A battery of 3.0, V is connected to a resisto | vedclass

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Question

$P_{R}=0.5 W$

$\Rightarrow i ^{2} R =0.5 W$

$Also , V = E - ir$

$2.5=3-$ ir

$\Rightarrow$ ir $=0.5$

Power dissipated across 'r'

Vedclass · Accepted Answer

$0.10$

Vedclass · Answer

$P_{R}=0.5 W$

$\Rightarrow i ^{2} R =0.5 W$

$Also , V = E - ir$

$2.5=3-$ ir

$\Rightarrow$ ir $=0.5$

Power dissipated across 'r' : $P_{r}=i 2 r$

$Now$ iR $=2.5$

ir $=0.5$

On dividing : $\frac{ R }{ r }=5$

$\operatorname{Now} \frac{ P _{ R }}{ P _{ r }}=\frac{ i ^{2} R }{ i ^{2} r } \Rightarrow \frac{ P _{ R }}{ P _{ r }}=\frac{ R }{ r } \Rightarrow \frac{ P _{ R }}{ P _{ r }}=5$

$\Rightarrow P _{ r }=\frac{ P _{ R }}{5}$

$\Rightarrow P _{ r }=\frac{0.50}{5} \Rightarrow P _{ r }=0.10 W$

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$.......W$

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