Download A2 Capacitors - Animated Science

Capacitors Multi / Long Ans HW Questions
97 minutes
74 marks
Q1.
A 1000 μF capacitor and a 10 μF capacitor are charged so that the potential difference
across each of them is the same. The charge stored in the 100 μF capacitor is Q1 and the
charge
stored in the 10 μF capacitor is Q2. What is the ratio
A
100
B
10
C
1
?
D
(Total 1 mark)
Q2.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)
Q3.A nuclear fusion device is required to deliver at least 1 MJ of energy using capacitors. If the
largest workable potential difference is 10 kV, what is the minimum capacitance of the
capacitors that should be used?
A
0.01 F
B
0.02 F
C
2F
D
100 F
(Total 1 mark)
Q4.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)
Q5.An initially uncharged capacitor of capacitance 20 μF is charged by a constant current of 80 μA.
Which line, A to D, in the table gives the potential difference across, and the energy stored in,
the capacitor after 50 s?
potential difference / V
energy stored / J
A
4.0 × 10–3
2.0 × 10–3
B
4.0 × 10–3
4.0 × 10–1
C
2. 0 × 102
2.0 × 10–3
D
2. 0 × 102
4.0 × 10–1
(Total 1 mark)
Q6.Which one of the following statements about a parallel plate capacitor is incorrect?
A
The capacitance of the capacitor is the amount of charge stored by the capacitor when the
pd across the plates is 1V.
B
A uniform electric field exists between the plates of the capacitor.
C
The charge stored on the capacitor is inversely proportional to the pd across the plates.
D
The energy stored when the capacitor is fully charged is proportional to the square of the
pd across the plates.
(Total 1 mark)
Q7.
In experiments to pass a very high current through a gas, a bank of capacitors of total
capacitance 50 μF is charged to 30 kV. If the bank of capacitors could be discharged completely
in 5.0 ms, what would be the mean power delivered?
A
22 kW
B
110 kW
C
4.5 MW
D
9.0 MW
(Total 1 mark)
Q8.
A 400 µF capacitor is charged so that the voltage across its plates rises at a constant rate
from
0 V to 4.0 V in 20 s. What current is being used to charge the capacitor?
A
5 µA
B
20 µA
C
40 µA
D
80 µA
(Total 1 mark)
Q9.
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)
Q10.
A 10 mF capacitor is charged to 10 V and then discharged completely through a small
motor. During this process, the motor lifts a weight of mass 0.10 kg. If 10% of the energy stored
in the capacitor is used to lift the weight, through what approximate height will the weight be
lifted?
A
0.05 m
B
0.10 m
C
0.50 m
D
1.00 m
(Total 1 mark)
Q11.
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 Ris decreased.
D
After charging to the same voltage, the initial discharge current will be unaffected
ifC is increased.
(Total 1 mark)
Q12.
The graph shows how the charge stored by a capacitor varies with the potential difference
across it as it is charged from a 6 V battery.
Which one of the following statements is not correct?
A
The capacitance of the capacitor is 5.0 µF.
B
When the potential difference is 2 V the charge stored is 10 µC.
C
When the potential difference is 2 V the energy stored is 10 µJ.
D
When the potential difference is 6 V the energy stored is 180 µJ.
(Total 1 mark)
Q13.
In the circuit shown, the capacitor C is charged to a potential difference V when the switch
S is closed.
Which line, A to D, in the table gives a correct pair of graphs showing how the charge and
current 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)
Q14.
In experiments to pass a very high current through a gas, a bank of capacitors of total
capacitance 50 µF is charged to 30 kV. If the bank of capacitors could be discharged completely
in 5.0 m s what would be the mean power delivered?
A
22 kW
B
110 kW
C
4.5 MW
D
9.0 MW
(Total 1 mark)
Q15.
A 1.0 µF capacitor is charged by means of a constant current of 10 µA for 20 s. What is
the energy finally stored in the capacitor?
A
4.0 × 10–4 J
B
2.0 × 10–3 J
C
2.0 × 10–2 J
D
4.0 × 10–2 J
(Total 1 mark)
Q16.
(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 yaxis) 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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(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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(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)
Q17.
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)
Q18.
The graph shows how the charge stored by a capacitor varies with the potential difference
across it as it is charged from a 6 V battery.
Which one of the following statements is not correct?
A
The capacitance of the capacitor is 5.0 μF.
B
When the potential difference is 2 V the charge stored is 10 μC.
