Question

In Fig, initially the capacitor is charged to a potential of 5 V and then connected to position across the inductor at position 2. The maximum current flowing in the LC circuit when the capacitor is connected across the inductor is :

• A. $10A$
• B. $20A$
• C. $30A$
• D. None of these

Answer 1

The correct option is A $10A$

Total energy stored in capacitor $E=\frac{1}{2}C{V}^2$, which must be equal to the energy stored in the indicator $E=\frac{1}{2}L{I}_{max}^2$

i.e. $\frac{1}{2}C{V}^2=\frac{1}{2}L{I}_{max}^2$

When we substitute the values of $V=5V,C=10mF,L=2.5mH$,

We get,

$\frac{1}{2}\times 10\times {10}^{-3}\times 5^2=\frac{1}{2}\times 2.5\times {10}^{-3}\times I_{max}^2$

$I_{max}$ = $10A$