Question

In circuit, initially capacitor and inductor does not have any energy. Then find current through the battery (in ampere) just after switch is closed

Answer 1

Just after switch is closed, there is no current through the coil and capacitor offers no resistance.

Net resistance $R=\frac{9\times 9}{9+9}=4.5\Omega$

Current through the battery is $i=\frac{V}{R}=\frac{18}{4.5}=4\text{A}$

Accepted answer: $4$