Question: Two cells of emfs $E_1 = 2.0\text{ V}$ and $E_2 = 6.0\text{ V}$ and internal resistances $r_1 = 0.1\text{ }\Omega$ and $r_2 = 0.4\text{ }\Omega$ respectively, are connected in parallel. The equivalent emf of the combination is to be determined.

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Step 1 (Formula / Governing Law & Given Data): [1 Mark]

For two cells connected in parallel with emfs $E_1, E_2$ and internal resistances $r_1, r_2$, the equivalent emf ($E_{eq}$) is given by the formula:

$$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$$

Given Data:

• $E_1 = 2.0\text{ V}$, $r_1 = 0.1\text{ }\Omega$
• $E_2 = 6.0\text{ V}$, $r_2 = 0.4\text{ }\Omega$

Step 2 (Step-by-Step Mathematical Working): [2 Marks]

Substitute the given values into the formula:

$$E_{eq} = \frac{\left( \frac{2.0}{0.1} \right) + \left( \frac{6.0}{0.4} \right)}{\left( \frac{1}{0.1} \right) + \left( \frac{1}{0.4} \right)}$$

Calculate the individual terms:

$$E_{eq} = \frac{20 + 15}{10 + 2.5}$$ $$E_{eq} = \frac{35}{12.5}$$

Perform the final division:

$$E_{eq} = 2.8\text{ V}$$

Step 3 (Final Concluded Answer): [1 Mark]

The equivalent emf of the combination is $2.8\text{ V}$.

Correct Option: (B)

💡 EL10 Concept Insight: Think of this like two water pumps connected to the same pipe; the stronger pump tries to push harder, but the weaker one acts like a slight "bottleneck," so the final pressure (voltage) ends up being a weighted average that leans closer to the stronger pump but is pulled down by the weaker one.