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A battery of e.m.f 6 V and internal resistance 1 Omega gives a p.d of 5.8 V when connected to a resistance . Find the external resistance . |
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Answer» Solution :e.m.f of battery E= 6V Internal resistance of battery R = `1 OMEGA` p.d across the resistance V= 5.8 V Let the external resistance be .R. E= V+ IR E`=V+V/R r` `r/R = (E-V)/(V)` `R= (Vr)/(E-V) = (5.8xx1)/(6-5.8) =(5.8)/(0.2) = 29 Omega` |
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