EMF produced by generator of a car

AI Thread Summary
The discussion centers on calculating the electromotive force (e.m.f) produced by a car generator at different RPMs. Initially, a calculation suggested an output voltage of 28.9 V at 2500 RPM based on a ratio, but this was deemed incorrect. Participants highlighted that the voltage regulator maintains a constant output of 12.7 V, regardless of RPM changes, and clarified that e.m.f values are typically expressed as RMS (root mean square) values, not functions of time. The conversation emphasizes the importance of understanding RMS values in electrical calculations. Ultimately, the participants reached a consensus on the nature of e.m.f and the role of the voltage regulator in this context.
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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