It is known that the complex impedance of each phase of the star connected load is Z = 12 + j16 Ω. When it is connected to a symmetrical three-phase power supply, the line voltage is 380V, and the line current is calculated If the three-phase load is changed to triangle connection, what is the line current?

It is known that the complex impedance of each phase of the star connected load is Z = 12 + j16 Ω. When it is connected to a symmetrical three-phase power supply, the line voltage is 380V, and the line current is calculated If the three-phase load is changed to triangle connection, what is the line current?

The complex impedance of each phase of star connected load is Z = 12 + J 16 Ω = 20 Ω. I line = u line / √ 3 / z = 380 △ 3 △ 10 = 22a
If the three-phase load is changed to triangle connection, the line current I line = √ 3U / z = √ 3 × 380 △ 10 = 66A

In a three-phase symmetrical circuit, the line voltage of the power supply is 380V, and the impedance of each phase is 4 + 3j ohm
Try to find the line current and phase voltage of the load connected into star and triangle respectively

Line voltage U1 = 380 V phase voltage U2 = U1 / root 3 = 220 V z = 4 + 3 J ohm = 5 ohm Star: line current = phase current = U2 / z = 220 / 5 = 44 a phase voltage = 220 V triangle: line current = root 3 * phase current = 1.732 * U1 / z = 1.732 * 380

The three-phase symmetrical power supply with line voltage of 380V is used for two groups of symmetrical loads. One group of loads is star connected with impedance z = 4 + 3j Ω in each phase, and the other group of loads is triangle connected with resistance R = 38 Ω in each phase. The circuit diagram is drawn and the total active power, reactive power and power factor of two groups of loads are calculated

The impedance mode of load 1 is | Z1 | = √ (3 ^ 2 + 4 ^ 2) = 5 Ω. The star load uses phase voltage to calculate the power. The phase voltage is up1 = 220 V, the phase current is IP1 = up1 / | Z1 | = 220 / 5 = 44a, the power factor of load cos φ 1 = 4 / 5 = 0.8, the active power P1 = 3up1 * IP1 * cos φ 1 = 3 * 220 *