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Unread 01-02-2014, 01:40
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Ether Ether is offline
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar

Quote:
Originally Posted by zbrozek View Post
Well, not at the same time, but it's the same method. Basically I find a high-enough frequency where the voltage across at the output of the meter (post-50R source resistor) doesn't change when I attach the motor.

Then I lower the frequency until the voltage is exactly half that. At that point the voltage across the resistor and the voltage across the motor are the same...
They're not the same.

Consider:

When the voltage across the motor is half the voltage across the resistor+motor series circuit,
the frequency will be f = R/(2*pi*L*sqrt(3)).

At that frequency, the magnitude of the motor impedance will be R/sqrt(3), so the voltage across the resistor will be sqrt(3) times the voltage across the motor.



For the voltages across the resistor and motor to be equal, the frequency would have to be f = R/(2*pi*L). At that frequency, the magnitude of the motor's impedance would be R, and the voltage across the motor (and the resistor) would each be 0.707 times the voltage across the motor+resistor series circuit.



Last edited by Ether : 02-02-2014 at 01:09. Reason: added calculation for equal voltages
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