In the circuit, switch ‘S’ is in the closed position for a very long time. If the switch is opened at time t =0, then i_L( t)  in Amperes, for t ≥0 is

To find the inductor current $i_L\left(t\right)$ for $t\ge 0$, we analyze the circuit in two stages: before and after the switch $S$ is opened.

1. Initial Condition ($t<0$):
Before $t=0$, the switch $S$ has been closed for a very long time. Under DC steady-state conditions, the inductor acts as a short circuit.
The closed switch $S$ acts as a short circuit across the branch containing the $4\Omega$ resistor and the $30\text{V}$ source, bypassing them completely.
Thus, the active circuit simply consists of the $10\text{V}$ source connected to the $1\Omega$ resistor and the short-circuited inductor.
The current through the inductor just before the switch opens is:
$i_L\left(0^{-}\right)=\frac{10\text{V}}{1\Omega }=10\text{A}$
Since the current through an inductor cannot change instantaneously, the current immediately after the switch opens is:
$i_L\left(0^{+}\right)=i_L\left(0^{-}\right)=10\text{A}$

2. Final Steady-State Condition ($t\to \infty$):
At $t=0$, the switch $S$ is opened. The branch containing the $4\Omega$ resistor and the $30\text{V}$ source is now connected in series with the rest of the loop.
Looking at the polarity of the $30\text{V}$ source, its negative terminal is on the left and its positive terminal is on the right, which series-aids the $10\text{V}$ source.
The total equivalent resistance of the loop is:
$R_{\text{eq}}=4\Omega +1\Omega =5\Omega$
The total equivalent DC voltage in the loop is:
$V_{\text{eq}}=10\text{V}+30\text{V}=40\text{V}$
As $t\to \infty$, the circuit reaches a new steady state, and the inductor again acts as a short circuit. The final current is:
$i_L(\infty )=\frac{V_{\text{eq}}}{R_{\text{eq}}}$

3. Time Constant ($\tau$):
The equivalent resistance in the loop with the switch open is $R_{\text{eq}}=5\Omega$, and the inductance is $L=0.5\text{H}$.
The time constant of this RL circuit is:
$\tau =\frac{L}{R_{\text{eq}}}$
The exponential decay factor is:
$\frac{1}{\tau }=\frac{1}{0.1}=10{\text{s}}^{-1}$

4. Current Response Equation:
The general transient response equation for a first-order RL circuit is given by:
$i_L\left(t\right)=i_L(\infty )+\left[i_L\left(0^{+}\right)-i_L(\infty )\right]e^{-t/\tau }$
Substituting the derived values:
$i_L\left(t\right)=8+(10-8)e^{-10t}=8+2e^{-10t}\text{A}$

Thus, the transient current equation for the inductor is 8 + 2e^-10t. Note that the correct option contains a minor typographical rendering error showing as 8 + 2e^-0t, which corresponds to 8 + 2e^-10t.