The circuit shown in Fig. is operating in steady-state with switch $${S_1}$$ closed. The switch $${S_1}$$ is opened at $$t\, = \,0$$.

(a) Find $${i_L}\left( {{0^ + }} \right)$$.
(b) Find $${e_1}\left( {{0^ + }} \right)$$.
(c) Using nodal equations and Laplace transform approach, find an expression for the voltage across the capacitor for all $$t\, = \,0$$.

The network $$N$$ in Fig. consists only of two elements: a resistor of $$1\Omega $$ and an inductor of L Henry. $$A$$ $$5$$ $$V$$ source is connected at the input at $$t\, = \,0$$ seconds. The inductor current is zero at $$t\, = \,0$$. The output voltage is found to be $$5{e^{ - 3t}}\,\,V,$$ for $$t\, = \,0$$.

(a) Find the voltage transfer function of the network.
(b) Find L, and draw the configuration of the network.
(c) Find the impulse response of the network.

In the circuit of figure, the switch $$'S'$$ has remained open for a long time. The switch closes instantaneously at $$ t = 0$$. n

(a) Find $${V_0}$$ for $$t \le 0$$ and as $$t \to \infty $$.
(b) Write an expression for $${V_0}$$ as a function of time for $$0 \le t \le \infty $$.
(c) Evaluate $${V_0}$$ at $$t = 25\,\,\mu $$sec.

The circuit shown in the figure, is initially in its steady state. Switch is opened at t = 0.
(1) Determine the initial voltage, V_C(0^-), across the capacitor, and the initial current, i_L(0^-) , through the indicator.
(2) Calculate the voltage, V_L(t) , across the inductance for t > 0.