Question

In the circuit shown below, initially the switch $S_1$ is open, the capacitor C1 has a charge of 6 coulomb, and the capacitor C2 has 0 coulomb. After $S_1$ is closed, the charge on C2 in steady state is Coulomb.

• A. 4

Answer 1

The correct option is A 4
$V_{C1}\left(0^{-}\right)=\frac{6}{1}=6Vand{V}_{C2}\left(0^{-}\right)=0V$

For t > 0


Transforming the circuit to the Laplace domain,



$V_{C2}\left(s\right)=\frac{\frac{6}{s}}{\frac{1}{s}+1k+\frac{1}{2s}}\times \frac{1}{2s}=\frac{6}{s[2+2sk+1]}$

Voltage across the capacitor C2 in steady state,

$V_{C2}\left(\infty \right)={lim}_{s\to 0}s.V_{C2}\left(s\right)={lim}_{s\to 0}\frac{6}{[2+2sk+1]}=\frac{6}{3}$

$V_{C2}\left(\infty \right)=2V$

$\therefore$ The charge on $C_2$ in steady state,

$Q=C_2{V}_{C2}\left(\infty \right)$

Q = (2)(2) = 4 C

Alternate method :

$Q_1\left(0^{-}\right)=6C$

$V_{c_1}\left(0^{-}\right)=\frac{Q_1\left(0^{-}\right)}{C}=\frac{6}{1}=6V$

The Initial voltage across the capacitor $C_1$ is 6V

and capacitor $C_2$ is 0V i.e, $V_{C_2}\left(0^{-}\right)=0V$

In steady state the voltage across two capacitors are equal i.e,

$V_{C_1}\left(\infty \right)=V_{C_2}\left(\infty \right)=\frac{V_{C_1\left(0^{-}\right)C_1+V_{C_2}}\left(0^{-}\right)C_2}{C_1+C_2}$

$V_{c_1}\left(\infty \right)=V_{C_2}\left(\infty \right)$

$\Rightarrow \frac{(6\times 1)+0}{1+2}=2V$

The steady state voltage across capacitor $C_2$ is 2 V

and change $Q_2=C_2{V}_2=2\times 2=4C$