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A current of `2 A` flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of `0.5 A` when connected across a `9 Omega` resistor. The internal resistance of the battery isA. `0.5Omega`B. `1//3 Omega`C. `1//4Omega`D. `1Omega` |
Answer» Correct Answer - 2 Let internal resistance of b attery be r then according to question `(E)/(2+r)=2` and `(E)/(9+r)=0.5` `rArr(2)/(0.5)=(9+r)/(2+r)rArrr=(1)/(3)Omega` |
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