## anonymous 5 years ago please help... 'n' cells of emf 'E' & internal resistance 'r' is connected in series to an external resistance 'R'.What is its total emf, total resistance & current intensity ?????

1. Owlfred

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2. anonymous

Well since you are connecting them in series, the net e.m.f adds up $$E_{net}=nE$$ Same is the case with the resistance of the e.m.fs $$r_{net}=nr$$ That gives $$I=\frac{E_{net}}{r_{net}+R}=\frac{nE}{nr+R}$$