Question

In the arrangement as shown in the figure, $S_{1}and{S}_2$ are already closed for a long time. Now at t = 0 the $S_2$ is closed and $S_1,S_3$ are opened. [$ϵ$ = 1 volt, L = 1 henry, R = 1 $\Omega$ and C = 1 Faraday]

• A. After $S_2$ is closed the maximum current through the indicator is 2A.
• B. After $S_2$ is closed the maximum current through the indicator is $\sqrt{2}A$
• C. Maximum charge on the capacitor is $2+\sqrt{2}$ coulomb
• D. Maximum charge on the capacitor is $\sqrt{2}$ coulomb

Answer 1

Answer: (B), (D)

The correct options are
B After $S_2$ is closed the maximum current through the indicator is $\sqrt{2}A$
D Maximum charge on the capacitor is $\sqrt{2}$ coulomb

$\frac{Q_i^2}{2c}+\frac{1}{2}L{i}_i^2=\frac{1}{2}L{i}_{max}^2$
$I_{max}=\sqrt{2}amp$
Also, $\frac{Q_i^2}{2c}+\frac{1}{2}L{i}_i^2=\frac{Q_{max}^2}{2c}$