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, Diffcult
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Heat generation in $r$ will be, $H=I_{r}^{2} r$
$H=\left(\frac{V}{R+\frac{R r}{R+r}} \times \frac{r}{R+r}\right)^{2} r=\left(\frac{V}{R(R+2 r)}\right)^{2} r$
Putting $\frac{d H}{d r}=0$ gives, $r=\frac{R}{2} .$ Hence, $f=\frac{1}{2}$
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