![JEE Main 2016 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Current_Electricity/648b5a64417cc3fb48d6603c/question_1__q_648b5a64417cc3fb48d6603c__cdn-question-pool.getmarks.app__3605a536-52e2-40a1-a35a-b57290688e40-image__c56fdcb473_final_ppt_sync.png) In the circuit shown, the resistance r is a variable resistance. If for $r=fR$ , the heat generation in r is maximum then the value of $f$ is

![JEE Main 2016 Physics, Current Electricity — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Current_Electricity/648b5a64417cc3fb48d6603c/solution_1__q_648b5a64417cc3fb48d6603c__cdn-question-pool.getmarks.app__48c4dc1d-26ba-4f5b-8deb-ec5298878fc0-image__70187e91ce_final_ppt_sync.png) Power in “r” $P={I}_{2}^{2} r={ [ I\times \frac{R}{ R+r} ]}^{2} r={I}^{2} \frac{ {R}^{2} r}{ { ( R+r )}^{2} }$ $P={ [ \frac{E}{ R+\frac{ R\times fR}{ R+fR} } ]}^{2} \frac{ {R}^{2} ( fR )}{ { ( R+fR )}^{2} } =\frac{ {E}^{2} {R}^{2} fR}{ {R}^{2} { [ 1+\frac{f}{ ( f+1 )} ]}^{2} {R}^{2} { ( 1+f )}^{2} }$ $P=\frac{ {E}^{2} f}{ R{ ( 1+2f )}^{2} }$ For $P( \mathrm{max} ) \frac{ dP}{ df} =0$ $\frac{ {E}^{2} }{R} =[ \frac{ { ( 1+2f )}^{2} -f{ 2( 1+2f )( 0+2 ) }}{ { ( 1+2f )}^{2} } ]=0$ ${ ( 1+2f )}^{2} =4f( 1+2f )$ $1+2f=4f$ $f=\frac{1}{2}$

JEE Main 2016 — Physics Electromagnetism

2016mcqmedium

JEE Main 2016 Physics, Current Electricity — question figure

In the circuit shown, the resistance r is a variable resistance. If for $r=fR$ , the heat generation in r is maximum then the value of $f$ is

Official previous-year question

Held on 9 Apr 2016 · Verified 6 Jul 2026.

Options

A
$\frac{1}{2}$
B
$1$
C
$\frac{1}{4}$
D
$\frac{3}{4}$

Did you get this right?

Solution

JEE Main 2016 Physics, Current Electricity — solution figure

Power in “r”

$P={I}_{2}^{2} r={ [ I\times \frac{R}{ R+r} ]}^{2} r={I}^{2} \frac{ {R}^{2} r}{ { ( R+r )}^{2} }$

$P={ [ \frac{E}{ R+\frac{ R\times fR}{ R+fR} } ]}^{2} \frac{ {R}^{2} ( fR )}{ { ( R+fR )}^{2} } =\frac{ {E}^{2} {R}^{2} fR}{ {R}^{2} { [ 1+\frac{f}{ ( f+1 )} ]}^{2} {R}^{2} { ( 1+f )}^{2} }$

$P=\frac{ {E}^{2} f}{ R{ ( 1+2f )}^{2} }$

For $P( \mathrm{max} ) \frac{ dP}{ df} =0$

$\frac{ {E}^{2} }{R} =[ \frac{ { ( 1+2f )}^{2} -f{ 2( 1+2f )( 0+2 ) }}{ { ( 1+2f )}^{2} } ]=0$

${ ( 1+2f )}^{2} =4f( 1+2f )$

$1+2f=4f$

$f=\frac{1}{2}$

Your note

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Solution

JEE Main 2016 Physics, Current Electricity — solution figure

Power in “r”

$P={I}_{2}^{2} r={ [ I\times \frac{R}{ R+r} ]}^{2} r={I}^{2} \frac{ {R}^{2} r}{ { ( R+r )}^{2} }$

$P={ [ \frac{E}{ R+\frac{ R\times fR}{ R+fR} } ]}^{2} \frac{ {R}^{2} ( fR )}{ { ( R+fR )}^{2} } =\frac{ {E}^{2} {R}^{2} fR}{ {R}^{2} { [ 1+\frac{f}{ ( f+1 )} ]}^{2} {R}^{2} { ( 1+f )}^{2} }$

$P=\frac{ {E}^{2} f}{ R{ ( 1+2f )}^{2} }$

For $P( \mathrm{max} ) \frac{ dP}{ df} =0$

$\frac{ {E}^{2} }{R} =[ \frac{ { ( 1+2f )}^{2} -f{ 2( 1+2f )( 0+2 ) }}{ { ( 1+2f )}^{2} } ]=0$

${ ( 1+2f )}^{2} =4f( 1+2f )$

$1+2f=4f$

$f=\frac{1}{2}$

JEE Main 2016 — Physics Electromagnetism

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JEE Main 2016 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism