elec-student
Member
- Location
- Massachusetts
the power expended in a three phase alternating current is equal to the power facter times?
Three phase alternating current
- Thread starter elec-student
- Start date
- Status
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Three phase alternating current
The magnitude of the voltage (in units of ?volts RMS?), and times the magnitude of the current (in units of ?amps RMS?), and times the square root of three.
The magnitude of the voltage (in units of ?volts RMS?), and times the magnitude of the current (in units of ?amps RMS?), and times the square root of three.
bphgravity
Senior Member
- Location
- Florida
Re: Three phase alternating current
Charlie's calculation is correct, however, this will still only give the volt-amps of the circuit and not the true power. E x I x pF will give the true wattage of a three phase system.
Charlie's calculation is correct, however, this will still only give the volt-amps of the circuit and not the true power. E x I x pF will give the true wattage of a three phase system.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Three phase alternating current
A closer look at the question would reveal that the "power factor" factor was included in my answer. I was completing the original sentence: "...power factor times ...." Also, don't forget the square root of three that enters into three phase calculations.
robbie
Member
- Location
- California
Re: Three phase alternating current
Elec-Student:
The power triangle might help you. The 'run' of the triangle would be your 'real' power in Watts. The 'rise' of the triangle would be the 'reactive' power in Vars and the 'hypotenuse' of the triangle would be the 'apparent' power in Volt-Amps. The cosine of the angle between (run)/(hypotenuse) is your power factor.
Elec-Student:
The power triangle might help you. The 'run' of the triangle would be your 'real' power in Watts. The 'rise' of the triangle would be the 'reactive' power in Vars and the 'hypotenuse' of the triangle would be the 'apparent' power in Volt-Amps. The cosine of the angle between (run)/(hypotenuse) is your power factor.
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