bloguetronica
- Joined Apr 27, 2007
- 1,467
That's what I use.average emf is used for flux calculations (remeber using them for derivation in transformers)
but then again u can always use the form factor cant u?
That's what I use.average emf is used for flux calculations (remeber using them for derivation in transformers)
but then again u can always use the form factor cant u?
The 230V AC is an average value and has 325V peak (650V from positive peak to negative peak). That is the way in Europe. EOS.The average voltage of 230V RMS is 2*Vpeak/pi=2*325/3,14=207V
Para 220 V CA, a tensão de pico VP ou A é, portanto, 220 V × √2 = 311 V (aprox.). O valor de pico-a-pico VP-P de 220V CA é ainda mais alta: 2 × 220 V × √2 = 622V (aprox.)
No! I live in Europe and I have measured the output of a 230V wall outlet in a university lab with an oscilloscope and the peak voltage is 325V and the peak to peak is of course 650V. Using rms integration you get 230V rms and using average integration you get 207V average. I have seen it with my eyes talked about it with my professors so I do not think I can be wrong about that!The 230V AC is an average value and has 325V peak (650V from positive peak to negative peak). That is the way in Europe. EOS.
The value of 0.707 was calculated and confirmed in a real life situation. In a transformer, measured 22V AC, but after rectification and capacitative filtering measured around 30V. Aprox the peak voltage with 1V subtracted, being the 1V the voltage across two diodes from the bridge. Not to offend you, but if you need more proff, go to Europe, buy a wall wart and open it (I don't know how it applies in the US or other countries). I cannot say more than that. Sorry if I was too harsh.
Yes. With pulse waveforms having insignificant risetimes, you can calculate the power at the two voltage levels and average them proportional to the duty cycle.So, if you have a current with that waveform going across through a resistor, the power dissipated in the resistor is the same when it is crossed by a current with a constant potential of 3.53V? Sounds a little strange, so with a 5Ohm resistor:
Power when crossed by 5V: 5W
Average power (square wave regime): 2.5W
Power when crossed by 3.53V: 3.53^2 / 5 = 2.492W
Makes sense, but that is not the average voltage. So RMS is related to power, that is, RMS voltage is the constant voltage that causes the same amount of dissipation as the waveform. Right?
I think your average calculation is wrong. The integral of sin(x) for x=0 to x=pi is 2. To get the average, divide by pi. The average is 0.6366. For 0 to 2pi, the average of abs(sin)x) will be the same.This confuses me, since some transformers say 230V RMS instead of 230V AC, and they are refering to the average voltage, because I know the peaks are about 325.22 (230 x sqrt (2)). Unless RMS and average voltage are the same for a 230V AC waveform. Also, this definition of average voltage for AC current exists on Malvino (Peak = Average x sqrt(0.5)). I calculated myself the average voltage of the module (I'm refering here to abs, because in Portugal we call it módulo, and I´m not sure about around here) of a sin function. For example the integer from x=0 to x=Pi of sin (x) is 0.707/2. So the integer from x=0 to x=2Pi of abs(sin(x)) would be 0.707, the value of sqrt(0.5).
Are you sure about that last part. I just calculated it and it is zero... I think (unless of course the wave is rectified)For 0 to 2pi, the average of abs(sin)x) will be the same.
It is abs(sin(x)) and not sin(x). The areas of sin(x) will subtract and the integral will equal 0. abs(sin(x)) corresponds more or less to a rectified wave.Are you sure about that last part. I just calculated it and it is zero... I think (unless of course the wave is rectified)
I know I answered this wrong the other time. But I can give you a hint:As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
I know I answered this wrong the other time. But I can give you a hint:As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
For now I can say that that is exactly what we learned at polytechnic yesterday. The first component (term) in a fourier series is in fact the DC component (another term for the average value). Using a low pass filter you try to eliminate higher harmonics (the rest of the components of the Fourier series) and what is left is the average value.Getting back to the original question, somewhat, isn't it true that a low-pass filter will produce approximately the average value of a rectified signal? (and not only for sinusoids)-
by Jake Hertz
by Ikimi .O