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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 sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?

1 Answer

6 votes

Answer:

3.067MHz

Step-by-step 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 = (13)/(4.239*10^(-6)) \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz

Hence, the frequency required is 3.067MHz

User David Namenyi
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