1-Phase Load, 3-Phase Supply

[zimmerDN]

How do you calculate the current on each line conductor for a resistive load?

The load is single phase, 575V Line-Line, 1000W Resistive

I = 1000W / 575V / 1.73 = 1.005A

or is it

I = 1000W / 575V = 1.739A because it's single phase?


Is it correct to call it a single phase load because it's only connected to 2 out of 3 phases at the supply?

 

[LionelHutz]

No sqrt(3) is used in single phase calculations.

 

[DRWeig]

Yes, it's called a single phase load as well. As Lionel implied, your second answer is correct.

Best to you,

Goober Dave

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[zimmerDN]

If there are two 1000W 1-phase resistive loads, across phase A & B and phase A & C.

Does it mean phase A will carry a load of I = 1000W / 575V = 1.739A * 2 = 3.478 amps

phase B & C will carry a load of I = 1000W / 575V = 1.739A = 1.739 amps

Thanks guys

 

[rhatcher]

Yes, that is correct.

 

[jghrist]

Not quite. Phase A will carry 1.739 * sqrt(3) = 3.012 amps
I_A = I_BA - I_AC vectorially

 

[LiteYear]

jghrist is correct in that the two load currents are 60° apart (from the point of view of phase A), so instead of 2*1.739 it's sqrt(3)*1.739 for phase A current.

All phase currents have to sum to zero as usual, so I_A = -I_B - I_C. But magnitude of phase B and C currents is as you describe - the load current of 1.739A.