 shows the screen when a 1.5V cell is connected to the y-input.
Figure 2
(a)
On the grid below show the appearance of the screen if the y-input is connected to
a2.5V dc supply.
(1)
(b)
The y-input is now connected to a sinusoidal ac voltage supply and the screen is
shown inFigure 3.
Figure 3
(i)
Explain why a vertical line is now seen on the screen.
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(2)
(ii)
Calculate the peak-to-peak voltage of the ac supply.
answer = ..................................... V
(2)
(iii)
Calculate the root mean square voltage of the supply.
answer = ..................................... V
(2)
(Total 7 marks)
Q23.
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)
(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.
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(2)
(Total 10 marks)
Q24.
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)
(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)
(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.
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(2)
(ii)
Explain without calculation why the lamp Y glows more brightly.
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(2)
(Total 11 marks)
Q25.
(a) A student is given a piece of metal wire and asked to investigate how the resistance
of the wire changes between a temperature of 0 °C and 100 °C.
(i)
Draw a labelled diagram of a suitable arrangement that would enable the student to
carry out the experiment.
(3)
(ii)
Describe the procedure the student would follow in order to obtain accurate and
reliable measurements of the resistance of the wire at different temperatures
between 0 °C and 100 °C.
The quality of written communication will be assessed in your answer.
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(6)
(b)
A certain metal has a critical temperature of –268 °C (5 K). Explain what is meant by
critical temperature.
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(2)
(Total 11 marks)
Q26.
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)
(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.
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(3)
(Total 8 marks)
Q27.
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)
(b)
State and explain the effect on the brightness of the lamps in the circuit shown in Figure
1if the battery has a significant internal resistance.
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(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.
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(2)
(ii)
State and explain which of the lamps will be brighter assuming that the resistance of
the lamps does not change significantly with temperature.
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(3)
(Total 12 marks)
Q28.
(a) An alternating current supply provides an output voltage of 12 V rms at a frequency
of 50 Hz. Describe how you would use an oscilloscope to check the accuracy of the rms
output voltage and the frequency of the supply.
The quality of your written communication will be assessed in your answer.
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(6)
(b)
The power supply in part (a) is connected to a 12 V 24 W lamp.
(i)
Calculate the rms current in the lamp.
answer = ...................................... A
(1)
(ii)
Calculate the peak current in the lamp.
answer = ...................................... A
(1)
(iii)
Calculate the peak power of the lamp.
answer = ...................................... W
(2)
(Total 10 marks)
Q29.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)
(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).
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(2)
(Total 8 marks)
Q30.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)
(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........................................................................................................
...............................................................................................................
...............................................................................................................
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(2)
(Total 11 marks)
Q31.An experiment can be performed to determine whether a particular component is an ohmic
conductor.
(a)
State what is meant by an ohmic conductor.
........................................................................................................................
(1)
(b)
(i)
Draw a suitable circuit diagram for such an experiment.
(2)
(ii)
For the circuit diagram you have drawn, describe a suitable experiment. Your
account should include details of:
•
what measurements you would take
•
how you would use your measurements
•
how you would reach a conclusion.
The quality of written communication will be assessed in your answer.
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(6)
(c)
(i)
State the principal property of a superconductor.
...............................................................................................................
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(1)
(ii)
State what is meant by critical temperature.
...............................................................................................................
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(1)
(iii)
Give one use of a superconductor.
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(1)
(Total 12 marks)
Q32.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.
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(2)
(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)
Q33.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)
(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.
...............................................................................................................
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(2)
(Total 12 marks)
Q34.The graph below shows how a sinusoidal alternating voltage varies with time when connected
across a resistor, R.
(a)
(i)
State the peak-to-peak voltage.
peak-to-peak voltage...........................................V
(1)
(ii)
State the peak voltage.
peak voltage...........................................V
(1)
(iii)
Calculate the root mean square (rms) value of the alternating voltage.
rms voltage...........................................V
(2)
(iv)
Calculate the frequency of the alternating voltage. State an appropriate unit.
frequency.........................................unit ...............
(3)
(b)
On the graph above draw a line to show the dc voltage that gives the same rate of energy
dissipation in R as produced by the alternating waveform.
(2)
(c)
An oscilloscope has a screen of eight vertical and ten horizontal divisions.
Describe how you would use the oscilloscope to display the alternating waveform in the
graph above so that two complete cycles are visible.
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(3)
(Total 12 marks)
M1.
(a)
reverse mode: current zero or just negative at 50 -500 V (1)
sharp downward curve (1)
forward mode:
current zero or just positive up to
0.7 V (1)
rapid increase of for small increase in V (1)
4
(b)
at low V, I increases proportionally (or Ohm's law obeyed) (1)
(as V increases) greater I heats filament/wire
(or temp of filament/wire increases) (1)
resistance increases (1)
rate of increase of I with V decreases [or ref. to gradient = 1/R] (1)
reference to same form of the curve in negative quadrant (1)
4
[8]
M2.
