Question 261 of 380

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

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

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Answer: (A) $\dfrac{1}{2}$

Power in “r”

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

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

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

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

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

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

$1+2f=4f$

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

JEE Main Physics practice question on Current Electricity (Electric Power and Heating Effect of Current), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.