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Chapter 21: Problem 29

A coil in a 60 -Hz circuit has a resistance of \(100 \Omega\) and an inductance of \(0.45 \mathrm{H}\). Calculate (a) the coil's reactance and (b) the circuit's impedance.

Short Answer

Expert verified
The coil's reactance is about 169.56 惟 and the circuit's impedance is about 196.72 惟.

Step by step solution

01

Understanding Reactance

The reactance of an inductor is defined by the formula \[ X_L = 2 \pi fL \]where \( f \) is the frequency of the AC source, and \( L \) is the inductance. With \( f = 60 \) Hz and \( L = 0.45 \) H, the reactance \( X_L \) can be calculated.
02

Calculating Reactance

Substitute the given values into the formula for inductive reactance:\[ X_L = 2 \pi \times 60 \times 0.45 \]Calculate the result:\[ X_L = 2 \times 3.1416 \times 60 \times 0.45 \approx 169.56 \Omega \]So, the coil's reactance is approximately \(169.56 \Omega \).
03

Understanding Impedance

The impedance \( Z \) of the circuit, which combines resistance \( R \) and inductive reactance \( X_L \), is given by the formula:\[ Z = \sqrt{R^2 + X_L^2} \]where \( R = 100 \Omega \) and \( X_L = 169.56 \Omega \).
04

Calculating Impedance

Substitute the known values into the impedance formula:\[ Z = \sqrt{100^2 + 169.56^2} \]\[ Z = \sqrt{10000 + 28711.38} \]\[ Z = \sqrt{38711.38} \approx 196.72 \Omega \]Thus, the circuit's impedance is approximately \( 196.72 \Omega \).

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Key Concepts

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

Inductive Reactance
Inductive reactance is an important concept when dealing with AC circuits that include inductors. This reactance arises because inductors resist changes in current. It is represented by the symbol \( X_L \) and calculated with the formula:
  • \( X_L = 2 \pi fL \)
Here, \( f \) denotes the frequency of the AC signal, and \( L \) is the inductance of the coil. These two factors determine how much the inductor will oppose the AC current. The unit of inductive reactance is Ohms (\( \Omega \)).
  • Higher frequency or inductance results in higher reactance.
  • Reactance increases linearly with frequency.
Circuit Impedance
Circuit impedance, denoted by \( Z \), is a measure of how much an entire circuit resists the flow of electrical current. It combines both resistance (\( R \)) and reactance (\( X_L \)). The formula to find impedance in a circuit containing resistance and inductive reactance is as follows:
  • \( Z = \sqrt{R^2 + X_L^2} \)
Resistance contributes to energy loss as heat, while reactance relates to energy storage in the magnetic field of the inductor. Both together make up the total opposition to current flow.
  • Resistance and reactance are orthogonal (they act independently).
  • Impedance is also expressed in Ohms (\( \Omega \)).
Coil Resistance
Coil resistance is the inherent opposition to current flow presented by the coil's conductor. In AC circuits, coil resistance contributes to total impedance along with reactance. It does not change with frequency; instead, it depends solely on the coil's material, size, and temperature. Coil resistance in the exercise is given by 100 惟. While resistance is constant, it is crucial in determining circuit performance as it directly affects power loss. Too much resistance can lead to significant heating problems in a circuit.
  • Resistance is one part of impedance, where the other part is reactance.
  • The lower the resistance, the more efficient the coil.
Frequency in Circuits
Frequency in AC circuits indicates how many cycles per second the AC signal completes, measured in Hertz (Hz). In our exercise, the frequency is 60 Hz, which is typical for many electrical systems. Frequency affects the reactance of components within the circuit. For inductors, higher frequency increases reactance, which in turn impacts the overall impedance. Conversely, low-frequency signals lead to lower reactance.
  • High frequencies can result in greater energy storage in inductive coils.
  • Frequency is a key parameter in designing AC circuits for efficient operation.

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Most popular questions from this chapter

What are the peak and rms voltages of a \(120-\mathrm{V}\) ac line and a \(240-\mathrm{V}\) ac line?

An inductor is connected to a variable-frequency ac voltage source. (a) If the frequency decreases by a factor of \(2,\) the rms current will be (1) \(2,(2) \frac{1}{2},(3) 4,(4) \frac{1}{4}\) times the original rms current. Why? (b) If the rms current in an inductor at \(40 \mathrm{~Hz}\) is \(9.0 \mathrm{~A},\) what is its rms current if the frequency is changed to \(120 \mathrm{~Hz} ?\)

A small welding machine uses a voltage source of \(120 \mathrm{~V}\) at \(60 \mathrm{~Hz}\). When the source is operating, it requires \(1200 \mathrm{~W}\) of power, and the power factor is \(0.75 .\) (a) What is the machine's impedance? (b) Find the rms current in the machine while operating.

The voltage across a \(10-\Omega\) resistor varies as \(V=(170 \mathrm{~V}) \sin (100 \pi t) .\) (a) Is the current in the resistor (1) in phase with the voltage, (2) ahead of the voltage by \(90^{\circ},\) or (3) lagging behind the voltage by \(90^{\circ} ?\) (b) Write the expression for the current in the resistor as a function of time and determine the voltage frequency.

A coil with a resistance of \(30 \Omega\) and an inductance of \(0.15 \mathrm{H}\) is connected to \(\mathrm{a} 120-\mathrm{V}, 60\) -Hz source. \((\mathrm{a})\) Is the phase angle of this circuit (1) positive, (2) zero, or (3) negative? Why? (b) What is the phase angle of the circuit? (c) How much rms current is in the circuit? (d) What is the average power delivered to the circuit?
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