Correct Answer: Option A

Explanation

Total resistance in series Ra = 4\(\Omega\) and another 4 \(\Omega\)

Forming two resistors in parallel: \(\frac{R_a \times R_b}{R_a + R_b}\) = \(\frac{4 \times 4}{4 + 4}\) = 2 (\Omega\)

and this is in series with the 1 (\Omega\) resistor = 3(\Omega\)

I = \(\frac{12}{3}\) = 4.0A

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