Class 12th Physics Physics Current Electricity Overview

A simple $L R$ circuit is connected to a battery at $t = 0$ . The time instant at which rate of energy storage in inductor is half of power delivered by battery

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <mi>I</mi> <mi>n</mi> <mi> </mi> <mn>2</mn> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <mi>I</mi> <mi>n</mi> <mi> </mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <mi>I</mi> <mi>n</mi> <mi> </mi> <mn>2</mn> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <mi>I</mi> <mi>n</mi> <mi> </mi> <mn>3</mn> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 3

Detailed Solution:

$i = \frac{v_{0}}{R} (1 - e^{\frac{t R}{L}})$

$U = \frac{1}{2} L i^{2}$

$\frac{d U}{d t} = L i \frac{d i}{d t}$

$2 L * \frac{R}{L} * \frac{v_{0}^{2}}{R^{2}} (1 - e^{\frac{t R}{L}}) e^{\frac{t R}{L}} = \frac{v_{0}^{2}}{R} (1 - e^{\frac{t R}{L}})$

$t = \frac{L}{R} I n 2$

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