Question

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 ?????

Answer 1

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Answer 2

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 $$I=\frac{E_{net}}{r_{net}+R}=\frac{nE}{nr+R} $$