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The time constant of a circuit is 10 sec, when a resistance of 10 ohm is connected in series in a previous circuit then time constant becomes 2 second, then the self inductance of the circuit is:

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5 H

25 H

15 H

2.5 H

answer is D.

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Detailed Solution

$λ = \frac{L}{R} = 10$

$L = 10 R \dots (1)$

$\frac{L}{R + 10} = 2$

$L = 2 R + 20 \dots$

from eq (1) and (2)

$10 R = 2 R + 20$

$R = \frac{5}{2}$

$and L = 25 H$

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