(a)
V0 = √2 Vrms = √2 × 4.2 V (1) (5.94 V)
Vp-p (= 2 × V0) = 2 × 5.94 = 11.8 V (1)
2
(b)
(i)
voltage sensitivity = 11.8/5.9 = 2.0 V div–1 (1)
(ii)
T (= 1/f = 1/2500) = 4.0 × 10–4 s (1)
time base = 4.0 × 10–4/8 = 5.0 10–5 s div–1 (1)
3
(c)
(i)
spot at (1.75/0.5) = 3.5 div (1)
(ii)
(use of sum of emf = sum of pd)
1.75 = I (3.5 + 10) (1)
I = 0.13 A (1)
(iii)
V (= RI = 10 × 0.13) = 1.3 V (1)
[or V = ε – Ir = 1.75 – 0.13 × 3.5 = 1.3 V]
spot at (1.3/0.5) = 2.6 div (1) (accept 2.5 to 2.75 div)
5
[10]
M3.
(a)
(i)
R (= V2/P) = 122/45 (1)
R = 3.2Ω (1)
(ii)
(resistive strips are in parallel)
1/RT = 1/R1 + 1/R2 … = 5/R (1)
RT (= 5 × 3.2) = 16Ω (1)
4
(b)
(using R = pl/A and A = wt)
thickness = pl/wR (1)
= 3 × 10–5 × 0.80/2.5 × 10–3 × 16 (1)
= 0.60 mm (1)
3
(c)
I (= P/V) = 45/12 = 3.75 A (1)
t (Q/I) = 1.44 × 10–5/3.75 = 3.84 × 104s (1)
3.84 × 104/60 × 60 = 10.7 hr (1)
3
[10]
M4.
(a)
(i)
positive current showing increasing gradient (1)
(starting) any value in the range 0.6 – 0.7 V (1)
negative current zero or near zero for negative pd (1)
(ii)
diode is in the forward (bias) direction
this occurs when the current is high
so the current by passes the ammeter
(1)(1)(1) any 3 lines
6
(b)
pd across diode = any value in the range 0.6 – 0.7 V (1)
pd across 4Ω = (12 V – diode pd) = value consistent with above in
the range 11.3 to 11.4 V (1)
for 6Ω resistor, I (= V/R = 12/6) = 2.0 A (1)
for 4Ω resistor, I (= pd across 4Ω resistor/4Ω) = value consistent
with above in the range 2.82 to 2.85 A (1)
ammeter current (= sum of above currents) = value consistent with
above in the range 4.82 to 4.85 A (1)
5
[11]
M5.
(a)
(ii)
(i)
work done (by the battery) per unit charge (1)
or (electrical) energy per unit charge
or pd/voltage when open circuit/no current
the resistance of the materials within the battery (1)
or hindrance to flow of charge in battery
or loss of pd/voltage per unit current
2
(b)
(i)
(use of E = V + Ir)
12 = V + 800 × 0.005 (1) (working/equation needs to be shown)
V = 12 – 4 = 8.0V (1)
(ii)
(use of P = I2r)
P = 8002 × 0.005 (1) (working/equation needs to be shown)
P = 3200 (1) W (1) or J s–1
5
(c)
car will probably not start (1)
battery will not be able to provide enough current (1)
or less current
or lower terminal pd/voltage
2
[9]
M6.
(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]
M7.
(a)
superconductivity means a material has zero resistivity/resistance (1)
resistivity decreases with temperature or idea of cooling (1)
becomes superconducting when you reach the critical/certain/
transition temperature (1)
3
(b)
(i)
(use of R = ρl/A)
0.075 = ρ × 1/(2.28 × 10–7) (1) (must see working or equation)
R = 1.7 × 10–8 (1) Ωm (1)
(ii)
max 3 from
the resistance decreases (to zero) (1)
copper still has resistance (1)
but this is in parallel with filaments (which have zero resistance) (1)
hence total resistance is zero (1)
current goes through filaments (1)
6
[9]
M8.
(a)
(use of E = V + Ir)
12 = V + 420 × 0.0095 (1)
V = 8.0(1)V (1)
2
(b)
ρ = RA/I = 1.6 × 10–3 × 7.9 × 10–5/0.75 (1)
R = 1.7 × 10–7 (1) Ωm (1)
3
[5]
M9.
