Physics - EMI/AC Question with Solution | TestHub

Initially, the capacitor $C$ is charged to $V_0$. The total energy stored in the circuit is $E=\frac{1}{2}C{V}_0^2$. When the capacitor is connected to the inductor $L$ (first stage), LC oscillations begin.

At time $t_1$, the current in the circuit is maximum. At this instant, the capacitor is fully discharged (charge $q=0$), and all the energy is stored in the inductor. The maximum current in the first stage, $I_{max,1}$, is given by energy conservation:$$\frac{1}{2}L{I}_{max,1}^2=\frac{1}{2}C{V}_0^2$$ $$I_{max,1}=V_0\sqrt{C/L}$$

At this moment ($t_1$), the switch $S_1$ is opened. The capacitor has zero charge, and the inductor $L$ has energy $E_{L,1}=\frac{1}{2}L{I}_{max,1}^2$.

The capacitor is then immediately connected to a new series combination of two identical inductors, each of inductance $L$. The total inductance of this new combination is $L^{\prime }=L+L=2L$.

The new LC circuit for oscillation consists of capacitance $C$ and inductance $L^{\prime }$. The total energy for this new oscillation is the energy that was stored in the inductor $L$ at time $t_1$, which is $E_{L,1}$.

The maximum current in this new circuit, $I_{max,2}$, will occur when the capacitor is again fully discharged (all energy is in the inductors). By energy conservation for the second stage:$$\frac{1}{2}{L}^{\prime }{I}_{max,2}^2=E_{L,1}$$ $$\frac{1}{2}(2L)I_{max,2}^2=\frac{1}{2}L{I}_{max,1}^2$$ $$2L{I}_{max,2}^2=L{I}_{max,1}^2$$ $$I_{max,2}^2=\frac{1}{2}{I}_{max,1}^2$$ $$I_{max,2}=\frac{I_{max,1}}{\sqrt{2}}$$ Substitute the expression for $I_{max,1}$:$$I_{max,2}=\frac{V_0\sqrt{C/L}}{\sqrt{2}}=V_0\sqrt{\frac{C}{2L}}$$