Question

A capacitor 1 $\mu$F is charged to potential of 1 V. It is connected in parallel to an inductor of inductance ${10}^{-3}$ H. The maximum current that will flow in the circuit has the value

• A. $\sqrt{1000}$ mA
• B. 1 mA
• C. 1 $\mu$ A
• D. 1000 magnitude

Answer 1

The correct option is A $\sqrt{1000}$ mA

Current in inductor increases as it get charged. so maximum current in inductor will occur at the time when all energy of capacitor will be stored in inductor.

So, energy stored in capacitor (U) = $\frac{1}{2}C{V}^2$

C = $1\mu F$, V = 1volt

$U=\frac{1}{2}{\times 10}^{-6}\times 1^2=\frac{{10}^{-6}}{2}J$

Let $i$ is the maximum current in the inductor.
Energy stored in inductor at the instant of maximum current $=E$

$E=\frac{1}{2}L{i}^2$, which will be equal to U.

$\frac{1}{2}C{V}^2=\frac{1}{2}L{i}^2$

$L={10}^{-3}H$

Therefore,$\frac{1}{2}\times {{10}^{-3}\times i}^2=\frac{{10}^{-6}}{2}$

Hence, $i=\sqrt{1000}\times {10}^{-3}Aor\sqrt{1000}mA$.