Q.
An inductor of inductance $L = 400\ mH$ and resistors of resistances $R_{1} = 4\Omega$ and $R_{2} = 2\Omega$ are connected to 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) $6e^{- 5t}\ V$
B) $\frac{12}{t}\ e^{- 3t}\ V$
C) $6(1 - e^{- t/0.2})\ V$
D) $12\ e^{- 5t}\ V$
Explanation
(d) $l_{1} = \frac{F}{R_{1}} = \frac{12}{2} = 6\ A$
$E = L\frac{dl_{2}}{dt} + R_{2} \times l_{2}$
$I_{2} = I_{0}(1 - e^{- t/t_{c}})$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ I_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6A\ \ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{c} = \frac{L}{R} = \frac{400 \times 10^{- 3}}{2} = 0.2$
$I_{2} = 6(1 - e^{- t/0.2})$
Potential drop across
$L = E - R_{2}L_{2} = 12 - 2 \times 6\left( 1 - e^{- bt} \right) = 12e^{- 5t}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.