(a)
(i)
use of 1.5 cycles (1)
conversion to time eg time for 1.5 cycles = 10 × 1.5 = 15ms (1)
calculation of frequency eg frequency = 1 / 0.010 = 100 ± 3Hz (1)
(ii)
peak voltage = 1.5 × 2 (1) = 3.0V (1)
(iii)
rms voltage = 3.0/√2 (1) (ce from (a) (i))
rms voltage = 2.12V (1)
7
(b)
vertical line is formed (1)
of length equal to twice the peak voltage (1)
because trace no longer moves horizontally
or spot moves just up and down (1)
max 2
[9]
M10.
(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
(ii)
the mark scheme for this part of the question includes anoverall assessment for the
Quality of Written Communication
QWC
descriptor
mark
range
good-excellent
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
modest-adequate
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
poor-limited
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
incorrect,inappropriateor No answer at all or answer refers to unrelated, incorrect or
noresponse
inappropriate physics.
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)
(b)
(i)
(use of I = V/R)
I = 12/8 (1)= 1.5A (1)
0
(ii)
I = (12 – 0.65 (1))/4 = 2.8 A (1) sig figs (1)
5
[14]
M11.
(i)
10.0 (V) (1)
1
(ii)
Vrms = 10.0/√2 = 7.1 (V) (1)
1
(iii)
time period = 3 × 2 = 6 (ms) (1)
1
(iv)
frequency = 1/0.006 or 1/6 (1)
frequency = 167 (1) (Hz)
2
[5]
M12.
(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) isdouble/more than that
passing through A or B (1)
energy given to charge passing through cells per second is doubleor more than in cells C
and D (1) or in terms of power
3
[9]
M13.(a)
(use of R = ρl/A)
R = 4.0 × 10–3 × 0.060 (1)/(π × 0.0122) (1)
R = 0.53 (Ω) (1)
2 significant figures (1)
4
(b)
the mark scheme for this part of the question includes an overallassessment for the
Quality of Written Communication
circuit must include:
voltmeter and ammeter connected correctly (1)
power supply with means of varying current (1)
2
QWC
goodexcellent
descriptor
(i)
Uses accurately appropriate grammar, spelling,
punctuation and legibility.
(ii)
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]
5-6
An excellent candidate will have a working circuit
diagram with correct description of measurements
(including range of results) and processing. An
excellent candidate uses a range of results and finds a
mean value or uses a graphical method, eg IV characteristics. They also mention precision eg use
of vernier callipers.
(i)
modestadequate
mark
range
Only a few errors.
(ii) Some structure to answer, style acceptable,
arguments or explanations partially supported by
evidence or examples.
An adequate candidate will have a working circuit and a
description with only a few errors, eg do not consider
precision. They have not taken a range of results and
fail to realise that the diameter needs to be measured in
3-4
several places.
poor-limited
(i)
Several significant errors.
(ii)
Answer lacking structure, arguments not
supported by evidence and contains limited
information.
1-2
Several significant errors, eg important measurement
missed, incorrect circuit, no awareness of how to
calculate resistivity.
incorrect,
inappropriate
or no
response
0
The explanation expected in a good answer should include a coherentaccount of the
procedure and include most of the following points.
•
length with a ruler
•
thickness/diameter with vernier callipers/micrometer
•
measure voltage
•
measure current
•
calculate resistance
•
use of graph, eg I-V or resistance against length
•
use of diameter to calculate cross-sectional area
•
mention of precision, eg vernier callipers or full scale readingsfor V and I
•
flat metal electrodes at each end to improve connection
6
[12]
M14.
(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]
M15.
(a)
a non-ohmic conductor does not have a constant resistance (1)
1
(b)
(i)
curve of decreasing gradient with increasing V (1)
attempt to make graph symmetric in two opposite quadrants (1)
2
(ii)
resistance increases as pd increases/current increases (1)
1
(c)
(i)
(use of P = V2/R)
36 = 144/R (1)
R = 4.0 (Ω) (1)
2
(ii)
reference to temperature change (1)
(resulting in) a lower resistance (1)
(hence) power rating would be greater (1)
3
[9]
M16.
(a)
(i)
working circuit including power supply and thermistor
(correct symbol) (1)
voltmeter and ammeter or ohm meter (1)
2
(ii)
The candidate’s writing should be legible and the spelling,punctuation and
grammar should be sufficiently accuratefor the meaning to be clear.
The candidate’s answer will be assessed holistically. The answerwill be assigned to
one of three levels according to the followingcriteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised,logical and coherent,
using appropriate specialist vocabularycorrectly. The form and style of writing is
appropriate toanswer the question.The candidate states that the thermistor is
connected in asuitable circuit with voltmeter and ammeter or ohmmeter.The
candidate gives details of how the thermistor is heatedin a beaker of water or a
water bath and a thermometer is usedto measure the temperature at small regular
intervals.The candidate states that the resistance is found at varioustemperatures
either directly with an ohmmeter or by dividingvoltage by current. The candidate
may mention that the watermust be stirred to ensure that the thermistor is at
thetemperature measured by the thermometer.The candidate may give some
indication of the range oftemperatures to be used.The candidate may refer to
repetition of whole experiment.The candidate may plot a graph of resistance against
temperature.The candidate may use a digital thermometer.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well
organised and not fully coherent. There is less use of specialist
vocabulary, or specialist vocabulary may be used incorrectly.
