Question

Four capacitors C1=1uF, C2=2uF, C3=3uF, C4=4uF are connected to a 12V battery as shown in the figure in which C1 and C2 are in series and are parallel to C3 and C4 which are in series. Find
a) the capacitance between the points A and B
b) the total charge in the circuit
c) the charge on each capacitor, and
d) the potential on each capacitor

Answer 1

The equivalent capacitance of a series combination of capacitors is given

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+...$$

The equivalent capacitance of a parallel combination of capacitors is given by

$$C=C_1+C_2+...$$

The circuit configuration of the problem is shown in the figure.

Here $C_1$ and $C_2$ are in series. The equivalent capacitance is

$$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}\\ \frac{1}{C_{12}}=\frac{1}{1}+\frac{1}{2}\\ C_{12}=0.667\mu F$$

Again $C_3$ and $C_4$ are in series. The equivalent capacitance is

$$\frac{1}{C_{34}}=\frac{1}{C_3}+\frac{1}{C_4}\\ \frac{1}{C_{34}}=\frac{1}{3}+\frac{1}{4}\\ C_{34}=1.71\mu F$$

Now $C_{12}\ and\ C_{34}$ are in parallel. The equivalent capacitance is

$$C=C_{12}+C_{34}\\ C=0.667+1.71=2.38\mu F$$

Therefore the capacitance between points A and B is $2.38\mu F$

The total charge in the circuit is given by

$$Q=CV=2.38\mu F\times 14V=33.32\mu C$$

The charge on the upper branch i.e, in the capacitors $C_1\ and \ C_2$ is

$$Q_1=Q_2=C_{12}V=0.667\mu F\times 14V=9.34\mu C$$

[ Since in series combination, same current flows through the capacitors the charge stored in the capacitors is also the same. ]

The charge on the upper branch i.e, in the capacitors $C_3\ and \ C_4$ is

$$Q_3=Q_4=C_{34}V=1.71\mu F\times 14V=23.94\mu C$$

Potential across $C_1$ is

$$V_1=\frac{Q_1}{C_1}=\frac{9.34\mu C}{1\mu F}=9.34V$$

The potential across $C_2$ is

$$V_2=\frac{Q_2}{C_2}=\frac{9.34\mu C}{2\mu C}=4.67V$$

The potential across $C_3$ is

$$V_3=\frac{Q_3}{C_3}=\frac{23.94\mu C}{3\mu F}=7.98V$$

The potential across $C_4$ is

$$V_4=\frac{Q_4}{C_4}=\frac{23.94\mu C}{4\mu F}=5.99V$$