Question

In the circuit shown in the figure, $E=50.0V,R=250\Omega$ and $C=0.500\mu F$. The switch $S$ 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 $150V$. What is the inductance $L$?

• A. $0.98H$
• B. $0.78H$
• C. $0.28H$
• D. $0.38H$

Answer 1

The correct option is C $0.28H$

In steady state when switch was closed,
$i_0=E/R=\left(1/5\right)A=0.2A$
After switch is opened, it becomes $L-C$ circuit in which peak value current is $0.2A$
$\therefore$ $\frac{1}{2}L{i}_0^2=\frac{1}{2}L{V}_0^2$
or $L=\frac{V_0^2}{i_0^2}\cdot C$
$=\frac{{\left(150\right)}^2}{{\left(0.2\right)}^2}\times 0.5\times {10}^{-6}=0.28H$