Question

In the circuit shown in figure, switch remains closed for long time. It is opened at time $t=0$. Match the following two columns at an instant $t=\left(\ln 2\right)$ second.

Answer 1

Steady state current through inductor
$i_0=\frac{9}{3}=3A$
Now this current decays exponentially across inductor and two resistors.
${\tau }_L=\frac{L}{R}=\frac{9}{6+3}=1s$
$t_{1/2}=\left(\ln 2\right){\tau }_L=\left(\ln 2\right)s$
Given time is half-lif time. Hence current will remain $1.5A$.
$i=i_0{e}^{-t/{\tau }_L}=3e^{-t}$
$(-\frac{di}{dt})=3e^{-t}$
In the beginning $(-\frac{di}{dt})=3A/s$
After one -half time $(-\frac{di}{dt})==1.5A/s$
(a) $V_L=L(-\frac{di}{dt})=9\times 1.5=13.5V$
(b)$V_{3\Omega }=iR=1.5\times 3=4.5V$
(c) $V_{6\Omega }=iR=1.5\times 6=9V$
(d) $V_{bc}=V_L-V_{3\Omega }=9V$