Solution: The original picture is as above. Three150Ω resistors are connected in a triangle, which can be equivalent to a Y-shaped connection. The equivalent resistance is:

R ' = R×R/(R+R+R)= R/3 = 150/3 = 50(ω)。 As shown in the figure below:

Because the power supply is symmetrical in three phases, the resistance is symmetrical in three phases and the inductance is symmetrical in three phases, Un' (phasor) =Un "(phasor) =Un (phasor) =0, and R' and j50Ω resistors run in parallel.

(1)UAB (phasor) = 380 ∠ 30 V. According to the symmetry, UBC (phasor) = 380 ∝ (30-120) = 380 ∝-90 V. Meanwhile, UAn' (phasor)

Therefore: I 1a (phasor) =UAn' (phasor) /R' = 220 ∠ 0/50 = 4.4 (a).

I2a (phasor) =UAn' (phasor)/J50 = 220 ∠ 0/50 ∠ 90 = 4.4 ∠-90 =-J4.4 (a).

According to KCl: IA (phasor) =I 1a (phasor) +I2a (phasor) = 4.4-j 4.4 = 4.4√2∞-45(A).

According to the symmetry: IB (phasor) =IA (phasor) ∞-120 = 4.4 ∠ 2 ∠ (-45-120) = 4.4 ∠ 2 ∠-160.

IC (phasor) =IA (phasor) ≈120 = 4.4 ∠ 2 ∠ (-45+120) = 4.4 ∠ 2 ∠ 75 (a).

(2) φ u = 0 and φ I =-45, so φ = φ u-φ I = 0-(-45) = 45.

So the power absorbed by the three-phase load: p = √ 3 UICOS φ = √ 3× 380× 4.4 √ 2× COS45 =1672 √ 3 (w).

Because only resistors consume power in the circuit, the absorbed power can be calculated by calculating the consumption of three-phase resistors: p = 3× 4.4× 220 = 3× (380/√ 3 )× 4.4 =1672 √ 3 (w).

—— Note: If other methods are used, the calculation results may be different, mainly 380/√ 3 = 2 19.5438+0 ≈ 220, and a big error may be obtained in the calculation process.