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Chapter 21: Problem 87

Calculate the peak output voltage of a simple generator whose square armature windings are 6.60 cm on a side; the armature contains 125 loops and rotates in a field of 0.200 T at a rate of 120 rev/s

Short Answer

Expert verified
The peak output voltage is approximately 82.0 volts.

Step by step solution

01

Understand the Problem Requirements

We are tasked with calculating the peak output voltage of a generator with specific properties. These include the dimensions of the armature loop, the number of loops, the magnetic field strength, and the rotational speed. The formula for the peak voltage in a rotating coil is needed.
02

Identify the Formula

The peak output voltage \( V_{peak} \) of the generator can be calculated using the formula: \[V_{peak} = NAB\omega\]where \( N \) is the number of loops, \( A \) is the area of one loop, \( B \) is the magnetic field strength, and \( \omega \) is the angular velocity in radians per second.
03

Calculate the Area of One Loop

The loop is square with sides of length 6.60 cm. Convert this to meters, and use the formula for the area of a square:\[A = (0.066 \, \text{m})^2 = 0.004356 \, \text{m}^2\]
04

Convert Rotational Speed to Angular Velocity

The rotational speed is given in revolutions per second (rev/s). Convert this to radians per second by using the conversion \( 1 \, \text{rev} = 2\pi \, \text{radians} \):\[\omega = 120 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 240\pi \, \text{rad/s}\]
05

Plug Values Into the Formula

Now substitute \( N = 125 \), \( A = 0.004356 \, \text{m}^2 \), \( B = 0.200 \, \text{T} \), and \( \omega = 240\pi \, \text{rad/s} \) into the formula for the peak voltage:\[V_{peak} = 125 \times 0.004356 \, \text{m}^2 \times 0.200 \, \text{T} \times 240\pi \, \text{rad/s}\]
06

Calculate the Result

Calculate the numerical result:\[V_{peak} = 125 \times 0.004356 \times 0.200 \times 240\pi \approx 82.0 \, \text{volts}\]

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Key Concepts

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

Generator
A generator is a device that converts mechanical energy into electrical energy, often using the principle of electromagnetic induction. In essence, it produces electricity by moving a conductor through a magnetic field or by changing the magnetic field around the conductor. Generators are fundamental to supplying electricity for various applications, from powering appliances to providing large-scale power in industries.
  • They can vary significantly in size depending on their intended application, from compact portable units to enormous industrial machines.
  • Although details may vary, all generators work on the same basic principle: inducing a voltage by moving a coil inside a magnetic field.
Understanding the operation of generators helps to grasp how they achieve this critical conversion, offering insights into how electricity is provided for everyday use.
Armature Windings
Armature windings are a key component of electric generators. These consist of coils of wire, often wound in configurations like circular loops or squares, as in our case, forming the moving part of the generator where voltage is induced.
  • The configuration and number of loops in the armature winding can greatly affect the generator's efficiency and output voltage.
  • In our example, the armature windings consist of 125 loops of square coils with each side measuring 6.60 cm, contributing to the calculation of peak output voltage.
  • The area of the loop is crucial, as it directly impacts the amount of voltage induced during rotation in a magnetic field.
This mechanism is central to ensuring optimal power generation by influencing the strength and consistency of the induced voltage.
Magnetic Field
The magnetic field is a critical factor in the operation of a generator. It is the region around a magnet where magnetic forces are exerted, and in a generator, this field interacts with the armature windings to induce voltage.
  • A stronger magnetic field can lead to a higher induced voltage, assuming other variables such as the number of loops and their rotation speed remain constant.
  • The given problem uses a magnetic field strength of 0.200 T (Tesla), a fundamental parameter in calculating the generator鈥檚 output voltage.
  • The magnetic field and its uniformity directly affect the efficiency and stability of the voltage generated.
Understanding how magnetic fields interact with conductors is key to mastering the principles behind how generators function effectively.
Angular Velocity
Angular velocity, represented by the symbol \( \omega \), describes how fast an object rotates or revolves relative to another point, usually the center of a circle. In generators, it measures the speed at which the armature rotates within the magnetic field.
  • Angular velocity is typically expressed in radians per second, and is crucial for calculating induced voltage.
  • In the given exercise, the armature rotates at 120 rev/s, converting to \( \omega = 240\pi \) rad/s. This conversion is essential for utilizing the formula \( V_{peak} = NAB\omega \).
  • Higher angular velocity generally increases the peak voltage output of a generator, assuming that the number of loops and magnetic field strength are constant.
This understanding is pivotal for anyone studying electric machines, as it affects both the performance and design of various types of generators.

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Most popular questions from this chapter

(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?

Conceptual Example 21-9 states that an overloaded motor may burn out due to high currents. Suppose you have a blender with an internal resistance of 3.0 \(\Omega\). (\(a\)) At 120 V, what is the initial current through the blender? (\(b\)) The blender is rated at 2.0 A for continuous use.What is the back emf of the blender? (\(c\))At what rate is heat dissipated in the blender during normal use? (\(d\)) If the blender jams and stops turning, at what rate is heat dissipated in the motor coils?

A high-intensity desk lamp is rated at 45 W but requires only 12 V. It contains a transformer that converts 120-V household voltage. (\(a\)) Is the transformer step-up or stepdown? (\(b\)) What is the current in the secondary coil when the lamp is on? (\(c\)) What is the current in the primary coil? (\(d\)) What is the resistance of the bulb when on?

(III) A 22.0-cm-diameter coil consists of 30 turns of circular copper wire 2.6 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of \(8.65 \times 10^{-3}\) T/s. Determine (\(a\)) the current in the loop, and (\(b\)) the rate at which thermal energy is produced.

A 130-mH coil whose resistance is is 15.8 \(\Omega\) connected to a capacitor \(C\) and a 1360-Hz source voltage. If the current and voltage are to be in phase, what value must \(C\) have?
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