A resistor R is connected across a battery of emf 12 V. It observed that terminal potential differenceis 9 V and the power delivered to resistor is 18 W. Internal resistance of the battery ‘r’ and the resistance R are respectively

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2Ω,5Ω

1.8Ω,3.2Ω

1.6Ω,5.6Ω

1.5Ω,4.5Ω

answer is D.

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Detailed Solution

P=V2R⇒R=V2P⇒R=9×918=4.5Ω&TPD=ERR+r⇒9=12×4.54.5+r ⇒2(4.5+r)=12⇒R=4.5Ω⇒9+2r=12⇒2r=3⇒r=1.5Ω

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