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In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of `10Omega` is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:- A. `1Omega`B. ` 3Omega`C. `10Omega`D. `5Omega` |
Answer» Correct Answer - 1 `r=(( l_(0)-l_(C))/(l_(C)))R=((55-50)/(50))10=1Omega` Here `l_(0)=55cm:l_(c)=50cm:R=10Omega` |
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