# Wye/Delta Load

Discussion in 'Homework Help' started by Three Phase, Mar 27, 2013.

1. ### Three Phase Thread Starter Member

Jul 11, 2007
13
0
A three source with a voltage of 120/208 Volts is connected to a wye load of three 10 ohms resistors and a delta connected load of three pure coils (pf=0) that draws 15 amps each. The line current to the source will be..... the correct answer is 28.6 amps.
My answer comes in much higher. On the delta load I find it's line current by multiplying 15 amps by 1.732 which equals to 25.98 amps. Then on the wye connected load I find the phase current by 120 volts by 10 ohms which equals to 12 amps. Phase current equals line current in the Wye configuration.
I add the line currents of the delta and wye loads.......25.98 amps + 12 amps = 37.98 amps. Help on this question will be greatly appreciated.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You need to do careful phasor additions. Draw the phasor diagram and the result is more readily deduced.

3. ### Three Phase Thread Starter Member

Jul 11, 2007
13
0
I'm still having difficulty.

4. ### panic mode Senior Member

Oct 10, 2011
1,329
305
post circuit and your attempt

5. ### Three Phase Thread Starter Member

Jul 11, 2007
13
0
Got it. The coils are pure inductors meaning their currents lag by 90 degrees. To find the true line current I must use the square root of (12 squared + 25.98 squared) = 28.617 amps.