C
When the potential difference is 2 V the energy stored is 10 μJ.
D
When the potential difference is 6 V the energy stored is 180 μJ.
(Total 1 mark)
Q19.
A 10 mF capacitor is charged to 10 V and then discharged completely through a small
motor. During the process, the motor lifts a weight of mass 0.10 kg. If 10% of the energy stored
in the capacitor is used to lift the weight, through what approximate height will the weight be
lifted?
A
0.05 m
B
0.10 m
C
0.50 m
D
1.00 m
(Total 1 mark)
Q20.
The graph shows how the charge stored by a capacitor varies with the pd applied across
it.
Which line, A to D, in the table gives the capacitance and the energy stored when the potential
difference is 5.0 V?
capacitance/μF
energy stored/μJ
A
2.0
25
B
2.0
50
C
10.0
25
D
10.0
50
(Total 1 mark)
Q21.
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)
Q22.
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)
Q23.
A capacitor of capacitance C stores an amount of energy E when the pd across it
is V.Which line, A to D, in the table gives the correct stored energy and pd when the charge is
increased by 50%?
energy
pd
A
1.5 E
1.5 V
B
1.5 E
2.25 V
C
2.25 E
1.5 V
D
2.25 E
2.25 V
(Total 1 mark)
Q24.
A 400 μF capacitor is charged so that the voltage across its plates rises at a constant
ratefrom 0 V to 4.0 V in 20 s. What current is being used to charge the capacitor?
A
5 μΑ
B
20 μΑ
C
40 μΑ
D
80 μΑ
(Total 1 mark)
Q25.
A student was required to design an experiment to measure the acceleration of a heavy
cylinder as it rolled down an inclined slope of constant gradient. He suggested an arrangement
that would make use of a capacitor-resistor discharge circuit to measure the time taken for the
cylinder to travel between two points on the slope. The principle of this arrangement is shown in
the figure below.
S1 and S2 are two switches that would be opened in turn by plungers as the cylinder passed over
them. Once opened, the switches would remain open. The cylinder would be released from rest
as it opened S1. The pd across the capacitator would be measured by the voltmeter.
(a)
Describe the procedure the student should follow, including the measurements he should
make, when using this arrangement. Explain how he should use the measurements taken
to calculate the acceleration of the cylinder down the slope.
The quality of your written communication will be assessed in this question.
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(6)
(b)
When the student set up his experiment using the arrangement shown in the figure above,
he used a 22 μF capacitor, C, and a 200 kΩ resistor, R. In one of his results, the initial pd
was 12.0 V and the final pd was 5.8 V. The distance between the plungers was 2.5 m.
(i)
From the student’s result, calculate the time taken for the cylinder to reach the
second plunger.
answer = ................................... s
(3)
(ii)
What value does this result give for the acceleration of the cylinder down the slope,
assuming the acceleration is constant?
answer = ............................m s–2
(2)
(Total 11 marks)
Q26.
Capacitors and rechargeable batteries are examples of electrical devices that can be
usedrepeatedly to store energy.
(a)
(i)
A capacitor of capacitance 70 F is used to provide the emergency back-up in a
lowvoltage 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)
(b)
Give two reasons why a capacitor is not a suitable source for powering a
cordlesstelephone.
Reason 1.....................................................................................................
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Reason 2......................................................................................................
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(2)
(Total 7 marks)
Q27.
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)
Q28.
The flash tube in a camera produces a flash of light when a 180 μF capacitor is discharged
across the tube.
(a)
The capacitor is charged to a pd of 100 V from an electronic charging unit in the camera,
as shown in the diagram above.Calculate,
(i)
the energy stored in the capacitor,
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(ii)
the work done by the battery.
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(2)
(b)
When a photograph is taken, switch S in the diagram above is automatically moved from
A to B and the capacitor is discharged across the flash tube. The discharge circuit has a
resistance of 1.5 Ω. Emission of light from the flash tube ceases when the pd falls below
30 V.
(i)
Calculate the duration of the light flash.
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(ii)
The capacitor in the circuit in the diagram above is replaced by a capacitor of
greater capacitance. Discuss the effect of this change on the photograph image of a
moving object.
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(4)
(Total 6 marks)
Q29.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)
Q30.
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)
M1.
A
[1]
M2.D
[1]
M3.B
[1]
M4.C
[1]
M5.D
[1]
M6.C
[1]
M7.
