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Chapter 22: Problem 71

When its coil rotates at a frequency of \(280 \mathrm{Hz},\) a certain generator has a peak emf of 75 V. (a) What is the peak emf of the generator when its coil rotates at a frequency of 45 Hz? (b) Determine the frequency of the coil's rotation when the peak emf of the generator is \(180 \mathrm{V}\).

Short Answer

Expert verified
(a) 12.05 V (b) 672 Hz

Step by step solution

01

Understand the Relationship

The emf induced in a generator is proportional to the frequency of the coil rotation. This can be expressed as: \( \text{emf} \propto \text{frequency} \). Therefore, when the frequency changes, the emf changes proportionally.
02

Set Up the Proportional Relationship

From Step 1, we express this relationship as \( \frac{\text{emf}_1}{\text{emf}_2} = \frac{f_1}{f_2} \), where \( \text{emf}_1 \) and \( f_1 \) are the known values (75 V and 280 Hz), and \( \text{emf}_2 \) and \( f_2 \) are the values we want to find in each part.
03

Solve for Part (a)

Given: \( f_1 = 280 \) Hz, \( \text{emf}_1 = 75 \) V, and \( f_2 = 45 \) Hz.Rearranging \( \frac{\text{emf}_1}{\text{emf}_2} = \frac{f_1}{f_2} \) to find \( \text{emf}_2 \), we get \( \text{emf}_2 = \frac{\text{emf}_1 \cdot f_2}{f_1} = \frac{75 \cdot 45}{280} \). Calculating this gives \( \text{emf}_2 \approx 12.05 \) V.
04

Solve for Part (b)

Given: \( f_1 = 280 \) Hz, \( \text{emf}_1 = 75 \) V, and \( \text{emf}_2 = 180 \) V.Rearrange \( \frac{\text{emf}_1}{\text{emf}_2} = \frac{f_1}{f_2} \) to solve for \( f_2 \). We get \( f_2 = \frac{f_1 \cdot \text{emf}_2}{\text{emf}_1} = \frac{280 \cdot 180}{75} \). Calculating this gives \( f_2 \approx 672 \) Hz.

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Key Concepts

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

Generator Basics
Generators play a crucial role in converting mechanical energy into electrical energy. They do this through a process called electromagnetic induction. Using a coil, typically made of wire, that spins in a magnetic field, a generator creates an electric current. As the coil rotates, it cuts through magnetic field lines, inducing a voltage, which is why a generator is so crucial in power production.
In simple terms:
  • The coil rotates in a magnetic field.
  • This rotation induces voltage (also known as electromotive force or emf).
  • The induced emf leads to the generation of electrical current.
The amount of voltage generated by a generator depends on several factors, but most importantly, the rate at which the coil spins, which is its frequency.
Understanding Frequency
Frequency, often measured in Hertz (Hz), represents how many times a rotating coil completes a full cycle in one second. It's a crucial aspect because the frequency directly influences how much voltage a generator can produce. Consider the following:
  • Higher frequency means more cycles per second.
  • More cycles result in more opportunities for the coil to cut through the magnetic field.
  • Hence, a higher frequency typically results in a higher emf.
In practical terms, changing the frequency of coil rotation changes the induced emf. If the coil rotates faster (higher frequency), it generates more voltage, and if it rotates slower, the emf decreases. For instance, in the provided exercise, reducing the frequency from 280 Hz to 45 Hz resulted in a significant drop in peak emf.
The Role of Coil Rotation
Coil rotation is the dynamic action within a generator that allows it to produce electricity. When discussing coil rotation, there's an important principle to keep in mind: the relationship between the speed of rotation and the electromagnetic force produced. This rotation speed is intrinsically tied to the generator's frequency, affecting how much voltage is produced. Key facts about coil rotation in generators:
  • The speed of rotation determines how quickly magnetic field lines are cut.
  • Faster rotation leads to higher frequency, and thus, more emf.
  • Slower coil rotation decreases the frequency, reducing the emf.
Understanding this relationship helps in manipulating a generator's performance. For example, to increase power, you might need to increase coil rotation speed, leading to higher frequency and thus higher emf, as illustrated when calculating a frequency increase to 672 Hz for a peak emf of 180 V.

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Most popular questions from this chapter

The coil of a generator has a radius of \(0.14 \mathrm{m} .\) When this coil is unwound, the wire from which it is made has a length of \(5.7 \mathrm{m}\). The magnetic field of the generator is \(0.20 \mathrm{T}\), and the coil rotates at an angular speed of \(25 \mathrm{rad} / \mathrm{s}\). What is the peak emf of this generator?

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of \(0.46 \mathrm{V}\) to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a handgun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

A generator is connected across the primary coil \(\left(N_{\mathrm{p}}\right.\) turns ) of a transformer, while a resistance \(R_{2}\) is connected across the secondary coil \(\left(N_{\mathrm{s}}\right.\) turns). This circuit is equivalent to a circuit in which a single resistance \(R_{1}\) is connected directly across the generator, without the transformer. Show that \(R_{1}=\left(N_{\mathrm{p}} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with Ohm's law as applied to the secondary coil.

A magnetic field has a magnitude of \(0.078 \mathrm{T}\) and is uniform over a circular surface whose radius is \(0.10 \mathrm{m}\). The field is oriented at an angle of \(\phi=25^{\circ}\) with respect to the normal to the surface. What is the magnetic flux through the surface?

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?
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