Interpret the circuit

From the given data and options, the standard arrangement is:

• battery of $12\text{ V}$,
• resistor $R_1=2\Omega$ in series with a parallel branch containing:
  • inductor $L=400\text{ mH}=0.4\text{ H}$,
  • resistor $R_2=2\Omega$.

We need the potential drop across the inductor as a function of time after closing the switch at $t=0$.

Initial condition at $t=0^+$

Just after switching on, current through an inductor cannot change instantaneously, so the inductor behaves like an open circuit.

Therefore, initially the circuit reduces to $R_1$ and $R_2$ in series: $R_1+R_2=2+2=4\Omega$

Initial current through the circuit: $I(0^+)=\frac{12}{4}=3\text{ A}$

Hence the voltage across $R_2$ is $V_{R_2}(0^+)=I(0^+)R_2=3\times 2=6\text{ V}$

Since $L$ is in parallel with $R_2$, the initial voltage across the inductor is also $V_L(0^+)=6\text{ V}$

Final condition at $t\to\infty$

In steady state, the inductor behaves like a short circuit.

Then the branch containing $L$ shorts the parallel combination, so $R_2$ gets bypassed. The circuit becomes only $R_1$ in series with the battery.

Thus the voltage across the inductor branch becomes $V_L(\infty)=0$

Time constant

For the transient of an inductor, the time constant is $\tau=\frac{L}{R_{\text{th}}}$ where $R_{\text{th}}$ is the Thevenin resistance seen by the inductor with the battery replaced by a short.

Looking into the inductor terminals:

• battery is shorted,
• $R_1$ and $R_2$ both connect from the inductor terminal to ground, so they are in parallel.

Therefore, $R_{\text{th}}=R_1\parallel R_2=\frac{2\cdot 2}{2+2}=1\Omega$

Hence, $\tau=\frac{0.4}{1}=0.4\text{ s}$

Voltage across the inductor

Since the inductor voltage decays exponentially from $6\text{ V}$ to $0$: $V_L(t)=V_L(0^+)e^{-t/\tau}$

Substituting values, $V_L(t)=6e^{-t/0.4}=6e^{-2.5t}\text{ V}$

Compare with options

• A: $\dfrac{12}{t}e^{-3t}$ — incorrect form
• B: $6(1-e^{-t/0.2})$ — rises from $0$ to $6$, incorrect
• C: $12e^{-5t}$ — wrong initial value and exponent
• D: $6e^{-5t}$ — correct initial value but wrong time constant

So the correct expression should be: $\boxed{6e^{-2.5t}\text{ V}}$

Comparison with stored answer

Stored correct answer is C, i.e. $12e^{-5t}\text{ V}$.

But our derivation shows the voltage across the inductor is $\boxed{6e^{-2.5t}\text{ V}}$ which does not match any option exactly.

Therefore, the stored answer appears inconsistent with the circuit analysis. If forced to choose the closest option by initial value behavior, none is fully correct.