C
[1]
M8.
D
[1]
M9.
D
[1]
M10.
A
[1]
M11.
B
[2]
M12.
D
[2]
M13.
D
[1]
M14.
C
[1]
M15.
C
[1]
M16.
(a)
(ii)
(iii)
(i)
straight line through origin (1)
(1)
energy (stored by capacitor) (1) (or work done
(in charging capacitor))
3
(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 × 103 × 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]
M17.
B
[1]
M18.
D
[1]
M19.
A
[1]
M20.
A
[1]
M21.
C
[1]
M22.
A
[1]
M23.
C
[1]
M24.
D
[1]
M25.
(a) The candidate’s writing should be legible and the spelling, punctuationand
grammar should be sufficiently accurate for the meaning to beclear.
The candidate’s answer will be assessed holistically. The answer will beassigned to one
of the 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 andcoherent, using
appropriate specialist vocabulary correctly. The form andstyle of writing is appropriate to
answer the question.
The candidate provides a comprehensive and logical description of thesequence of
releasing the ball and taking measurements of initial and finalvoltages. They should
identify the correct distance measurement and showa good appreciation of how to use
these measurements to calculate thetime and acceleration from them. Time should be
found from capacitordischarge, using known C and R values. Repeated readings would
beexpected in any answer worthy of full marks, but five marks may beawarded where
repetition is omitted.
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 provides a comprehensive and logical description of the
sequence of releasing the ball and taking measurements of the initial and
final voltages. They are likely to show some appreciation of the use of suvat
equations to calculate the acceleration, although they may not recognise the
need to measure a distance.
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 only be partly appropriate.
The candidate is likely to have recognised that initial and final voltages
should be measured, but may not appreciate the need for any other
measurement. They may present few details of how to calculate the
acceleration from the voltage measurements.
The explanation expected in a competent answer should include a
coherent selection of the following points.
Measurements
•
initial pd across C (V0) from voltmeter (before releasing roller)
•
distance s along slope between plungers
•
final pd across C (V1) from voltmeter
•
measurements repeated to provide a more reliable result
Analysis
(b)
•
time t is found from V1 = V0e-t/RC, giving t = RC ln (V0/V1)
•
from s = ut + ½ at2 with u = 0, acceleration a = 2s/t2
•
repeat and find average a from several results
(i)
RC = 22 × 10–6 × 200 × 103 [or = 4.4 (s)] (1) (4.40)
5.8 = 12.0 e–t/4.40 (1)
gives t = 4.40 ln (12.0/5.8) = 3.2 (3.20) (s) (1)
3
(ii)
(1)
= 0.49 (0.488) (m s–2) (1)
2
[11]
M26.
(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
(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]
M27.
(a)
E
V2 (or E = 1/2CV2) (1)
pd after 25 s = 6 V (1)
2
(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]
M28.
(a)
(i)
E (= ½ CV2 = 0.5 × 180 × 10–6 × 1002) = 0.90J (1)
(ii)
W (= QV = CV2 = 180 × 10–6 × 1002) = 1.8 J (1)
2
(b)
(i)
(V = V0e–t/RC) gives 30 = 100 e–t/RC (1)
t = (–RC ln (30/100) = –1.5 × 180 × 10–6 × –1.204 s)
= 3.3 × 10–4 s (1)
(ii)
image would be less sharp (or blurred) because the discharge
would last longer and the image would be photographed
as it is moving (1)
image would be brighter because the capacitor stores more
energy and therefore produces more light (1)
4
[6]
M29.(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]
(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)]
M30.
2
(b)
time constant (= RC) = 470 × 103 × 330 × 10–6 = 155 s (1)
1
(c)
= 2970 × e–60/155
= 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]
E2.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.
E3.This question was a direct test of energy storage by a capacitor using ½ CV2. The facility was 73%
and the question discriminated reasonably well.
E4.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 “half-life”, by using t = RC ln 2.
E5.Capacitors were the topic tested by this question. The question needed knowledge of how to
apply Q = I t for a constant current, C = Q / V and energy stored = ½ CV 2. Three quarters of the
candidates succeeded.
E6.The Capacitors were the topic tested by this question which needed knowledge of how to apply Q
= I t for a constant current, C = Q/V and energy stored = ½ CV2.
E8.
This question involved the charging of a capacitor by a constant current. The two-stage
calculation, using Q = CV and I = Q/t , caused no difficulty for 69% of the candidates. Incorrect
responses were almost all divided between distractors A and B, with very few selecting
distractor C.
