Question

A series $L-R$ circuit is connected to a battery of emf $E$ at $t=0$ as shown in the figure. The rate of increment of energy in the inductor with the time is best represented by-

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Energy stored in an inductor is,

$U=\frac{1}{2}L{i}^2$

Where, $i$ is the current flowing through the inductor at time $t.$

The rate of increment of energy in the inductor is,

$\frac{dU}{dt}=\frac{d}{dt}\left[\frac{1}{2}L{i}^2\right]$

$\frac{dU}{dt}=\frac{L}{2}\times 2i\left(\frac{di}{dt}\right)$

$\frac{dU}{dt}=Li\left(\frac{di}{dt}\right)$

At time $t=0$ inductors behaves as an open circuit i.e., $i=0.$

$\frac{dU}{dt}=0\text{at}t=0$

Now, at $t=\infty$ steady state is achieved i.e., $\frac{di}{dt}=0$

$\frac{dU}{dt}=0\text{at}t=\infty$

Thus, the rate of increment of energy in the inductor must be maximum between $0<t<\infty$.




<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.