Physics - Current Electricity Question with Solution | TestHub

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

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

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

$1+2f=4f$

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