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The circuit shown here is in its steady state. Time constant of the circuit is

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0.1 ms

0.2 ms

0.3 ms

information is insufficient of determine time constant

answer is B.

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Detailed Solution

The inductance, $L_{1} = 3 m H$ and so $l_{2} = 6 m H$
$L = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = \frac{3 \times 6}{3 + 6} = 2 m H$
And $R = 6 + \frac{12 \times 6}{12 + 6} = 10 Ω$
Now $τ = \frac{L}{R} = \frac{2}{10} = 0.2 m s$

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