Question

A Radio Inductor. You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.80 mA when a sinusoidal voltage with amplitude 12.0 V is applied across the inductor. What frequency is required?

Short answer

The required frequency of the ac source is 2.36 MHz.

Step-by-step solution

Given Data

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

The amplitude of voltage, V = 12.0 V

The amplitude of current, I = 1.80 mA

The inductance of the coil, L = 0.450 mH

Determination of frequency

Equation for voltage across the inductor –

$$\begin{gathered} V_L=1\omega L \\ {V}_L=I\left(2\mathrm{\pi f}\right)L \\ f=\frac{V_L}{2\mathrm{\pi IL}} \\ f=\frac{12.0V}{2\mathrm{\pi }\left(1.80*{10}^{-3}\mathrm{A}*\right)\left(0.450*{10}^{-3}\mathrm{H}\right)} \\ f=2.36*{10}^{6}Hz \end{gathered}$$

Therefore, the required frequency of the ac source is 2.36 Mega-Hertz.