Question

In the circuit shown in the figure, $E=50\text{V},R=250\Omega ,L=20\text{mH}$ and $C=0.5\mu \text{F}$. The switch is closed for a long time and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of $\text{V}$(Integer only)

Answer 1

In steady state, the inductor will behave as a short circuit while the capacitor will behave as an open circuit.

The equivalent circuit diagram in steady state is given by


Therefore, the current through the inductor in steady state is,

$i_0=\frac{E}{R}=\frac{50}{250}=0.2\text{A}$

After the switch is opened, it becomes an $LC$ circuit in which peak current is $0.2\text{A}$.

By energy conservation, we have

$\frac{1}{2}L{i}_0^2=\frac{1}{2}C(V_{max}{)}^2$

$\Rightarrow 20\times {10}^{-3}\times (0.2{)}^2=0.5\times {10}^{-6}\times (V_{max}{)}^2$

$\Rightarrow (V_{max}{)}^2=1600$

$\Rightarrow V_{max}=40\text{V}$

Correct answer: $40$