# 3-phase circuits

Discussion in 'Homework Help' started by MareBear, Jan 25, 2009.

1. ### MareBear Thread Starter Member

Oct 29, 2008
22
0
im doing the homework my professor assigned on friday for 3-phase circuits and im going over my notes but i dont know how he got the phase current. he divided phase voltage (254) by 10+j15 and got 14.09 A. how did he get 14.09, when i put it in my calculator i get it in polar form. am i missing something?

2. ### Wendy Moderator

Mar 24, 2008
21,438
2,958
Try drawing the problem as a vector, and solve it using trig.

3. ### MareBear Thread Starter Member

Oct 29, 2008
22
0
i got it, thanks

4. ### mhean_ee Member

Aug 25, 2008
10
0
You can also compute for 14.09A by changing first the impedance of 10+15j in a polar form. Which will be resulted to 18.02776 angle 56.31 degrees. but since the voltage will be cosidered at 0' angle, your impedance lags, which makes up a -56.31 degrees.
Angle -56.31 degrees only denoted the magnitude or direction (which can explain by vector). The 254V now can be divided by 18.02776. Which resulted to 14.09A.