Question

A $L-R$ circuit is connected to a battery of constant EMF $E$. The switch is closed at time $t=0$. If $E_{ind}$ denotes the magnitude of induced EMF across inductor and $i$ the current in the circuit at any time $t$. Then which of the following graph shows the variation of $E_{ind}$ with $i$?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We know that current in an $L-R$ circuit is,

$i=\frac{E}{R}(1-e^{\frac{-t}{\tau }}).........\left(1\right)$

Where, $\tau =\frac{L}{R}$

EMF induced across the inductor is,

$E_{ind}=-L\frac{di}{dt}$

Using $\left(1\right)$,

$E_{ind}=-L\frac{d}{dt}\left[\frac{E}{R}(1-e^{\frac{-t}{\tau }})\right]$

$=-L\times (\frac{E}{R}\times (-e^{\frac{-t}{\tau }})\times (-\frac{1}{\tau }))$

$=-\frac{EL}{R\tau }{e}^{\frac{-t}{\tau }}$

$=-\frac{EL}{R}\times \frac{R}{L}{e}^{\frac{-t}{\tau }}[\because \tau =\frac{L}{R}]$

$=-E{e}^{\frac{-t}{\tau }}$

If considering magnitude of $E_{ind}$:

$\left|\begin{array}{c}{E}_{ind}\end{array}\right|=E{e}^{\frac{-t}{\tau }}$

As we have, $i=\frac{E}{R}(1-e^{\frac{-t}{\tau }})$

$\Rightarrow i=\frac{E}{R}-\frac{E_{ind}}{R}$

$\therefore E_{ind}=E-iR$

This equation is similar to the equation of a straight line with a negative slope.


Hence, $\left(\text{A}\right)$ is the correct answer.

Alternate solution: We can apply KVL in the given circuit and we will get, $E-E_{ind}-iR=0$ $\therefore E_{ind}=E-iR$