Question

An inductor of self inductance $100mH$ and a resistor of resistance $50\Omega$, are connected to a $2V$ battery. The time required for the current to come to half its steady value is

• A. $2$ millisecond
• B. $2\ln$ $\left(0.5\right)$ millisecond
• C. $2\ln \left(3\right)$millisecond
• D. $2\ln \left(2\right)$ millisecond

Answer 1

The correct option is D $2\ln \left(2\right)$ millisecond

The time constant of the circuit is

$\tau =\frac{L}{R}=\frac{100\times {10}^{-3}}{50}=2\times {10}^{-3}s=2$millisecond

Current at time $t$ is given by $I=I_0{e}^{-t/\tau }$
where $I_0$ is the steady current. Therefore, time for $I$ to fall to $I_0/2$ is

$e^{-t/\tau }=\frac{1}{2}$ or $e^{-t/\tau }=2,t=\tau ln\left(2\right)=2ln\left(2\right)$ millisecond