Question

Only switch $S_1$ closed in Figure. What is the steady-state reading of the voltmeter?

Answer 1

a. As explained earlier that a capacitor is an open circuit to DC in steady state. The effective circuit through which current will pass initially after closing switch $S_1$ is shown in figure. When steady state is reached, the capacitor will not allow current. Thus, there is no current from battery. Reading of voltmeter is zero.
b. Without current, the resistors act as conducting wires. Hence, potential difference across capacitor is 12 V. Charge on capacitor, $3\mu F$, is $Q=(3\times 12)\mu C=36\mu C$
c. When all the switches are closed, the branch containing capacitor will not allow current after steady state is reached. The equivalent circuit from point of view of current is redrawn as in figure.
Let the current in the single loop circuit be $I$. We begin at left lower corner and traverse the circuit anticlockiwise while applying $KVL$. We get $-20I+20-10I-12=0$
$I=\frac{4}{15}A$
Power received by 12V battery is $P=VI=12\times \frac{4}{15}=\frac{16}{5}W$
d. $V_c-10I-12=V_d$
or $V_c-V_d=\frac{44}{3}V$ or $Q=CV=\frac{88}{3}\mu C$