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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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