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\(.......W\)

\(\Rightarrow i ^{2} R =0.5 W\)

\(Also , V = E - ir\)

\(2.5=3-\) ir

\(\Rightarrow\) ir \(=0.5\)

Power dissipated across 'r' : \(P_{r}=i 2 r\)

\(Now\) iR \(=2.5\)

ir \(=0.5\)

On dividing : \(\frac{ R }{ r }=5\)

\(\operatorname{Now} \frac{ P _{ R }}{ P _{ r }}=\frac{ i ^{2} R }{ i ^{2} r } \Rightarrow \frac{ P _{ R }}{ P _{ r }}=\frac{ R }{ r } \Rightarrow \frac{ P _{ R }}{ P _{ r }}=5\)

\(\Rightarrow P _{ r }=\frac{ P _{ R }}{5}\)

\(\Rightarrow P _{ r }=\frac{0.50}{5} \Rightarrow P _{ r }=0.10 W\)