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# Load currents in an alternator

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1.  Lord of the Lightning
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Reputation ## Load currents in an alternator

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?

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  Reply With Quote

2. Originally Posted by 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?

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:  Reply With Quote

3.  Lord of the Lightning
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Reputation  Originally Posted by Kalbi_Rob See previous thread on this question:

Thanks! Unfortunately still not able to figure it out.  Reply With Quote

4. Originally Posted by JeremyBrown87 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.  Reply With Quote

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