time constant

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?


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.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.