An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductan

Question

An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH . The current at a time when the potential difference across the capacitor is 5 V is:

TestHub · Accepted Answer

The energy stored in the capacitor is given by, $U_{c} = \frac{1}{2} C V^{2}$ . The energy stored in the inductor is given by. $U_{L} = \frac{1}{2} L I^{2}$ . In $L C$ circuit there is no loss of energy in the form of heat so, using energy conservation, $U_{c i} + U_{L i} = U_{c f} + U_{L f}$

$\Rightarrow \frac{1}{2} \times 0.2 \times 1 0^{- 6} \times 1 0^{2} + 0 = \frac{1}{2} \times 0.2 \times 1 0^{- 6} \times 5^{2} + \frac{1}{2} \times 0.5 \times 1 0^{- 3} I^{2}$
$I = 3 \times 1 0^{- 1} A = 0.17 A$

Physics - EMI/AC Question with Solution | TestHub

Options:

Doubts & Discussion