Question

A capacitor $C_1=12\mu F$ charged with voltage 4 V is connected across uncharged capacitor $C_2=4\mu F$ and resistor 2 $\Omega$. At $t=0$, switch S is closed. At $t=t_0$, voltage across $C_1$ is 3 V, then current in circuit (in A) at $t=t_0$ is (Answer upto two decimal places )

Answer 1

$Q=CV=12\times 4=48\mu C$
When $V_1=3V\Rightarrow q_1=12\times 3=36\mu C$
$\because Q=q_1+q_2\therefore q_2=12\mu C$
$V_2=\frac{q_2}{C_2}=\frac{12}{4}=3V$
$V_1=\frac{q_1}{C_1}=\frac{36}{12}=3V$
By KVL,
$V_1=V_2+iR$
3 = 3 + $iR$
$\therefore i=0A$