Question 254 of 380

A cell of internal resistance $r$ drives current through an external resistance $R.$ The power delivered by the cell to the external resistance will be maximum when:

A$R=2r$B$R=r$C$R=1000r$D$R=0.001r$

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Answer: (B) $R=r$

Current $i=\dfrac{E}{r+R}$
Power generated in $R$
$P={i}^{2}R$
$P=\dfrac{{E}^{2}R}{(r+R{)}^{2}}$
for maximum power $\dfrac{dP}{dR}=0$
${E}^{2}\left[\dfrac{(r+R{)}^{2}\times 1-R\times 2(r+R)}{(r+R{)}^{4}}\right]=0$
$\Rightarrow \text{when,}r=R \text{then,} \dfrac{dP}{dR}=0$

JEE Main Physics practice question on Current Electricity (Electric Power and Heating Effect of Current), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.