Here is my equivalent balanced three phase circuit Y-Y with neutral:

enter image description here

Rla + jXla is my line impedance by phase. L1 and L2 are my loads. The exercise is saying that L1 absorbs 180 W and 240 VAr, L2 absorbs 600 VAr and has 0.6 lagging power factor. The line impedance is 5 + j8. The line voltage is 240 V. Its asking me what is the voltage VAN across the loads.


VAN.IL1* = SL1 = 180 + j240 VA
VAN.IL2* = SL2 = 450 + j600 VA
VAN.IL* = SL = SL1 + SL2 = 630 + j840 VA
IL = IL1 + IL2
VaA = IL.(Ra+jXLa)
VAN = Van - VaA
Van = 240/sqrt(3) with angle of -30 degrees
SL/IL* = Van - IL.(Ra+jXLa)
(630 + j840)/IL* = (120 - j69.282) - IL(5 + j8)

When I throw this equation at wolfram alpha, It says there's no solution:

Link for wolfram alpha's analysis

What am I doing wrong?

  • \$\begingroup\$ Yeah it's tricky for sure. You don't know the impedance of L1 - all you know is what power it takes and what reactive power it takes when the supply is somewhat less than 240V (this level being dictated by the other load and the series line impedance). 2 or 3 pages of maths methinks. \$\endgroup\$
    – Andy aka
    Commented Feb 26, 2014 at 22:35

2 Answers 2


You need:

  • an equation for the current through your line, in terms of VAN. (Hint: the voltage across the line is 240 - VAN. Use Ohm's law.)
  • an equation for the current through the combined load of L1 and L2, in terms of VAN. (Hint: Consider L1 and L2 as a single load absorbing 630W and 840var. Use S=V*I)

These two currents are equal so you should be able to solve for VAN.

  • \$\begingroup\$ Isolating VAN I have this: VAN = Van - (Ra+jXLa).(SL/VAN)* --- Wolframalpha analysis: --- adf.ly/eBn3r -- same no solution... I don't know how, cause every quadratic equation has solution... \$\endgroup\$ Commented Feb 28, 2014 at 14:29

As @Li-Wen Yip indicated in the answer, you can match the two equations below: $$i=(630+j480)/VAN$$ $$i=(240-VAN)/(5+j8)$$ You end up in the equation: $$VAN^2-240VAN-(3570+j9420)=0$$

You can use Bhaskara's formula to solve this equation. Apparently, there are two valid solutions.


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