Q. In an $L - R$ circuit shown in above figure switch S is closed at time $t = 0$. If $e$ denotes the induced emf across inductor and $i$, the current in the circuit at any time $t,$ then which of the following graphs, figure shows the variation of $e\ with\ i$?
A) [image:7_image1.emf]
B) [image:7_image2.emf]
C) [image:7_image3.emf]
D) [image:7_image4.emf]
Explanation
(c) In $L - R$ circuit, current at any time $t$ is given by
$\ \ i = \frac{E}{R}\left( 1 - e^{- \frac{R}{L}t} \right) = \frac{E}{R} - \frac{E}{R}e^{- \frac{R}{L}t}$
$\ \ \ \ \frac{di}{dt}\ = \frac{E}{R}e^{- \frac{R}{L}t}\ \left( \frac{R}{L} \right) = \frac{E}{L}e^{- \frac{R}{L}t}$
Induced emf $= L\frac{di}{dt} = Ee^{- \frac{R}{L}t}$
From Eq. (i), $iR = E - Ee^{- \frac{R}{L}t}$
Using Eq. (ii), $iR = E - e\ \ \ \ \ \ \ \ or\ \ \ \ \ e = E - iR$
Therefore, graph between $e\ and\ i$ is a straight line with negative slope and positive intercept. The choice (c) is correct.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.