Download When fully charged the 2.0 mF capacitor used as a backup

1
When fully charged the 2.0 mF capacitor used as a backup for a memory unit has a potential
difference of 5.0 V across it. The capacitor is required to supply a constant current of 1.0 μA and
can be used until the potential difference across it falls by 10%. For how long can the capacitor
be used before it must be recharged?
A
10 s
B
100 s
C
200 s
D
1000 s
(Total 1 mark)
2
(a)
The graph shows how the current varies with time as a capacitor is discharged through a
150 Ω resistor.
(i)
Explain how the initial charge on the capacitor could be determined from a graph of
current against time.
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(1)
(ii)
The same capacitor is charged to the same initial potential difference (pd) and then
discharged through a 300 kΩ resistor. Sketch a second graph on the same axes
above to show how the current varies with time in this case.
(3)
Page 1 of 74
(b)
In an experiment to show that a capacitor stores energy, a student charges a capacitor
from a battery and then discharges it through a small electric motor. The motor is used to
lift a mass vertically.
(i)
The capacitance of the capacitor is 0.12 F and it is charged to a pd of 9.0 V.
The weight of the mass raised is 3.5 N.
Calculate the maximum height to which the mass could be raised.
Give your answer to an appropriate number of significant figures.
maximum height ................................................. m
(4)
(ii)
Give two reasons why the value you have calculated in part (i) would not be achieved
in practice.
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(2)
(Total 10 marks)
Page 2 of 74
3
The specification for a pacemaker requires a suitable charge to be delivered in 1.4 ms. A
designer uses a circuit with a capacitor of capacitance 3.0 μF and a 2.5 V power supply to deliver
the charge. The designer calculates that a suitable charge will be delivered to the heart as the
capacitor discharges from a potential difference (pd) of 2.5 V to a pd of 1.2 V in 1.4 ms.
(a)
(i)
Calculate the charge on the capacitor when it is charged to a pd of 2.5 V.
charge .................................................. C
(1)
(ii)
Draw a graph showing how the charge, Q, on the capacitor varies with the pd, V, as
it discharges through the heart.
Include an appropriate scale on the charge axis.
(3)
Page 3 of 74
(b)
Calculate the energy delivered to the heart in a single pulse from the pacemaker when the
capacitor discharges to 1.2 V from 2.5 V.
energy ................................................... J
(3)
(c)
(i)
Calculate the resistance of the heart that has been assumed in the design.
resistance ................................................. Ω
(3)
(ii)
Explain why the rate of change of pd between the capacitor plates decreases as the
capacitor discharges.
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(2)
(Total 12 marks)
4
(a)
When an uncharged capacitor is charged by a constant current of 4.5 μA for 60 s the pd
across it becomes 4.4 V.
(i)
Calculate the capacitance of the capacitor.
capacitance ......................................... F
(3)
Page 4 of 74
(ii)
The capacitor is charged using the circuit shown in Figure 1. The battery emf is 6.0 V
and its internal resistance is negligible. In order to keep the current constant at
4.5 μA, the resistance of the variable resistor R is decreased steadily as the charge
on the capacitor increases.
Figure 1
Calculate the resistance of R when the uncharged capacitor has been charging for
30 s.
resistance ........................................ Ω
(3)
Page 5 of 74
(b)
The circuit in Figure 2 contains a cell, an uncharged capacitor, a fixed resistor and a
two-way switch.
Figure 2
The switch is moved to position 1 until the capacitor is fully charged. The switch is then
moved to position 2.
Describe what happens in this circuit after the switch is moved to position 1, and after it has
been moved to position 2. In your answer you should refer to:
•
the direction in which electrons flow in the circuit, and how the flow of electrons
changes with time,
•
how the potential differences across the resistor and the capacitor change with time,
•
the energy changes which take place in the circuit.
The terminals of the cell are labelled A and B and the capacitor plates are labelled P and
Q so that you can refer to them in your answer.
The quality of your written communication will be assessed in your answer.
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(6)
(Total 12 marks)
Page 6 of 74
5
A capacitor of capacitance 330 µF is charged to a potential difference of 9.0 V. It is then
discharged through a resistor of resistance 470 kΩ.
Calculate
(a)
the energy stored by the capacitor when it is fully charged,
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(2)
(b)
the time constant of the discharging circuit,
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(1)
(c)
the p.d. across the capacitor 60 s after the discharge has begun.
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(3)
(Total 6 marks)
6
(a)
A capacitor is marked ‘2200 μF 15 V’.
(i)
Explain what is meant by a capacitance of 2200 μF.
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(2)
Page 7 of 74
(ii)
What is the significance of the 15 V marking?
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(1)
(b)
An egg-timer is designed to produce a sound when an egg has been boiled for a sufficient
time. The time which elapses before the alarm sounds is controlled by the circuit shown
below. The circuit is operated from a 6.0 V cell of negligible internal resistance.
The time is set by means of the variable resistor R.
The capacitor is charged by moving the two-way switch to position S1 for a short time. The
timing is then started automatically when the two-way switch is moved to position S2. An
alarm rings when the potential difference between terminals XY reaches 2.0 V.
(i)
In one setting the time constant of the circuit when the capacitor is discharging is 3.0
minutes. Sketch a graph to show how the potential difference between the terminals
X and Y varies with time for the first 6.0 minutes after the switch moves to the
position S2.
(2)
(ii)
How long after timing commences will the alarm sound for the setting in part (i)?
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(1)
Page 8 of 74
(iii)
Calculate the resistance of the variable resistor when the time constant is
3.0 minutes.
(2)
(iv)
The system is designed to measure cooking times up to 5.0 minutes. Determine the
maximum value of the resistance R that is needed.
(2)
(v)
State how a suitable capacitor would be connected to increase the measurable
cooking time.
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(1)
(Total 11 marks)
7
The graph shows the variation of potential difference V with time t across a 470 μF capacitor
discharging through a resistor.
The resistance of the resistor is approximately
A
900 Ω
B
1300 Ω
C
1900 Ω
D
4700 Ω
(Total 1 mark)
Page 9 of 74
8
A capacitor is first charged through a resistor and then discharged through the same resistor.
The magnitude of which one of the following quantities varies with time in the same way during
both charging and discharging?
A
Energy stored
B
Current
C
Potential difference
D
Charge
(Total 1 mark)
9
(a)
(i)
A label on a capacitor shows it to have a capacitance of 0.020 F. Explain what this
tells you about the capacitor.
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(1)
(ii)
Sketch on Figure 1 the graph that shows how the charge on the 0.020 F capacitor
varies with the potential difference across it over the voltage range given. Insert an
appropriate scale on the charge axis.
(2)
(iii)
Explain how your graph could be used to obtain the energy stored for a given
potential difference.
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(2)
Page 10 of 74
Figure 1
(iv)
Figure 2
Show on Figure 2 how two similar capacitors could be connected to a supply to store
more energy for the same potential difference.
(1)
(b)
Figure 3 shows one 0.020 F capacitor connected to a 20 V supply. By means of the
changeover switch S, the capacitor is disconnected from the supply and connected to a
small motor. The motor lifts an object of mass 0.15 kg through a height of 0.80 m, after
which the energy left in the capacitor is negligible.
acceleration of free fall, g = 9.8 m s–2
Figure 3
Page 11 of 74
Calculate:
(i)
the initial energy stored by the capacitor;
(2)
(ii)
the efficiency of the energy conversion.
(3)
(Total 11 marks)
10
(a)
A capacitor, initially charged to a pd of 6.0V, was discharged through a 100 kΩ resistor.
A datalogger was used to record the pd across the capacitor at frequent intervals. The
graph shows how the pd varied with time during the first 40 s of discharge.
(i)
Calculate the initial discharge current.
answer = ........................... A
(1)
(ii)
Use the graph to determine the time constant of the circuit, giving an appropriate unit.
answer = ...............................
(4)
Page 12 of 74
(iii)
Hence calculate the capacitance of the capacitor.
answer = .................................. µF
(1)
(iv)
Show that the capacitor lost 90% of the energy it stored originally after about 25 s.
(3)
(b)
In order to produce a time delay, an intruder alarm contains a capacitor identical to the
capacitor used in the experiment in part (a). This capacitor is charged from a 12 V supply
and then discharges through a 100 kΩ resistor, similar to the one used in the experiment.
(i)
State and explain the effect of this higher initial pd on the energy stored by this
capacitor initially.
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(2)
(ii)
State and explain the effect of this higher initial pd on the time taken for this capacitor
to lose 90% of its original energy.
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(1)
(Total 12 marks)
Page 13 of 74
11
(a)
Explain what is meant by a capacitance of 1 farad (F).
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(1)
(b)
A parallel plate capacitor was made from two circular metal plates with air between them.
The distance between the plates was 1.8 mm. The capacitance of this capacitor was found
to be 2.3 × 10–11 F.
