In the circuit shown in Fig., the capacitance of the capacitor is $C=5 \mu \mathrm{F}$. When the steady state is reached, the charge on the capacitor is

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Aicharya · Accepted Answer

In the steady state, no current flows through the branch containing the capacitor. Charge on the capacitor is $Q=C V$, where $V$ is the potential difference between the capacitor plates [see Fig.]
The $30 \Omega$ resistance carries no current and hence can be ignored. The resistance of circuit is
$\begin{aligned} R &= 4+\frac{10 \times 15}{10+15}=10 \, \Omega \\ \text{Current } \, I &= \frac{12 \, \text{V}}{10 \, \Omega}=1.2 \, \text{A} \end{aligned}$ Applying Kirchhoff's voltage law,
$V_{B}-V_{A}=-12+4 \times 1.2=-12+4.8=-7.2 \mathrm{~V}$
$\therefore$$V=V_{A}-V_{B}=7.2 \mathrm{~V}$
Hence $\quad Q=C V=5 \mu \mathrm{F} \times 7.2 \mathrm{~V}=36 \mu \mathrm{C}$
So the correct choice is (d).

In the circuit shown in Fig., the capacitance of the capacitor is \(C=5 \mu \mathrm{F}\). When the steady…

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