Speed of a DC generator is proportional to the generated voltage?

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SUMMARY

The speed of a DC generator is directly proportional to the Electromotive Force (EMF) generated, which in turn affects the output voltage, particularly in separately excited generators. When the speed is halved, the EMF and consequently the output voltage also halve, assuming all other factors remain constant. For a shunt wound generator, the terminal voltage must account for the field winding resistance, which influences the overall output voltage. The relationship between speed, EMF, and voltage is crucial for understanding generator performance.

PREREQUISITES
  • Understanding of Electromotive Force (EMF) in DC generators
  • Knowledge of shunt wound generator operation
  • Familiarity with Ohm's Law and voltage calculations
  • Basic principles of magnetic flux in electrical machines
NEXT STEPS
  • Study the relationship between rotor speed and EMF in DC generators
  • Learn about the effects of field winding resistance on terminal voltage in shunt generators
  • Explore mathematical modeling of DC generator performance
  • Investigate the differences between separately excited and shunt wound generators
USEFUL FOR

Electrical engineers, students studying electrical machines, and professionals involved in generator design and analysis will benefit from this discussion.

ajskelly
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Homework Statement
If the speed of a DC generator is reduced by half, will this reduce the Voltage and the EMF by half as speed is directly proportional to EMF? Resistance is staying constant,
Relevant Equations
E=V+IaRa
I'm confused as I'm aware speed is directly proportional to the EMF, but is it proportional to the voltage as well? If the speed of a DC generator is halved and everything else is remaining constant the EMF will have so will the output voltage half
 
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ajskelly said:
If the speed of a DC generator is halved and everything else is remaining constant the EMF will have so will the output voltage half
Yes, provided it is a separately excited generator.
 
How about a shunt wound generator, for example...

the output voltage is 200V
the Ia is 50A
the Ra is 0.03ohms

E=V+IaRa
E=200+(50X0.03)
E=201.5

The speed of the generator is now reduced by half. As the speed of the generator is directly proportional to the EMF, half the speed will mean half the EMF so...

E/2 = 201.5/2 = 100.75

Now if I wanted to find the current which would have changed, can I use

E=V+IaRa

so can I say,

100.75= 100+(IaX0.03)

100 being the 100V that was found by dividing the orignal voltage by 2 due to the speed being halved.

Hope this makes sense!

Thanks for your help
 
ajskelly said:
As the speed of the generator is directly proportional to the EMF
ajskelly said:
I'm aware speed is directly proportional to the EMF
I prefer 'emf is directly proportional to the speed'.

Also, emf is proportional to the magnetic flux produced by the generator's field winding.
This means emf is directly proportional to the product of field flux and rotor speed.
As the terminal voltage changes, the field flux also changes.
Hence, in case of shunt generator, you will need to do some math considering field winding resistance to find the terminal voltage.
 
Sorry I forgot to thank you cnh1995 for your assistance,
Cheers!
 

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