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Chapter 31: Problem 10

31.10. A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

Short Answer

Expert verified
The required frequency is approximately 1632.53 Hz.

Step by step solution

01

Identify the given values and formula

In this problem, we have the following given values:- Inductance of the inductor \(L = 0.450\, \text{mH} = 0.450 \times 10^{-3}\, \text{H}\)- Current amplitude \(I_0 = 2.60\, \text{mA} = 2.60 \times 10^{-3}\, \text{A}\)- Voltage amplitude \(V_0 = 12.0\, \text{V}\)The relationship between voltage, current, and inductance in an AC circuit is given by:\[ V_0 = I_0 \times X_L \]where \(X_L\) is the inductive reactance, given by \(X_L = 2\pi f L\), and \(f\) is the frequency.
02

Calculate inductive reactance

Rearrange the formula for inductive reactance to solve for it:\[ X_L = \frac{V_0}{I_0} \]Substitute the given values:\[ X_L = \frac{12.0 \, \text{V}}{2.60 \times 10^{-3} \, \text{A}} = 4615.38 \, \text{Ohms} \]
03

Solve for frequency

Now, use the formula for inductive reactance to solve for frequency \(f\):\[ X_L = 2 \pi f L \]Rearrange to find \(f\):\[ f = \frac{X_L}{2 \pi L} \]Substitute \(X_L = 4615.38 \, \text{Ohms}\) and \(L = 0.450 \times 10^{-3}\, \text{H}\):\[ f = \frac{4615.38}{2 \pi \times 0.450 \times 10^{-3}} \approx 1632.53 \, \text{Hz} \]
04

Conclude the solution

The frequency required for a current amplitude of 2.60 mA through the inductor with a voltage amplitude of 12.0 V is approximately 1632.53 Hz.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

AC Circuits
AC circuits are an essential part of modern electrical systems. They are characterized by alternating current (AC), which means the current changes direction periodically. In contrast to direct current (DC), AC circuits offer several advantages, such as easier voltage transformation and reduced power losses over long distances.
AC circuits are important in various applications, from household electrical systems to complex radio receivers like the one in this problem. They often include components like inductors, capacitors, and resistors.
  • An inductor in an AC circuit resists changes in current by establishing a magnetic field, which leads to inductive reactance.
  • Inductive reactance is the resistance to current changes from inductors in an AC circuit. It impacts how the current flows and interacts within the circuit.
Understanding AC circuits helps us comprehend how waves of voltage and current work together to power our electronics.
Frequency Calculation
Frequency in AC circuits refers to the number of times the current changes direction per second, measured in Hertz (Hz). Calculating frequency is crucial for tuning radio receivers, ensuring devices operate effectively at the desired electromagnetic spectrum portion.
In this exercise, we calculate frequency using the relationship between voltage, current, and inductive reactance. The formula for calculating frequency in relation to inductive reactance is:

- Frequency Calculation Formula: \[ f = \frac{X_L}{2 \pi L} \]

Here, \( X_L \) stands for inductive reactance, \( L \) is the inductance, and \( f \) is the frequency. By substituting known values, we can find the precise frequency needed for specific inductor current amplitudes.
A correct frequency calculation ensures that electronic devices like radios function at their optimal capability, receiving the right signals at the desired frequencies.
Inductance and Reactance
Inductance and reactance are integral to understanding AC circuits effectively. Inductance, represented by the symbol \( L \), measures an inductor's ability to resist changes in electrical current. It is influenced by factors such as the coil's number of turns, the coil's shape, and the material around which the coil is wound.
Reactance, often referred to as inductive reactance in this context, is the opposition that inductance provides to alternating current. This is expressed by the formula:

- Inductive Reactance Formula: \[ X_L = 2\pi f L \]
As seen in the equation, inductive reactance depends on both the frequency (\( f \)) of the AC signal and the inductance (\( L \)) of the coil. Reactance varies with frequency; it increases as the frequency rises, meaning at higher frequencies the inductor offers more opposition to the current flow.
By understanding inductance and reactance, engineers and technicians can design electronic circuits that perform optimally, ensuring that devices like radios pick up the necessary signals without distortions.

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Most popular questions from this chapter

31.37. A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}\) (mns) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

31\. 46. A circuit consists of a resistor and a capacitor in series with an ac source that supplies an rms voltage of 240 \(\mathrm{V}\) . At the frequency of the source the reactance of the capacitor is 50.0\(\Omega\) . The rms current in the circuit is 3.00 A. What is the average power supplied by the source?

31.36. In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=4.00 \mu \mathrm{F}\) . The voltage amplitude of the source is 120 \(\mathrm{V}\) . (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 \(\mathrm{A}\) . What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

31\. 72 . An \(L-R-C\) series circuit 220 \(\mathrm{W}\) from a \(120-\mathrm{V}(\mathrm{rms})\) 50.0-Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacitor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

31.11. Kitchen Capacitance. The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 \(\mathrm{V}\) and frequency 60.0 \(\mathrm{Hz}\) applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?
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