Question

A Circuit contains a resistor of resistance $R$ and a capacitor $C_1$ of capacitance $C$, a second Capacitor $C_2$ of capacitance $2C$, and a battery that provides an emf of $V$, as shown above. Initially both switches are open and both capacitors are uncharged. At time t=0, Both switches are closed. After the circuit reaches steady state, what is the charge $Q_2$ on capacitor $C_2$?

• A. $2CV$
• B. $Q_2=\frac{2V}{9C}$
• C. $Q_2=\frac{V}{2C}$
• D. $Q_2=\frac{2V}{3C}$
• E. $Q_2=\frac{V}{C}$

Answer 1

The correct option is A $2CV$

When the circuit reaches steady state, no current will flow through the circuit so potential drop across $R_1$ will be zero. Thus, the potential difference across capacitors $C_1,C_2$ are same i.e V because they are in parallel.

Hence, the charge on $C_2$ is $Q_2=C_{2}V=2CV$