
How much is the alternator phase current?
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?

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

Originally Posted by
Kalbi_Rob
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 wyeconnected alternator that has phase voltages of 277 V is feeding two loads. The first load is wyeconnected inductors that have 40 ohms of XL. The second load is a set of deltaconnected capacitors that have XC of 60 ohms. How much is the alternator phase current?

Originally Posted by
clayton
Ahh, well wouldn't that make a whole lot of sense. I guess I assumed people could read my mind
A wyeconnected alternator that has phase voltages of 277 V is feeding two loads. The first load is wyeconnected inductors that have 40 ohms of XL. The second load is a set of deltaconnected 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 Wyeconnected inductors appears to be 6.93A. I have to ask if you might have misunderstood the question.

Originally Posted by
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?
1. Convert Xc from delta connection to Y connection. In Y connection, Xc'= 60*60/(60+60+60)=20
2. Calculate balance 3phase circuit for two loads, parallel Xc' & XL. Total load 1/X= (12)j/40 > X=40j (reactance)
3. 3Phase balance current I = 277/40 = 6.925 Ans

Originally Posted by
Jimmy Lee
1. Convert Xc from delta connection to Y connection. In Y connection, Xc'= 60*60/(60+60+60)=20
2. Calculate balance 3phase circuit for two loads, parallel Xc' & XL. Total load 1/X= (12)j/40 > X=40j (reactance)
3. 3Phase balance current I = 277/40 = 6.925 Ans
Aww, that's the equation I forgot
Z=1/(sqrt((1/R)^2+((1/X_{L})(1/X_{C}))^2))

Originally Posted by
Kalbi_Rob
Aww, that's the equation I forgot
Z=1/(sqrt((1/R)^2+((1/X
_{L})(1/X
_{C}))^2))
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/X_{L})(1/X_{C}))^2)
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