The form and style of writing is less appropriate.
The candidate states that the thermistor is connected in a
suitable circuit with voltmeter and ammeter or ohmmeter.
The candidate gives details of how the thermistor is heated
in a beaker of water and a thermometer is used to measure
the temperature.
The candidate states that the resistance is found at various
temperatures either directly with an ohmmeter or by dividing
voltage by current.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organisedand may not be
relevant or coherent. There is little correctuse of specialist vocabulary.The form and
style of writing may be only partly appropriate.
The candidate changes temperature at least once andmeasures V and I or R.
The explanation expected in a competent answershould include a coherent
selection of the followingpoints concerning the physical principles involved
andtheir consequences in this case.
Max 6
(b)
(i)
pd = 6.0 – 1.6 = 4.4 (V) (1)
1
(ii)
current = 4.4/1200 = 3.7 × 10–3 (A) (1) (not 3.6)
1
(iii)
resistance = 1.6/3.7 × 10–3 = 440 or 430 (Ω) (1)
2 sfs (1)
2
(c)
less current now flows or terminal pd/voltage lower (1)
(or voltage across cell/external circuit is lower)
(hence) pd/voltage across resistor will decrease (1)
2
[14]
M17.
(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
(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]
M18.
(a)
current = 0.40 A (1)
1
(b)
(i)
resistance = 12/0.2 = 60 Ω (1)
1
(ii)
power = 12 × 0.2 = 2.4 W (1)
1
(c)
resistance of filament increases or more collisions/scattering (1)
as temperature of filament increase or filament gets hot/heats
(until reaches thermal equilibrium) (1)
2
(d)
(i)
voltage of supply now shared by lamps
or resistance increased (1)
hence current reduced (1)
2
(ii)
current through the lamps unchanged/stays the same (1)
as both connected directly to the supply
or correct resistance argument (1)
2
(e)
resistance of lamps will be lower when first switched on (1)
hence initial current will be larger (1)
sudden rapid change in temperature (1)
max 2
[11]
M19.
(a) the square root of the mean of the squares of all the values of the
voltage in one cycle (1)
or the equivalent dc/steady/constant voltage that produces the same
heating effect/power (1)
1
(b)
(i)
peak voltage = 230 × √2 (1)
peak voltage = 325 V (or 324 V) (1)
2
(ii)
average power = 230 × 0.26 = 60 W (1)
1
(c)
shape and symmetrical with consistent values of x at y = 0 and consistent
ymax (must be at least one cycle) (1)
appropriate scale y-axis (1)
correct peak values (to within one 2 mm square) (1)
correct period (accept 0.02 or 20) (1)
4
[8]
M20.
(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]
M21.
(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]
M22.
(a)
1
(b)
(i)
the voltage reverse/changes direction/sign
this makes the spot move up and down or correctexplanation of lack of horizontal
movement
2
(ii)
length of line = 8 divisions
peak to peak = 8 × 0.5 = 4.0 V
2
(iii)
(peak = 2.0 V)
rms = 2.0/√2 = 1.4 V
2
[7]
M23.
(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
(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]
M24.
(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
(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 lampincreases
2
[11]
M25.
(a)
(i)
circuit with ammeter and voltmeter correct or ohmmeter
some means of heating eg water bath
thermometer in water bath
3
(ii)
The candidate’s writing should be legible and the spelling, punctuation
and grammar should be sufficiently accurate for the meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be assigned
to one of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and coherent,
using appropriate specialist vocabulary correctly. The form and style of writing is
appropriate to answer the question.
The candidate states that resistance is measured using an ohmmeter or voltmeter
ammeter method. The wire is heated in a beaker of water and the temperature
measured with a thermometer. Ice is added to the water and the water is stirred as
the water is heated. Details of how resistance is calculated and how results are
presented e.g. graph of resistivity against temperature.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and not fully
coherent. There is less use of specialist vocabulary, or specialist vocabulary may be
used incorrectly. The form and style of writing is less appropriate.
The candidate states that resistance is measured using an ohmmeter or voltmeter
ammeter method. The wire is heated in a beaker of water and the temperature
measured with a thermometer. Ice is added to the water. Details of how resistance
is calculated.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not be
relevant or coherent. There is little correct use of specialist vocabulary. The form
and style of writing may be only partly appropriate.
