The power loss of circuit is \$16kW\$ when the power factor, PF, is 0.6 lag. Now if I improve the PF from \$0.6\$ lag to \$0.8\$ lag, what is the power loss now?

The solution provide me the formula below

\$P_{L2}=P_{L1}\times \frac{I^2_2}{I^2_1}=P_{L1}\times \frac{(cos\theta_1)^2}{(cos\theta_2)^2}=16\times \frac{0.6^2}{0.8^2}=9kW\$

And I have two thinking about this question, however, I don't know where am I wrong, because the answer are both wrong!

Method 1

If we draw a relation between apparent power and real power

\$\frac{6}{10}=\frac{x}{16k},x=0.6 \times 16k=9.6k VA\$

now if we improve the PF to \$0.8\$,that is \$\frac{8}{10}=\frac{9.6k VA}{y},\$ so \$y=\frac{9.6k VA}{0.8}=12k\$, so the power loss is \$12kW\$ when the PF is 0.8

enter image description here

Method 2

As the question shown,the power loss is 16kW when PF is 0.6, that is, \$IVcos\theta=IV0.6=16k\$,so \$IV=\frac{16k}{0.6}\$

Now the PF become 0.8, that is, power loss\$=IVcos\theta=IV \times 0.8=\frac{16k}{0.6}\times 0.8=16k\times\frac{4}{3}=\frac{64}{3}W\$

Now as we know, the solution of my both thinking are wrong, does anyone know where is my calculation or thinking wrong?

  • \$\begingroup\$ Have you stated the question exactly as you found it? Is the question about a circuit that is described I'm more detail in a previous question? \$\endgroup\$
    – Gone Off
    May 6, 2020 at 15:34
  • \$\begingroup\$ I think you need to be much clearer about "power loss" - for instance, how do you distinguish between power loss through cabling and power delivered to a load (which can also be called a loss in terms of electrical energy that disappears into heat or mechanical work). \$\endgroup\$
    – Andy aka
    May 6, 2020 at 15:44
  • \$\begingroup\$ @Andyaka The power loss from the wire \$\endgroup\$
    – Shine Sun
    May 8, 2020 at 4:52

2 Answers 2


In both the methods the answer is 21.333 There is a mistake in the calculation of the first method because of 0.6 = 16k/x and not x/16k

Now coming to the solution

As far as I know, there is no direct formula linking power factor and power loss. Either there is something missing from the question or missing in the assumption.

If the power factor increases the True power of the system should increase and the apparent power should decrease and the loss depends on the circuit given.


The solution provided is explained by the following reasoning:

You are given the losses, initial power factor and final power factor. If this is a power transmission problem, the losses are a relatively small percentage of the total power. Losses are the current squared multiplied by the resistance of the current path from the source to the load. If the power factor of the load is increased, the power remains unchanged but the volt-amperes is reduced as shown by the power triangle below. Since the voltage is not changed, the current ie reduced in proportion to the volt-amperes. The losses are reduced in proportion the the ratio of the square of the final current to the square of the initial current. There may be a small error in the calculation depending on the effect on the load caused by the reduced voltage drop in the transmission line. Since the loss is a small percentage of the total power, and the exact value is unknown, that effect is neglected and the simple calculation is sufficient.

enter image description here

  • \$\begingroup\$ but it seems that according to the explanation from your figure,no matter how much the real power or apparent power change,the loss power won't change \$\endgroup\$
    – Shine Sun
    May 8, 2020 at 4:36
  • \$\begingroup\$ The figure doesn't show the change is loss, only that is small compared to the useful power. The loss is proportional the the square of the current because it is due to the load current flowing through the distribution system resistance. The load current is proportional to the load VA. The loss is therefore calculated using the formula that you say was provided to you. \$\endgroup\$
    – Gone Off
    May 8, 2020 at 15:25

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