Question

In the circuit shown below, the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q_1\mu C$. Then $S$ is switched to position $Q$. After the long time, the charge on the capacitor is $q_2\mu C.$

The magnitude of $q_2$ (in $\mu C$) is

• A. 0.66
• B. 0.667
• C. 0.666
• D. 0.67

Answer 1

Answer: (A), (B), (C), (D)

At steady state current through capacitor $=0$
$I=\frac{2}{3}\text{A}$
Potential difference across capacitor = Potential difference across $1\Omega$ resistor
$V_A-V_B=I\left(1\right)=\frac{2}{3}V$
Charge on capcitor $=q_2=C(V_A-V_B)$
$y=q_2=1\times \frac{2}{3}\mu C=0.67\mu C$