The permittivity of free space ε0 = 8.9 ×10–12 F m–1
The relative permittivity of air = 1.0
Calculate:
(i)
the radius of the plates used in the capacitor;
(3)
(ii)
the energy stored when the potential difference between the capacitor plates is 6.0 V.
(2)
Page 14 of 74
(c)
A student charged the capacitor and then tried to measure the potential difference between
the plates using an oscilloscope. The student observed the trace shown in the diagram
below and concluded that the capacitor was discharging through the oscilloscope.
Calculate the resistance of the oscilloscope.
(3)
(Total 9 marks)
Page 15 of 74
12
A student used a voltage sensor connected to a datalogger to plot the discharge curve for a 4.7
μF capacitor. She obtained the following graph.
Use data from the graph to calculate
(a)
the initial charge stored,
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(2)
(b)
the energy stored when the capacitor had been discharging for 35 ms,
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(3)
(c)
the time constant for the circuit,
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(3)
(d)
the resistance of the circuit through which the capacitor was discharging.
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(2)
(Total 10 marks)
Page 16 of 74
13
A capacitor of capacitance 15 μF is fully charged and the potential difference across its plates is
8.0 V. It is then connected into the circuit as shown.
The switch S is closed at time t = 0. Which one of the following statements is correct?
A
The time constant of the circuit is 6.0 ms.
B
The initial charge on the capacitor is 12 μC.
C
After a time equal to twice the time constant, the charge remaining on the
capacitor is Q0e2, where Q0 is the charge at time t = 0.
D
After a time equal to the time constant, the potential difference across the
capacitor is 2.9 V.
(Total 1 mark)
14
(a)
As a capacitor was charged from a 12 V supply, a student used a coulomb meter and a
voltmeter to record the charge stored by the capacitor at a series of values of potential
difference across the capacitor. The student then plotted a graph of pd (on the y-axis)
against charge (on the x-axis).
(i)
Sketch the graph obtained.
(ii)
State what is represented by the gradient of the line.
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Page 17 of 74
(iii)
State what is represented by the area enclosed by the line and the x-axis of the
graph.
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(3)
(b)
The student then connected the capacitor as shown in the diagram below to carry out an
investigation into the discharge of the capacitor.
The student used a voltage sensor, datalogger and computer to obtain values for the pd
across the capacitor at various times during the discharge.
(i)
At time t = 0, with switch S2 open, switch S1was moved from position A to position B.
Calculate the pd across the capacitor when t = 26 s.
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(ii)
At time t = 26 s, as the discharge continued, the student closed switch S2. Calculate
the pd across the capacitor 40 s after switch S1 was moved from position A to
position B.
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Page 18 of 74
(iii)
Sketch a graph of pd against time for the student’s experiment described in parts
(b)(i) and (b)(ii).
(7)
(Total 10 marks)
15
A capacitor of capacitance C discharges through a resistor of resistance R.
Which one of the following statements is not true?
A
The time constant will increase if R is increased.
B
The time constant will decrease if C increased.
C
After charging to the same voltage, the initial discharge current will increase if R
is decreased.
D
After charging to the same voltage, the initial discharge current will be unaffected if
C is increased.
(Total 1 mark)
Page 19 of 74
16
A 1000 μF capacitor, initially uncharged, is charged by a steady current of 50 μA. How long will it
take for the potential difference across the capacitor to reach 2.5 V?
A
20 s
B
50 s
C
100 s
D
400 s
(Total 1 mark)
17
A 680 µF capacitor is charged fully from a 12 V battery. At time t = 0 the capacitor begins to
discharge through a resistor. When t = 25 s the energy remaining in the capacitor is one quarter
of the energy it stored at 12 V.
(a)
Determine the pd across the capacitor when t = 25s.
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(2)
(b)
(i)
Show that the time constant of the discharge circuit is 36 s.
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(ii)
Calculate the resistance of the resistor.
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(4)
(Total 6 marks)
Page 20 of 74
18
When a 220 μF capacitor is discharged through a resistor R, the capacitor pd decreases from 6.0
V to 1.5 V in 92 s.
What is the resistance of R?
A
210 kΩ
B
300 kΩ
C
420 kΩ
D
440 kΩ
(Total 1 mark)
19
The graph shows how the charge on a capacitor varies with time as it is discharged through a
resistor.
What is the time constant for the circuit?
A
3.0 s
B
4.0 s
C
5.0 s
D
8.0 s
(Total 1 mark)
Page 21 of 74
20
When switch S in the circuit is closed, the capacitor C is charged by the battery to a pd V0. The
switch is then opened until the capacitor pd decreases to 0.5 V0, at which time S is closed again.
The capacitor then charges back to V0.
Which graph best shows how the pd across the capacitor varies with time, t, after S is opened?
(Total 1 mark)
Page 22 of 74
21
Figure 1 shows a circuit that is used in a defibrillator in which a short pulse of charge is used to
revive a patient who suffers a cardiac arrest in which their heart stops beating.
Figure 2 shows how the charge on the capacitor varies with time when the capacitor is charging.
Figure 1
Figure 2
(a)
(i)
Use Figure 2 to determine the initial charging current.
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initial charging current ....................... A
(2)
Page 23 of 74
(ii)
Calculate the emf of the supply used to charge the capacitor.
Assume that the supply has negligible internal resistance.
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emf of the supply ...................................... V
(2)
(iii)
Explain why the current that charges the capacitor falls as the capacitor charges.
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(3)
(b)
For the system to work successfully, the capacitor has to deliver 140 J of energy to the
heart in a pulse that lasts for 10 ms.
(i)
Show that the charge on the capacitor when it is storing this much energy is about 85
mC.
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(2)
(ii)
Calculate the average power supplied during the pulse.
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average power ....................................... W
(1)
Page 24 of 74
(c)
The circuit designer suggests that the capacitor can be used successfully after a charging
time equal to 1.5 time constants of the charging circuit shown in Figure 1.
Explain with a calculation whether or not the designer’s suggestion is valid.
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(3)
(Total 13 marks)
22
When a capacitor discharges through a resistor it loses 50% of its charge in 10 s. What is the
time constant of the capacitor-resistor circuit?
A
0.5 s
B
5s
C
14 s
D
17 s
(Total 1 mark)
Page 25 of 74
23
A 2.0 mF capacitor, used as the backup for a memory unit, has a potential difference of 5.0 V
across it when fully charged. The capacitor is required to supply a constant current of 1.0 µA and
can be used until the potential difference across it falls by 10%. How long can the capacitor be
used for before it must be recharged?
A
10 s
B
100 s
C
200 s
D
1000 s
(Total 1 mark)
24
Capacitors and rechargeable batteries are examples of electrical devices that can be used
repeatedly to store energy.
(a)
(i)
A capacitor of capacitance 70 F is used to provide the emergency back-up in a low
voltage power supply.
Calculate the energy stored by this capacitor when fully charged to its maximum
operating voltage of 1.2 V. Express your answer to an appropriate number of
significant figures.
answer = ...................................J
(3)
(ii)
A rechargeable 1.2 V cell used in a cordless telephone can supply a steady current of
55 mA for 10 hours. Show that this cell, when fully charged, stores almost 50 times
more energy than the capacitor in part (a)(i).
(2)
Page 26 of 74
(b)
Give two reasons why a capacitor is not a suitable source for powering a cordless
telephone.
Reason 1.....................................................................................................
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Reason 2......................................................................................................
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(2)
(Total 7 marks)
25
(a)
A particular heart pacemaker uses a capacitor which has a capacitance of 4.2 μF.
Explain what is meant by a capacitance of 4.2 μF.
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(2)
(b)
Capacitor A, of capacitance 4.2 μF, is charged to 4.0 V and then discharged through a
sample of heart tissue. This capacitor is replaced by capacitor B and the charge and
discharge process repeated through the same sample of tissue.
The discharge curves are shown in the figure below.
Page 27 of 74
(i)
By considering the discharge curve for capacitor A, show that the resistance of the
sample of heart tissue through which the discharge occurs is approximately 150 Ω.
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(4)
(ii)
State and explain whether capacitor B has a larger or smaller capacitance than that
of capacitor A.
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(2)
(c)
Capacitor A was charged to a potential difference of 4.0V before discharging through the
sample of heart tissue.
Determine how much energy it passed to the sample of heart tissue in the first 0.90 m s of
the discharge.
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energy ...................................... J
(3)
(Total 11 marks)
Page 28 of 74
26
(a)
Define the capacitance of a capacitor.
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(2)
(b)
The circuit shown in the figure below contains a battery, a resistor, a capacitor and a
switch.
The switch in the circuit is closed at time t = 0. The graph shows how the charge Q stored
by the capacitor varies with t.