The candidate states that resistance is measured using an ohmmeter or voltmeter
ammeter method. The wire is heated in a beaker of water and the temperature
measured with a thermometer.
The explanation expected in a competent answer should include a coherent
selection of the following points concerning the physical principles involved
and their consequences in this case.
•
resistance measured calculated
•
water bath used
•
ice added to water
•
water stirred
•
temperature measured with thermometer
•
resistance calculated
•
graph drawn
6
(b)
the temperature at or below which a material
becomes a superconductor or has zero resistance/resistivity
2
[11]
M26.
(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]
M27.
(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
(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]
M28.
(a) The candidate’s writing should be legible and the spelling, punctuation and
grammar should be sufficiently accurate for the meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be assigned to one
of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and coherent, using
appropriate specialist vocabulary correctly. The form and style of writing is appropriate to
answer the question.
The candidate states that the power supply is connected to the input of the oscilloscope.
The time base is switched off and the y gain adjusted until a complete vertical line is seen
on the screen. The length of the line is measured and this is converted to peak to peak
voltage using the calibration. The peak voltage is divided by root two to get the rms
voltage and this is compared with the stated value. The time base is now switched on and
adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is
measured and this is converted to time using the time base setting. Frequency is the
reciprocal of this time.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and not fully
coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used
incorrectly. The form and style of writing is less appropriate.
The candidate states that the power supply is connected to the input of the oscilloscope.
The y gain adjusted. The length of the line/height of peak is measured. The peak voltage
is divided by root two to get the rms voltage. The time base is now switched on and
adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is
measured and this is converted to time using the time base setting.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not be relevant or
coherent. There is little correct use of specialist vocabulary. The form and style of writing
may be only partly appropriate.
The candidate states that the power supply is connected to the input of the oscilloscope.
The length of the line/height of peak is measured. The time base is now switched on and
adjusted until a minimum of one cycle is seen on the screen. The length of one cycle is
measured and this is converted to time.
The explanation expected in a competent answer should include a coherent
selection of the following points concerning the physical principles involved and
their consequences in this case.
•
power supply connected to oscilloscope input
•
time base initially switched off
•
y gain adjusted to get as long a line as possible
•
length of line used to find peak to peak voltage
•
rms voltage found
•
time base switched on and adjusted to get several cycles on the screen
•
use the time base setting to find period
•
use period to find frequency
•
compare vales with stated values
6
(b)
(i)
(use of P = IV)
I = 24/12 = 2.0 (A)
1
(ii)
peak current = √2 × 2.0 = 2.8 (A)
1
(iii)
peak power = √2 × 12 × √2 × 2.0
= 48 (W)
2
[10]
M29.(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]
M30.(a)
(i)
1/R total = 1/(40) +1/(10+5)
R total = 10.9 kΩ
= 0.09167
3
(ii)
I = 12 / 10.9 k = 1.1 mA
1
(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]
M31.(a)
a component with constant resistance OR V ∝ I
1
(b)
(i)
circuit using correct symbols with means of varying current / voltage
correct voltmeter and ammeter
ignore symbol for component
unless it is a variable resistor
2
(ii)
The candidate’s writing should be legible and the spelling, punctuation and
grammar should be sufficiently accurate for the meaning to be clear.The
candidate’s answer will be assessed holistically. The answer will be assigned to one
of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marksThe information conveyed by the
answer is clearly organised, logical and coherent, using appropriate specialist
vocabulary correctly. The form and style of writing is appropriate to answer the
question.
Candidate draws an appropriate circuit diagram with correctly positioned ammeter
and voltmeters. Candidate has a means of varying the current. Sets current to
different values and measures pd. Mentions wide range. Has a sensible way of
varying current (e.g. variable resistor / potential divider). Plots a graph of pd against
current. Relates constant gradient to a constant resistance.
Level 5 / 6meaning of line through originreverse current
readingssuitable range with suggested values
Intermediate Level (Modest to adequate): 3 or 4 marksThe information conveyed
by the answer may be less well organised and not fully coherent. There is less use
of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form
and style of writing is less appropriate.Candidate draws an appropriate circuit
diagram with correctly positioned ammeter and voltmeters. Candidate has a means
of varying the current. Varies current and measures pd. Plots a graph of pd against
current. Relates constant gradient to a constant resistance.
Level 3 / 4Draw best fit line or state R constantRelate straight line
on graph to ohmic conductor
Low Level (Poor to limited): 1 or 2 marksThe information conveyed by the answer
is poorly organised and may not be relevant or coherent. There is little correct use of
specialist vocabulary. The form and style of writing may be only partly appropriate.
The candidate measures resistance at least twice to see if constant. Has some
means of varying current.
