For the circuit shown, with $R_{1} = 1.0 Ω, R_{2} = 2.0 Ω, E_{1} = 2 V$ and $E_{2} = E_{3} = 4 V$ , the potential difference between the points ‘a’ and ‘b’ is approximately (in V)

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3.3

2.7

3.7

2.3

answer is D.

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Detailed Solution

Applying parallel combination of batteries $\frac{\frac{E_{1}}{1 + 1} + \frac{E_{2}}{2} + \frac{E_{3}}{1 + 1}}{\frac{1}{1 + 1} + \frac{1}{2} + \frac{1}{1 + 1}}$

$\frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{5 \times 2}{3} = \frac{10}{3} = 3.3 V o l t$

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