(b)
(i)
When the capacitor is fully charged, the charge stored is 13.2 μC. The electromotive
force (emf) of the battery is 6.0 V. Determine the capacitance of the capacitor.
answer = ................................. F
(2)
Page 29 of 74
(ii)
The time constant for this circuit is the time taken for the charge stored to increase
from 0 to 63% of its final value. Use the graph to find the time constant in
milliseconds.
answer = ................................. ms
(2)
(iii)
Hence calculate the resistance of the resistor.
answer = ................................. Ω
(1)
(iv)
What physical quantity is represented by the gradient of the graph?
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(1)
(c)
(i)
Calculate the maximum value of the current, in mA, in this circuit during the charging
process.
answer = ................................. mA
(1)
(ii)
Sketch a graph on the outline axes to show how the current varies with time as the
capacitor is charged. Mark the maximum value of the current on your graph.
(2)
(Total 11 marks)
Page 30 of 74
27
A capacitor of capacitance C discharges through a resistor of resistance R. Which one of the
following statements is not true?
A
The time constant will decrease if C is increased.
B
The time constant will increase if R is increased.
C
After charging to the same voltage, the initial discharge current will increase if R is
decreased.
D
After charging to the same voltage, the initial discharge current will be unaffected if C is
increased.
(Total 1 mark)
28
The figure below shows part of the discharge curve for a capacitor that a manufacturer tested for
use in a heart pacemaker.
The capacitor was initially charged to a potential difference (pd) of 1.4 V and then discharged
through a 150 Ω resistor.
(a)
Show that the capacitance of the capacitor used is about 80 μF.
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(3)
Page 31 of 74
(b)
Explain why the rate of change of the potential difference decreases as the capacitor
discharges.
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(3)
(c)
Calculate the percentage of the initial energy stored by the capacitor that is lost by the
capacitor in the first 0.015 s of the discharge.
energy lost .........................................................%
(3)
(d)
The charge leaving the capacitor in 0.015 s is the charge used by the pacemaker to
provide a single pulse to stimulate the heart.
(i)
Calculate the charge delivered to the heart in a single pulse.
charge .........................................................C
(1)
Page 32 of 74
(ii)
The manufacturer of the pacemaker wants it to operate for a minimum of 5 years
working at a constant pulse rate of 60 per minute.
Calculate the minimum charge capacity of the power supply that the manufacturer
should specify so that it will operate for this time.
Give your answer in amp-hours (Ah).
minimum capacity .........................................................Ah
(2)
(Total 12 marks)
29
In the circuit shown the capacitor C charges when switch S is closed.
Which line, A to D, in the table gives a correct pair of graphs showing how the charge on the
capacitor and the current in the circuit change with time after S is closed?
charge
current
A
graph 1
graph 1
B
graph 1
graph 2
C
graph 2
graph 2
D
graph 2
graph 1
(Total 1 mark)
Page 33 of 74
30
A voltage sensor and a datalogger are used to record the discharge of a 10 mF capacitor in
series with a 500 Ω resistor from an initial pd of 6.0 V. The datalogger is capable of recording
1000 readings in 10 s.
After a time equal to the time constant of the discharge circuit, which one of the rows gives the pd
and the number of readings made?
Potential difference / V
Number of readings
A
2.2
50
B
3.8
50
C
3.8
500
D
2.2
500
(Total 1 mark)
31
A voltage sensor and a datalogger are used to record the discharge of a 10 mF capacitor in
series with a 500 Ω resistor from an initial pd of 6.0 V. The datalogger is capable of recording
1000 readings in 10 s. Which line, A to D, in the table gives the pd and the number of readings
made after a time equal to the time constant of the discharge circuit?
potential difference/V
number of readings
A
2.2
50
B
3.8
50
C
3.8
500
D
2.2
500
(Total 1 mark)
Page 34 of 74
32
The graph below shows how the charge stored by a capacitor varies with time when it is
discharged through a fixed resistor.
(a)
Determine the time constant, in ms, of the discharge circuit.
time constant ............................... ms
(3)
(b)
Explain why the rate of discharge will be greater if the fixed resistor has a smaller
resistance.
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(2)
(Total 5 marks)
Page 35 of 74
33
The voltage across a capacitor falls from 10 V to 5 V in 48 ms as it discharges through a resistor.
What is the time constant of the circuit?
A
24 ms
B
33 ms
C
69 ms
D
96 ms
(Total 1 mark)
34
This question is about capacitor charging and discharging.
A student designs an experiment to charge a capacitor using a constant current. The figure
below shows the circuit the student designed to allow charge to flow onto a capacitor that has
been initially discharged.
The student begins the experiment with the shorting lead connected across the capacitor as in
the figure above. The variable resistor is then adjusted to give a suitable ammeter reading. The
shorting lead is removed so that the capacitor begins to charge. At the same instant, the stop
clock is started.
The student intends to measure the potential difference (pd) across the capacitor at 10 s intervals
while adjusting the variable resistor to keep the charging current constant.
The power supply has an emf of 6.0 V and negligible internal resistance. The capacitor has a
capacitance of 680 µF. The variable resistor has a maximum resistance of 100 kΩ.
(a)
The student chooses a digital voltmeter for the experiment. A digital voltmeter has a very
high resistance.
Explain why it is important to use a voltmeter with very high resistance.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
Page 36 of 74
(b)
Suggest one advantage of using an analogue ammeter rather than a digital ammeter for
this experiment.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(c)
Suggest a suitable full scale deflection for an analogue ammeter to be used in the
experiment.
full scale deflection = ............
(2)
(d)
The diagram shows the reading on the voltmeter at one instant during the experiment. The
manufacturer gives the uncertainty in the meter reading as 2%.
Calculate the absolute uncertainty in this reading.
uncertainty = ............V
(1)
Page 37 of 74
(e)
Determine the number of different readings the student will be able to take before the
capacitor becomes fully charged.
number = ............
(3)
(f)
The experiment is performed with a capacitor of nominal value 680 µF and a manufacturing
tolerance of ± 5 %. In this experiment the charging current is maintained at 65 µA. The data
from the experiment produces a straight-line graph for the variation of pd with time. This
shows that the pd across the capacitor increases at a rate of 98 mV s–1.
Calculate the capacitance of the capacitor.
capacitance = ............µF
(2)
(g)
Deduce whether the capacitor is within the manufacturer’s tolerance.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(h)
The student decides to confirm the value of the capacitance by first determining the time
constant of the circuit when the capacitor discharges through a fixed resistor.
Describe an experiment to do this. Include in your answer:
•
•
•
a circuit diagram
an outline of a procedure
an explanation of how you would use the data to determine the time constant.
Page 38 of 74
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 15 marks)
Page 39 of 74
35
Switch S in the circuit is held in position 1, so that the capacitor C becomes fully charged to a pd
V and stores energy E.
The switch is then moved quickly to position 2, allowing C to discharge through the fixed resistor
R. It takes 36 ms for the pd across C to fall to
What period of time must elapse, after the
switch has moved to position 2, before the energy stored by C has fallen to
A
51 ms
B
72 ms
C
432 ms
D
576 ms
?
(Total 1 mark)
Page 40 of 74
Mark schemes
1
2
D
[1]
(a)
(i)
determine area under the graph
[or determine area between line and time axis] ✓
1
(ii)
as seen
line starts at very low current (within bottom half of first square) ✓
either line continuing as (almost) horizontal straight line to end ✓✓
or
very slight exponential decay curve ✓
which does not meet time axis ✓
OR suitable verbal comment that shows appreciation of difficulty of
representing this line on the scales involved ✓✓✓
Use this scheme for answers which treat the information in the
question literally.
3
as intended
line starts at half of original initial current ✓
slower discharging exponential (ie. smaller initial gradient)
than the original curve ✓
correct line that intersects the original curve
(or meets it at the end) ✓
Use this scheme for answers which assume that both resistance
values should be in Ω or kΩ.
½ initial current to be marked within ±2mm of expected value.
3
(b)
(i)
energy stored (= ½ CV2) = ½ × 0.12 × 9.02 ✓ ( = 4.86 (J) )
4.86 = 3.5 Δh ✓
gives Δh = (1.39) = 1.4 (m) ✓
to 2SF only ✓
SF mark is independent.
Students who make a PE in the 1st mark may still be awarded the
remaining marks: treat as ECF.
4
Page 41 of 74
(ii)
energy is lost through heating of wires or heating the motor
(as capacitor discharges) ✓
Allow heating of circuit or I2 R heating.
energy is lost in overcoming frictional forces in the motor
(or in other rotating parts) ✓
Location of energy loss (wires, or motor, etc) should be indicated in
each correct answer.
