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A battery of emf `epsilon` and internal resistance r sends currents `I_(1)` and `I_(2)`, when connected to external resistance `R_(1)` and `R_(2)` respectively. Find the emf and internal resistance of the battery. |
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Answer» In first case, current is `I_(1)=epsilon/R_(1)+r or epsilon = I_(1)(R_(1)+r)`....(i) In second case, current, `I_(2)=epsilon/(R_(2)+r)` or `epsilon = I_(2)(R_(2)+r)`....(ii) From (i) and (ii), we have `I_(1)(R_(1) +r)=I_(2)(R_(2) +r)` or `r (I_(1)- I_(2))=I_(2)R_(2) - I_(1)R_(1) or r =(I_(1)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))` Putting this value of r in equation (i), we get, `epsilon= I_(1)[R_(1) +(I_(2)R_(2) - I_(1)R_(1))/(I_(1)-I_(2))]` `=I_(1)[(I_(1)R_(1)-I_(2)R_(1)+I_(2)R_(2)- I_(1)R_(1))/(I_(1)-I_(2))]` `=(I_(1)I_(2)(R_(2)-R_(1)))/(I_(1)-I_(2))` |
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