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Question

# 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

A
3.2 W
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B
8.2 W
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C
2 W
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D
4 W
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Solution

## The correct option is C 2 WFrom the given diagram, we can easily observe that both cells are in parallel combination with same polarity. So, equivalent emf of cell is given by Eeq=E1r1+E2r21r1+1r2 ⇒Eeq=21+2111+11=2 V And equivalent internal resistance will be req=r1r2r1+r2=1×11+1=0.5 Ω So, simplified circuit is given as So, current in circuit will be i=EeqR+req=2R+0.5 So, power developed through R is given below P=i2R=(2R+0.5)2R For, maximum power, dPdR=0, we get R=0.5 Ω ∴P=(20.5+0.5)20.5=2 W Hence, option (c) is correct. 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

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