Answer:Iph=7A, IL=12.12A, P=65.65W, Q=1538.6VAR, Theta=87.55ْ

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SUMMARY

The discussion focuses on calculating the phase current (Iph), line current (IL), real power (P), reactive power (Q), and phase angle (Theta) for a delta-connected three-phase AC network with inductors. The calculations yield Iph = 7A, IL = 12.12A, P = 65.65W, Q = 1538.6VAR, and Theta = 87.55°. The user highlights the contradiction of having real power dissipated by a purely reactive load, leading to the conclusion that a purely inductive load should not dissipate real power, thus P should be 0.

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JasonHathaway
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Hi everyone,1. Homework Statement

Three inductors with 0.1 H each are delta connected to three phase ac newtork with 220V/50Hz. Determine Iph, IL, P, Q and Theta.

Homework Equations


Showed at the attempt.

The Attempt at a Solution


[/B]
XL=2*pi*f*L=31.4 ohms
Iph=Uph/XL=220/31.4=7 A
IL=1.73*7=12.12 A

Q=(Iph)^2 XL=Uph Iph sin(theta)

Since I don't have theta and I need to determine it in thte first place, I shall use he first law.
Q=(7)^2 31.4=1538.6 VAR
QT=3*Q=4615 VAR

Then, I shall substitute the value of Q in the second law to get theta.
sin^-1 (theta)=Q/Uph Iph=1538.6/220*7=87.55 ْ

P=Uph Iph cos(theta)=65.65 W
PT=3*P=196.95 W
 
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Are you sure that you should be seeing real power being dissipated by a purely reactive load? You might want to reconsider what the phase angle must be for such a load.
 
I found out that a purley inductor doesn't disspate any power, which leads to P=0. Thanks :)
 

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