In the circuit shown above, the switch is closed after a long time. The current i_S (0⁺) through the switch is

A t = 0, steady state is reached

\({I_2}\left( {{0^ - }} \right) = \frac{{12}}{{8 + 4}}\)

= 1 A

\({V_1}\left( {{0^ - }} \right) = \left( {\frac{6}{{6 + 3}}} \right){V_L}\)

\( = \left( {\frac{6}{9}} \right) \times 4\)

\( = \frac{8}{3}V\)

\({V_2}\left( {{0^ - }} \right) = \frac{3}{{6 + 3}} \times 4\)

\( = \frac{4}{3}V\)

At t = 0⁺, we can draw the circuit as follows

\({I_1} = \frac{{\frac{8}{3}}}{4} = \frac{2}{3}A\)

I₁ + I_s = 1 A

\({I_s} = \left( {1 - \frac{2}{3}} \right)A\)

\( = \frac{1}{3}A\)