CAIE A-Level Physics 19 Capacitance

Practise analysing capacitance, series or parallel combinations, stored energy, discharge curves, time constants and exponential decay.

Exam points

define capacitance and calculate charge, voltage and equivalent series/parallel capacitance
calculate capacitor energy from graph area and energy formulae
analyse discharge graphs, time constants and exponential decay

Covered syllabus points

19.1 Capacitors and capacitance 19.2 Energy stored in a capacitor 19.3 Discharging a capacitor

[Maximum number: 9]

(a)

Define capacitance.

[ 1 ]

(b)

Three capacitors of capacitances $C_{1}, C_{2}$ and $C_{3}$ are initially uncharged. They are then connected in series to a battery, as shown in Fig. 7.1.

Fig. 7.1

The battery applies a potential difference V across the three capacitors.
Show that the combined capacitance C of the capacitors is given by

\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} .

[ 2 ]

$V=V_{1}+V_{2}+V_{3}$
either $\quad Q / C=Q / C_{1}+Q / C_{2}+Q / C_{3}$ or $V / Q=V_{1} / Q+V_{2} / Q+V_{3} / Q$
and so $1 / C=1 / C_{1}+1 / C_{2}+1 / C_{3}$

(c)

A battery of e.m.f. 12 V and negligible internal resistance is connected to a network of two capacitors and a resistor, as shown in Fig. 7.2.

Fig. 7.2

The capacitors have capacitances of $200 \mu \mathrm{~F}$ and $600 \mu \mathrm{~F}$ . The switch has two positions, A and B.

[ 6 ]

(i)

The switch is moved to position A.

Calculate
1. the combined capacitance of the two capacitors,

combined capacitance =μF [1]

2. the charge on the $600 \mu \mathrm{~F}$ capacitor,

charge =

3. the potential difference across the $600 \mu \mathrm{~F}$ capacitor.

potential difference =

[ 3 ]

1. $1 / C_{\top}=(1 / 200)+(1 / 600)$ $C_{\mathrm{T}}=150 \mu \mathrm{~F}$
2. Q=C V

\begin{aligned} & =150 \times 10^{-6} \times 12 \text { or } 600 \times 10^{-6} \times 3.0 \text { or } 200 \times 10^{-6} \times 9.0 \\ & =1.8 \times 10^{-3} \mathrm{C} \end{aligned}

3. $V=Q / C=1.8 \times 10^{-3} / 600 \times 10^{-6}$ or $V=[200 /(200+600)] \times 12$

=3(.0) \vee

(ii)

The switch is now moved from position A to position B.

Calculate the potential difference across the $600 \mu \mathrm{~F}$ capacitor when it has discharged 50 % of its initial energy.
potential difference = V

[ 3 ]

energy $=1 / 2 C V^{2}$ or energy =1 / 2 Q V and C=Q / V
$\frac{1}{2} \times C \times 3^{2}=2 \times \frac{1}{2} \times C \times V^{2}$ $V=2.1 \mathrm{~V}$
A1

[Maximum number: 10]

The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1.

Fig. 5.1

The capacitor is connected to an 8.0 V power supply and two resistors R and S as shown in Fig. 5.2.

Fig. 5.2

The resistance of R is $25 \mathrm{k} \Omega$ and the resistance of S is $220 \mathrm{k} \Omega$ .
The switch can be in either position X or position Y .

(a)

The switch is in position X so that the capacitor is fully charged.

Calculate the energy E stored in the capacitor.

[ 3 ]

(energy stored =) area under line or $\frac{1}{2}$ QV
$=1 / 2 \times 8.0 \times 1.2 \times 10^{-4}$

C1

$=4.8 \times 10^{-4} \mathrm{~J}$

A1

(b)

The switch is now moved to position Y.

[ 5 ]

(i)

Show that the time constant of the discharge circuit is 3.3 s .

[ 2 ]

(t=) R C

C1

( t= ) $220 \times 10^{3} \times\left(1.2 \times 10^{-4} / 8.0\right)=3.3 \mathrm{~s}$

A1

(ii)

The fully charged capacitor in (a) stores energy E.

Determine the time t taken for the stored energy to decrease from E to E / 9.

[ 3 ]

$E \propto V^{2}$

C1

(so time to) $V_{\mathrm{o}} / 3$ $V=V_{0} e^{-t / R C}$

C1

$\begin{aligned} \frac{V_{o}}{3}=V_{o} e^{-t / 3.3} \frac{1}{3}=e^{-t / 3.3} \end{aligned}$

C1

$\mathrm{t}=3.6 \mathrm{~s}$

A1

(c)

A second identical capacitor is connected in parallel with the first capacitor.

State and explain the change, if any, to the time constant of the discharge circuit.

[ 2 ]

[Maximum number: 6]

A capacitor of capacitance $470 \mu \mathrm{~F}$ is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1.

Fig. 5.1

The two-way switch S is initially at position X.
P and Q are identical long straight wires, each with a resistance of $5.6 \mathrm{k} \Omega$ . These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter.

At time t=0, switch S is moved to position Y so that the capacitor discharges through wire P .

(a)

Calculate the charge $Q_{0}$ on the capacitor at time t=0.

Q_{0}=

[ 2 ]

Q=C V

C1

$\begin{aligned} Q_{0} =24 \times 470 \times 10^{-6} =0.011 \mathrm{C} \end{aligned}$

A1

(b)

Calculate the current $I_{0}$ in wire P at time t=0.

I_{0}=

[ 1 ]

$\begin{aligned} I_{0} =24 / 5600 =4.3 \times 10^{-3} \mathrm{~A} \end{aligned}$

A1

(c)

Calculate the time constant $\tau$ of the discharge circuit.

\tau=

[ 1 ]

$\tau=R C$

C1

$\begin{aligned} =5600 \times 470 \times 10^{-6} =2.6 \mathrm{~s} \end{aligned}$

A1

(d)

On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges.

Fig. 5.2

[ 2 ]

line with negative gradient throughout passing through ( $0, I_{0}$ )

B1

exponential decay curve asymptotic to t-axis

B1

All question bank results loaded

Covered syllabus points

19. Capacitance question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

19. Capacitance question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

19. Capacitance question 3

Question (a)

Question (b)

Question (c)

Question (d)