Question

An inductor of inductance $L =400\, mH$ and resistors of resistance $R 1=2 \,\Omega$ and $R_{2}=2 \Omega$ are connected to a battery of emf $12\, V$ 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{d i}{d t}=\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}$
$=12 e-\frac{2 t}{400 \times 10^{-3}} $
$=12 e^{-5 t} V$