The steady state current in the circuit shown below is

Question

Professor PC Thomas Classes · Answer

(c) : In steady state capacitor acts as an open circuit. So, the circuit will become \ $ \begin{aligned} & R_{e q}=2+3=5 \Omega \ & V=10 \mathrm{~V} \ & I=\frac{V}{R_{e q}}=\frac{10}{5}=2 \mathrm{~A} \end{aligned} $