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Q. Three similar cells, each of emf 2 V and internal resistance $r\,\Omega$ send the same current through an external resistance of $2\,\Omega$ when connected in series or in parallel. The strength of the current flowing through the external resistance is (A)

KCETKCET 1999Current Electricity

Solution:

Using I = $\frac{nE}{R+nr} $ we get I = $\frac{3 \times 2}{2 + 3r} $ (in series)
and I = $\frac{mE}{mR + r} = \frac{3 \times 2}{3 \times2 + r} $ (in parallel)
Then 2 + 3r = 6 + r $i.e.$ r =2 $\Omega$ current = $\frac{6}{2 + 3r} = \frac{6}{8}$ = 0.75 A.