Question

A capacitor of capacitance $2\mu F$ is charged to 50 V. The charging battery is then disconnected and a coil of inductance 5 mH is connected across it. Assuming that the coil has negligible resistance, the peak value of the current in the circuit will be

• A. 1A
• B. 2A
• C. 3A
• D. 4A

Answer 1

The correct option is A 1A

$\frac{1}{2}L{I}_0^2=\frac{1}{2}C{V}^2$ ($\because$ there is no loss of energy due to joule heating as R = 0). Hence
$I_0=V\sqrt{\frac{C}{L}}=50\times \sqrt{\frac{2\times {10}^{-6}}{5\times {10}^{-3}}}=1A$.