# 3 PHASE CIRCUIT PROBLEM - Haaalllppp

1. Mar 16, 2013

### JGrecs

1. The problem statement, all variables and given/known data

Three load each having a resistance of 16 ohms and an inductive reactance of 12 ohms are wye connected to a 240V three phase supply. Determine:

a) the impedance per phase
b) the current per phase
c) the kVA total
d) the Power Factor
e) the kW total

2. Relevant equations

impedance? (Z= √R^2 + XL^2)??
kVA= 3 X VPhase X IPhase
kVA= 1.732 X VLine X ILine
PF= kW/kVA
p= 3 X VPhase X IPhase X PF
P= 1.732 X VLine X ILine X PF

3. The attempt at a solution

A) ?

B) VLine= 240V

IPhase= VPhase/RPhase

VPhase= 240V/1.732
VPhase= 138.5V

IPhase= 138.5V/16Ω
IPhase= 8.6A

C) kVA= 3 X VPhase X IPhase

kVA= 3 X 138.5V X 8.6A
= 3.5kVA

D) ?

E) ?

Not sure how to take these ones on. Any tips? formulas to use?

2. Mar 16, 2013

### JGrecs

http://1-media-cdn.foolz.us/ffuuka/board/wsg/image/1339/18/1339188551986.gif

3. Mar 16, 2013

### Staff: Mentor

I did some browsing and came across this site:

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE302

with the following graphic:

which seems to indicate that the voltage specification for a Wye generating system specifies the line-to-neutral (or phase) voltage of the generator. If that is true for this problem, then the line-to-line voltage will be in the neighborhood of 416V.

A circuit diagram would look something like:

For the impedance you can use the reactance as you've shown. You can determine the power factor for the load from its real and reactive components.

An alternative is to leave the impedance in complex form and do the calculations in complex form too (phasors). This would automatically take care of all the "1.732" and pf factors and give you the kVA, kW and kVAR values all at once. It helps if your calculator handles complex numbers

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