Question

An L-R circuit connected to a battery of a constant emf E. The switch S is closed at time t = 0. If ‘e’ denotes the induced emf across the inductor which of the graphs represents the variation of e with i?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$i=\frac{1}{2}(1-e^{\frac{-t}{t}})$ where $\tau =\frac{L}{R};\therefore \frac{di}{dt}=\frac{E}{R}\times (-\frac{1}{\tau })e^{\frac{-t}{t}}=\frac{E}{L}{e}^{\frac{-t}{t}}$
Induced emf is $e=-L\frac{di}{dt}=E{e}^{\frac{-t}{t}}$ --(1) From (2) , $e^{\frac{-t}{t}}=\frac{e}{E}$ Using this in (1) we have $i=\frac{E}{R}(1-\frac{e}{E})=\frac{E}{R}-\frac{e}{R}ore=E-iR$
Equation (3) shows that the graph of e against i is a straight line with negative slope and positive intercept.