The circuit in the figure shows two cells connected in opposition to each other. Cell $E_1$ has an $emf$ of $6 \ V$ and an internal resistance of $2 \ \Omega$. Cell $E_2$ has an $emf$ of $4 \ V$ and an internal resistance of $8 \ \Omega$. Find the potential difference between the points $A$ and $B$.

(N/A) The equivalent internal resistance of the cells in series is:
$r = r_1 + r_2 = 2 + 8 = 10 \ \Omega$
Since $E_1 > E_2$,the equivalent $emf$ of the circuit is:
$E = E_1 - E_2 = 6 - 4 = 2 \ V$
Using Ohm's law,the current $I$ in the circuit is:
$I = \frac{E}{r} = \frac{2}{10} = 0.2 \ A$
The current flows from the higher potential cell $(E_1)$ to the lower potential cell $(E_2)$.
To find the potential difference $V_A - V_B$,we apply Kirchhoff's voltage law from $A$ to $B$:
$V_A - E_2 - I r_2 = V_B$
$V_A - V_B = E_2 + I r_2$
$V_A - V_B = 4 + (0.2 \times 8)$
$V_A - V_B = 4 + 1.6 = 5.6 \ V$