Question

The ratio of time constants in charging and discharging of inductor in the circuit shown in figure is

• A. $1:1$
• B. $3:2$
• C. $2:3$
• D. $1:3$

Answer 1

The correct option is B $3:2$

When the key is closed the potential of the battery will act directly across the series combination of $L$ and $2R$.

Therefore, the time constant of charging is,

${\tau }_c=\frac{L}{2R}$

When the switch is open the inductor will discharge. The equivalent circuit of discharging is,


$\therefore$ The net resistance of the discharging circuit is

$R_{net}=3R$

$\therefore$ Time constant of discharging is,

${\tau }_d=\frac{L}{3R}$

Now,

${\tau }_c:{\tau }_d=\frac{L}{2R}:\frac{L}{3R}$

$\therefore {\tau }_c:{\tau }_d=3:2$

Hence, $\left(\text{B}\right)$ is the correct answer.