Question

Initially capacitor was uncharged and switch was open. Switch is closed at t=0. Ammeter and voltmeterare ideal. [All units are in SI]

Answer 1

After long time, current through the circuit will be zero since capacitor will behave as open circuit. Therefore voltmeter will read $10V$.

Just after closing the switch, voltage across capacitor will be $0V$. Hence current will be $i=\frac{20+10}{4+2}=5A$.

Reading of voltmeter just after closing the switch will be $10-2i=0V$.

After long time, entire voltage will appear across the capacitor, hence, $Q=CV=2\times 30=60C$