I've solved this circuit but i got some issues about the alternatives.
English question at this picture:
Switch \$S_1\$ had been closed for long time and after that \$S_2\$ was opened. Therefore, at \$t=0\$, the switch \$S_2\$ has closed.
The current \$i_1(t)\$ is equal for \$t\geq 0\$
My attempt:
\$L\frac{di_1(t)}{dt} + i_1(t).1 = 2\Rightarrow\$
\$i(t) = C.e^{-t} + 2\$
\$i_1(t=0^-) = i_1(t=0^+) = 2V/2\Omega = 1A\$
\$C = -1\$
\$i_1(t) = -e^{-t} + 2\Rightarrow \boxed{i_1(t) = 2\left(1-\frac{e^{-t}}{2}\right)}\$
\$\tau = 1s\$
Is this alternatives about that question all wrong respecting time constant?
Best Answer
Let us re-check your solution through laplace method.
At t>=0,
$$\frac{2}{s} = L(sI(s) - I(0)) + I(s)R_{eq} $$ $$\implies \frac{2}{s} = sI(s)-1+I(s)$$ $$\implies \frac{2}{s}+1 = I(s)(s+1)$$ $$\implies \frac{2+s}{s(s+1)} = I(s)$$ $$\implies I(s) = \frac{2}{s}+\frac{-1}{s+1}$$ Taking inverse laplace, $$I(t) = 2 - e^{-t} = 2(1-\frac {e^{-t}}{2})$$
Your solution is therefore correct.