[or any other well-expressed sensible reason that is valid
e.g. capacitor will not drive motor when voltage becomes low ✓ ]
Don’t allow losses due to sound, air resistance or resistance (rather
than heating of) wires.
max 2
[10]
3
(a)
(i)
7.5 × 10−6 (C) or 7.5 µ (C)
B1
1
(ii)
Suitable scale and charge from (i) correctly plotted at 2.5 V
Large square = 1 or 2 µC or
With false origin then large square = 0.5 µC
B1
Only a Straight line drawn through or toward origin
C1
Line must be straight, toward origin and only drawn between
2.5 V and 1.2 V (± 1 / 2 square on plotted points)
A1
3
(b)
Attempted use of E= ½ CV2 Or attempted use of E=½ QV
C1
9.38 ( µ J) − 2.16 ( µ J) seen
or E = ½ × 3 × 10−6 × 2.52 ‒ ½ × 3 × 10−6 × 1.22 seen
or E = ½ × 3 × 10−6 × (2.52 ‒ 1.22) seen
or E =½ × 7.5 × 10−6 × 2.5 ‒ ½ × 3.6 × 10−6 × 1.2 seen
C1
7.2 × 10−6 (J) c.a.o
A1
3
Page 42 of 74
(c)
(i)
Use of V = V0
or equivalent with
Q = Q0
C1
R=−
or R = −
or R =
C1
636 or 640 (Ω)
A1
3
(ii)
Current decreases (I = V / R) / describes rate of flow of
electrons decreasing / rate of flow of charge decreases
M1
Charge lost more slowly so pd falls more slowly because
V∝Q or Q=CV where C is constant
A1
MAX 2
[12]
4
(a)
(i)
Q(= It) 4.5 × 10–6 × 60 or = 2.70 × 10–4 (C) ✓
✓ = 6.1(4) × 10–5 = 61 (μF) ✓
3
(ii)
since VC was 4.4V after 60s, when t = 30s VC = 2.2 (V) ✓
[ or by use of Q = It and VC = Q / C ]
∴ pd across R is (6.0 – 2.2) = 3.8 (V) ✓
= 8.4(4) × 105 (Ω) ✓ (=844 kΩ)
In alternative method,
Q = 4.5 × 10–6 × 30 = 1.35 × 10–4 (C)
VC = 1.35 × 10–4 / 6.14 × 10–5 = 2.2 (V)
(allow ECF from wrong values in (i)).
3
Page 43 of 74
(b)
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 gives a coherent and logical description of the flow of electrons taking
place during the charging and discharging processes, indicating the correct directions
of flow and the correct time variations. There is clear understanding of how the pds
change with time during charging and during discharging. The candidate also gives a
coherent account of energy transfers that take place during charging and during
discharging, naming the types of energy involved. They recognise that the time
constant is the same for both charging and discharging.
A High Level answer must contain correct physical statements
about at least two of the following for both the charging and the
discharging positions of the switch:• the direction of electron flow in the circuit
• how the flow of electrons (or current) changes with time
• how V R and / or VC change with time
• energy changes in the circuit
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 has a fair understanding of how the flow of electrons varies with time,
but may not be entirely clear about the directions of flow. Description of the variation
of pds with time is likely to be only partially correct and may not be complete. The
candidate may show reasonable understanding of the energy transfers.
An Intermediate Level answer must contain correct physical
statements about at least two of the above for either the charging
or the discharging positions of the switch.
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 is likely to confuse electron flow with current and is therefore unlikely
to make effective progress in describing electron flow. Understanding of the variation
of pds with time is likely to be quite poor. The candidate may show some
understanding of the energy transfers that take place.
A Low Level answer must contain a correct physical statement
about at least one of the above for either the charging or the
discharging positions of the switch.
Incorrect, inappropriate or no response: 0 marks
Page 44 of 74
No answer, or answer refers to unrelated, incorrect or inappropriate physics.
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.
Charging
• electrons flow from plate P to terminal A and from terminal B to plate Q
(ie. from plate P to plate Q via A and B)
• electrons flow in the opposite direction to current
• plate P becomes + and plate Q becomes –
• the rate of flow of electrons is greatest at the start, and decreases to zero
when the capacitor is fully charged
• VR decreases from E to zero whilst VC increases from zero to E
• at any time VR + VC = E
• time variations are exponential decrease for VR and exponential increase
for VC
• chemical energy of the battery is changed into electric potential energy
stored in the capacitor, and into thermal energy by the resistor (which passes
to the surroundings)
• half of the energy supplied by the battery is converted into thermal energy and
half is stored in the capacitor
Discharging
• electrons flow back from plate Q via the shorting wire to plate P
• at the end of the process the plates are uncharged
• the rate of flow of electrons is greatest at the start, and decreases to zero
when the capacitor is fully discharged
• VC decreases from –E to zero and VR decreases from E to zero
• at any time VC = – VR
• both VC and VR decrease exponentially with time
• electrical energy stored by the capacitor is all converted to thermal energy
by the resistor as the electrons flow through it and this energy passes to
the surroundings
• time constant of the circuit is the same for discharging as for charging
Any answer which does not satisfy the requirement for a Low Level
answer should be awarded 0 marks.
max 6
[12]
5
(a)
Q (= CV = 330 × 9.0) = 2970 (μC) (1)
E (= ½QV) = ½ x 2.97 × 10–3 × 9.0 = 1.34 × 10–2J (1)
[or E (= ½CV2) = ½ × 300 × 10–6 × 9.02 (1) = 1.34 × 10–2J (1)]
2
(b)
time constant (= RC) = 470 × 103 × 330 × 10–6 = 155 s (1)
1
Page 45 of 74
= 2970 × e–60/155
(c)
= 2020 (μC)
(allow C.E. for time constant from (b))
(1)
(allow C.E. for Q)
[or V = V0e–t/RC (1) = 9.0 e–60/155 (1)
= 6.11 V (1)]
3
[6]
6
(a)
(i)
2200 × 10–6 farads (C V–1 ) or 2200 μC V–1
or idea of capacitance measuring charge (or coulomb) per volt
or C = Q / V with terms defined
C1
the capacitor ‘stores’ 2200 μC of charge for a potential
difference of 1 volt
A1
(2)
(ii)
15 V is
the maximum safe voltage between the terminals of
the capacitor.
or
the maximum voltage that should be used across
the capacitor
or
the voltage at which the capacitor breaks down / insulator
conducts
B1
(1)
(b)
(i)
correct curvature starting at 6 V at time = 0
points plotted correctly at 3 and 6 minutes with reasonable curve
(2.2 V and 0.8 V)
B1
or at 3 V at 2.1 minutes and 1.5 V at 4.2 minutes if
‘half life’ calculated and used
allow ±0.5 small square
B1
(2)
Page 46 of 74
(ii)
time alarm rings read correctly from the graph at 2 V
(about 200 s but use candidate's graph condone any shape graph)
B1
(1)
(iii)
time constant = RC or (R =
) or time to halve = 0.69CR
C1
82 kΩ
A1
(2)
(iv)
cooking time ∝ CR ∝ R
or quotes V = V0 e–t / CR or 2 = 6 e–300 / CR
C1
resistance = 120 kΩ (124 kΩ)
A1
(2)
(v)
connect it in parallel (with the other capacitor)
or
replace capacitor with one of higher value (not just use a larger capacitor)
B1
(3)
[11]
7
8
9
C
[1]
B
[1]
(a)
(i)
0.02 C of charge produce a p.d. of 1 V between the two terminals
or 0.02 C of charge per unit p.d.
