Question

A $1.0\mu F$ capacitor is charged to $50V$ potential difference and then discharged through a $10mH$ inductor of negligible resistance. The maximum current in the inductor will be

• A. $0.5A$
• B. $1.6A$
• C. $1.0A$
• D. $0.16A$

Answer 1

The correct option is A $0.5A$

Energy stored in Capacitor$=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}C{V}^2$ at maximum current energy stored in inductor.

$\frac{1}{2}{CV}^2=\frac{1}{2}L{I}_0^2$

Where $I_0=$ Maximum Current.

$I_0^2=\frac{{CV}^2}{L}=\frac{{10}^{-6}\times 50\times 50}{10\times {10}^{-3}}=25\times {10}^{-2}$

$I_0=0.5A$