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# A Coil of Inductance 5.0 Mh and Negligible Resistance is Connected to the Oscillator of the Previous Problem. Find the Peak Currents in the Circuit for ω = 100 S−1, 500 S−1, 1000 S−1. - Physics

Sum

A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for ω = 100 s−1, 500 s−1, 1000 s−1.

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#### Solution

Given:
Inductance of the coil, L = 5.0 mH = 0.005 H
(a) At  ω = 100 s−1:
Reactance of coil (X_L) is given by,
X_L = omegaL
Here, omega= angular frequency
therefore X_L = 100xx0.005 = 0.5 Ω
peak current , l_0 = 10/0.5 = 20 A

(b) At omega = 500 s^-1
Reactance, X_L = 500 xx 5/1000
= 2.5 Ω

peak current, l_0 = 10/2.5 = 4A\

(c) ω = 1000 s−1:
Reactance, X_L = 1000 xx 0.005 = 5 Ω
Peak current, l_0 = 10/5 = 2A

Concept: Peak and Rms Value of Alternating Current Or Voltage
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 17 Alternating Current
Q 9 | Page 330
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