Medium
Four cells, each of emf $E$ and internal resistance $r$, are connected in series across an external resistance $R$. By mistake, one of the cells is connected in reverse. Then the current in the external circuit is
- A$\frac{2 E}{4 r+R}$
- B$\frac{3 E}{4 r+R}$
- C$\frac{3 E}{3 r+R}$
- D$\frac{2 E}{3 r+R}$
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