Explain AC voltage measurements - diyAudio
 Explain AC voltage measurements
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 27th June 2013, 02:36 PM #1 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders 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? __________________ regards Andrew T. Sent from my desktop computer using a keyboard
 27th June 2013, 02:57 PM #2 diyAudio Member     Join Date: Sep 2006 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. __________________ . .Circlophone your life !!!! . . ♫♪ My little cheap Circlophone© ♫♪
 27th June 2013, 03:08 PM #3 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders 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. __________________ regards Andrew T. Sent from my desktop computer using a keyboard
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Join Date: May 2004
Location: Melbourne
Quote:
 Originally Posted by AndrewT 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.
__________________
Glenn.

 28th June 2013, 10:45 AM #5 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders 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?" __________________ regards Andrew T. Sent from my desktop computer using a keyboard
 28th June 2013, 11:00 AM #6 diyAudio Moderator     Join Date: May 2003 Location: Palatiw, Pasig City just get a step down traffo with 30 volt secondary with the correct current(ampere) rating and be done with it... __________________ the best advertisement for a good audio design is the number of diy'ers wanting to build it after all the years....never the say so of so called gurus....

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