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 $t=\ln 2$ second.

$\begin{array}{cc}\text{Column\_I}& \text{Column-II}\\ \text{(a) Potential difference across inductor}& \text{(p) 9 V}\\ \text{(b) Potential difference across}3\Omega \text{resistance}& \text{(q) 4.5 V}\\ \text{(c) Potential difference across}6\Omega \text{resistance}& \text{(r) 6 V}\\ \text{(d) Potential difference between points b and c}& \text{(s) None of these}\end{array}$

• A. $\left(\text{a}\right)\to \text{p};\left(\text{b}\right)\to \text{q};\left(\text{c}\right)\to \text{r};\left(\text{d}\right)\to \text{s}$
• B. $\left(\text{a}\right)\to \text{s};\left(\text{b}\right)\to \text{q};\left(\text{c}\right)\to \text{p};\left(\text{d}\right)\to \text{p}$
• C. $\left(\text{a}\right)\to \text{r};\left(\text{b}\right)\to \text{q};\left(\text{c}\right)\to \text{r};\left(\text{d}\right)\to \text{p}$
• D. $\left(\text{a}\right)\to \text{p};\left(\text{b}\right)\to \text{q};\left(\text{c}\right)\to \text{q};\left(\text{d}\right)\to \text{p}$

Answer 1

The correct option is B $\left(\text{a}\right)\to \text{s};\left(\text{b}\right)\to \text{q};\left(\text{c}\right)\to \text{p};\left(\text{d}\right)\to \text{p}$

At steady state, inductor is short-circuited.

Steady state current through inductor
$i_0=\frac{4.5\times 6}{6+3}=3\text{A}$

Now this current decays exponentially across inductor and two resistors.

${\tau }_d=\frac{L}{R}=\frac{9}{6+3}=1\text{s}$

$t_{\frac{1}{2}}=\left(\ln 2\right){\tau }_d=\left(\ln 2\right)\text{s}$

Given time is half-time. Hence current will remain $1.5\text{A}$.

$i=i_0{e}^{\frac{-t}{{\tau }_L}}=3e^{-t}$

$\Rightarrow (-\frac{di}{dt})=3e^{-t}$

In the beginning , $(-\frac{di}{dt})=3\frac{\text{A}}{\text{s}}$

After one half-time , $(-\frac{di}{dt})=1.5\frac{\text{A}}{\text{s}}$

(a) $V_L=L(-\frac{di}{dt})=9\times 1.5=13.5\text{V}$
(b) $V_{3\Omega }=iR=1.5\times 3=4.5\text{V}$
(c) $V_{6\Omega }=iR=1.5\times 6=9\text{V}$
(d) $V_{bc}=V_L-V_{3\Omega }=9\text{V}$