Correct Answer: Option D

Explanation

Effective resistance in the circuit
= 3 + 2 + 1 = 6Ω
Current through the I = \(\frac{E}{(R+r)}\)
∴ I = \(\frac{2}{6}\)
=> V1 = I x R1 = \(\frac{2}{6}\times\frac{3}{1}\) = 1V
= V2 = I x R2 = \(\frac{2}{6}\times\frac{2}{1}\) = \(\frac{2}{3}\)V
∴ we have 1V and \(\frac{2}{3}\)V

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