Physics - Current Electricity Question with Solution | TestHub

Power in “r”

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

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

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

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

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

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

$1+2f=4f$

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