How much is the alternator phase current?

[clayton]

I have been stuck on this question and just can't seem to figure out how to get the correct answer of 6.93.

Would you be able to show the work needed for this?

[Kalbi_Rob]

I have been stuck on this question and just can't seem to figure out how to get the correct answer of 6.93.

Would you be able to show the work needed for this?

Can you assist us by stating the question you need answering?

[clayton]

Can you assist us by stating the question you need answering?

Ahh, well wouldn't that make a whole lot of sense. I guess I assumed people could read my mind

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?

[Kalbi_Rob]

Ahh, well wouldn't that make a whole lot of sense. I guess I assumed people could read my mind

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?

I can't help you, I only get 20.781 A. Now the current flowing through the Wye-connected inductors appears to be 6.93A. I have to ask if you might have misunderstood the question.

[J2019L]

I have been stuck on this question and just can't seem to figure out how to get the correct answer of 6.93.

Would you be able to show the work needed for this?

1. Convert Xc from delta connection to Y connection. In Y connection, Xc'= 60*60/(60+60+60)=20
2. Calculate balance 3-phase circuit for two loads, parallel Xc' & XL. Total load 1/X= (1-2)j/40 --> X=40j (reactance)

3. 3-Phase balance current I = 277/40 = 6.925 --Ans

[Kalbi_Rob]

1. Convert Xc from delta connection to Y connection. In Y connection, Xc'= 60*60/(60+60+60)=20
2. Calculate balance 3-phase circuit for two loads, parallel Xc' & XL. Total load 1/X= (1-2)j/40 --> X=40j (reactance)

3. 3-Phase balance current I = 277/40 = 6.925 --Ans

Aww, that's the equation I forgot

Z=1/(sqrt((1/R)^2+((1/XL)-(1/XC))^2))

334

[J2019L]

Aww, that's the equation I forgot

Z=1/(sqrt((1/R)^2+((1/XL)-(1/XC))^2))

334
To solve this question, need to have math and electric concept. In the complex number, R represents real part and Xc, XL represent complex parts. So, R is the number on the real axis and Xc and XL are the number on the complex axis. Xc and XL are the number on complex axis, but XL is positive (to up side) Xc is negative (to down side) on the complex axis. Therefore, it is showed minus between (1/XL -1/Xc). And in this question R = 0, The equation becomes simplified Z = 1/sqrt((1/XL)-(1/XC))^2)