Problem 39
Question
(II) Show that the rms output of an ac generator is \(V_{\mathrm{rms}}=N A B \omega / \sqrt{2}\) where \(\omega=2 \pi f\).
Step-by-Step Solution
Verified Answer
The RMS voltage is \( V_{\mathrm{rms}} = \frac{N A B \omega}{\sqrt{2}} \) where \( \omega = 2 \pi f \).
1Step 1: Understand the problem
We need to show that the RMS (Root Mean Square) output voltage of an AC generator is given by the formula \( V_{\mathrm{rms}} = \frac{N A B \omega}{\sqrt{2}} \) where \( \omega = 2 \pi f \). The variables \( N \), \( A \), \( B \), and \( \omega \) represent the number of turns, the area of the coil, the magnetic field strength, and the angular frequency respectively.
2Step 2: Recall the induced EMF formula
The instantaneous induced EMF (Electromotive Force) for a coil rotating in a magnetic field is given by \( V = NAB\omega \sin(\omega t) \), where \( \omega t \) is the angular position of the coil at time \( t \).
3Step 3: Calculate RMS Voltage
The RMS value of a sinusoidal voltage \( V = V_0 \sin(\omega t) \), where \( V_0 \) is the peak voltage, is given by \( V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} \). Here, \( V_0 = NAB\omega \). Substitute \( V_0 \) to get \( V_{\mathrm{rms}} = \frac{NAB\omega}{\sqrt{2}} \).
4Step 4: Verify the expression for \(\omega\)
From the problem, \( \omega \) is given as \( 2\pi f \), where \( f \) is the frequency. This is consistent with the standard formula for angular frequency. Therefore, substituting \( \omega = 2 \pi f \) into the expression does not change its form, proving it represents the RMS voltage correctly.
Key Concepts
RMS VoltageElectromotive Force (EMF)Angular FrequencyMagnetic Field
RMS Voltage
RMS Voltage, or Root Mean Square Voltage, is an important concept when dealing with alternating current (AC) systems. It helps you determine the effective voltage that performs equivalent work as a direct current (DC) circuit. Instead of dealing with varying values throughout a cycle, RMS provides a single steady value.
In a sinusoidal waveform, the RMS voltage comes from the peak voltage, understood with the formula:
In a sinusoidal waveform, the RMS voltage comes from the peak voltage, understood with the formula:
- For a sine wave: \( V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} \).
- The \( V_0 \) or peak voltage is the maximum voltage attained.
Electromotive Force (EMF)
Electromotive Force (EMF) is a term used to denote the energy provided per charge in a circuit. In the context of an AC generator, it is the voltage induced when the coil rotates through a magnetic field. This concept is central to generating electrical energy.
The induced EMF in a rotating coil within a magnetic field is expressed as:
The induced EMF in a rotating coil within a magnetic field is expressed as:
- \( V = N A B \omega \sin(\omega t) \)
- Where \( N \) is the number of turns, \( A \) is the area of the coil, \( B \) is the magnetic field strength, and \( \omega t \) is the angular position over time \( t \).
Angular Frequency
Angular Frequency, symbolized as \( \omega \), gives insight into how fast the coil in an AC generator rotates. It is vital for understanding the time-based behavior of the system.
Angular frequency is defined by the formula:
Angular frequency is defined by the formula:
- \( \omega = 2\pi f \)
- Where \( f \) is the frequency of rotations in hertz (cycles per second).
Magnetic Field
The Magnetic Field, represented by \( B \), is an essential element in the generation of EMF within an AC generator. It provides the necessary force to induce voltage as the coil moves through it.
Characteristics of the magnetic field in this scenario include:
Characteristics of the magnetic field in this scenario include:
- The magnetic field is uniform and allows for consistent EMF generation.
- Its strength is directly proportional to the induced EMF, which implies stronger fields generate higher voltages.
Other exercises in this chapter
Problem 37
(I) A simple generator is used to generate a peak output voltage of 24.0 \(\mathrm{V}\) . The square armature consists of windings that are 5.15 \(\mathrm{cm}\)
View solution Problem 38
(II) A simple generator has a 480-loop square coil \(22.0 \mathrm{~cm}\) on a side. How fast must it turn in a \(0.550-\mathrm{T}\) field to produce a \(120-\ma
View solution Problem 40
(II) A 250-loop circular armature coil with a diameter of \(10.0 \mathrm{~cm}\) rotates at \(120 \mathrm{rev} / \mathrm{s}\) in a uniform magnetic field of stre
View solution Problem 41
(I) The back emf in a motor is \(72 \mathrm{~V}\) when operating at \(1200 \mathrm{rpm} .\) What would be the back emf at \(2500 \mathrm{rpm}\) if the magnetic
View solution