Level 1 / 2Take several readings of V and I and plot graph or
calculate R
The explanation expected in a competent answer should include a coherent
selection of the following points concerning the physical principles involved
and their consequences in this case.
method for varying currentcurrent varied in regular stepspd and current
measureresistance calculatedgraph drawnsignificance of gradient of the graph
discussed
6
(c)
(i)
a material with zero resistivity / resistance
not negligible
1
(ii)
material becomes superconducting at / below critical temperature
accept reverse argument
1
(iii)
any correct usage e.g. powerful magnets, mri, maglev trains / bullet train / (high
power) transmission lines / particle accelerators / LHC
1
[12]
M32.(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)
V = 6.7 × 10−5 × 5000
= 0.33 (0.33 − 0.35) V
OR
V = 5 / 90 × 6
= 0.33( V)
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]
M33.(a)
(i)
(use of V=Ir)
V= 4.2 × 1.5
= 6.3 (V)
1
(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
(v)
R= 5.7 / 1.35 =4.2 Ω
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
answer
s in (a)
but not
for first
value
2.0:
a(iii)2×2
R: a(iv)2×a(v)
3
(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]
M34.(a)
(i)
128 V
1
(ii)
64 V
CE from (i)
1
(iii)
Vrms= 64 / √2
=45.3 V
CE from (ii)
2
(iv)
frequency = 1 / 0.01 = 100
Hz
do not accept kHz for unit mark unless correct for candidate value
if use 10 s instead of 10 ms then can score second two marks
3
(b)
horizontal line
through y = 45 (44 − 48) x =0
CE from (a)(iii)+ / - half square
straight line must extend to at least to 6.0 ms
2
(c)
connect to y-input
adjust / change time base
so that each division is 2.0 ms OR 20 ms across screen
reference to y-gain / sensitivity
if inappropriate numbers quoted for y gain then lose last mark
3max
[12]
E1.
A question on l - V characteristics has appeared many times over the last few years and the
answers should really be familiar to candidates who practise past papers. In general, the
characteristics for the silicon semiconductor diode were drawn reasonably well in part (a) and
most candidates were able to gain at least two of the allocated four marks. Points which lost
marks, and to which attention should be given, did not show a sufficiently sharp downward
curve in reverse bias, or show the current in the forward bias as being zero or just positive for a
short distance. Many candidates drew the forward characteristic as a current increasing
smoothly from zero. This was not acceptable.
Answers to explaining the characteristic for the filament lamp in part (b) were vague, e.g. a
statement that the current increased the temperature would not gain a mark unless it was stated
clearly that it was the temperature of the filament that increased. Another error which occurred
frequently was after reasoning correctly that the resistance of the filament increased, then going
on to state that this caused the current to decrease, rather than causing the rate of increase of
current to decrease. Reference to the curve in the negative quadrant was very often superficial.
E5.
Most candidates were able to explain what is meant by internal resistance but were less
clear about the meaning of the emf of a battery. Most appreciated that it was connected to
energy but their answers were far from convincing.
Part (b) (i) produced some good responses although a number calculated the potential
difference across the internal resistance as opposed to the terminal potential difference which
meant that the answer 4.0 V was commonly seen.
Part (b) (ii) was generally answered well and the unit J s–1 or W did not seem to present too
many problems.
Part (c) is an application mentioned in the specification and there was good evidence that this is
something that has been considered by most centres. The most common answer was that the
current would be reduced. The effect of a reduced current was sometimes not clearly expressed
and candidates tended to say things like it would take longer to start or that it would be more
difficult to start rather than making a definite statement about the car not starting.
E6.
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.
E7.
For a number of candidates this question proved to be quite difficult. In part (a) there was
much confusion about superconductivity with a significant proportions of candidates stating that
the resistivity is very small or close to zero rather than making a clear statement that it is zero.
The conditions necessary for superconductivity were also not clearly expressed and many
candidates seemed to think that this was a high temperature property. The term critical
temperature was seldom seen.
The calculation in part (b) (i) was done well but part (b) (ii) caused candidates more problems.
Full marks were rare and although some candidates did appreciate that the copper still had
resistance, they did not see this as effectively a question about parallel resistors.
E8.
Part (a) caused similar problems to the question on emf and internal resistance in the
January examination. A common, incorrect approach was to calculate the potential difference
across the internal resistance and quote this as the value of terminal pd.
Part (b) proved to be much more accessible and the calculation only caused a few candidates
problems. The unit for resistivity does confuse a significant proportion of candidates and this is
often quoted as Ω m–1 or Ω/m.
E9.
