Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part of the circuitry for a radio receiver) to be 3.00 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. Part A What frequency is required? ff = nothing Hz Request Answer Provide Feedback

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1), which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π * (3.00*10⁻³) A * (4.9*10⁻³) H = 130 Khz