Question

In the circuit of figure, the battery emf is $50 V$, the resistance is $200 \Omega$, and the capacitance is $1.0 \mu F$. The switch $S$ is closed for a long time interval, and zero potential difference is measured across the capacitor.

Answer 1

When the switch has been closed for a long time, battery, resistor and coil carry constant current $I_{i}=\frac{\varepsilon}{R}$. When the switch is opened, current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the $L C$ loop. We interpret the problem to mean that the voltage amplitude of these oscillations is $\Delta V$, in $\frac{1}{2} C(\Delta V)^{2}=\frac{1}{2} L I_{i}^{2}$.
Then, $L=\frac{C(\Delta V)^{2}}{I_{i}^{2} .}=\frac{C(\Delta V)^{2} R^{2}}{\varepsilon^{2}}=\frac{\left(1.0 \times 10^{-6}\right)(100)^{2}(200)^{2}}{(50)^{2}}$ $=0.16 H =160 mH$