Question

Ratio of charge on $C_1$ when switch S is open and when S is closed is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Charge on $C_1$ when switch was open, $q_i=12\times 3$

By $V=iR\Rightarrow 12=i\times 6\Rightarrow i=2A$
$\therefore$ V across $C_1$ is $=2\times 2=4V$
So, charge on $C_1$ after switch is closed = $q_f=3\times 4$
$\frac{q_i}{q_f}=\frac{12\times 3}{3\times 4}=3$