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

Answer: (A)

Answer 2

Rate of increment of energy in inductor

= $\frac{dU}{dt}$ = $\frac { \mathrm { d } } { \mathrm { dt } }$ $\left\lbrack \frac{1}{2}Li^{2} \right\rbrack$ = Li $\frac{di}{dt}$

Current in the inductor at time t is:

i = i₀ (1 –$e^{\frac{- t}{\tau}}$) and $\frac{di}{dt}$ = $\frac{i_{0}}{\tau}$ $e^{\frac{- t}{\tau}}$

$\frac{dU}{dt}$ = $\frac { \mathrm { Li } _ { 0 } ^ { 2 } } { \tau }$ $e^{\frac{- t}{\tau}}$ (1 –$e^{\frac{- t}{\tau}}$)

$\frac{dU}{dt}$ = 0 at t = 0 and t = 

Hence E is best represented by