# What is the time constant L/R of this circuit?

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?

## 1 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.