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Can you explain the theory behind why the inductor in phase R and phase B become in series when phase Y is disconnected? And why do the voltage of phase R and phase B become to be equal to line voltage of 400V if phase Y and Neutral are short circuited?

Can you explain the theory behind why the inductor in phase R and phase B become in series when phase Y is disconnected? And why do the voltage of phase R and phase B become to be equal to line voltage of 400V if phase Y and Neutral are short circuited?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Can you explain the theory behind why the inductor in phase R and phase B become in series when phase Y is disconnected? And why do the voltage of phase R and phase B become to be equal to line voltage of 400V if phase Y and Neutral are short circuited?

Problem 9.5 Three equal inductors connected in star take 5 kW at 0.7 Pf when connected to a 400 V, 50 Hz three-phase, three-wire supply. Calculate the line currents (i) if one of the inductors is disconnected, and (ii) If one of the inductors is short circuited. Solution B. Y. Fig. 9.57 (a) Total power when they are connected to 400 V supply P- J3 V, I, cos o= 5000 W 5000 IPh== = 10.31 A V3 x 400 x 0.7 400 =D22.4 2 Impedance/phase = 3 x 10.31 RP. = ZP, cos 0= 22.4 x 0.7 = 15.68 2 XP = ZPh sin 0 = 22.4 x 0.714 16 2 (i) If phase Y is disconnected from the circuit, the other two inductors are connected in series across the line voltage of 400 V as shown in Fig. 9.57(a). 400 = 8.928 A 2x Zph ly 0 (ii) If phase Y and N are short circuited as shown in Fig. 9.57(b), the phase voltages VRy and VEy will be equal to the line voltage 400 V.
Transcribed Image Text:Problem 9.5 Three equal inductors connected in star take 5 kW at 0.7 Pf when connected to a 400 V, 50 Hz three-phase, three-wire supply. Calculate the line currents (i) if one of the inductors is disconnected, and (ii) If one of the inductors is short circuited. Solution B. Y. Fig. 9.57 (a) Total power when they are connected to 400 V supply P- J3 V, I, cos o= 5000 W 5000 IPh== = 10.31 A V3 x 400 x 0.7 400 =D22.4 2 Impedance/phase = 3 x 10.31 RP. = ZP, cos 0= 22.4 x 0.7 = 15.68 2 XP = ZPh sin 0 = 22.4 x 0.714 16 2 (i) If phase Y is disconnected from the circuit, the other two inductors are connected in series across the line voltage of 400 V as shown in Fig. 9.57(a). 400 = 8.928 A 2x Zph ly 0 (ii) If phase Y and N are short circuited as shown in Fig. 9.57(b), the phase voltages VRy and VEy will be equal to the line voltage 400 V.
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