CAIE A-Level Physics 19.1.4 Capacitors in Series and Parallel

Practise calculating capacitance, charge and potential difference in series or parallel networks, applying common-charge or common-potential conditions.

Exam points

calculate equivalent capacitance, charge or potential difference for capacitors in series
calculate equivalent capacitance, charge or potential difference for capacitors in parallel
apply common-charge and common-potential conditions to identify charge distribution

[Maximum number: 3]

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.

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 =

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

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19.1.4—The capacitance formulas for capacitors in series and in parallel question 1