E9.
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).
E10.
Energy stored by a capacitor was the subject tested in this question, which had been used
before in an Advanced level examination. In June 2003 the examination facility was 56%, as
opposed to 67% when it was used before. The discrimination index was very similar on both
occasions.
E11.
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.
E12.
This question involved calculations of capacitance, charge stored and energy stored by a
capacitor, using data from a graph. When the question was pre-tested, 56% gave the correct
response, but this advanced to 73% in the examination. Presumably it was confusion
between E= (1/2) QV and E = QV that caused 18% of the candidates to select distractor C.
E13.
This question was about capacitors. It dealt with the time-dependent variations of charge
and current. It had a facility of 63%. In this question, the incorrect responses were almost
equally divided between the three distractors.
E14.
This question concerned the power delivered from a bank of charged capacitors. It had
been used in a previous A level examination, when the facility was 60%. In 2006 the facility
dropped to 50% but the discrimination was unchanged. The most popular incorrect choice was
distractor D, selected by no less than 30%; this is almost certainly because these candidates
forgot about the ½ in ½ CV2.
E15.
This question was about capacitors. It dealt with energy storage. Generally it was probably
inability to handle powers of 10 that caused over one in five of the candidates to choose
distractor B. This question had a facility of 63%.
E16.
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
att = 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.
E17.
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.
E19.
This question had been used in an earlier examination. Its facility of 58% this time was a
slight improvement on that achieved previously. Either arithmetic errors, or failure to account for
the 10% efficiency, were probably responsible for almost a quarter of the candidates choosing
distractor C (0.50 m) rather than the correct 0.05 m.
E20.
This was the easiest question in the test, with a facility of 86%. The candidates were
obviously competent when applying the equations C = Q/V and E = ½ CV2 to find the
capacitance and energy stored from data on the graph of charge against pd.
E21.
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.
E22.
This question, about factors affecting time constant in an RC circuit, was the easiest of the
three with a facility of 75%.
E23.
This question was more demanding, with a facility of 49%. Here the energy and voltage of
a capacitor had to be considered when the charge is increased by half; distractors A and B each
attracted over 20% of the responses.
E24.
This question involved finding the current when a capacitor is charged using a constant
current, by combining Q = CV and Q = I t. 68% of candidates chose the correct alternative.
E25.
Part (a), which was about a student’s proposed experimental arrangement for measuring
the acceleration of a cylinder down a slope, was also a test of candidates’ quality of written
communication. Answers were rather better than those on the quality of written communication
questions in the 2010 unit 4 examinations. Perhaps this is because candidates now have a
clearer idea about what is expected in this type of question and are making greater efforts to
address these requirements, or perhaps this question may have been more accessible because
it involved an experimental technique. The main errors saw candidates not answering the
specific aspects posed in the question; what procedure should be followed, what measurements
should be taken, and how would these measurements be used to calculate the acceleration?
Less able candidates in particular, ended up writing much of their answer about how the circuit
would work. Common misconceptions over measurements included the need to measure the
mass of the cylinder, or the angle of the slope. Some candidates thought the time should be
measured with a stopwatch rather than by the intended discharge circuit, whilst others thought
there would be value in repeating the measurements for a different angle of slope. The best
answers were usually brief and to-the-point, showing excellent understanding of what this
experiment would involve.
The need to measure the distance from S1 to S2 was sometimes overlooked. The omission of
any reference to the calculation of acceleration was a serious error and tended to limit the mark
that examiners could award. Among the answers making an attempt to show how a could be
determined, a frequent error was to suggest that this could be done by dividing the average
speed by time rather than by dividing the final speed (ie twice the average speed) by time. The
more successful answers were those using s = ut + ½ at2, with u taken to be zero.
Many candidates achieved full marks in parts (b) (i) and (ii), showing confidence in solving an
exponential equation to determine the time and then applying an appropriate uniform
acceleration equation to calculate a. Some of the less convincing attempts to solve the
exponential equation omitted the essential minus sign in t = –RC ln (V/V0). A small proportion of
the candidates used log10 rather than loge, therefore ending up with the incorrect answer.
Many attempts at part (b) (ii) received no credit because the acceleration was found by dividing
the average speed by the time of descent – the error already noted in part (a).
E26.
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.
E27.
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.
E29.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 ln2
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.
E30.
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
viaQ = 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.