EMF produced by generator of a car

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SUMMARY

The discussion centers on calculating the electromotive force (e.m.f) produced by a car generator at different RPMs. The initial calculation of output voltage using the ratio method yielded 28.9 V, which was deemed incorrect due to the constant output voltage regulated at 12.7 V regardless of RPM changes. Participants clarified that the e.m.f values discussed are likely root mean square (rms) values, not functions of time, and emphasized the importance of understanding rms calculations in this context.

PREREQUISITES
  • Understanding of electromotive force (e.m.f) and its calculation
  • Familiarity with root mean square (rms) values in electrical engineering
  • Knowledge of the formula for e.m.f: ε = N B A ω sin(ωt)
  • Basic concepts of voltage regulation in automotive systems
NEXT STEPS
  • Research the principles of voltage regulation in automotive generators
  • Study the derivation and application of rms values in AC circuits
  • Learn about the significance of sine functions in alternating current (AC) calculations
  • Explore practical examples of e.m.f calculations in different rotational speeds
USEFUL FOR

Electrical engineers, automotive technicians, and students studying electromagnetism or electrical engineering principles will benefit from this discussion.

songoku
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Homework Statement
The generator of a car idling at 1100 rpm produces 12.7 V. What will the output be at a rotation speed of 2500 rpm, assuming nothing else changes?
Relevant Equations
ε = N B A ω sin ωt
First I assumed the question asks about max e.m.f so I just used ratio:

output voltage = (2500 / 1100) x 12.7 = 28.9 V, but the answer is wrong.

Then I tried to use the ratio of the whole formula:
$$\frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{NBA \omega_{1} \sin(\omega_{1}t)}{NBA \omega_{2} \sin(\omega_{2}t)}$$

But I can't calculate because I don't know the value of ##t##.

What is the approach to solve this question? Thanks
 
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Assume 12.7 V is the peak voltage
 
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Gordianus said:
Assume 12.7 V is the peak voltage
I did and this is my attempt:

songoku said:
First I assumed the question asks about max e.m.f so I just used ratio:

output voltage = (2500 / 1100) x 12.7 = 28.9 V, but the answer is wrong.
But the answer is wrong. Maybe I need to take something else into consideration?

Thanks
 
You don't say what the right answer is. Perhaps it's a tricky question and you're supposed to know the voltage regulator keeps the output voltage constant at 12.7 V, regardless the rpms.
 
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songoku said:
Homework Statement:: The generator of a car idling at 1100 rpm produces 12.7 V. What will the output be at a rotation speed of 2500 rpm, assuming nothing else changes?
Relevant Equations:: ε = N B A ω sin ωt

First I assumed the question asks about max e.m.f so I just used ratio:

output voltage = (2500 / 1100) x 12.7 = 28.9 V, but the answer is wrong.
I believe your answer of 28.9V is correct. Sometimes 'official' answers are wrong. It happens occasionally.

(I'm assuming you have given the complete question, so that, for example, 12.7V and 28.9V are both rms values or both peak values.)

songoku said:
Then I tried to use the ratio of the whole formula:$$\frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{NBA \omega_{1} \sin(\omega_{1}t)}{NBA \omega_{2} \sin(\omega_{2}t)}$$But I can't calculate because I don't know the value of ##t##.
##\varepsilon_1## and ##\varepsilon_2## are rms (or peak) values . They aare not functions of time. So you can replace each of the two sine terms by a single number.

{Minor edits made.]
 
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Gordianus said:
You don't say what the right answer is. Perhaps it's a tricky question and you're supposed to know the voltage regulator keeps the output voltage constant at 12.7 V, regardless the rpms.
I also don't know what the right answer is. The question is just like that, no other information, no other sentences and no diagram given

Steve4Physics said:
I believe your answer of 28.9V is correct. Sometimes 'official' answers are wrong. It happens occasionally.

(I'm assuming you have given the complete question, so that, for example, 12.7V and 28.9V are both rms values or both peak values.)
Yes I have given the whole question

Steve4Physics said:
##\varepsilon_1## and ##\varepsilon_2## are rms (or peak) values . They are not functions of time. So you can replace each of the two sine terms by a single number.

{Minor edits made.]
Sorry I don't understand why ##\varepsilon## is not a function of time. From the graph of ##\varepsilon## and also the equation, wouldn't ##\varepsilon## changes with respect to time?

Thanks
 
songoku said:
Sorry I don't understand why ##\varepsilon## is not a function of time. From the graph of ##\varepsilon## and also the equation, wouldn't ##\varepsilon## changes with respect to time?
Emf is a function of time: ##\varepsilon (t) =NBA \omega \sin {(\omega t)}##.

Note the key values of emf during a cycle:
Minimum value = ##-NBA \omega##
Maximum (peak) value = ##+NBA \omega##
Average value = zero.
Rms average value ##\varepsilon_{rms} =\frac {NBA \omega}{\sqrt 2}##
None of these values are functions of time.

When you are told that the emf is 12.7V this is probably the rms value. Make sure you understand what an rms value means.

The ratio you are trying to calculate is in fact: ##\frac {\varepsilon_{2, rms}}{\varepsilon_{1, rms}}##. This is not a function of time.
 
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Steve4Physics said:
Emf is a function of time: ##\varepsilon (t) =NBA \omega \sin {(\omega t)}##.

Note the key values of emf during a cycle:
Minimum value = ##-NBA \omega##
Maximum (peak) value = ##+NBA \omega##
Average value = zero.
Rms average value ##\varepsilon_{rms} =\frac {NBA \omega}{\sqrt 2}##
None of these values are functions of time.

When you are told that the emf is 12.7V this is probably the rms value. Make sure you understand what an rms value means.

The ratio you are trying to calculate is in fact: ##\frac {\varepsilon_{2, rms}}{\varepsilon_{1, rms}}##. This is not a function of time.
Ah ok I understand

Thank you very much for the help and explanation Gordianus and Steve4Physics
 
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