The voltage and the current between A and B points shown in the circuit are :

The circuit contains 4 parallel horizontal branches acting as sources between the left vertical wire and right vertical wire:

Branches 1, 2, and 4: Each has an EMF $$E = 27\text{ V}$$ and internal resistance $$r = 3\ \Omega$$.

Branch 3 features two cells connected in series aiding each other ($$14\text{ V} + 13\text{ V} = 27\text{ V}$$) with a net branch resistance of $$3\ \Omega$$.

$$E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3} + \frac{E_4}{r_4}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4}}$$

$$E_{\text{eq}} = \frac{\frac{27}{3} + \frac{27}{3} + \frac{27}{3} + \frac{27}{3}}{\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \frac{4 \times \left(\frac{27}{3}\right)}{4 \times \left(\frac{1}{3}\right)} = 27\text{ V}$$

$$\frac{1}{r_{\text{eq}}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{4}{3}\ \Omega^{-1} \implies r_{\text{eq}} = \frac{3}{4}\ \Omega$$

$$R_L = 3\ \Omega + 3\ \Omega = 6\ \Omega$$

$$i_{AB} = \frac{E_{\text{eq}}}{R_L + r_{\text{eq}}}$$

$$i_{AB} = \frac{27}{6 + \frac{3}{4}} = \frac{27}{\frac{27}{4}} = 4\text{ A}$$

$$V_{AB} = i_{AB} \times R_L$$

$$V_{AB} = 4\text{ A} \times 6\ \Omega = 24\text{ V}$$