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A condenser of capacity C is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is
A
$\left(\frac{C(V_1 - V_2)^2}{L}\right)^{\frac{1}{2}}$
B
$\frac{C(V_1^2 - V_2^2)}{L}$
C
$\frac{C(V_1^2 + V_2^2)}{L}$
D
$\left(\frac{C(V_1^2 - V_2^2)}{L}\right)^{\frac{1}{2}}$
Detailed Solution
The inductor is ideal (no resistance), so the total energy of the LC circuit is conserved.
Initially all the energy is in the capacitor: $U_i = \frac{1}{2}CV_1^2$
When the potential difference has fallen to $V_2$, the energy is shared: $\frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$
$\frac{1}{2}CV_1^2 = \frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$
$LI^2 = C(V_1^2 - V_2^2)$
$I = \left(\frac{C(V_1^2 - V_2^2)}{L}\right)^{\frac{1}{2}}$
Initially all the energy is in the capacitor: $U_i = \frac{1}{2}CV_1^2$
When the potential difference has fallen to $V_2$, the energy is shared: $\frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$
$\frac{1}{2}CV_1^2 = \frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$
$LI^2 = C(V_1^2 - V_2^2)$
$I = \left(\frac{C(V_1^2 - V_2^2)}{L}\right)^{\frac{1}{2}}$