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Calculating Average Power

Posted Feb 18th 2011

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Andy Wolin
1Score: 

3 years ago:  

1/2*(V^2/R) = 100^2/2*30 = 166.67 W

Since the peak voltage is given (and not the rms value), V^2/R must be multiplied by 1/2.  The Average power is also known as the Real power which uses the units of Watts.

Richard Percifield

3 years ago:  

The above equation is only true for a purely resistive impedance having a power factor of 1. Thus, if you have a low inductance heater this would be true. For an inductive load like a motor or capacitive load like a cheap switcher found in many CFLs, the power dissipated by the load will not be as the equation as above. The complex impedance would have the voltage and current waveforms out of phase leaving you a power factor of less than 1.

Andy Wolin
1Score: 

3 years ago:  

I think I see what you are saying, if the load was a 30ohm resistor in series with a 3h inductor, then we could not use that same formula because the 100v source is now across a resistive and inductive load.  The procedure would be to find the voltage across each element (by adding impedances and then finding i for example), at which point the average power could be calculated for the resistor (i^2*r) and we know that the average power of the inductor would be 0.

Richard Percifield

3 years ago:  

Hello Andy,

Unfortunately, the real world doesn’t fit the “textbook” equations. There are no perfect inductors, capacitors, or resistors. Coil ceramic resistors have inductance and capacitance. Inductors have both capacitance and resistance. And finally Capacitors have inductance and resistance. Many a design has been placed in unknown operation land because of these imperfections.

To your question it is easiest to look at the system as a lumped element impedance in the form of Z= Rres + jXinductive. This makes the impedance frequency dependent, and will result in a vector quantity. In your example the result for 60Hz would be Z= 30 + j1130.97 ohms, or 1131.37 Ohms at 88.4deg. The voltage would lead the current by 88.4 degrees. Of course this is only a first order calculation, since coil resistance and resistor inductance have not been used in the calculation. Given the low frequency, the capacitive effects would be minimal and can be ignored (If it was a 100MHz signal then we would have to look at the capacitance). The power factor would be the cos(88.4)= 0.0279, a low power factor indeed! Most of the energy is not reaching the resistor, and the P(ave) of the resistor would be 0.0279 of what is available in VA.

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