Question

In figure, battery B supplies 12 V. Find the charge on each capacitor when $S_2$ is also closed.

(Take $C_1=1.0\mu F,C_2=2.0\mu F,C_3=3.0\mu F$ and $C_4=4.0\mu F$)

Answer 1

When $S_2$ is closed, the corresponding equivalent circuit will be as shown in figure.

Now $C_1,C_2$ are in parallel so the equivalent capacitance for this is $C_{12}=C_1+C_2=1+2=3\mu F$

Similarly , $C_{34}=C_3+C_4=3+4=7\mu F$

As $C_{12}$ and $C_{34}$ are in series so they have same charge i.e $Q_{12}=Q_{34}$

Thus, $Q_{12}=C_{12}{V}_1$ and $Q_{34}=C_{34}{V}_2$

So, $V_1/V_2=C_{34}/C_{12}=7/3\mu F$

From circuit, $V_1+V_2=12$

or $\left(7/3\right)V_2+V_2=12$

or $V_2=3.6V$

So, $V_1=12-3.6=8.4V$

Now, $q_1=C_1{V}_1=1\times 8.4=8.4\mu C$ ; $q_2=C_2{V}_1=2\times 8.4=16.8\mu C$; $q_3=C_3{V}_2=3\times 3.6=10.8\mu C$ and $q_4=C_4{V}_{=}4\times 3.6=14.4\mu C$