Alternating Current Ques 23
23. A coil of $40$ henry inductance is connected in series with a resistance of $8$ $ ohm$ and the combination is joined to the terminals of a $2$ volt battery. The time constant of the circuit is
[2004]
(a) $20$ seconds
(b) $5$ seconds
(c) $1 / 5$ seconds
(d) $40$ seconds
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Answer:
Correct Answer: 23.(b)
Solution:
- (b) Time constant is $L / R$
Given, $L=40$ $ H$ & $R=8 $ $\Omega$
$\therefore \tau=40 / 8=5 $ $sec$.
BETA
