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xz_007 [3.2K]
3 years ago
14

You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sin

usoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Physics
1 answer:
Kobotan [32]3 years ago
4 0

Answer:

<h2><em>3.067MHz</em></h2>

Explanation:

The formula for calculating the voltage across an inductor is expressed as

V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor V_l = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz

<em>Hence, the frequency required is 3.067MHz</em>

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