JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 26)
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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 :
$${1 \over 4}$$
$${1 \over 2}$$
$${3 \over 4}$$
1
Explanation
Here equivalent resistance
Req = 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 = $${1 \over 2}$$
Req = 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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