N batteries each of emf E and internal resistance r are first connected in series and then in parallel to an external resistance R. if current through R is same in both cases then.A. `R=r^(2)//N`B. `R-Nr//2`C. `R=Nr`D. `R=r`

N batteries each of emf E and internal resistance r are first connecte

1.	N batteries each of emf E and internal resistance r are first connected in series and then in parallel to an external resistance R. if current through R is same in both cases then.A. `R=r^(2)//N`B. `R-Nr//2`C. `R=Nr`D. `R=r`
Answer» Correct Answer - 4 In serie s curren t `=(NE)/(Nr+R)` In parallel, current `=(E)/((r//N)+R)=(NE)/(r+NR)` since the currents in the two cases are equal, `(NE)/(Nr+R)=(NE)/(r+NR)` or `Nr+R=r+NR` or `Nr-r=NR-R` or `r(N-1)=R(N-1)` or r=R `therefore R=r`.

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N batteries each of emf E and internal resistance r are first connected in series and then in parallel to an external resistance R. if current through R is same in both cases then.A. `R=r^(2)//N`B. `R-Nr//2`C. `R=Nr`D. `R=r`

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