Q.
An inductor of inductance $L = 400mH$ and resistors of resistances $R_{1} = 2\Omega$ 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) $6e^{- 5t}V$
B) $\frac{12}{t}e^{- 3t}V$
C) $6\left( 1 - e^{\frac{- t}{0.2}} \right)V$
D) $12e^{- 5t}V$
Explanation
(d) $E$(across $BC$) $= L\frac{dI_{2}}{dt} + R_{2}I_{2}$
[image:6_image62.emf]
$I_{2} = I_{0}(1 - e^{- t/t_{0}})$
$I_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6\ A$
$\tau = t_{0} = \frac{L}{R} = \frac{400 \times 10^{- 4}}{2\Omega} = 0.2\ S$
$\therefore I_{2} = 6(1 - e^{- t/0.2})$
Potential drop areas $L = E - R_{2}I_{2}$
$= 12 - 2 \times 6(1 - e^{- t/0.2})$
$= 12e^{- t/0.2} = 12e^{- 5t}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.