Explanation

(b) Emf’s $E_{1}$ and $E_{2}$ are opposing each other. Since $E_{2} > E_{1}$ so, current will move from right to left.
Current in circuit
$i = \frac{E_{2} - E_{1}}{R + r_{1} + r_{2}}$
$\ \ \ = \frac{4 - 2}{5 + 1 + 2} = \frac{2}{8} = 0.25A$
The potential drop between points A and C is
$V_{A} - V_{C} = E_{1} + ir_{1}$
$= 2 + (0.25 \times 1)$
$= 2.25V$