Question

In the circuit shows in the figure $C_1=2C_2$. Capacitor $C_1$ is charged to a potential of $V$. The current in the circuit just after the switch is closed, is:

• A. zero
• B. $\frac{2V}{R}$
• C. Infinite
• D. $\frac{V}{2R}$

Answer 1

The correct option is D $\frac{V}{2R}$

Since$,$ capacitor is charged to a potential of $V$ volts it will act as a battery of $V$ volts$.$

Let the current in the circuit be $=I$

Now potential accross capacitor $c_2$ is given by $,$ $\frac{C_1}{2}=\frac{Q}{V_1}$

$=V_1=\frac{2Q}{C_1}$ $-------\left(1\right)$

Now$,$ same charge $Q$ is flowing through both $C_1$ and $\frac{C_1}{2}$

for capacitor$,$ $C_1,$ $C_1=\frac{Q}{V}$

$=Q=C_{1}V$ $-------------\left(2\right)$

$($ since$,$ it is charged to potential $v$$)$

from $\left(1\right)$ and $\left(2\right)$

$V_1=\frac{2C_{1}V}{C_1}=V_1=2V$

Appling kirchoff voltage law$:$

$V-IR-2V-IR=0$

$-V-2IR=0$ $=|I|=\frac{V}{2R}$

Hence$,$

Option $\left(D\right)$ is correct