Question

Current through the battery at the instant when the switch S is closed is (in A):

Answer 1

At t = 0 , the capacitor will act as short-circuit. So as a result curretn will choose the path of least resistance and will not flow in the branch having $3\Omega$. Hence effective resistance will be only $2\Omega$. Therefore, $I=\frac{V}{R}=\frac{10}{2}=5A$