Question

Rate of increment of energy in an inductor with time, in series LR circuit getting charge with battery of e.m.f. $E$ is best represented by [inductor has initially zero current]-

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Rate of increment of energy in indcutor
$=\frac{dU}{dt}=\frac{d}{dt}\left[\frac{1}{2}L{i}^2\right]=Li\frac{di}{dt}$
Current in the inductor at time t is:
$i=i_0(1-e^{\frac{-t}{\tau }})$ and $\frac{di}{dt}=\frac{i_0}{\tau }{e}^{\frac{-t}{\tau }}$
$\frac{dU}{dt}=\frac{L{i}_0^2}{\tau }{e}^{\frac{-t}{\tau }}(1-e^{\frac{-t}{\tau }})$
$\frac{dU}{dt}=0$ at t = 0 and $t=\infty$
Hence, E is best represented by (A).