B1
(2)
(ii)
straight line through the origin
M1
correct gradient (possible check point 0.2 C at 10 V)
and graph line up to 20 V
A1
(2)
Page 47 of 74
(iii)
area between graph line and charge axis
(allow area under graph)
not area of the graph
not area under graph / 2
M1
from 0 to the required voltage or up to the required voltage
A1
or energy = ½QV or ½CV2
M1
read corresponding Q from the graph
A1
(only allow second mark if graph is straight line through the origin)
or C determined from gradient of graph and V given
(2)
(iv)
sketch showing two capacitors in parallel connected to a supply
B1
(1)
(b)
(i)
energy stored = 0.5 CV2
C1
4.0 J (condone 1 sf answer)
A1
(2)
(ii)
(useful) energy output = mgh
or
efficiency = useful energy out / energy input(in same time)
or
efficiency = useful power out / power input
C1
energy output = 0.15 × 9.8 × 0.8 = 1.18 J
C1
efficiency = 0.294 or 29.4%
e.c.f. from (b)(i)
A1
(allow 29% – 30%)
(3)
[11]
Page 48 of 74
10
(a)
(i)
initial discharge current
= 6.0 × 10–5 (A) (1)
1
(ii)
time constant is time for V to fall to (1/e) [or 0.368] of initial value (1)
pd falls to (6.0/e) = 2.21 V when t = time constant (1)
reading from graph gives time constant = 22 (± 1) (1)
unit: s (1) (ΩF not acceptable)
[alternatively accept solutions based on use of V = V0e–t/RC
eg 1.5 = 6.0 e–30/RC (1) gives RC =
(1) = 22 (1) s (1)]
4
(iii)
capacitance of capacitor C =
= 2.2 × 10–4 (F) = 220 (µF) (1)
1
Page 49 of 74
(iv)
energy
V2 (or energy = ½ CV2) (1)
= 0.10 gives =
(0.10)1/2 (1) (= 0.316)
V2 = 0.316 × 6.0 = 1.90 (V) (1)
reading from graph gives V2 = 1.90 V when t = 25 s (1)
[alternatively accept reverse argument:
ie when t = 25 s, V2 = 1.9 V from graph (1)
final energy stored = ½ × 2.2 × 10–4 × 1.92
= 3.97 × 10–4 (J) and initial energy stored = 3.96 × 10–3 (J) (1)
which is 10 × greater, so 90% of initial energy has been lost (1)]
[alternatively, using exponential decay equation:
use of V = V0e–t/R with t = 25 s and RC = 22 s gives V = 1.93 V (1)
energy
V2 (or energy = ½ CV2) gives
fraction of stored energy that is lost =
= 0.103 (1)
= 0.90 (1)]
3
(b)
(i)
initial energy stored is 4 × greater (1)
because energy
V2 (and V is doubled) (1)
2
(ii)
time to lose 90% of energy is unchanged because time constant
is unchanged (or depends only on R and C) (1)
1
[12]
11
(a)
1 coulomb of charge is stored for a p.d. of 1 V between
the plates
(or equivalent statement) Condone I coulomb per volt
B1
1
Page 50 of 74
(b)
(i)
Correct substitution in C =
(ignore powers of 10)
C1
Plate area = 4.65 × 10–3 m2 or C =
data
with correct
A1
Radius = (their area /3.14)1/2; 0.038(4 or 5) m if correct
B1
3
(ii)
E = ½ CV2 or correct numerical substitution or
E = ½ QV & Q = VC
C1
4.1(4) × 10–10 J
A1
2
(c)
Time constant = RC or Time to halve = 0.69 RC
or V = V0 e– t/RC
C1
Time to fall to 1/e (0. 19 ms) or time to halve (0. 13 ms)
or V0 = 6 V and correct coordinates of point on line
(0.6 ms max)
C1
8.1 - 8.6 MΩ
A1
3
[9]
12
(a)
Q = CV (1)
(= 4.7 × 10-6 × 6.0) = 28 × 10-6 C or 28 μC (1)
2
Page 51 of 74
(b)
E = ½CV2 (1)
= ½ × 4.7 × 10-6 × 2.02 (1)
= 9.4 × 10-6 J (1)
[or E = ½QV (1)
= ½ × 9.4 × 10-6 × 2.0 (1)
= 9.4 × 10-6 J (1)]
3
(c)
time constant is time taken for V to fall to
(1)
∴V must fall to 2.2 V (1)
time constant = 32 ms (1)
[or draw tangent at t = 0 (1)
intercept of tangent on t axis is time constant (1)
accept value 30 - 35 ms (1)]
[or V = V0 exp(-t / RC) or Q = Q0 exp(-t / RC) (1)
correct substitution (1)
time constant = 32 ms (1)]
3
(d)
time constant = RC (1)
R=
= 6800 Ω (1)
(allow C.E. for value of time constant from (c))
2
[10]
13
14
D
[1]
(a)
(i)
straight line through origin (1)
(ii)
(1)
(iii)
energy (stored by capacitor) (1) (or work done
(in charging capacitor))
3
Page 52 of 74
(b)
(i)
RC = 5.6 × 103 × 6.8 × 10–3 (1) (= 38.1 s)
V(= V0 e–t/RC) = 12 e–26/38.1 (1)
= 6.1 V (1) (6.06 V)
[or equivalent using Q = Q0e–t/RC and Q = CV]
(ii)
(RC)’ = 2.8 × 10 3 × 6.8 × 10–3 (1) (= 19.0 s)
V (= 6.06 e–14/19) = 2.9(0) V (1)
(use of V’ = 6.1 V gives V = 2.9(2) V)
(iii)
7
[10]
15
16
17
B
[2]
B
[1]
(a)
E
V2 (or E = 1/2CV2) (1)
pd after 25 s = 6 V (1)
2
Page 53 of 74
(b)
(i)
use of Q = Q0 e−t/RC or V = V0 e−t/RC (1)
(e.g. 6 = 12e−25/RC) gives e
=
and
= 1n 2 (1)
(RC = 36(.1) s)
[alternatives for (i):
V = 12 e−25/36 gives V = 6.0 V (1) (5.99 V)
or time for pd to halve is 0.69RC
RC =
(ii)
R=
(1) = 36(.2) s]
(1) = 5.3(0) × 104 Ω(1)
4
[6]
18
19
20
B
[1]
C
[1]
A
[1]
Page 54 of 74
21
(a)
(i)
tangent drawn at t = 0
M1
coordinates correct and manipulated correctly
0.015 to 0.020 (A) 15 mA – 20 mA
or V = 4000 V as in (ii) then I = 18 mA
A1
2
(ii)
V = 220 × their (i) condoning powers of 10
C1
about 4000 V (3300 – 4400 V)
A1
or use of V = Q/C; V = 100 mC/25 µF
C1
4000 V
A1
2
(iii)
more charge leads to increased potential difference across
the capacitor
M1
pd = VR + VC
or if VC increases then VR decreases
M1
(if VR falls) so I falls
A1
3
(b)
(i)
use of energy = ½ Q2/C or use of C = Q/V and ½ QV
C1
0.083(7) or 0.084 C
condone 0.083 C
A1
2
(ii)
power = 14 kW
B1
1
Page 55 of 74
(c)
time constant = 5.5 s
M1
sensible attempt to find the charge after 8.3 s – by
calculation or reading from graph
M1
about 78 mC and needs to be 85 mC/has not reached
85 mC so designer’s suggestion is not valid
A1
3
[13]
22
23
24
C
[1]
D
[1]
(a)
(i)
energy stored by capacitor (= ½ CV2)
= ½ × 70 × 1.22
(= 50.4) = 50 (J)
to 2 sf only
3
(ii)
energy stored by cell (= I V t) = 55 × 10–3 × 1.2 × 10 × 3600
(= 2380 J)
=
(ie about 50)
2
Page 56 of 74
(b)
capacitor would be impossibly large (to fit in phone)
capacitor would need recharging very frequently
[or capacitor could only power the phone for a short time]
capacitor voltage [or current supplied or charge] would fall
continuously while in use
max 2
[7]
25
(a)
ratio of charge to potential
C1
4.2 μC per volt etc
A1
2
(b)
(i)
method: time for voltage to half/tangent at
origin/use of decay equation/1/e value
B1
appropriate reading from graph (T½ = 440 or 450 μs)
B1
substitution into correct equation
B1
R correct for method (151/152/155 Ω)
B1
4
(ii)
B smaller than A M0
B discharges faster/A discharges slower
B1
reference to decay equation/calculation for B
B1
2
Page 57 of 74
(c)
E = ½ CV2 or ½ QV seen
C1
both 4.0 (V) and 0.9 (V)/16.8 (μC) and 3.8 (μC) seen
C1
31.9 (μJ)
A1
3
[11]
26
(a)
per unit potential difference
charge (stored)
[or C = Q/V where Q = charge (stored by one plate)
V = pd (across plates)
]
2
(b)
(i)
=
(or 2.2 μF)
= 2.2 × 10–6 (F)
2
(ii)
when t = time constant Q = 0.63 × 13.2 = 8.3 (μC)
[or = 0.63 × 13(.0) (from graph) = 8.2 (μC)]
reading from graph gives time constant = 15 (± 1) (ms)
2
(iii)
resistance of resistor
=
= 6820 (Ω)
1
(iv)
gradient = current
1
(c)
(i)
maximum current =
=
= 0.88 (mA)
[or value from initial gradient of graph: allow 0.70 – 1.00 mA for this approach]
1
Page 58 of 74
(ii)
curve starts at marked lmax on l axis and has decreasing negative gradient
line is asymptotic to t axis and approaches ≈ 0 by t = 60 ms
2
[11]
27
28
A
[1]
(a)
time to halve = 0.008 s or two coordinates correct
C1
C = T1/2/(0.69 × 150) or eg 0.4 = 1.4 e–0.015/150C
A1
77 μF (consistent with numerical answer)
A1
3
(b)
max 3 from
as capacitor discharges:
pd decreases
B1
current through resistor decreases (since I
V)
B1
rate at which charge leaves the capacitor decreases (since I = ∆Q/∆t)
B1
rate of change of charge is proportional to rate of change of pd
(since V Q)
B1
condone quicker discharge when pd is larger
B1
3
Page 59 of 74
(c)
energy stored ∝ V 2 or use of ½ CV 2
or initial energy = 78.4 (or 75.5) μJ
or final energy using V = 0.38–0.4 0 V
(answer in range 5.6 – 6.4 μJ)
C1
fraction remaining = (0.4/1.4)2 or 0.072 – 0.081
or energy lost = 72 μJ
C1
91.8 to 92.8% lost
A1
3
(d)
(i)
charge = 77 μC to 82 μC
B1
1
(ii)
charge required = 77 × 10–6 × 5 × 3.15 × 107 (= 12128 C)
or 1A–h =3600 C
C1
3.36(3.4) Ah
A1
2
[12]
29
30
31
D
[1]
D
[1]
D
[1]
Page 60 of 74
32
(a)
(Q = Q 0e−t /RC gives )1.0 = 4.0e−300 / RC
from which
[Alternative answer:
time constant is time for charge to decrease to Q0 /e [or 0.37 Q0 ]
4.0/e = 1.47
reading from graph gives time constant = 216 ± 10 (ms)
In alternative scheme, 4.0/e = 1.47 subsumes 1st mark. Also, accept
T½ = 0.693 RC (or = ln 2 RC) for 1st mark.
