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.
You need to do careful phasor additions. Draw the phasor diagram and the result is more readily deduced.
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.