Question

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. $\frac{2E}{4r+R}$
• B. $\frac{3E}{4r+R}$
• C. $\frac{3E}{3r+R}$
• D. $\frac{2E}{3r+R}$

Answer 1

The correct option is A $\frac{2E}{4r+R}$

Equivalent resistance $r_{eq}=r+r+r+r=4r$

Equivalent emf $E_{eq}=E+E+E-E=2E$

Let current through the circuit i.e figure (2), be $i$

$\therefore$ $E_{eq}=i(r_{eq}+R)$

$2E=i(4r+R)$ $⟹i=\frac{2E}{4r+R}$