Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 $$\Omega$$. The internal resistance of the two batteries are 1 $$\Omega$$ and 2 $$\Omega$$ respectively. The voltage across the load lies between,

We have two real batteries, each represented by an ideal emf in series with its own internal resistance. For the first battery $$E_1 = 12\ \text{V}$$ with $$r_1 = 1\ \Omega$$. For the second battery $$E_2 = 13\ \text{V}$$ with $$r_2 = 2\ \Omega$$. Both positive terminals are joined together, and this common point is connected to one end of the external load resistor $$R_L = 10\ \Omega$$. The other end of every element is on the common return line, which we shall take as the reference (zero-volt) point.

Let the unknown potential of the junction (the point common to the two positive terminals and to the load) be $$V$$. The direction of current we choose is from each battery towards the junction and then through the load to the return line. According to Ohm’s law, the current contributed by each battery is the difference between its emf and the node voltage divided by its internal resistance:

$$I_1 = \dfrac{E_1 - V}{r_1}, \qquad I_2 = \dfrac{E_2 - V}{r_2}.$$

The current leaving the junction through the load is

$$I_L = \dfrac{V - 0}{R_L} = \dfrac{V}{R_L}.$$

Kirchhoff’s current law states that the algebraic sum of currents at a node is zero (total current entering equals total current leaving). Stating the law explicitly,

$$I_1 + I_2 = I_L.$$

Substituting the algebraic expressions just written, we obtain

$$\dfrac{E_1 - V}{r_1} \;+\; \dfrac{E_2 - V}{r_2} \;=\; \dfrac{V}{R_L}.$$

Now we put the numerical values:

$$\dfrac{12 - V}{1} \;+\; \dfrac{13 - V}{2} \;=\; \dfrac{V}{10}.$$

Removing the denominators step by step, we first leave the left-hand side untouched and write the right-hand side explicitly,

$$(12 - V) \;+\; \dfrac{13 - V}{2} = \dfrac{V}{10}.$$

To combine the terms on the left, we bring them to a common denominator 2:

$$\dfrac{2(12 - V)}{2} \;+\; \dfrac{13 - V}{2} = \dfrac{V}{10},$$

so

$$\dfrac{24 - 2V + 13 - V}{2} = \dfrac{V}{10}.$$

Simplifying the numerator,

$$\dfrac{37 - 3V}{2} = \dfrac{V}{10}.$$

Next we clear both denominators by multiplying every term by 10 × 2 = 20:

$$20 \times \dfrac{37 - 3V}{2} = 20 \times \dfrac{V}{10}.$$

This gives

$$(37 - 3V)\;10 = 2V.$$

Expanding the product on the left:

$$370 - 30V = 2V.$$

We collect all terms containing $$V$$ on one side:

$$370 = 32V.$$

Finally, dividing by 32, we obtain the node (terminal) voltage

$$V = \dfrac{370}{32}\ \text{V} = 11.5625\ \text{V}.$$

This value clearly lies between 11.5 V and 11.6 V.

Hence, the correct answer is Option C.