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Question

A battery of emf 12 V having 10 Ω internal resistance is connected with 100 Ω resistor. If the resistance of a resistor increases by 1%, while total power loss remains constant, what should be the percentage change in emf ?

A
0.5%
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B
0.45%
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C
0.55%
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D
0.62%
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Solution

The correct option is B 0.45%
Given:
E=12 V; R=100 Ω; r=10 Ω

Current flowing through the circuit is,

I=ER+r

So, the total power dissipated from the circuit will be

Pt=I2R+I2r=I2(R+r)

Pt=E2(R+r)2(R+r)

Pt=E2(R+r)

Taking log of both sides,

logPt=2logElog(R+r)

On differentiating, we get

dPtPt=2dEEd(R+r)(R+r)

Since, Pt and r are constants, so dPt=0, and dr=0

2dEE=dR(R+r) .........(1)

Given that, dR=1% of R=0.01×100=1 Ω

dRR+r=1100+10

So, equation (1) becomes

2×dEE=1110dEE=1220

dEE×100=1220×100=0.45

Thus, the percentage change in emf is 0.45%.

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