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The p.d between the terminals A & B is

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3.6 V

1.8 V

answer is D.

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Detailed Solution

$i = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}}{1 + R (\frac{1}{r_{1}} + \frac{1}{r_{2}})} = \frac{\frac{5}{20} + \frac{2}{10}}{1 + 10 (\frac{1}{20} + \frac{1}{10})}$
V = iR = i x10 =1.8v

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