3
(b)
current is larger (for given V)(because resistance is lower)
[or correct application of I = V / R]
current is rate of flow of charge
[or correct application of I =Δ Q / Δt]
larger rate of flow of charge (implies greater rate of discharge)
[or causes larger rate of transfer of electrons from one plate back to the other]
[Alternative answer:
time constant (or RC) is decreased (when R is decreased)
explanation using Q = Q 0e−t / RC or time constant explained
]
Use either first or alternative scheme; do not mix and match.
Time constant = RC is insufficient for time constant explained.
max 2
[5]
33
34
C
[1]
(a)
Capacitor must not lose charge through the meter ✓
1
(b)
Position on scale can be marked / easier to read quickly etc ✓
1
(c)
Initial current =
= 60.0 μA ✓
100 μA or 200 μA ✓ (250 probably gives too low a reading)
Give max 1 mark if 65 μA (from 2.6) used and 100 μA meter chosen
2
(d)
0.05 V ✓
1
Page 61 of 74
(e)
Total charge = 6.0 x 680 x 10-6 (C) (= 4.08 mC) ✓
Time = 4.08 x 10-3 / 60.0 x 10-6 = 68 s ✓
Hence 6 readings ✓
3
(f)
Recognition that total charge = 65 t μC and final pd = 0.098 t
so C = 65μ / 0.098✓
660 μF ✓
Allow 663 μF
2
(g)
(yes) because it could lie within 646 – 714 to be in tolerance ✓
OR
it is 97.5 % of quoted value which is within 5% ✓
1
Page 62 of 74
(h)
Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals ✓
Plot ln V or ln I versus time ✓
gradient is 1 / RC ✓
OR
Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals✓
Use V or I versus time data to deduce half-time to discharge ✓
1 / RC = ln 2 / t½ quoted ✓
OR
Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals ✓
Plot V or I against t and find time T for V or I to fall to 0.37 of initial value ✓
T = CR ✓
Either A or V required
For 2nd mark, credit use of datalogger for recording V or I.
4
[15]
35
B
[1]
Page 63 of 74
Examiner reports
1
2
This question was similar but a little more demanding, because its facility was 67%. The 15% of
students who gave distractor B may have had difficulty in combining mF with μA, because they
arrived at an answer of 100 s instead of 1000 s.
AQA apologises for the unfortunate typographical error which crept in to the resistor values in
part (a) of this question. Both values ought to have been given in kΩ. The majority of students
actually answered the question as it had been intended to appear, and so the mark scheme that
would have applied to the intended question in part (ii) was used when marking their work. The
students who answered the question as it appeared in the paper were not disadvantaged,
because an alternative mark scheme which gave full credit for completely correct responses was
adopted for them. The main weaknesses in either approach were a lack of appreciation of the
effect of the resistor value on the initial current, and doubt as to whether increasing resistance
would speed up the decay or slow it down. In the case of the question as it had been intended,
the curve crosses the original curve within the time scale of the graph; this was rarely spotted
and so the award of all three marks was quite unusual. The determination of the initial charge on
the capacitor from the area under the curve was not as well known as expected. There were
many references to the intercept on the current axis, to the initial gradient, and some to “the initial
area under the curve”.
Most students wrote correct and complete answers in part (b)(i). A small number of students
mixed up mass and weight, leading to the unnecessary introduction of g into the calculation. The
more able ones who had done this then realised that g was self-cancelling. It was pleasing to see
that an appropriate number of significant figures was generally quoted in the final answers. The
reasons looked for in the answers to part (b)(ii) were those which cause the greatest energy loss
as the weight is raised by an electric motor: losses caused by heating of the connecting wires or
the motor and energy lost in overcoming frictional forces as the motor rotates. By comparison,
the energy lost in overcoming air resistance (for example) is trivial and was therefore discounted.
Examiners expected to see that the locations of the energy losses (wires, motor, circuit, etc) were
identified in acceptable answers.
3
(a)
(b)
(i)
This calculation was very well done with over 90% of candidates able to complete it
successfully. Mistakes seen were mostly power of ten errors on the value of the
capacitor (3 × 10−3 often quoted).
(ii)
The majority of candidates were able to achieve at least 2 marks. Of course, many
candidates attempted to draw discharge curves instead of the required straight line.
Candidates should treat these graphs with caution and be aware of the Q∝ V is
always directly proportional irrespective of charging and discharging.
There were lots of mistakes in this calculation. The most common error was the use of E =
CV2 where candidates substituted ΔV = 1.3 V into this formula, effectively calculating
1.32 instead of calculating 2.52 × 1.22. Another common mistake seen was treating Q as a
constant in the equation E=
QV , that is E =
× 7.2 × 10−6×ΔV
Page 64 of 74
(c)
(i)
A common mistake seen here was the use of the wrong formula, a significant number
of candidates chose to use
= 0.69 RC even though the fall in voltage was not
quite 50%. Other candidates selected the correct formula but then had difficulty in
their rearranging of the formula; many made the mistake of misplacing C in the
rearrangement of the equation: R =
(ii)
4
instead of R =
Candidates found this explanation difficult, with only the best candidates able to
deliver a detailed, coherent response. Most answers were limited to less charge less
potential difference arguments rather than dealing with the rate aspect of the
question.
The capacitance calculation in part (a)(i) rewarded most students with full marks. Answers to part
(a)(ii) made a distinct contrast, because relatively few students were able to progress. Correct
answers were rare. The circuit in Figure 1 is one in which the current is maintained constant by
reducing the resistance as the capacitor is charged. Consequently the large number of attempted
solutions that introduced exponential decay equations were totally inappropriate. An
understanding of the principle that in a series circuit the sum of the voltages across components
is equal to the applied voltage was essential. Many of the efforts progressed as far as
establishing that the pd across the capacitor at 30 s would be 2.2V, but then went on to find what
is effectively “the resistance of the capacitor” by dividing 2.2V by the current.
The final question in this examination, part (b), concerned a C-R circuit is which R is constant
and charging / discharging are exponential processes. Apart from testing this subject content, the
question was also used to assess the communications skills of the students. The guidance given
in the bullet points helped most students to organise their answers systematically. A very good
spread of marks was seen, ranging from students who clearly knew everything that happens
during charging and discharging to ones who understood little or nothing about capacitors. A
large number of correct statements about the factors listed in the bullet points for both charging
and discharging constituted a high level answer (5 / 6 marks). Fewer correct statements about
either charging or discharging put answers into the intermediate level (3 / 4 marks) whilst even
fewer correct statements put answers into the low level (1 / 2 marks). Contributing also to the
overall assessment was examiners’ consideration of the incorrect statements made in the
answers, and how satisfactorily the answers had been had been written. There were many
instances of answers in which it was stated that electrons passed directly from plate Q across the
gap to plate P – these tended to condemn the knowledge of the student concerned. A common
misapprehension concerning this circuit was that the reduction in current is caused by an
increase in the resistance of the capacitor rather than by a decrease in the net potential
difference as the capacitor charges or discharges. A large proportion of the students chose to
ignore the advice given to refer in their answers to points A, B, P and Q in the circuit. This
omission usually made their answers somewhat more difficult to assess.
Page 65 of 74
5
The mathematical competence of the majority of candidates in this question was much better
than has been seen in several recent papers and full marks were frequently awarded. Previous
reports have emphasised that ½ CV2 is a safer route to the energy of a capacitor than ½ QV, and
in part (a) the message appeared to have got through to the candidates. In part (b) the main
problems appeared to be with the meaning of micro in μF and of kilo in kΩ; the unit of time
constant was expected to be shown as s and not ΩF.
The exponential decay equation was usually used correctly in part (c), where approaches via
Q = Q0 e–t/RC and V = V0 e–t/RC were equally valid. Only a tiny minority of the candidates
attempted any other approach and almost all of them were wrong.
