question_answer

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)

Correct Answer: A

Solution :

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