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Q. A battery of e. m. f. $E$ and internal resistance $r$ is connected to an external resistance $R$ the condition for maximum power transfer is

WBJEEWBJEE 2008Current Electricity

Solution:

We know that the current in the circuit
$I=\frac{E}{R+r}$
and power delivered to the resistance $R$ is
$P=I^{2} R=\frac{E^{2} R}{(R+r)^{2}}$
It is maximum when $\frac{d P}{d R}=0$
$\frac{d P}{d R}=E^{2}\left[\frac{(r+R)^{2}-2 R(r +R)}{(r +R)^{4}}\right]=0$
or $(r+ R)^{2}=2 R(r+ R)$ or $R=r$