Question

A $8\text{μH}$ inductor is shorted across a $2\mu F$ capacitor. The capacitor is initially charged to $20V$ . The maximum value of the current in the circuit is

• A. $10.0A$
• B. $7.5A$
• C. $12.0A$
• D. $8.2A$

Answer 1

Answer: (A)

we are given :
capacitance $C=2\mu F$
voltage $V_0=20V$
Inductance $L=8\mu H$
Maximum current flows in a circuit at Resonance.
At resonance,
$\omega L=\frac{1}{\omega C}$
${\omega }^2=\frac{1}{LC}\omega =\frac{1}{\sqrt{LC}}$
The maximum amount of current will be
$I_m=\omega Q_0$
where charge on the capacitor $Q_0=C{V}_0$
$\Rightarrow \omega Q_0=\frac{C{V}_0}{\sqrt{LC}}=\frac{2\mu F\cdot 20V}{\sqrt{(8\mu H)(2\mu F)}}=10A$
$I_m=10A$