In the circuit shown, the resistance r is a variable resistance. If for r = f R , the heat generation in r…

In the circuit shown, the resistance r is a variable resistance. If for $\mathrm{r}=\mathrm{f}\mathrm{R}$ , the heat generation in r is maximum then the value of $\mathrm{f}$ is

1. $\frac{1}{2}$
2. $1$
3. $\frac{1}{4}$
4. $\frac{3}{4}$

Solution

Power in “r”

$P=I_2^{2}r={\left[I\times \frac{R}{R+r}\right]}^{2}r=I^2\frac{R^{2}r}{{\left(R+r\right)}^2}$

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

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

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

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

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

$1+2f=4f$

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

Asked in: JEE Main 2016 (09 Apr Online)

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