Correct Answer: Option C

Explanation

E.m.f = 1.5

r = \(\frac{1 \times\ 1 \times 1}{(1 \times\ 1) + (1 \times\ 1) + (1 \times\ 1)}\)

r = \(\frac{1}{3}\)\(\Omega\)

I = \(\frac{E.m.f}{r + R}\)

I = \(\frac{1.5}{\frac{1}{3}\ + \frac{2}{3}}\)

I = 1.5A

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