# Three phase ac newtork

Discussion in 'Homework Help' started by SilverKing, Jan 31, 2015.

1. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Hi everyone,

I need some help in solving this problem:

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.

This how I tried to solve it:
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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

______________________

Is that right?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If the load comprises purely inductive components where will any real load power be dissipated?

Mar 31, 2012
17,748
4,796

4. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
t_n_k
So, any purly inductor would disspate zero power? I didn't know that before. Thanks

WBahn

???

5. ### WBahn Moderator

Mar 31, 2012
17,748
4,796
Yep. If the voltage and current are 90 degrees out of phase, then no real power is delivered or received. Energy flows from the source and is temporarily stored in the magnetic field of the inductor (or electric field of a capacitor) and then returned to the source. This happens twice each cycle.