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#### macaber

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Good day, I have come across a question during studying that I am struggling with, any help would be great.

Three coils with a resistance of 8 ohms and an impedance of 12 ohms are connected in wye to a 480V supply. What will each wattmeter indicate if three single phase wattmeters are used to measure the power?

#### backstay

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macaber said:
Good day, I have come across a question during studying that I am struggling with, any help would be great.

Three coils with a resistance of 8 ohms and an impedance of 12 ohms are connected in wye to a 480V supply. What will each wattmeter indicate if three single phase wattmeters are used to measure the power?
I'll take a stab at 28.8 kW

#### bkmichael65

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Voltage squared divided by resistance. Backstay gave you the correct answer

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#### bkmichael65

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Yeah, now that I look at it again. That would be the wattage in a resistant load.

#### Paulusgnome

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1) A 480V supply will have line voltage = 480/sqrt(3) = 277V, so this will be the voltage across each coil.
2) The current drawn by a 12 ohm impedance connected across a 277V supply is 277V/12ohms = 23A.
3) The power dissipated in the 8 ohms of resistance (1 coil) is I^2R = 23A x 23A x 8ohms = 4250W.
4) Total power dissipated in all 3 coils is 3 x 4250W x 3 = 12.75kW.

#### Zog

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Measured in Henreys, so the question is misleading.

#### 8V71

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Measured in Henreys, so the question is misleading.
Sorta like inductance, huh?

#### Zog

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Sorta like inductance, huh?
Doh! My bad, I brain farted :whistling2:

#### dronai

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Ok, now add another inductor of 6L onto each phase.

#### oliquir

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you can read all the current, voltage, power factor,...

#### macaber

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1) A 480V supply will have line voltage = 480/sqrt(3) = 277V, so this will be the voltage across each coil.
2) The current drawn by a 12 ohm impedance connected across a 277V supply is 277V/12ohms = 23A.
3) The power dissipated in the 8 ohms of resistance (1 coil) is I^2R = 23A x 23A x 8ohms = 4250W.
4) Total power dissipated in all 3 coils is 3 x 4250W x 3 = 12.75kW.
Thanks Paul, this is the answer I was looking for, I was forgetting to calculate the current drawn first before calculating the active power, much appreciated!

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