Question

A cell having EMF E and internal resistance r is connected to a load R. Find the value of R such that power dissipated by the cell is maximum

Answer 1

Current$=I=\frac{E}{R+r}$

Power dissipated by cell is $P=I^{2}R=\frac{E^{2}R}{(R+r{)}^2}$

For maximum power by cell-

$\frac{dP}{dR}=0$

$⟹E^2\frac{(R+r{)}^2-2R(R+r)}{(R+r{)}^4}=0$

$⟹R=r$

That is $R=$ internal resistance.