Question

A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage of 120 V rms. The load draws a total of 10 kW at a power factor of 0.85 (lagging). Calculate the rms value of the phase currents and the magnitude of the per-phase load impedance. Draw a phasor diagram showing all tlme voltages and currents.

Answer 1

Following are the solution to the given question:

Step-by-step explanation:

Line voltage:

`V_L=√(3)V_(ph)=√(3)(120) \ v`

Power supplied to the load:

`P_(L)=√(3)V_(L)I_(L) \cos \phi`

`10* 10^3=√(3)(120 √(3)) I_(L)\ (0.85)\\\\I_(L)= 32.68\ A`

Check wye-connection, for the phase current:

`I_(ph)=I_L= 32.68\ A`

Therefore,

Phasor currents:
`32.68 \angle 0^(\circ) \ A \ ,\ 32.68 \angle 120^(\circ) \ A\ ,\ and\ 32.68 -\angle 120^(\circ) \ A`

Magnitude of the per-phase load impedance:

`Z_(ph)=(V_(ph))/(I_(ph))=(120)/(32.68)=3.672 \ \Omega`

Phase angle:

`\phi = \cos^(-1) \ (0.85) =31.79^(\circ)`

Please find the phasor diagram in the attached file.