Load currents in an alternator

[JeremyBrown87]

16. A wye-connected alternator that has phase voltages of 277 V is feeding two loads. The first load is wye-connected inductors that have 40 ohms of XL. The second load is a set of delta-connected capacitors that have XC of 60 ohms. How much is the alternator phase current?
Your Answer: 14.93 A
Correct Answer: 6.93 A

So I didn't realized that the One load was Wye and other was delta, how does this change the total Z eqn? I tried to answer in a couple ways but couldn't come up with correct answer

[Kalbi_Rob]

16. A wye-connected alternator that has phase voltages of 277 V is feeding two loads. The first load is wye-connected inductors that have 40 ohms of XL. The second load is a set of delta-connected capacitors that have XC of 60 ohms. How much is the alternator phase current?
Your Answer: 14.93 A
Correct Answer: 6.93 A

So I didn't realized that the One load was Wye and other was delta, how does this change the total Z eqn? I tried to answer in a couple ways but couldn't come up with correct answer

See previous thread on this question:

https://testguy.net/threads/5955-How...-phase-current

[JeremyBrown87]

See previous thread on this question:

https://testguy.net/threads/5955-How...-phase-current

Thanks! Unfortunately still not able to figure it out.

[Kalbi_Rob]

Thanks! Unfortunately still not able to figure it out.

So first you want to convert Xc Delta connected load to Xc Wye load(simplify the math): Xc(Wye)=(Xa*Xb)/(Xa+Xb+Xc)=(60*60)/(60+60+60)=20 (this is just a single phase version of all three equations to be simplified, see https://www.allaboutcircuits.com/tex...y-conversions/).

Next, you have to add two reactances in parallel: Z=1/(sqrt((1/R)^2+((1/XL)-(1/XC))^2)), where R=0 thus Z=1/(sqrt((1/0)^2+((1/40)-(1/20))^2))=40
Taken from Impedance of a Parallel RLC Circuit found on this website: https://www.electronics-tutorials.ws...l-circuit.html

Then calculate the current of the alternator knowing that current is equal on a Wye connection between phase and line, I=277/40=6.93A

There is another much more difficult (easier math but higher level) way to calculate this using rectangular formulas and Siemens using Laplace equations.