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

Answer:

Following are the solution to the given question:

Explanation:

Line voltage:

$V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v$

Power supplied to the load:

$P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi$

$10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A$

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:

$Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega$

Phase angle:

$\phi = \cos^{-1} \ (0.85) =31.79^{\circ}$

Please find the phasor diagram in the attached file.