6
(a)
(b)
(i)
This is the basic definition in the work on capacitors and the number of weak or
incorrect responses was very disappointing if not disturbing. Many stated that 2200
μF meant that 2200 μF of charge could be stored on the capacitor (some did add ‘per
volt’). Some defined capacitance as charge per volt without reference to the value of
the capacitance.
(ii)
It was not always clear that candidates knew that this was the maximum potential
difference that can exist between the terminals of a capacitor without causing
damage. Many candidates thought it was the voltage at which the capacitance would
be 2200 μF or that it was the voltage at which 2200 μC of charge would be stored.
(i)
Some candidates demonstrated excellent graph drawing skills but sketches were
frequently poor. The starting voltage and the direction of the curvature was most
often correct but many candidates took no care in producing realistic values after
times of one and two time constants. Many associated time constant with the time to
halve.
(ii)
Most read the value correctly from their graph. There were those, however, who were
unable to read the graph accurately giving answers only to the nearest minute.
(iii)
This was generally well done but, again, confusion between time constant and half life
was a frequent cause of lost marks. Neglecting to convert minutes to seconds was
also a common error.
(iv)
This part was rarely correct. Most candidates assumed a time constant of 5 minutes.
It was necessary for candidates to take one of two approaches. They could calculate
the time constant necessary for the voltage to fall from 6 V to 2 V and hence
determine the resistance. More simply they could use the fact that the cooking time
would be proportional to the time constant and, hence, the resistance.
(v)
This part was usually completed successfully. A statement that a larger capacitor
would be used without stating how it would be connected was inadequate as this did
not answer the question.
Page 66 of 74
9
(a)
(b)
10
(i)
‘The capacitor stores 0.02 F of charge per volt’ was not an uncommon answer. Many
simply defined capacitance and some defined the farad. Candidates are required to
apply their knowledge to the context of the question.
(ii)
Failure to include a unit on the axis was a common error. There were many curves
seen, sometimes even when part (i) was correct, which demonstrated poor
understanding.
(iii)
Many correctly stated that the area under the graph would be determined but did not
say what area would be used for any given voltage. Some quoted ½ CV2 but did not
link this with the graph as required by the question.
(iv)
Most were successful in this part but there were many who drew capacitors in series
and some who did not use a capacitor symbol or label their ‘boxes’ as capacitors. A
few, although knowing the symbol, had no idea how to draw them in either a series or
parallel circuit.
(i)
Only very weak candidates failed to gain some marks in part (b). In this part the main
error was thinking that the answer was given by mgh. This led to problems for some
candidates in part (ii).
(ii)
The majority of candidates obtained the correct answer. Even those who were
confused in (i) often recovered and used energies correctly in this part.
Most candidates were able to calculate the initial discharge current successfully in part (a) (i).
The common approaches to finding the time constant in part (a) (ii) were reading from the graph
at the point where the pd had fallen to 6.0/e, or solving the exponential equation V = V0e–t/RC for
corresponding V and t values. It was expected that candidates would know that a time constant
is measured in s; the unit ΩF was not accepted. The principal difficulty experienced by some
candidates in part (a) (iii) was not spotting that the capacitance value had to be expressed in μF.
Answers of 2.2 × 10–4 μF were clearly wrong and caused this mark to be lost.
A wide variety of approaches could be adopted when answering part (a) (iv). Most candidates
attempted to answer the question ‘in reverse’, by showing that after 25 s the energy lost would be
90% of the original; this was acceptable. Some lost the third mark when using this method by
failing to link the two energies (calculated correctly) to the 90% value for the loss. A neat and
concise solution was seen in a few cases, where candidates reasoned that, since E
V2, the
percentage loss of energy would be [1 – (V/V0)2] × 100.
In part (b) (i) one mark was awarded for a correct consequence of the increased charging pd,
V2 was quite well known, and it was
and one mark for an explanation. The fact that E
expected that candidates would realise that doubling the pd would quadruple the energy stored:
to state just that the energy stored would ‘increase’ was too simplistic to deserve this mark.
Candidates who resorted to E = ½ QV almost invariably reached the wrong conclusion, because
they thought that the energy stored would double.
In part (b) (ii) the one available mark was given for a correct consequence together with an
acceptable explanation. Relatively few candidates were able to state that the time taken for 90%
of the energy to be lost would be unchanged because the time constant had not altered.
Page 67 of 74
11
(a)
Most of the candidates were able to provide an acceptable explanation. Some weaker
candidates defined capacitance in general and others thought that the farad was a charge
unit.
(b)
(i)
Failure to use correct powers of 10 for the separation and an incorrect formula for
area were not uncommon but most were able to gain 2 of the 3 available marks.
(ii)
This was generally well done. Some did not square V having quoted the formula
correctly and others forgot the ½.
(c)
12
Most candidates knew what was to be done but many made errors somewhere along the
way. When quoting formulae candidates need to be absolutely clear as to whether the ‘T’
they use is the period or the time to halve. As a start point T = RC and T = 0.69 RC were
both common and unless they proceeded correctly this gained no compensation mark.
There were many who misread the scales. Use of 0.13 s for the time to halve was common
as was 0.65 ms and 1.25 s.
It was satisfying to see so many excellent answers to a question on a subject area that has
caused problems in the recent past, and also on those sections testing parts of the specification
dealing with the mathematics of exponential discharge, which have been re-introduced at A level.
Part (a) only seemed to trouble those candidates who had not learnt Q = CV, together with those
who did not know that 1 μF = 10-6 F. When finding the stored energy in part (b), many more
candidates realised that E = ½CV2 is a safer approach than E = ½QV, but the latter equation
also provided a large number of correct answers.
Three alternative routes were possible when answering part (c). Most candidates preferred to
start from the exponential decay equation (either in terms of V or in terms of Q), substituted
values, took logs and proceeded to a solution. It was pleasing that so many succeeded. The
most elegant solutions came from the candidates who knew that the charge stored falls to (1 / e)
of the initial charge in a time equal to the time constant. Solutions that made use of the gradient
of the initial section of the graph were exceedingly rare. Part (d) was usually well rewarded in
most scripts, with candidates working from their knowledge of the time constant as RC.
13
This question required candidates to be familiar with capacitor discharge and the concept of time
constant. Almost half of them chose the correct response, but 27% of them thought that the
remaining charge would be Q0 e2 after a time of 2RC. Perhaps this was caused by misreading
(Q0e2) as (Q0/e2).
Page 68 of 74
14
This question was often well answered, with marks of 9 or 10 frequently being awarded. Part (a)
(ii) proved troublesome for most. Although almost all candidates recognised that V = Q/C would
lead to a straight line through the origin, relatively few were sufficiently alert to spot that the
gradient was 1/C; a far more popular choice was C. Most knew that the area represented energy
(or work done).
In part (b), the two resistors in parallel posed a problem for some, but there were many correct
solutions to (b) (i). The principal errors in (b) (ii) were to take the wrong resistance value (11.2 kΩ
instead of 2.8 kΩ), or to use the wrong time (40 s instead of 14 s), or both. The sketch graphs in
(b) (iii) were often drawn well, even by some candidates who had not been successful with the
previous calculations. Examiners were expecting the exponential decay curve to start at t = 0 and
to become steeper after a discontinuity at t = 26 s. Some candidates drew a linear decay graph,
whilst others showed an exponential curve passing continuously through t = 26s.
15
16
17
When pre-tested, the facility of this question on capacitor discharge, was only 48%. In the
examination no fewer than 75% gave the correct response, a remarkable improvement.
Charging a capacitor by the use of a constant current was the subject in this question, where
77% of the candidates had no difficulty in combining Q = CV and Q = It to achieve the desired
result. This question showed the largest advance in facility over the pre-testing value of any of
the questions used in this test.
Candidates with a sound knowledge of capacitors and capacitor discharge had little difficulty in
gaining all six marks. However, it did seem that some centres had not been able to cover these
areas fully (if at all) in time for the January examination; candidates from such centres were
frequently unable to make anything of the complete question.
Almost inevitably, misunderstanding of E = ² QV in part (a) led many candidates to believe that
the pd at 25 s would be 3 V. These candidates were then unable to arrive at a time of 36 s for the
time constant in part (b), but could still access both marks in part (b) (ii). Many excellent
responses were seen in part (b) (i), where familiarity with logarithmic solutions to exponential
relationships was almost essential. Examiners gave no credit in part (a) to those candidates who
attempted an exponential solution by using the 36 s given in part (b); a successful solution had to
come from the energy information. Similarly, only one of the two marks in part (b) (i) was
available for those who turned the question on its head by showing that V would be 6 V after 25
s, if the time constant were 36 s.
Page 69 of 74
18
19
This question which required application of a knowledge of time constant by correctly using the
equation V = V0e-t/RC, discriminated between the strongest and weakest students better than any
other question in this test. The facility of this question was 71%.
