Question

In the circuit shown, the switch $S_1$ is closed for long time and at $t=0$s, the switch $S_2$ is closed and $S_1$ is opened simultaneously. What is the maximum charge (in the unit of $\mu C$) on the $4\mu F$ capacitor?

• A. $q_0=3\mu C$
• B. $q_0=6\mu C$
• C. $q_0=3.4\mu C$
• D. $q_0=6.8\mu C$

Answer 1

The correct option is A $q_0=3\mu C$

After a long time when switch $S_1$ is closed, the voltage entirely drop across the 20 $\Omega$ resistor.

The current in the circuit will be $i=\frac{V}{R}=\frac{1.5}{20}=0.075amp$

Energy in the inductor is $U=\frac{1}{2}L{i}^2$

When $S_1$ is opened and $S_2$ is closed, max energy in the capacitor will be stored when the charge on the capacitor is maximum.

$\therefore \frac{1}{2}L{i}^2=\frac{1}{2}\frac{q_{max}^2}{C}$

$\Rightarrow q_{max}=\sqrt{LC{i}^2}=0.075\times \sqrt{0.4\times {10}^{-3}\times (4\times {10}^{-6})}$
$=4\times {10}^{-5}\times 75\times {10}^{-3}=3\times {10}^{-6}=3\mu C$