In the given figure, the value of $$V_0$$ will be ______ V.

We need to determine the output voltage $$V_0$$ for the given parallel combination of practical voltage sources (batteries with internal resistances).

1. Identify the Circuit Parameters

From the circuit diagram,we have three branches connected in parallel to the common output node $$V_0$$:

• Branch 1: Voltage $$E_1 = 2\text{ V}$$, Resistance $$r_1 = 1\text{ k}\Omega$$
• Branch 2: Voltage $$E_2 = 4\text{ V}$$, Resistance $$r_2 = 1\text{ k}\Omega$$
• Branch 3: Voltage $$E_3 = 6\text{ V}$$, Resistance $$r_3 = 1\text{ k}\Omega$$

All the negative terminals of the batteries are tied together to the ground ($$0\text{ V}$$).

2. Apply Millman's Theorem

Millman's Theorem states that when multiple voltage sources are connected in parallel, the equivalent voltage ($$V_0$$) across the common terminals is given by:

$$V_0 = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$

Since all the internal resistances are equal ($$r_1 = r_2 = r_3 = 1\text{ k}\Omega$$), the expression simplifies directly to the arithmetic mean of the voltages:

$$V_0 = \frac{\frac{2}{1} + \frac{4}{1} + \frac{6}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}}$$

$$V_0 = \frac{2 + 4 + 6}{1 + 1 + 1}$$

$$V_0 = \frac{12}{3} = 4\text{ V}$$

Conclusion

The value of $$V_0$$ is 4 V.