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A battery of emf epsiand internal resistance r, when connected across an external resistance of 12 Omega , produces a current of 0.5 A. When connected across a resistance of 25Omega, it produces a current of 0.25 A. Determine the emf and internal resistance of the cell. |
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Answer» Solution :We KNOW that current DRAWN `I = (epsi)/(R + r)` As per question when external resistance `R_1 = 12OMEGA` current `I_1`=0.5 A, and for resistance `R_2 = 25Omega`, current `I_2 = 0.25 A` . ` therefore0.5 = (epsi)/(12 + r) `....(i)and`0.25 = (epsi)/(25+ r)`.....(ii) From (i) and (ii) , we GET`r = 1 Omega ` and `epsi = 6.5 V` |
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