JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Current Electricity

The current from the cell is $i=\frac{E}{r+R}$, and the power delivered to the external resistor is $P=i^{2}R=\frac{E^{2}R}{(r+R{)}^2}$. To find the $R$ that maximises this, set $\frac{dP}{dR}=0$, which leads to $(r+R{)}^2-2R(r+R)=0$, so $(r+R)-2R=0$ and $R=r$. This is the maximum power transfer theorem: a source hands over the most power to its load when the load resistance equals the internal resistance.