CAIE A-Level Physics 19.1 Capacitors and Capacitance

Practise defining capacitance with C = Q/V and deriving or using equivalent capacitance for series and parallel networks.

Exam points

define capacitance and apply C=Q/V to charge, voltage and capacitance
derive and use equivalent capacitance for series and parallel networks

Covered syllabus points

19.1.1—Capacitance, as applied to both isolated spherical conductors and to 19.1.2—C = Q / V 19.1.3—C = Q / V, formulas for the combined capacitance of capacitors in series and 19.1.4—The capacitance formulas for capacitors in series and in parallel

[Maximum number: 6]

(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.

[ 3 ]

(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

[Maximum number: 2]

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 .

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

Q_{0}=

Q=C V

C1

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

A1

All question bank results loaded

Covered syllabus points

19.1 Capacitors and capacitance question 1

Question (a)

Question (b)

Question (c)

Question (i)

19.1 Capacitors and capacitance question 2