Question

An inductor of inductance $L=400mH$ and resistors of resistance $R1=2\Omega$ and $R_2=2\Omega$ are connected to a battery of emf $12V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is

• A. $\frac{12}{t}{e}^{3t}V$
• B. $6\left(1-e^{-t/0.2}\right)V$
• C. $12e^{-5t}V$
• D. $6e^{-5t}V$

Answer 1

Answer: (C)

Growth in current in $L{R}_2$ branch when switch is closed is given by
$i=\frac{E}{R_2}\left[1-e^{-R_{2}t/L}\right]$
$\Rightarrow \frac{di}{dt}=\frac{E}{R-2}\frac{R_2}{L}\cdot e^{-R_{2}t/L}=\frac{E}{L}{e}^{-\frac{R_{2}t}{L}}$
Hence, potential drop across
$L=\left(\frac{E}{L}{e}^{R_{2}t/L}\right)$
$L=E{e}^{R_{2}t/L}$
$=12e-\frac{2t}{400\times 10^{-3}}$
$=12e^{-5t}V$