Question

In the circuit shown, initially $C_1$ and $C_2$ are uncharged. After closing the switch,

• A. $\text{The charge on}{C}_2\text{is greater than that on}{C}_1.$
• B. $\text{The charges on}{C}_1\text{and}{C}_2\text{are the same}.$
• C. $\text{The potential drops across}{C}_1\text{and}{C}_2\text{are the same}.$
• D. $\text{The potential drop across}{C}_2\text{is greater than that across}{C}_1.$

Answer 1

The correct option is B $\text{The charges on}{C}_1\text{and}{C}_2\text{are the same}.$


In series combination, charges on the capacitors will be same.

Applying KVL around the circuit,

$12-\frac{q}{8}+6-\frac{q}{4}=0$
$\Rightarrow q=48\mu C$

$V_{C_2}=\frac{48}{8}=6\text{V}$
& $V_{C_1}=\frac{48}{4}=12\text{V}$

Hence, option (b) is correct.