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(I) At what frequency will a 32.0-mH inductor have a reactance of 660 \(\Omega\)?

Short Answer

Expert verified
The frequency is approximately 3282 Hz.

Step by step solution

01

Understanding Inductive Reactance

The inductive reactance, denoted by \(X_L\), is the opposition that an inductor presents to the change of current through it. It can be calculated using the formula \(X_L = 2 \pi f L\), where \(f\) is the frequency and \(L\) is the inductance.
02

Identifying Known Values

From the given problem, we know that the inductance \(L\) is 32.0 mH (which is equivalent to 0.032 H) and the inductive reactance \(X_L\) is 660 \(\Omega\).
03

Rearranging the Reactance Formula

To find the frequency \(f\), rearrange the formula for inductive reactance:\[ X_L = 2 \pi f L \] to \[ f = \frac{X_L}{2 \pi L} \].
04

Calculating the Frequency

Substitute the known values into the rearranged equation: \(f = \frac{660}{2 \pi \times 0.032}\). Calculate this to find the frequency.
05

Final Calculation

Compute \(f = \frac{660}{2 \pi \times 0.032} = \frac{660}{0.20106} \approx 3282.34\text{ Hz}\). The frequency is approximately 3282 Hz.

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Key Concepts

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

Frequency Calculation
When we speak about calculating frequency for inductive reactance, it refers to determining how often the alternating current (AC) cycles per second. This is important because frequency influences how inductors behave in an electrical circuit.
Understanding the frequency in a circuit helps in fine-tuning electronic devices and systems.
Frequency, typically measured in Hertz (Hz), tells us the number of cycles per second. In the context of inductors, it plays a critical role as it interacts with the inductance to produce reactance.

To find the frequency where a specific inductive reactance is reached, we use the equation:
  • Start with the formula for inductive reactance: \(X_L = 2 \pi f L\)
  • To isolate frequency \(f\), rearrange the equation to: \(f = \frac{X_L}{2 \pi L}\)
  • By substituting the values given for reactance and inductance into this formula, you can compute the exact frequency.
This calculation is crucial in designing circuits that rely on precise frequencies for operation, such as in communication systems.
Inductance
Inductance is a fundamental concept in electromagnetism that describes how effectively an inductor can produce an electromotive force (EMF) as the current changes.
It is measured in Henry (H). It's fascinating how inductance quantifies an inductor's ability to store energy in a magnetic field.
In practical terms, the higher the inductance, the greater the tendency of the inductor to resist changes in current.

For example, if you have an inductor with a 32.0 mH (millihenry) inductance:
  • Convert this to Henry by recognizing that 1 mH = 0.001 H, so 32.0 mH = 0.032 H.
  • This conversion is crucial when applying formulas that involve inductance because standard units ensure that calculations are accurate and meaningful.
  • Inductors with large inductance values are generally used in applications involving high power or sudden changes in currents.
By having a good grasp of inductance, you can better understand how inductors affect AC circuits, aiding in both analysis and circuit design.
Reactance Formula
The reactance formula for inductors is vital for calculating how they impede current flow at various frequencies.
Reactance itself is not a fixed resistance; it varies with frequency, making it unique to AC circuits.

The inductive reactance formula is expressed as:
  • \(X_L = 2 \pi f L\), where:
  • \(X_L\) is the inductive reactance measured in Ohms (Ω).
  • \(f\) is the frequency of the AC signal, measured in Hertz (Hz).
  • \(L\) is the inductance, measured in Henry (H).
By using this formula:
  • You can determine how many Ohms of reactance an inductor presents at a certain frequency.
  • This relationship highlights that as frequency increases, inductive reactance also increases, indicating that faster changing currents face greater opposition.
  • Understanding this formula allows electronics engineers to design circuits with desired characteristics by selecting appropriate inductances for specific frequencies.
This knowledge is particularly useful in RF (radio frequency) engineering, where precise reactance values are crucial.

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Most popular questions from this chapter

(I) Determine the inductance \(L\) of a 0.60-m-long air-filled solenoid 2.9 cm in diameter containing 8500 loops.

(II) An ac voltage source is connected in series with a 1.0-\(\mu\)F capacitor and a 650-\(\Omega\) resistor. Using a digital ac voltmeter, the amplitude of the voltage source is measured to be 4.0 V rms, while the voltages across the resistor and across the capacitor are found to be 3.0 V rms and 2.7 V rms, respectively. Determine the frequency of the ac voltage source.Why is the voltage measured across the voltage source not equal to the sum of the voltages measured across the resistor and across the capacitor?

The \(\textbf{Q factor}\) of a resonant ac circuit (Section 21-15) can be defined as the ratio of the voltage across the capacitor (or inductor) to the voltage across the resistor, at resonance. The larger the \(Q\) factor, the sharper the resonance curve will be and the sharper the tuning. (\(a\)) Show that the \(Q\) factor is given by the equation \(Q = (1/R) \sqrt{L/C}\). (\(b\)) At a resonant frequency \(f_0 =\) 1.0 MHz, what must be the values of L and R to produce a Q factor of 650? Assume that \(C =\) 0.010 \(\mu\)F.

(II) Determine the total impedance, phase angle, and rms current in an \(LRC\) circuit connected to a 10.0-kHz, 725-V (rms) source if \(L =\) 28.0 mH, \(R =\) 8.70 k\(\Omega\), and \(C =\) 6250 pF.

(II) When a car drives through the Earth's magnetic field, an emf is induced in its vertical 55-cm-long radio antenna. If the Earth's field \((5.0 \times 10^{-5})\) points north with a dip angle of 38\(^\circ\), what is the maximum emf induced in the antenna and which direction(s) will the car be moving to produce this maximum value? The car's speed is 30.0 m/s on a horizontal road.

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