In this question, candidates were required to determine the value of the time constant of an RC
circuit from data on a Q - t graph. The answer could be found in various ways. Perhaps the
quickest was to read off the time at which the charge had fallen to Q0/e, 0.036 µC, but many
candidates would have resorted to substituting values into Q = Q0e-t/RC. Almost 70% of responses
were correct. The remaining five questions were all set on section 3.4.5 of the specification,
magnetic fields.
20
This question was about capacitor discharge through a resistive circuit. It was answered
correctly by 66% of the candidates. The most popular incorrect response was distractor B, which
showed the correct shape for the discharging section but had the wrong shape for the charging
section.
Page 70 of 74
21
Candidates were penalised heavily for poor technique in this part (a) (i). It was expected that, at
this level candidates, would draw a tangent to the curve and not simply read off coordinates in
the first few small squares of the graph to find the initial rate of change of charge. Some who
appreciated that a tangent would be useful drew one in the wrong place.
Most candidates used the V = IR approach in part (a) (ii). Some candidates incorporated the 510
Ω heart resistance into the resistance of the charging circuit.
Part (a) (iii) was not done well and very few candidates gave completely convincing arguments
and many who seem to have no idea where to start. There were many candidates who wrote
about charge build up on the capacitor increasing the resistance of the capacitor. Others
appreciating the rise in pd as charge accumulates on the capacitor stated that this reduced the
emf of the supply. Appreciation of the reduction in the pd across the resistor as the pd across the
capacitor increases was rare. Arguments such as ‘charge accumulating repels other charges’
gained some credit. There was a significant proportion who wrote that as charge builds up there
is less space on the capacitor for more charge or that because charge has been added to the
capacitor there is less available in the circuit to provide further charge.
A surprising number failed to obtain the correct answer to part (b) (i). Many used ½ QV or ½ CV2
assuming V to be the value of the emf that had been calculated in (a) (ii).
Those who failed to obtain 14000 W in part (b) (ii) usually failed because of problems with
powers of 10.
Only the more able candidates made progress with part (c). Most of these obtained the time
constant correctly. Fewer went on to determine the charge after 8.3 s using the graph and to
compare this with 85 mC required or to state that the graph showed that about 10.5 s was
needed to achieve the 85 mC. There was a small minority of candidates who showed that the
energy stored would be insufficient after 8.3 s although this was not necessary.
22
23
This question was the best discriminator on this paper. Its facility was 59%. Satisfactory answers
required the equation 0.5 = e-10/RC to be solved for RC. Not even confusion with ‘half life’ could
account for the 26% of candidates who chose option B, 5 s. The remaining questions all tested
various aspects of electromagnetism.
The constant current idea also cropped up in this question, which was rather more demanding.
Combining Q = CV with Q = I t, and appreciating that Q will fall by 10% when V falls by 10%,
ought to have brought a satisfactory outcome. Just about half of the candidates succeeded with
this, and wrong answers were mostly split between distractors B and C.
Page 71 of 74
24
The data used in this question is realistic. A low voltage 70 F capacitor is available for back-up
purposes, and there is a rechargeable cell with the specification quoted. Part (a)(i) was readily
answered by the application of ½ CV2. The choice of an inappropriate number of significant
figures, typically three, caused the loss of a mark. Candidates should realise that a final value
should only be quoted to two significant figures when the data in the question is given to no more
than two significant figures.
Part (a)(ii) was answered poorly, usually because the calculation was approached from the
capacitor energy equation (½ QV), instead of that giving the energy delivered by a cell (QV).
Examiners were ready to penalise the candidates who, having started from the wrong principle,
introduced a mysterious factor of two in order to show that the energy stored was 50× greater,
rather than 25× greater.
In part (b) candidates’ responses were often inadequate because of incompleteness, and/or an
inability to express ideas sufficiently clearly. It was expected that satisfactory answers would
relate to the use of the capacitor in a cordless telephone. ‘A capacitor discharges quickly’ is an
incomplete answer; ‘a capacitor would need recharging frequently’ or ‘a capacitor would only
power the phone for a short time’ were much more explicit in the context of the question. Other
acceptable answers were that a 70 F capacitor would be too large to fit in the telephone, or that
the voltage supplied by it would decrease continuously whilst in use.
25
For part (a), most candidates knew the definition of capacitance, but frequently omitted the 4.2
μC per V aspect.
There was a variety of techniques used in part (b) (i) but most answers were complete.
In part (b) (ii), nearly all candidates recognised that B had a smaller capacitance than A but most
answers only gained a single mark for stating that the discharge happens faster in B without
explaining why this meant that the capacitance was smaller.
Few candidates correctly calculated the change in energy in part (c) – most used ½ QV but did
not calculate the values of Q and V before and afterwards.
Page 72 of 74
26
Good definitions of capacitance were usually seen in part (a), leading to an easily gained couple
of marks. Vague statements such as ‘capacitance is a measure of the ability to store charge’
went unrewarded.
Part (b) caused very few problems, apart from those arising from careless arithmetic or
misunderstood powers of 10. In part (b) (iv) it was expected that students would identify the
gradient with current; ‘rate of charging’ seemed too obvious for the mark to be given.
The maximum value of the current could be found directly in part (c) (i) by applying I = V/R,
where V is the emf of the battery and R is the resistance of the resistor. No doubt it was the
previous part of this question that directed so many students to base their response on the initial
slope of the graph. This was equally acceptable, and a wide tolerance was placed on answers
arrived at by this technique. The sketch graphs in part (c) (iii) were often too careless to deserve
full credit. This exponential decay curve should start at an intersection with the current axis (with
Imax marked as required) and should be asymptotic to the time axis. More able students realised
that the gradient of this Q-t graph in part (b) was practically zero at t = 60 ms, and that the current
should therefore be very close to zero at this time. There were many answers showing a current
that increased with time, and many others that had a constant negative gradient.
27
28
This question, about factors affecting time constant in an RC circuit, was the easiest of the three
with a facility of 75%.
Most students made progress with part (a) and a majority of the students gained full marks.
Part (b) was not done well. Few could provide a complete response and some confused charging
and discharging.
Few could cope with all the steps necessary to arrive at the correct answer to part (c).
There were surprisingly few correct responses to part (d) (i) and part (d) (ii) was not done well.
The number of seconds in a year is on the Data and Formulae Booklet but many tried
unsuccessfully to calculate it. Few seemed familiar with the unit Ah (ampere–hour).
29
This question was about charge-time and current-time graphs for a capacitor being charged
through a resistor. The facility of the question, at 63%, hardly changed from the value obtained
when used previously. Even the barest acquaintance with capacitors should have indicated to
students that the charge must increase (graph 2) and also that, because charge cannot flow
permanently through a capacitor, the current must fall to zero (graph 1). However, 16% of
responses were for distractor C, where both charge and current increase with time.
Page 73 of 74
31
32
This question to be rather more demanding than those who faced this question when it appeared
in a previous test, thus causing the facility to decrease from 69% to 64%. Students who were
successful here had to understand the meaning and significance of the time constant of an RC
circuit, and in particular that the pd would decrease to 1/e of its initial value after a time equal to
one time constant.
Part (a) required the evaluation of the time constant of an RC circuit from data on a graph of
charge against time. This proved to be an easy test, and marks were high. The most economical
solution followed from recognising that the charge falls to (1/e) of its initial value in a time equal to
the time constant, or from appreciating that Q0 becomes Q0/2 in a time equal to ln 2 RC. More
extensive answers that relied on a solution of Q = Q0e-t / RC were less common; in these it was
essential for candidates to show their working correctly for full marks to be accessible. A few
candidates knew that the time constant is equal to the time at which the capacitor would have
discharged completely had the initial current been maintained. Therefore they drew a tangent to
the curve at t = 0, continued the line to the time axis and then determined the required value by
reading off the time.
Careless use of the language of physical quantities was sometimes an obstacle to progress in
part (b). Loose terms such as “the current flows more quickly” (when the resistance is less)
should be avoided: the candidate should have stated that the current is greater, or that more
charge passes per second. The key to success in this part was to understand the meaning of a
rate of change. Those who stated that the current is larger, and that current is the rate of flow of
charge, readily scored both marks. Answers which stated that the time constant would be
decreased were also accepted but it was then necessary to make reference to the implications,
from Q = Q0e-t/RC, for the second mark to be awarded.
33
35
This question has been used to test candidates in earlier years. The former one showed an
improvement in facility and the latter one a deterioration. This question, which had a facility this
time of 76%, was a calculation on capacitor discharge. It could be answered either by full solution
of the exponential decay equation, or having recognised that the situation is equivalent to “halflife”, by using t = RC ln 2.
This question tested students’ understanding of exponential decay as well as energy storage.
The “half value period” of the RC circuit was 36ms; in this time the pd would decrease to V / 2
and the energy stored would fall to E / 4. In a further 36ms the energy stored would fall to E / 16.
62% of the responses were correct. Distractor C was selected by 17% of the students.
Page 74 of 74