Question

In the circuit shown aside, the switch S is closed for long time and at t = 0 s, S is opened. The current at t s (> 0) through the resistance R is:

• A. 1.5 (1 + e${}^{-3t}$) A
• B. 3.0 (1 - e${}^{-3t}$) A
• C. 3.0 (1 + e${}^{-3t}$) A
• D. 1.5 (1 - e${}^{-3t}$) A

Answer 1

The correct option is A 1.5 (1 + e${}^{-3t}$) A

When switch is closed $t=0s$
$1H$ is shorted in steady state.

Therefore, current through $1 H$,

$I_o=\frac{3}{1}=3A$

When S is opened, at $t=0s$, current through $R=I_o=3A$

As $t=\infty$, $I_f=\frac{3}{2}=1.5A$

Now time constant $=\frac{R_{eq}}{L}=\frac{3}{1}=3s^{-1}$

At time t, let $i\left(t\right)=a+b{e}^{-3t}$

at $t=0$ $3=a+b$

at $t=\infty$ $a=1.5$

Therefore $i\left(t\right)=1.5(1+e^{-3t})A$