Alternating Current - Result Question 28
28. 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: 28. (b)
Solution:
- (b) Time constant is $L / R$
Given, $L=40 H & R=8 \Omega$
$\therefore \tau=40 / 8=5 sec$. (d) At resonance $L \omega=\frac{1}{C \omega}, \omega=\frac{1}{\sqrt{L C}}$
Current through circuit $i=\frac{E}{R}$
Power dissipated at Resonance $=i^{2} R$
