Q. A $8\,\text{μH}$ inductor is shorted across a $2\,μF$ capacitor. The capacitor is initially charged to $20\,V$ . The maximum value of the current in the circuit is
NTA AbhyasNTA Abhyas 2022
Solution:
we are given :
capacitance $C=2μF$
voltage $V_{0}=20V$
Inductance $L=8μH$
Maximum current flows in a circuit at Resonance.
At resonance,
$ωL=\frac{1}{ωC}$
$\omega ^{2 }=\frac{1}{LC}\omega =\frac{1}{\sqrt{LC}}$
The maximum amount of current will be
$I_{m}=\omega Q_{0}$
where charge on the capacitor $Q_{0}=CV_{0}$
$\Rightarrow \omega Q_{0}=\frac{C V_{0}}{\sqrt{L C}}=\frac{2 \mu F \cdot 20 V}{\sqrt{\left(\right. 8 \mu H \left.\right) \left(\right. 2 \mu F \left.\right)}}=10A$
$I_{m }=10A$
capacitance $C=2μF$
voltage $V_{0}=20V$
Inductance $L=8μH$
Maximum current flows in a circuit at Resonance.
At resonance,
$ωL=\frac{1}{ωC}$
$\omega ^{2 }=\frac{1}{LC}\omega =\frac{1}{\sqrt{LC}}$
The maximum amount of current will be
$I_{m}=\omega Q_{0}$
where charge on the capacitor $Q_{0}=CV_{0}$
$\Rightarrow \omega Q_{0}=\frac{C V_{0}}{\sqrt{L C}}=\frac{2 \mu F \cdot 20 V}{\sqrt{\left(\right. 8 \mu H \left.\right) \left(\right. 2 \mu F \left.\right)}}=10A$
$I_{m }=10A$