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

The correct option is 8+2e^-10t.

We are given a circuit with a switch $S$ that has been closed for a very long time and is opened at $t=0$. We want to find the inductor current $i_L\left(t\right)$ for $t\ge 0$.

Step 1: Analyze the circuit for $t<0$ (Switch closed)
Before $t=0$, the switch $S$ has been closed for a very long time, which means the circuit has reached steady state.
- Since the switch $S$ is closed, it acts as a short circuit, bypassing the parallel branch containing the $4\Omega$ resistor and the $30\text{V}$ battery.
- At steady state, the inductor acts as a short circuit.
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, we have:
$i_L\left(0^{+}\right)=i_L\left(0^{-}\right)=10\text{A}$

Step 2: Analyze the circuit for $t>0$ (Switch open)
When the switch $S$ is opened, the branch with the switch is disconnected. The loop now contains:
- The $10\text{V}$ DC source.
- The $4\Omega$ resistor in series with the $30\text{V}$ DC source.
- The $1\Omega$ resistor in series with the $0.5\text{H}$ inductor.
Looking at the polarity of the $30\text{V}$ source shown in the diagram, its positive terminal is on the right and negative on the left, opposing the direction of the $10\text{V}$ source or adding to it depending on the loop. Tracing the clockwise loop:
The total DC voltage in the loop is:
$V_{\text{total}}=10\text{V}+30\text{V}=40\text{V}$
At steady state ($t\to \infty$), the inductor again acts as a short circuit. The steady-state current $i_L(\infty )$ is:
$i_L(\infty )=\frac{10\text{V}+30\text{V}}{4\Omega +1\Omega }=\frac{40}{5}=8\text{A}$

Step 3: Determine the time constant ($\tau$)
The equivalent resistance of the loop with the independent sources deactivated is:
$R_{\text{eq}}=4\Omega +1\Omega =5\Omega$
The time constant $\tau$ of the RL circuit is:
$\tau =\frac{L}{R_{\text{eq}}}=\frac{0.5}{5}=0.1\text{s}$

Step 4: Find the expression for $i_L\left(t\right)$
The transient response of the inductor current is given by:
$i_L\left(t\right)=i_L(\infty )+\left[i_L\right(0^{+})-i_L(\infty \left)\right]e^{-t/\tau }$
Substituting the values we found:
$i_L\left(t\right)=8+(10-8)e^{-t/0.1}$
$i_L\left(t\right)=8+2e^{-10t}\text{A}$