# EMF and Peak Voltage of a Generator

• David Truong
In summary, the conversation discusses a simple electric generator with a 30-turn coil and a magnetic field of 0.80 T at a frequency of 60 Hz. The peak output voltage is calculated using the equation emf = NBωAsinωt, with a known angular frequency of 377 rad/s. The time (t) is assumed to be 1 second since frequency is measured in Hz. The peak output voltage is found to be 1.75 V. To maintain the same output voltage at a frequency of 50 Hz, the number of turns in the coil must be adjusted. The amplitude of the signal is dependent on the angular frequency (w) and is equal to 1 at peak voltage.
David Truong

## Homework Statement

A simple electric generator contains a 30-turn coil of area 6.6 x 10^-4 meters squared. The coil spins in a magnetic field of 0.80 T at a frequency of 60 Hz.

a) What is the peak output voltage?
b) How must the number of turns be changed to maintain the same output voltage, but operate at a frequency of 50 Hz.

## Homework Equations

ω = 2πf
emf/peak voltage= NBωAsinωt

## The Attempt at a Solution

ω = 2π(60) = 377 rad/s

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)t

My problem is that I do not know how/where to obtain the time (t) to finish the calculation. My first assumption is to use t = 1 second, since frequency is in Hz, which is revolutions per second.

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)(1) = 1.75 V

Is this correct? If so, then I am sure I can do part b) on my own. Thanks for the help!

David Truong said:
emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)t

My problem is that I do not know how/where to obtain the time (t) to finish the calculation. My first assumption is to use t = 1 second, since frequency is in Hz, which is revolutions per second.

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)(1) = 1.75 V

Is this correct? If so, then I am sure I can do part b) on my own. Thanks for the help!
If a signal is Asin(wt), what is its amplitude? Does it depend on w or t?

So I believe I figured it out. At peak voltage, the amplitude sinwt is at its peak too, which is equal to 1. Thanks for the help!

## 1. What is EMF and how is it related to a generator?

EMF stands for electromotive force, which is the electrical potential difference that drives an electric current. In a generator, EMF is created as a result of the relative motion between a conductor and a magnetic field, which induces a voltage in the conductor.

## 2. How is the peak voltage of a generator determined?

The peak voltage of a generator can be calculated by multiplying the peak magnetic flux density by the number of turns in the generator's coil and the frequency of the alternating current being generated.

## 3. What factors affect the EMF and peak voltage of a generator?

The EMF and peak voltage of a generator can be affected by the strength of the magnetic field, the speed of rotation, the number of turns in the coil, and the frequency of the alternating current.

## 4. What are the units of measurement for EMF and peak voltage?

The units of measurement for EMF and peak voltage are volts (V). However, for larger generators, the unit of measurement may be kilovolts (kV) or megavolts (MV).

## 5. How is EMF and peak voltage controlled in a generator?

EMF and peak voltage can be controlled in a generator by adjusting the speed of rotation, the strength of the magnetic field, and the number of turns in the coil. Additionally, voltage regulators can be used to maintain a consistent output voltage.

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