calculation of Electrical Power
- Thread starter berad
- Start date
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kingpb
Senior Member
- Location
- SE USA as far as you can go
- Occupation
- Engineer, Registered
S = (V)^2/z
S = P + jQ
S = P + jQ
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
190121-0925 EST
berad:
Consider only two terminal devices as the load.
For any type of load the instantaneous power is p = e*i , where p = instantaneous power, and can be +, -, or 0.
Any other power is some sort of an average. And to perform simple calculations the load needs to be made up of linear components that are invariant in all ways with time.
For a simple resistive load P = V*I where P = average real power in watts, V = Vrms voltage in volts * Irms current in amperes. Applies to AC sine waves or DC. These measurements V and I do not need to be time correlated, but must be invariant with time. So the same multimeter could be used to make the two measurements.
If the load has reactive components, inductance, and/or capacitance, then a power factor multiplier between 0 and 1 must be included in the equation. A perfect capacitor has a PF = 0, and some real world capacitors are close to this.
A real electromechanical wattmeter measures real average power by instantaneous multiplication being done with two interacting magnetic fields, and averaged by the mechanical inertia of the meter movement, and calibration by a non-linear scale.
.
berad:
Consider only two terminal devices as the load.
For any type of load the instantaneous power is p = e*i , where p = instantaneous power, and can be +, -, or 0.
Any other power is some sort of an average. And to perform simple calculations the load needs to be made up of linear components that are invariant in all ways with time.
For a simple resistive load P = V*I where P = average real power in watts, V = Vrms voltage in volts * Irms current in amperes. Applies to AC sine waves or DC. These measurements V and I do not need to be time correlated, but must be invariant with time. So the same multimeter could be used to make the two measurements.
If the load has reactive components, inductance, and/or capacitance, then a power factor multiplier between 0 and 1 must be included in the equation. A perfect capacitor has a PF = 0, and some real world capacitors are close to this.
A real electromechanical wattmeter measures real average power by instantaneous multiplication being done with two interacting magnetic fields, and averaged by the mechanical inertia of the meter movement, and calibration by a non-linear scale.
.
√3VLILcosphi
Works for me.
But do you want AC single phase or three phase, resistive, reactive, or DC?
kwired
Electron manager
- Location
- NE Nebraska
Too generic of a question, you have more details on what you are inquiring about?
WarrMann
Senior Member
- Location
- Atlanta, GA
Paper and pencil, calculator, smart phone app, or even the digits on your hand. They have all worked well for me over the years
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