Two identical batteries, each of emf 2 V and internal resistance r=1 Ω are connected as shown. The maximum power that can be developed across R using these batteries is
Alternate solution: P=i2R=(2R+0.5)2R From maximum power theorem, Rext=rint ∴R=req=r1r2r1+r2=0.5 Ω So, P=(20.5+0.5)20.5=2 W |