Download A2 Unit G485: Fields, particles and frontiers of physics

Sample student answers
A2 Unit G485: Fields, particles and frontiers of physics
Module 5: Capacitors and exponential decay
Question 1
Total marks: 10
This question is about capacitor discharge in the circuit shown in the diagram.
The switch S is moved from X to Y. The 4700 μF capacitor discharges through the 1100 Ω resistor.
The graph below shows p.d. against time.
(a)
Use the information from the diagram and the graph to show that
(i)
the initial charge on the capacitor is about 0.03 C
(ii)
the initial rate of discharge is about 5 mA
(iii)
the time constant of the circuit is about 5 s.
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Marks available:
(i) 1 (ii) 1 (ii) 1
Student answer:
(a) (i) Q = CV
= 4700 × 10−6 × 6
= 0.028 C
(ii) I = V/R = 6/1100
= 5.5 mA
(iii) T = CR
= 4700 × 10−6 × 1100 = 5.2 s
Examiner comments:
(a) Good. All calculations show equation, substitution, answer and units.
(b)
Explain why the rate of fall of voltage is proportional to the rate of fall of charge and hence
proportional to the current in the circuit.
Marks available: 2
Student answer:
(b) In a capacitor, the voltage is proportional to the charge. Therefore the rate of fall of voltage
is proportional to the rate of fall of charge, but the rate of fall of charge is the current.
Examiner comments:
(b) Good answer
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(c)
A series of models of the discharge of a capacitor C through a resistor R are considered.
(i)
In the simplest model, the current is assumed to remain constant at its initial value
throughout the discharge. Show that this model predicts that the capacitor would
fully discharge in time RC.
(ii)
A better model calculates the change of charge ΔQ in successive time intervals Δt
using the equation
ΔQ = −Q Δt
RC
The graph below shows the graph produced when Δt is set at 4.0 s.
To improve the model, Δt is reduced to 2.0 s. This graph gives the charge remaining at 2.0 s as
0.017 C. Use this value to show that the loss of charge during the next 2 s will be about 6.5 × 10−3 C.
(iii)
(iv)
Sketch a copy of the graph above, and draw a line on the graph to represent the
loss of charge from the capacitor between 2.0 s and 4.0 s.
Explain why reducing the time interval Δt leads to a more accurate model of the
discharge.
Marks available:
(i) 2 (ii) 1 (iii) 1
(iv) 1
Student answer:
(c) (i) Q = It
It = CV = CIR
t = CR
(ii) Loss of charge = (0.017/5.2) × 2.0 = 6.5 × 10−3 C
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(iii)
(iv)
This makes the rate of decay constant for a smaller period of time.
Examiner comments:
(c) (i) The answer is correct, but a few words would have helped in the explanation:
Q = It
but Q = CV
therefore It = CV
and V = IR
therefore It = CIR
and t = CR.
(iv) You could have added that this keeps the graph closer to a continuous change.
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Module 5: Capacitors and exponential decay
Question 2
Total marks: 7
The diagram shows two capacitors, C1 and C2, of capacitances 10 μF and 4 μF respectively, a cell of
e.m.f. 4.0 V and a two-way switch. The capacitors are initially uncharged.
(a)
With the switch connected to X, calculate the charge stored on C1.
Marks available: 1
Student answer:
(a) Q = CV
= 10 × 10−6 × 4.0
= 4 × 10−5 C
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(b)
The switch is now connected to Y. Calculate:
(i)
the total capacitance of the combination of C1 and C2
(ii)
the potential difference across C1
(iii)
the charge stored on C1.
Marks available:
(i) 1 (ii) 1 (iii) 1
Student answer:
(b) (i) The capacitors will be in parallel.
C = C 1 + C2
= 10 + 4
= 14 μF
(ii) V = Q/C = 40 × 10−6/(14 × 10−6)
= 2.9 V
(iii) Q = 10 × 10−6 × 2.9
= 2.9 × 10−5 C
Examiner comments:
(b) These calculations could do with some explanations.
(i) When they are connected, all the positive charge remains on the top two plates and all
the negative charge is on the bottom plates. Therefore they are in parallel.
(ii) Total voltage = total charge/total capacitance
(iii) Same voltage across all components when in parallel
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(c)
The next diagram shows C2 replaced by a resistor R of resistance 24 MΩ. The switch is
connected to X to recharge C1 and then the switch is reconnected to Y. Calculate the charge
stored on C1 after the switch has been connected to Y for 30 s.
Marks available: 3
Student answer:
(c) Q = Q0e−t/CR
= 40 × 10−6 × exp[−30/(10 × 10−6 × 24 × 106)]
= 3.53 × 10−5 C
Examiner comments:
(c) Make sure you know how to use your calculator to do this.
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