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Definition of Relations

Given L1=1 ...

Question

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:

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

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

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1, 1, 1

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

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Solution

The correct option is A $1, 1, \frac{9}{2}$

For (i): $L_{e q} = L_{1} + L_{2} = 1 + 2 = 3 m H$ and $R_{e q} = R_{1} + R_{2} = 1 + 2 = 3 Ω$
Time constant $= \frac{L_{e q}}{R_{e q}} = \frac{3}{3} = 1 m s$
For (ii): $L_{e q} = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = \frac{1 (2)}{1 + 2} = 2 / 3 m H$ and $R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{1 (2)}{1 + 2} = 2 / 3 Ω$
Time constant $= \frac{L_{e q}}{R_{e q}} = \frac{2 / 3}{2 / 3} = 1 m s$
For (iii): $L_{e q} = L_{1} + L_{2} = 1 + 2 = 3 m H$ and $R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{1 (2)}{1 + 2} = 2 / 3 Ω$
Time constant $= \frac{L_{e q}}{R_{e q}} = \frac{3}{2 / 3} = 9 / 2 m s$

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