Gazza

It is all about the Power Triangle. So the answer is, it depends on what you mean by power. If the current and voltage are in phase, all of the power will be "real" so real power = apparent power or KW = KVA. If there is an inductor or a capacitor in the circuit, the voltage and current will be out of phase, therefore the real power will be the same but there will be a component of "active" power which has a phasor +/- 90 degrees out of phase with the real power. When you determine the "apparent Power" it will now be more than the "real" power or KVA>KW The long story short is study the power triangle. Audioguru, there are many types of AC power supplies, what about a UPS or a VFD?

Like the last poster said, there a few factors involved. However, I would say that could definitely damage them.

Im glad some one gets it.

The resistance has NOTHING to do with the turns ratio of a transformer. What you are seeing is something totaly different. PIn=POut therefore if you have a transformer that is 5:1 with a primary of 120V and a 60ohm load. The secondary voltage is equal to 120/5= 24V The current is = 24/60 = 400 mA Now, if you reduce the load to 30 ohms The secondary voltage is equal to the primary voltage divided by the turns ratio, it does not change 120/5 = 24V Current = 24/30 = 800mA The secondary current is a function of the load, and is not used to determine the turns ratio.

You can do it either way as long and you keep your signs consistent. Example (XT = Xc-XL) The reactance in the circuit is equal to the absolute value of Xc-XL If the value of XT is positive, the reactance is capacitive, if it is negative, it is inductive. If you reverse the formula, the opposite is now true, positive values are inductive and negative numbers are positive.

Resistance has absolutely nothing to do with the turns ratio. What do you mean you use 120Hz?

Not to mention the frequency of the voltage is 50Hz in Europe and 60Hz in the USA.

I don't think you know what you are talking about. Once you have the turns ratio which is equal to a=V2/V1 or a=N2/N1 You can find the voltage on the opposite side of the transformer. If you have a 240 to 120V transformer, the turns ratio is equal to 120/240 = .5 Therefore if you feed the primary side of the transformer with 120 volts, the secondary voltage will be .5= v2 / 120 v2 = 120 x .5 v2 = 60 Since current is inversely proportional to the voltage and power in must equal power out, the opposite is true, if you have 1 amp of current at 120V you must have twice as much current at 60V. The iron core is to help mitigate losses in the transformer, I suggest you look into the equivalent circuit of a transformer.

I am going to have to say it will have a drastic effect. Anything with an inductor or a capicitor in it will be affected by a change in frequency.

My bad, I read the question too fast. You are 100% correct. If the turns ratio is .5, the output voltage will be half of the input voltage. Which is the case here. My appologies.

Do you have a wiring diagram of the transformer? If it has a 50% tap then yes it can, if not, it cannot.

Allen Bradley was bought out by Rockwell Automation. The SLC500 is a PLC from the 80's and it uses RSLogix500 for is programming. You also need RSLinx to communicate from the software to the PLC. The software is about $7000 for the package, a SLC is at least $1000. Good find!

The easiest way to do that is to use an oscilloscope, measure the number of divisions between the start and the finish of one sine wave. Multiply that by the S/div and invert :) This doesn't make any sense, there will only be harmonics present if the load is non-linear. Even if there were any harmonics, the only way to "short" them out is to tune a filter circuit to the frequency of the harmonic, it will then act like a short to those frequencies. Even then, the filter exists between line and ground, not a dead short between the terminals. Shorting out the transformer's secondary with a wire is a sure way to set your transformer on fire.

Hey Gruthos, You are right, you need to vary the frequency to change the speed of your motor. I would use a single phase v/hz drive for this. The cap on the top of your motor is a starting cap for single phase induction motors You motor is using approx. 1.5HP, (10.2 x 110)/746 I would just buy one from www.automationdirect.com, http://web2.automationdirect.com/adc/Shopping/Catalog/AC_Drives_-z-_Motors/GS1_(120_-z-_230_VAC_V-z-Hz_Control)/GS1-22P0 That one costs about $150, You just have to make sure you can single phase the output, I dont have enough time to look right now.

wow I didn't read that first post properly, you can short circuit the seconday of a transformer for testing purposes but you have to be really careful about the current getting too high. However I have only ever heard of short circuiting a transformers secondary to find copper losses in the transformer. What were you trying to test? The resonant frequency of a wire?

Can you post a picture of the circuit?

Hey Azza, I think this is a great idea, I am all for anything that makes our energy sucking houses more efficient. Although 150 is quite a bit of money, it if works well and it lasts for more than a year, I think it is great.

Cool idea, If it were me, I would use a cheap PLC from www.automationdirect.com, you can get a cheep one for $129. I would probably monitor the water flow with some flow sensors and write a bit of logic to decide what pumps to use. Flow sensor - http://cgi.ebay.ca/Flow-Sensor-Air-Water-0-5-Vdc-Output-Omega-FLR1010_W0QQitemZ200071635353QQihZ010QQcategoryZ67003QQssPageNameZWDVWQQrdZ1QQcmdZViewItem PLC - http://web4.automationdirect.com/adc/Shopping/Catalog/PLC_Hardware/DirectLogic_05/PLC_Units Software - http://web4.automationdirect.com/adc/Shopping/Catalog/Software_Products/Directsoft_PLC_Programming_Software/Directsoft_Software - there is a free version I am a Power Electronics guy, so I tend to use a sledge hammer to solve these problems, I am sure one of the electronics guys will have a smaller idea :)

Hi there sskendre, Welcome to the board! I am not exactaly sure what you are looking for, protecting a load or device for UV and OV is application specific, meaning you need to be a bit more specific :)

is it a rectifier?

You need to step back about 50 yards, harmonics are not just added to the fundamental. PT=(PF + P 3rd harmonic2 + P 5th Harmonic 2 + P 7th Harmonic 2.........)1/2 I really don't mean to be rude but I think you are completely missing the point here, harmonics do not affect the value of the fundamental portion of a signal, they are additive but in a bad way. They are adding "distortion" to the signal. The fundamental frequency is the only part of the signal that you want to use. The harmonics are undesirable side effects created by non-linear loads. Here is a bad analogy: I have a rock, the rock grows moss due to the environmental conditions surrounding it, I scrape the moss off. Rock = fundamental moss = harmonics environment = type of load/power grid

You are missing the point. Harmonics are created in any load that is non-linear. There is no magic crystal. Just non-linear loads. I promise

Hey Kevin, I am not 100% sure that I am getting your question. As far as this quote is concerned maybe I can help a bit. When you are talking about the power grid in North America, the fundamental frequency of the power is 60hz. If you have a non-linear load (i.e. a rectifier) harmonics are created. Depending on the nature of the load, you will get different harmonics. Here is an example: If you have a fundamental frequency of 60hz feeding a 3 phase 6pulse rectifier. You will have 5th and 7th harmonics present, they will have frequencies of 300 and 420 Hz respectively. If you try to tune a filter to 480 Hz, you will get rid of the 7th harmonic, however the waveform will still be distorted due to the 5th. If you could filter out all harmonics (which is very difficult and costly) you would see a pure sine wave at 60hz. If you are passing a purely fundamental wave through an inductor or capacitor, ELI the ICE man has all of your answers. When a circuit is inductive: Voltage leads current. When the circuit is capacitive: Current leads voltage