QJEEM16Ph359d: JEE Main 2016 Physics - Class 12 JEEM16Ph359d Solution: In the circuit shown, the resistance r is a variable…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

Here equivalent resistance

R_eq = R +

{{r \times R} \over {r + R}}

= R +

{{f{R^2}} \over {fR + R}}

= R +

{{fR} \over {f + 1}}

=

{{\left( {2f + 1} \right)R} \over {\left( {f + 1} \right)}}

Circuit current,

{\rm I}

=

{V \over {{{\mathop{\rm R}\nolimits} _{eq}}}}

=

{{V\left( {f + 1} \right)} \over {{\mathop{\rm R}\nolimits} \left( {2f + 1} \right)}}

Current in the resistance r,

I =

{{\rm I} \over {f + 1}}

=

{V \over {R\left( {2f + 1} \right)}}

Heat generated in r,

H =

{\rm I}_2^2\,r

=

{{{V^2}f} \over {R{{\left( {2f + 1} \right)}^2}}}

For maximum H,

{{dH} \over {df}}

= 0

\Rightarrow

{{{V^2}} \over R}\left[ {{1 \over {{{\left( {2f + 1} \right)}^2}}} - {{4f} \over {{{\left( {2f + 1} \right)}^3}}}} \right]

= 0

\Rightarrow

2f + 1 = 4f

\Rightarrow

2f = 1

\Rightarrow

f =

{1 \over 2}

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