Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is :

Option 1)
$ln\; 2$
Option 2)
$2\; ln\; 2$
Option 3)
$\frac{1}{2}\; ln\; 2$

Option 4)
$\frac{2}{ ln\; 2}\;$

Answers (1)

best_answer

For L-R circuit

$i=\frac{E}{R}(I-e^{-t/t})$ where $L = \frac{L}{R}$

and rate of magnetic energy stored in inductor = $i\times L\times \frac{di}{dt}.......\; \; \; \; (1)$

and rate of dissipiation of energy across Resistance = $i^{2}R\; \; \; \; \; ......(2)$

from question $(1)=(2)$

So $i^{2}R=i\; L\; \frac{di}{dt}$

$R\times\frac{E}{R}(I-e^{-\frac{Rt}{L}})=\left ( L\times\frac{E}{R}\times\frac{R}{L}e^{-\frac{Rt}{L}} \right )$

$\Rightarrow 2\; e^{-\frac{Rt}{L}}=1$

$\frac{R}{L}t=ln\; 2$

$t=\frac{L}{R}ln\; 2$

$t=\frac{20}{10}ln\; 2$

$t=2\; ln\; 2$

Option 1)

$ln\; 2$

Option 2)
$2\; ln\; 2$
Option 3)

$\frac{1}{2}\; ln\; 2$

Option 4)

$\frac{2}{ ln\; 2}\;$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions

anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question