Question

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)

• A. 0.75
• B. 1
• C. 1.5
• D. 2

Answer 1

Answer: (A)

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