When we have a fault, say on one line out of three, the reactive power demands increase, which leads to voltage instability if the source can't provide it.

My question is, what causes reactive power demand to increase? Reactive power depends on frequency and frequency in power systems is always fairly constant. This concept is quite confusing to me.

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  • \$\begingroup\$ Reactive power depends on power factor as much as it does on frequency. If you have a power factor of unity you'll have no reactive power. If you have a power factor of 0 all of your power is reactive. \$\endgroup\$ – Tom Carpenter Jul 12 '18 at 10:17
  • \$\begingroup\$ @tom carpenter yes, but having more load on any line should not increase its reactive power demand or should it? Here as described in picture above, a fault on one line causes load on others and thereby increases its reactive power demand. \$\endgroup\$ – Jason Jul 12 '18 at 10:21

The reactive power loss in any short, lossless transmission line is given by,

\$\Delta Q= \frac{P_{flow}^2+Q_{flow}^2}{V^2}\times X\$

where \$P_{flow}\$ and \$Q_{flow}\$ are the active and reactive power flow in the line.

Now let us analyse


The power flowed from bus1 to bus2 is divided on 6 lines. Which means that the power flow in each line is 1/6 of total power (\$P_{total}\$), assuming that all lines having same parameters and length. This \$\frac{1}{6} P_{total}\$ flowing in each line will not lead to much voltage drop and reactive power losses in each line.


When three lines are tripped, the total power \$P_{total}\$ will be redistributed among all lines. Again, assuming that all lines having same parameters and length, the share of power flow in each line \$\frac{1}{3} P_{total}\$ which means double of power flow, and hence (from equation above) almost \$2^2=4\$ times more reactive power losses compared to pre-fault case.

Moreover, the voltage drop across the lines will increase as total impedance of the lines is significantly increased (doubled in case of same parameters and length) after fault occurred compared to pre-fault case.

  • \$\begingroup\$ Hi Hazem, thanks for your help. Its really very helpful and insightful. There is a little bit confusion in my mind and i would be delighted if you help me with it. When we lose three lines, the impedance should decrease because each line had its own resistance and inductance. And moreover after fault when power distribution became 1/3 the inductance of lines were still the same for each line so why does the reactive power demand increased? I mean active power flow doubled but reactive power flow should have stayed the same for each line. Can you correct me where i am confusing it please? \$\endgroup\$ – Jason Jul 12 '18 at 10:52
  • \$\begingroup\$ NO. When we lose three lines, total impedance increases. Remember these are parallel lines (impedances), tripping any one of them would increase the total impedance. So total reactance will be two times of pre-fault not 1/3. Regarding second confusion, check the equation in the answer. When \$P_{flow}\$ increases, reactive power losses increases (and do not forget the square), which you are translating to "reactive power demand". \$\endgroup\$ – Hazem Jul 12 '18 at 11:14
  • \$\begingroup\$ Thank you alot Hazem. I am clear now. It was very helpful. Please suggest me some books or sources that could be helpful in grasping these concepts more in detail. Thanks once again \$\endgroup\$ – Jason Jul 12 '18 at 11:35
  • \$\begingroup\$ You are welcome @Jason. I'm glad that the answer helped you. If you want to build a strong background in power system I think Stevenson and Saadat books are the best sources to start with. \$\endgroup\$ – Hazem Jul 12 '18 at 11:59

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