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InterviewSolution
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Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `R_(1)-R_(2)`B. `(R_(1)+R_(2))/2`C. `(R_(1)-R_(2))/2`D. `R_(1)+R_(2)` |
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Answer» Correct Answer - A According to question `E-Ir_(1)=0&I=(E+E)/(r_(1)+r_(2)+R)` `therefore (E)/(r_(1))=(2E)/(r_(1)+r_(2)+R)` `r_(1)+r_(2)=R=2r_(1) implies R=r_(1)-r_(2)` |
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