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Two cell with the same e.m.f. E and difference internal resistance `r_(1) and r_(2)` are connected in series to an external resistance `R` a value of `R` be selected such that the potential difference as the flow cell E should be zero when A. `R = r_(1)+r_(2)`B. `R = r_(1)- r_(2)`C. `R = r_(1)//r_(2)`D. `R = r_(1)=r_(2)` |
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Answer» Correct Answer - b Current is circuit, `I= (E+E)/(r_(1)+ r_(2) +R) = (22E)/(r_(1)+ r_(2) +R) `….(i) The potential difference first cell is zero, Then `E - I r_(1) = 0 or I= E//r_(1)`…..(ii) From (i) and (ii) `(2E)/(r_(1) +r_(2)+R) = (E )/(r_(1))` On solving `R = (r_(1)- r_(2))` |
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