# EMF produced by generator of a car

• songoku
In summary, the conversation discusses a question about the output voltage of a car generator at different rotation speeds. The initial attempt to solve the question using a ratio resulted in an incorrect answer. The conversation then explores using the entire formula for emf, but notes that the value of time is unknown. It is suggested that the emf is not a function of time and the ratio to be calculated is actually the rms values of the emf. The conversation concludes with a better understanding of the problem and appreciation for the assistance provided.

#### songoku

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

Assume 12.7 V is the peak voltage

• songoku
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.

• songoku
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'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.

• songoku
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'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.

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.

• songoku
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

• Steve4Physics