Question

Learning Goal: To find voltages and power in a balanced three-phase circuit. Three identical AC voltage sources, each with amplitude 170 V, are connected in a Y configuration to three identical resistances, each 50 , also in a Y configuration. There is also a fourth, neutral line in the circuit (see diagram). The voltage sources are each 120° out of phase from the other two sources R RE R -

Answer 1

Here we have given that,

Vmax = 170 V

R = 50 ohm each

Angle = 120° which means this a balanced Y connection

Now

Part A

Irms = Vl/sqroot3×R =  1.96299091524 A

Part B

Average power,  

Pav = 3l²R = 578 W

Part C

In this case the voltage would become twice so that,

Irms = 340/saqroot2 × 50 = (4.80832611207) A

Pav = Irms²R = 1156 W

Part D

In this the condition of single phase occurs so that,

Irsm = 170/sqroot2 × 50 = (2.40416305603) A

And

Pav = Irms²R = 289 W