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A $3 v o l t$ battery with negligible internal resistance is connected in a circuit as shown in the figure. The current $I$ , in the circuit will be

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$1 A$

$\frac{1}{3} A$

$1.5 A$

$2 A$

answer is B.

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Detailed Solution

In the given circuit, resistance of $3 Ω$ is in parallel with series combination of two $3 Ω$ resistance.

$R_{p} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 Ω$

Using ohm’s law $V = I R \Rightarrow I = \frac{V}{R} = \frac{3}{2} = 1.5 A$

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