In part (a) (i) the majority of candidates were able to relate the time-base setting to time
period and from this determine the frequency. Many however, did not use the whole trace and
did not recognise that there were one and half cycles across the ten divisions. Instead, they
tried to judge the number of divisions occupied by one cycle and consequently obtained a value
for frequency of less than 100 Hz.
Part (a) (ii) & (iii) were answered very well with only a minority of candidates confusing peak
voltage with peak to peak voltage.
Part (b) was less well done and it was rare for candidates to score full marks. It was not
uncommon for candidates to state that two horizontal lines were produced when the time base
is switched off. Some also confused this situation with what would occur if a source of direct
current had been used and stated that the trace or spot is deflected upwards.
E10.
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.
E11.
This was the most accessible question in the paper and candidates are clearly familiar with
the use of an oscilloscope and the idea of peak and rms voltages. The only consistent error was
the failure to convert the time period to seconds when calculating the frequency.
E12.
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.
E13.Part (a) proved straightforward and many candidates were able to calculate the resistance of the
putty correctly. A minority of candidates did confuse resistance with resistivity and did not
rearrange the equation from the data sheet. This question assessed significant figures and it
was clear that there are still many candidates who do not appreciate that their final answer
should reflect the precision of the data and in this case they should give their answer to two
significant figures.
Part (b) assessed quality of written communication and this question proved quite challenging
for the majority of candidates. It was extremely rare for candidates to obtain full marks and most
answers were either modest and/or limited. The circuit diagrams seen were often penalised for
careless errors such as incorrect symbols or the wrong positioning of meters.
It was rare for candidates to include a means of obtaining more than one result such as varying
the length of the putty or using a variable resistor. Descriptions were often vague and hard to
follow. Many candidates did not address the issue of precision in a convincing way and failed to
describe how they would make all the measurements needed. It is clear from this paper and
from previous papers that candidates find describing experiments difficult and would benefit
from some practice of this skill.
E14.
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.
E15.
This question was answered well although a minority stated that non-ohmic conductors did
not follow Ohm’s Law without explaining the consequence of this. The I –V characteristics of the
filament lamp seemed to be quite familiar but some answers were spoilt by carelessly drawn
graphs that were either horizontal at the end or were not noticeably symmetric in the two
quadrants.
The calculation in part (c) was well done but the deduction required for part (c) (ii) proved quite
discriminating and only the most able candidates obtained all three marks. Many identified why
the power might be 9.0 W, but were unable to explain why, in practice, the power rating is
slightly different.
E16.
A significant proportion of candidates found part (a) difficult. It seems that many candidates
were not familiar with an appropriate experiment that enables the variation of a thermistors
resistance with temperature to be investigated. While many drew a correct circuit diagram, few
were able to explain suitable techniques for varying the temperature of the thermistor. It was
common to see answers that confused this experiment with one that investigated the I –
Vcharacteristics of a filament lamp. This led candidates to suggest that it was appropriate to
investigate resistance change by using increasing current to change the thermistor resistance.
There was also much confusion as to how resistance was to be determined and the use of a
graph of current against voltage was a regular response. Candidates incorrectly stated that they
would determine the resistance of the thermistor by measuring the gradient of the graph. This
would not be an appropriate method as the graph would be a curve and the gradient of the
curve is not the resistance. It was extremely rare for water baths to be used for heating the
thermistor and also many did not explain how they would measure the temperature of the
thermistor. Many candidates did not address the issue of precision in a convincing way and
failed to describe how they would make all the measurements needed. It is clear from this paper
and from previous papers that candidates find describing experiments difficult and would benefit
from some practice of this skill.
In contrast parts (b) and (c) were generally answered well and full marks were frequently seen.
Less able candidates found it difficult to explain clearly the effect on the voltmeter reading if the
battery did have an appreciable internal resistance. There is some evidence that candidates did
not understand the meaning of the term negligible.
E17.
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 Ωresistor 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.
E18.
The majority of candidates were able to provide correct answers to parts (a) and (b)
although a minority did use the maximum current of 0.40 A rather than the normal working
current of 0.20 A.
Parts (c), (d) and (e) required more qualitative answers and proved to be much more
challenging.
Part (c) was well answered, although some candidates were let down by a looseness in their
technical language. An example of this occurred when a significant minority referred to the
‘filament increasing in heat’ rather than temperature.
Part (d) was not answered well and many candidates were confused by the context of the
question and gave answers that generally related to series and parallel circuits. This meant that
in (i) they stated the current was the same everywhere and therefore did not change, and in (ii)
they stated that the current split and therefore decreased. In both these cases, they did not
consider the effect the second bulb had on the resistance of the circuit.
Part (e) was more successfully answered by a significant proportion of the candidates, with
many realising the significance of the rapid increase in temperature.
E19.
