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Old 27th June 2013, 01:36 PM   #1
AndrewT is offline AndrewT  Scotland
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Default Explain AC voltage measurements

I connect a 1uF capacitor in series with a small AC motor.
When I measure the voltage across the combination I see 245Vac.
The voltage reading across the motor is 21.2Vac
The voltage reading across the capacitor is 258Vac.

Why?

If I square the 21.2 and subtract that from the square of the 258, I get the sqrt(answer) = ~245.

How are the phase voltages behaving?
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Old 27th June 2013, 01:57 PM   #2
Elvee is online now Elvee  Belgium
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The motor has a strong inductive component in its impedance, the capacitor is, ..well, capacitive, which means the voltage vectors generated by the common current are 180°out of phase (in the reactive parts of the components at least).

In other words, the capacitor (over)compensates the motor (in a rather excessive and imperfect way).

With a higher value of capacitor, compensation would be more "perfect", which could make things dangerous: the combination would behave like a resistor having the ohmic losses of the circuit as value, leading to a high current and high voltages across both the motor and capacitor.
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Old 27th June 2013, 02:08 PM   #3
AndrewT is offline AndrewT  Scotland
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It's a 30Vac motor, so I don't want a high voltage across it.
It will run on very low voltage so I don't mind that "only" 21.2Vac appears across it.
Using a 20r resistor (dropping 1.394Vac) in the circuit I see that ~71mAac is passing around the circuit.

I am going to substitute a 300r for the motor. That will bring the phasors to 90degrees.
Maybe that will show more of what I had expected.

I now see that the almost 180degrees of opposition leads to the diagonal being equal to the supply voltage.
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Old 28th June 2013, 09:30 AM   #4
glennb is offline glennb  Australia
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Quote:
Originally Posted by AndrewT View Post
I connect a 1uF capacitor in series with a small AC motor.
When I measure the voltage across the combination I see 245Vac.
The voltage reading across the motor is 21.2Vac
The voltage reading across the capacitor is 258Vac.
Why?
If I square the 21.2 and subtract that from the square of the 258, I get the sqrt(answer) = ~245.
How are the phase voltages behaving?
You have a series C L R circuit. The current magnitude through all components is the same. Work out the impedance (Z) of each component at the operating frequency. Vcomponent = I * Zcomponent.

At resonance, where Zl = Zc, the series resonant current becomes I = Vin / Rmotor. This can lead to a very very high voltage across each of the C and L, Vl = I * Zl, Vc = I * Zc.
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Old 28th June 2013, 09:45 AM   #5
AndrewT is offline AndrewT  Scotland
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If my arithmetic is right the phase of the motor current is at ~140degrees to the phase of the capacitor current.
I used the cosine rule to determine 50.3degrees.
That makes the cap vertical @ 258Vac and then down 21.2Vac from there at -50.3degrees gives the supply voltage of 245Vac.
It makes sense to me now.
I had the motor phasor as horizontal thinking of it as a resistive load. That misled me to thinking "how can the cap voltage be higher than the supply voltage?"
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Old 28th June 2013, 10:00 AM   #6
AJT is offline AJT  Philippines
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just get a step down traffo with 30 volt secondary with the correct current(ampere) rating and be done with it...
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