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InterviewSolution
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A cell develops the samepower across two resistance R_(1) and R_(2) separately. The internal resistance of the cell is |
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Answer» Solution :Let r be the internal RESISTANCE of the cell and E its EMF. When connected across the resistance `R_(1)` in the CIRCUIT, current passing through the resistance is `i=(E)/(R_(1)+r)""THEREFORE""P_(1)=i^(2)R_(1)=((E)/(R_(1)+rr))^(2)R_(1)` SIMILARLY `P_(2)-((E)/(R_(2)+r))^(2)R_(2)" Given that "P_(1)=P_(2)` Substituting the values, we get `r=sqrt(R_(1)R_(2))` |
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