Part (a) of this question on the meaning of root mean square voltage was not answered
well, with the majority responding by quoting a formula rather than by referring to an equivalent
direct voltage. In contrast part (b) was answered extremely well, with full marks being a
common outcome.
In part (c), candidates were required to draw a graph of an alternating voltage and although
some made a commendable attempt at this, there were a significant proportion of careless
answers with waves of varying amplitude and time period being a common occurrence. There
was also a tendency not to label the axes with appropriate values in spite of candidates being
instructed to do this in the rubric.
E20.
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.
E21.
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.
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.
E22.
Previous papers have suggested that the majority of candidates have a good
understanding of the use of an oscilloscope as a voltmeter and this proved to be the case here
too. The vast majority were able to successfully complete both parts of the question and only a
few confused peak to peak with peak voltage. Most candidates provided evidence that they
understood why a vertical line is produced if an alternating voltage is applied when the time
base is switched off. However, some answers were spoilt by a lack of precision in explanations
of how the voltage was varying – references to alternating currents rather than voltages were
common.
E23.
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’.
E24.
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 R2 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.
E25.
Part (a) (i) required students to draw a suitable arrangement for investigating the variation
in resistance of a piece of wire with temperature. Many students drew an appropriate circuit and
misplaced voltmeters were far less in evidence in circuit diagrams. However, the majority did
not show how the temperature of the wire was to be varied, in spite of being given the hint that
the temperature range should be varied between 0°C and 100°C.
The description of the experiment required in part (a) (ii) was answered better than has
previously been the case. Nevertheless, a significant proportion of students did not provide
suitable methods for temperature variation. Suggestions such as ‘vary the temperature using a
thermistor’ or ‘direct use of a Bunsen burner’ were not uncommon. It was clear that some
students did not appreciate that the wire could be safely placed in a water bath.
Part (b) on critical temperature was answered well.
E26.
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.
E27.
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.
E28.
Part (a) required students to describe the use of an oscilloscope to measure peak voltage
and frequency of an alternating current supply. This was answered well by a good proportion of
students and many were confident in their description of the use of the time base to determine
frequency and the y-gain to measure peak voltage. In a number of good quality answers
students mentioned switching off the time base and measuring peak to peak voltage so as to
find an accurate rms voltage. It was evident however, that a minority of students were unfamiliar
with the use of an oscilloscope and consequently gave very vague answers which scored few
marks.The calculation of rms and peak current were well done with the only common error
occurring when students assumed that the 12 V quoted in the question referred to peak voltage.
Those doing this were not heavily penalised as their answers were carried forward in the
subsequent calculations.
E29.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.
E30.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.
E31.In this question candidates needed to understand what is meant by an ohmic conductor and to
describe an experiment to determine whether a particular component exhibited the necessary
properties to be classed as this type of conductor. A significant proportion of candidates could
not state what is meant by an ohmic conductor and responses such as “obeys Ohm‘s Law” and
“something with resistance” were quite common.
A high proportion of correct circuit diagrams were seen in part (b)(i) and incorrectly placed
voltmeters were rare. The main omission was a means of varying the current in the circuit. The
experimental design caused problems and less than 15% of candidates wrote answers that
qualified them for the top band. Some described the wrong experiment such as investigating the
effect of temperature on the resistance of a thermistor. Those that did describe a correct
experiment tended not to fully develop their answer. It was expected that a top band answer
would have details of how a graph of results might be interpreted and in many cases this was
only done in a very superficial manner. Candidates who did refer to a straight line indicating
direct proportion often failed to mention that for this to be true the line must go through the
origin. There was also a tendency for them to state that the gradient of the graph is equal to the
resistance. While this is true for a straight line graph going through the origin, it is not generally
true and it was clear from a significant proportion of responses that candidates are not aware of
this. It was good to see evidence of candidates planning their answers and given that they find
these types of questions a challenge, this is a practice to be encouraged.
The remaining part of the question required knowledge of superconductors and although a high
proportion of candidates failed to give a convincing use of superconductors the important
property of a superconductor and the significance of critical temperature proved to be well
understood.
E32.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.
E33.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.
E34.This question on alternating currents was generally very well done. There were few major
problems with part (a) although a minority did leave the rms voltage in surd form thus not
completing the calculation.
Part (b) was also well answered and most candidates drew their line with care, using a ruler.
Part (c) however, was answered poorly and it is apparent that a significant proportion of
candidates were not clear on how to use an oscilloscope. Reference to the time base was seen
more often than reference to y – gain or y – sensitivity but frequently neither of these was
mentioned. It was also quite common for candidates to assume that the vertical scale only
needed to cover the peak voltage and not the peak to peak voltage i.e. eight divisions covered
64 V rather than 128 V.
On the other hand a far greater proportion was able to deduce that each horizontal division
needed to represent 2.0 ms.