Jump to content
Electronics-Lab.com Community


  • Posts

  • Joined

  • Last visited


About Gazza

  • Birthday 08/13/1980

Gazza's Achievements


Newbie (1/14)



  1. 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?
  2. Like the last poster said, there a few factors involved. However, I would say that could definitely damage them.
  3. 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.
  4. 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.
  5. Resistance has absolutely nothing to do with the turns ratio. What do you mean you use 120Hz?
  6. Not to mention the frequency of the voltage is 50Hz in Europe and 60Hz in the USA.
  7. 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.
  8. 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.
  9. 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.
  10. Do you have a wiring diagram of the transformer? If it has a 50% tap then yes it can, if not, it cannot.
  11. 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!
  12. 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.
  13. 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.
  14. 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?
  • Create New...