Correct Answer: Option B

Explanation

6Ω and 4Ω are parallel.

\(\frac{1}{R_T}\) = \(\frac{1}{R_1} + \frac{1}{R_2}\)

\(\frac{1}{R_T}\) = \(\frac{1}{6} + \frac{1}{4}\) 

\(R_T\) = \(\frac{12}{5}\) = 0.97Ω

  V = IR

 = 0.97 x 2.4 = 2.3V

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