DC generator question -- what is happening at no load?

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dave pallamino
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In the case of a d.c. generator, if V=E–IaRa when current flows through the armature, why does V = E at no load?
Why does the resistance element become disregarded at no load? Is this because the resistance can't cause the voltage to drop at no load?
 
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How is the armature current related to the load current? What can you say about the armature current under no load condition?

If you are having a hard time visualizing "no load". Then imagine a resistance load and you start increasing the resistance. When you get to the order of a Megohm (like what a voltmeter may have), how much current is through the armature, and what is the votage approaching?
 
dave pallamino said:
V=E–IaRa

Have you taken first year algebra ?

dave pallamino said:
Why does the resistance element become disregarded at no load?

Is multiplication by zero 'disregarded" ?

dave pallamino said:
Is this because the resistance can't cause the voltage to drop at no load?
Bravo !
Most anything when multiplied by zero yields a zero result.
:smile:

Always make your words and your math come into agreement.
It's important to form that habit because
our minds will believe word constructs that are physically impossible
and that's why they call math "the language of science" .
:oldwink:
old jim
 
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dave pallamino said:
Why does the resistance element become disregarded at no load? Is this because the resistance can't cause the voltage to drop at no load?
Yes!
Correct!
You got it!
Good insight!

Carry on with your studies, you passed the first exam.

Cheers,
Tom
 
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