# Power Factor Correction increases circuit total power?

I'm learning the concept of power factor and power factor correction, but I am having trouble to understand some points. Every book/article about power factor says something similar to:

Power factor can be an important aspect to consider in an AC circuit, because any power factor less than 1 means that the circuit’s wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same amount of (true) power to the resistive load.

So, I understand that when we are talking about a resistive load that is forced to receive the same amount of true power, increasing the load's power factor will indeed be a good improvement to the circuit, since the current that flows through the load will be reduced at the same time that the original true power doesn't change.

However, I draw a circuit where the power applied to the load is variable and, in that case, increasing the power factor made the circuit's total power be increased too. The circuit is below:

Original circuit:

Circuit with power factor correction:

So, it is easy to see that fixing the power factor made the circuit dissipates more true power.

However, the power that was being lost in the "transmission" was reduced. I understand that reducing the power loss in the transmission is maybe the most important point about power factor correction, but I always read that correcting the power factor is important to make the circuit more efficient and, also, reduce the price charged by electricty.

In that case, wouldn't the price charged be even higher? Even though the power is now being directed to the load instead of being lost in the transmission, I can't see how correcting the power factor would help when talking about reducing the eletricity bill.

I'm not sure if my question is clear, but any help is appreciated. Thank you very much!

• just to be clear, you are referring to displacement power factor
– user16222
Commented Nov 21, 2015 at 20:18
• @JonRB yes, sorry. I'm not very familiar with the subject. Commented Nov 21, 2015 at 20:20
• btw. this site has a built in circuit editor Commented Nov 21, 2015 at 20:26

Improving the power factor reduces the losses in the wires bringing power to your load. That in turn might increase the power delivered to your load -- so would changing the voltage supply.

If your load has an automatic adjustment, it will accommodate. If it is something like a water heater, it will also adjust -- the heating cycles will become shorter than before because they will operate at a higher power level. Ultimately, the wasted power (in the wires) will be the savings.

Note in most home and office installations, you don't pay for apparent power -- just real power. In industrial installations, you may pay for apparent power. Regulations in some countries require loads above a certain level (typ. 50 W) to have power factor correction. This would typically affect computer power supplies and televisions.

• By "automatic adjustment", you mean that the load will adjust its internal impedance to reduce the true power to the original value? Commented Nov 22, 2015 at 14:44
• No, I mean a regulator -- e.g. a thermostat. Your load (e.g. water heater) generally requires a certain average power, but the thermostat cycles on/off to regulate the temperature and deliver this average in chunks. When switched on, if the mains power is 'high', it will take a shorter time to deliver a certain amount of energy, so the on-time will be shorter. Many loads are like this -- refrigerators & most electronic equipment -- they require constant average power and will adjust their load to (on average) consume the same power. Incandescent lights are not -- brightness isn't regulated. Commented Nov 22, 2015 at 16:49

I ran the calculations and your only accounting for the real power in both cases. The power your paying for is apparent power, that consist of both real power and complex power.

In the first case the complex power is $\tilde{S}=\tilde{V}\tilde{I}^*=1132.3 + 587.5i W$. The apparent power is $\sqrt{1132.3^2 + 587.5^2} = 1275.5W$ This would be the power your paying for.

In the second case the complex power is $1149.8+18.3i$ as you can see the complex component is much smaller. This gives and apparent power of $\sqrt{1149.8^2+18.3^2}=1150W$ This would also be the power your paying for.

This is a reduction in power of 125.5W

Big companies (not homes) get charged for apparent power instead of real power because of the added current needed for apparent power. The power company needs a bigger infrastructure to provide that extra current capability to the customer. Adding more capability to the grid costs money and the power company passes the cost to big companies by charging them for apparent power instead of real power.

The example you gave is just one motor and the power factor was not that significant. Large factories for example can have hundreds or thousands of motors running at the same time. This can cause the power factor to be very significant making the current needs extremely high compared to what would be required to just feed the real power.

Feeding the reactive power back to the source sounds good in theory but the source does not want it back. What we are referring to as the "source" is usually is a very large generator. Returning excessive reactive power to these generators may damage them because the reactive power causes the generator to motor instead of generating interfering with the rotation of the generator.

• You are mixing up real power [Watt] and apparent power [VA] which makes thigs very confusing. Also your average household is charged for actual used W (hours), not for VA (hours). Only large consumers are charged for VA. Commented Nov 22, 2015 at 10:37
• @jippie Correct only large consumers are charged for VA, they are also the ones using power factor correction. Average households only pays for W because they normally only have a few devices that can skew the power factor and they normally don't correct it. Commented Nov 22, 2015 at 11:39
• @vini_i I thought that a big apparent power would increase the bill because it would normally cause a bigger current flowing through the transmission. (When comparing with a circuit that dissipates the same real power, but less apparent power). However, in the example circuit, reducing the power lost in the transmission didn't compensate the new power added to the load. Could you explain why a bigger apparent power would result in a bigger bill in the example? The extra reactive power will be fed back to the source, so I don't see a reason why this would be charged Commented Nov 22, 2015 at 14:48
• @felipeek i updated my answer. Commented Nov 22, 2015 at 16:01