Thank you for reporting, we will resolve it shortly
Q.
A battery of $3.0 V$ is connected to a resistor dissipating $0.5 W$ of power. If the terminal voltage of the battery is $2.5 V ,$ the power dissipated within the internal resistance is :
$P_{R}=0.5 W$
$\Rightarrow i ^{2} R =0.5 W$
Also, $V=E-$ ir
$2.5=3-$ ir
$\Rightarrow $ ir $=0.5$
Power dissipated across 'r' $: P _{ r }= i ^{2} r$
Now $iR =2.5$
ir $=0.5$
On dividing : $\frac{ R }{ r }=5$
Now $\frac{ P _{ R }}{ P _{ r }}=\frac{ i ^{2} R }{ i ^{2} r } $
$\Rightarrow \frac{ P _{ R }}{ P _{ r }}=\frac{ R }{ r } $
$\Rightarrow \frac{ P _{ R }}{ P _{ r }}=5$
$\Rightarrow P _{ r }=\frac{ P _{ R }}{5}$
$\Rightarrow P _{ r }=\frac{0.50}{5} $
$\Rightarrow P _{ r }=0.10 W$
