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Given $L_{1} = 1 m H, R_{1} = 1 Ω, L_{2} = 2 m H, R_{2} = 2 Ω$
Neglecting mutual inductance, the time constants (in ms) for circuits (i), (ii), and (iii) are

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$1, \frac{9}{4}, 1$

$\frac{9}{4}, 1, 1$

$1, 1, \frac{9}{2}$

1,1,1

answer is A.

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Detailed Solution

$G i v e n, L_{1} = 1 m H, L_{2} = 2 m H, R_{1} = 1 Ω, R_{2} = 2 Ω$

In the first circuit,
$L = L_{1} + L_{2} a n d R = R_{1} + R_{2}$

$τ_{1} = \frac{L}{R} = \frac{3 m H}{3 Ω} = 1 m s$

In the second circuit,

$L = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = \frac{2 \times 10^{- 6}}{3 \times 10^{- 3}} = \frac{2}{3} \times 10^{- 3}$

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2}{3} Ω$

$τ_{2} = \frac{\frac{2}{3} \times 10^{- 3}}{2 / 3} = 1 m s$

In the third circuit,

$L = L_{1} + L_{2} = 3 m H$

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2}{3} Ω$

$τ_{3} = \frac{L}{R} = \frac{3 \times 10^{- 3}}{2 / 3} = \frac{9}{2} m s$

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