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An ac emf V = 200 sin(100*pi*t) volt is connected across a choke coil of negligible resistance. To produce a current of amplitude 1 A, what must be the inductance of the choke?
- 2/pi H
- 200 H
- 1/pi H
- sqrt(2)/pi H
Correct answer: 2/pi H
Solution
For a pure inductor, peak current I0 = V0/XL where XL = omega L. From V = 200 sin(100 pi t): V0 = 200 V, omega = 100 pi rad/s. With I0 = 1 A, XL = V0/I0 = 200/1 = 200 ohm. Then L = XL/omega = 200/(100 pi) = 2/pi H.
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