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A 20 {H} inductor coil is connected to a 10 ohm resistance in

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A $20 H$ 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

Name: Video solution 1: A 20H 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
Uploaded: 2024-01-13T10:20:29.000Z
Duration: 21 min
Description: Video solution 1: A 20H 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

\frac{2}{l n 2}

\frac{1}{2} ln 2

2 ln 2

ln 2

Views: 5,311 students

Updated on: Jan 13, 2024

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

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20 H

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

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Text solutionVerified

Given circuit is a series

L - R

circuit

In an

L

R

circuit, current increases is

i = \frac{E}{R} (1 - e^{- \frac{R}{L} t})

Now, energy stored in inductor is

U_{L} = \frac{1}{2} L i^{2}

where,

L =

self inductance of the coil and energy dissipated by resistor is

U_{R} = i^{2} R

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

iL \frac{d i}{d t} = i^{2} R \Rightarrow \frac{d i}{d t} = \frac{R}{L} i \Rightarrow \frac{E}{R} \cdot \frac{R}{L} e^{- \frac{R}{L} t} = \frac{R}{L} \cdot \frac{E}{R} (1 - e^{- \frac{P _{f}}{L}}) \Rightarrow 2 e^{- \frac{R _{t}}{L}} = 1 \Rightarrow e^{- \frac{R _{2}}{L}} = \frac{1}{2}

Taking log on both sides, we have

\Rightarrow \Rightarrow \frac{- R}{L} t = ln (\frac{1}{2}) \Rightarrow \frac{R}{L} t = ln 2 t = \frac{L}{R} ln 2 = \frac{20}{10} ln 2 \Rightarrow t = 2 ln 2

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Question Text	A $20 H$ 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
Updated On	Jan 13, 2024
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	320
Avg. Video Duration	13 min

A $20 H$ 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

Text solutionVerified

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A 20H 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

Text solutionVerified

Filo tutor solutions (2)

Questions from JEE Mains 2019 - PYQs

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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A $20 H$ 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