Electromagnetic Induction and Alternating Current 4 Question 3

3. A 20H inductor coil is connected to a 10ohm 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

(Main 2019, 10 Jan I)

(a) 2ln2

(b) 12ln2

(c) 2ln2

(d) ln2

Show Answer

Answer:

Correct Answer: 3. (c)

Solution:

  1. Given circuit is a series LR circuit In an LR circuit, current increases as

i=ER1eRLt

Now, energy stored in inductor is

UL=12Li2

where, L= self inductance of the coil and energy dissipated by resistor is

UR=i2R

Given, rate of energy stored in inductor is equal to the rate of energy dissipation in resistor. So, after differentiating, we get

iLdidt=i2Rdidt=RLiERRLeRLt=RLER1eRLt2eRLt=1eRLt=12

Taking log on both sides, we have

RLt=ln12RLt=ln2t=LRln2=2010ln2t=2ln2



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक