How does a Low Power Factor draw more current?

Question

In the course I'm reading, there's this example to show that a low power factor leads to more current being drawn, and thus more power lost in the transmission lines.

Assume a certain load requires 1 kW to function correctly.

If the power factor were 0.9, and the power supply gives 110V, we would need 10A. If the power factor were 0.6, and the power supply gives 110V, we would need 15A.

However, something here just doesn't make sense. If we consider the power supply as an ideal voltage source, then the current has nothing to do with the given power.

It seems that he just assumed the load would draw the required power to function properly no matter what.

On the other hand, it seems to me that the current would stay the same, being affected by the impedance of the load, leading to a lower real power and thus a malfunctional load.

Where did I go wrong?

Answer 1

However, something here just doesn't make sense. If we consider the power supply as an ideal voltage source, then the current has nothing to do with the given power.

• The in-phase current has everything to do with the power. \$ P = VI \$.

• The out-of-phase current does not contribute to the power but does contribute to the overall current drawn from the supply. The increased current causes an increased voltage drop along the lines giving poorer voltage regulation at the load. This is a bad thing!

It seems that he just assumed the load would draw the required power to function properly no matter what.

• It will draw the required power and this will be measured in kW and billed in kWh.
• It will also draw reactive power measured in kVAr and billed in kVArh. Typically the utility will charge for excessive kVArh.

On the other hand, it seems to me that the current would stay the same, being affected by the impedance of the load, leading to a lower real power and thus a malfunctional load.

No.

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Links:

See my answer to How do capacitors improve power factor without destroying the capacitor?. It may help.

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Simulation

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A complicated struggle with CircuitLab's AC simulation capabilities.

Note that the values of L3, L4 and C2 have been chosen to have an impedance of 1 Ω at 50 Hz. The parallel R5-L3 pair should be out-of-phase by 45°. L4 and C2 should cancel out as their impedences are the same at 50 Hz. The 1 mΩ resistors are to help the simulator settle down after the initial transients.

Figure 2. The current waveforms for Figure 1. Note that the brown should be superimposed on the blue. The simulator is struggling.

• The voltage plot has been omitted for clarity but the blue curve is in phase with it and will be taken as the reference.
• The current for circuit (a), the resistor, is plotted in blue. As expected 1 V into 1 Ω gives 1 A peak.
• The current for circuit (b), R5 and L3, is plotted in orange. Notice that the orange curve crosses the blue curve exactly at its peak (90° into the blue cycle). This occurs because the inductor current is lagging the voltage by 90° and is zero at this point. As the inductor current continues to rise the resistor current is falling. The peak occurs at 45° into the cycle where both currents are at \$ \frac {1}{\sqrt 2} I_{PEAK} \$. It follows that the peak of this waveform will be \$ \sqrt 2 \$ times that of the blue.
• Finally note that if we add a capacitor of equal impedance to the inductor as in (c) the brown curve that we are back in phase again and the current has reduced. (I was expecting this to overlap the blue curve. I suspect the simulator is struggling.)

In summary:

• The generator is constant (AC) voltage output.
• The generator provides the current required by the load.
• If the load is reactive (not purely resistive) the current increases while the useful power transferred remains the same.
• By balancing the inductive and capacitive elements in the load the power factor can be brought back to unity.

Answer 2

Assume you had an ideal capacitor connected to your AC mains. It draws a current because it has an impedance of $$Z_C=\frac{1}{2\pi\cdot f\cdot C}$$ For example, a 1µF cap on a 230V, 50Hz mains draws $$I_C=\frac{U}{Z_C}=230\,\mathrm{V} \cdot 2\pi\cdot 50\,\mathrm{Hz}\cdot 10^{-6}\mathrm{F}=72\mathrm{mA}$$

This current leads to pure reactive power. The voltage is out of phase 90°, so there is no actual power, cos phi is zero, sin phi is 1.

$$Q_C=230\,\mathrm{V}\cdot 72\,\mathrm{mA}=16.6\,\mathrm{var}$$

Now connect a 40W incandescent light bulb in parallel. It has no reactive power:

$$I_B=\frac{P}{U}=\frac{40\,\mathrm{W}}{230\,\mathrm{V}}=174\,\mathrm{mA}$$

What is the current, in sum? The answer is

$$I=I_C + I_B=246\,\mathrm{mA}$$

And the appearant power?

$$S=U\cdot I=230\,\mathrm{V}\cdot 246\,\mathrm{mA}=56.6\,\mathrm{V\!A}$$

If you had the cap and the light bulb in a blackbox, know the voltage and could only measure the current, what would you tell it is?

Answer 3

If we consider the power supply as an ideal voltage source, then the current has nothing to do with the given power.

It is ok to consider power supply as an ideal voltage source, but the current has everything to do with the given power. The characteristics of the load dictate what the current and power factor will be with a given supply voltage. An ideal voltage source will do whatever is necessary to accommodate that demand without changing the voltage.

A load that is composed of resistors capacitors and inductors with a given applied voltage will draw a specific current at a specific power factor. The power factor will not change unless something is added or subtracted from the load. That change could be a malfunction or it could be a normal change in the load.

An induction motor load can be modeled as some inductors, a transformer and some resistors. The mechanical load is modeled as a variable resistor. If the mechanical load changes, the power factor changes due to the change of the variable resistor. There is also a small change in power factor due to a change in the proportion of current in the inductive components of the circuit. That is a normal change not a malfunction.

The an increase in the reactive portion of the load does not cause the generator to work harder except to the extent that the losses increase due to the increased current. Reactive power is energy circulated continuously between the load and the source. The prime mover needs to supply only the losses associated with that energy circulation.

Answer 4

This is a short answer, so I'm sure another one will come along soon with nice diagrams, but:

In an ideal situation, the current and voltage wave forms are in sync (power factor 1). This gives a power of X. If we take those same exact waveforms and start to shift them out of sync, the power level will decrease. If we need the same power (x), with the warriors out of sync, we need to increase either the volatge or the current. We usually want to keep voltages constant, so instead we increase the current. This means that, as wave forms go out of sync (I.e. the power factor decreases), we need to up the current to keep the power the same.

Edit: oops, a minute late

Answer 5

At any moment in time, the amount of power being transferred through the wires between a source and a load will be the instantaneous voltage times the instantaneous current. If a device has a power factor of 1, all power that is transferred will be consumed by the load. If a device has a power factor of zero, all of the power that is transferred to the load during one part of a cycle will be stored and then transferred back to the source during another part. If a device has a power factor between zero and one, some of the power that goes through the wires will drive the load, but some will be stored up and fed back through the wires.

Power which flows from the source to the load and back on each cycle won't do any useful work, but the current will still flow through the wires in the process of transferring that power. Transferring a given amount of "useful" power will require transporting enough current to carry that, plus additional current to transfer the power that keeps going back and forth.

Answer 6

Here's an analogy: If you're working behind a cash register the most efficient transaction is when everyone pays the exact amount in cash. That's power factor 1.

If everyone needs change the transactions are less efficient even though the accounting ends up the same. That's power factor less than one.

On the transmission line, the extra current isn't consumed, it "resonates" on the line where it generates parasitic I^2R losses and makes the transmission line less efficient.

Answer 7

Instantaneous power is voltage times current. The voltage defines the potential difference a current has to be put through: it's like some water current: the current does not eat or deliver energy as such until you look at the height difference you want to cross.

With a pure resistive load, the current is proportional to the voltage. Now we are talking about AC here, so the voltage and current and instantaneous power change all the while. When they are in synch, the total energy needed for a longer amount of time is maximized. When they aren't in synch, the load is not purely resistive (which would make it dissipate power). Instead, there is inertia involved which then stores energy momentarily and gives it off later. The voltage source no longer only delivers power but has moments where its current and voltage have different signs and thus it receives energy back that it has put out previously.

In the extreme, a load factor of 0, voltage and current are out of synch by 90 degrees and no energy is actually consumed but only pumped back and forth. The water equivalent is a pail of water on a swing: it passes back and forth and changes height, but its direction and its height are out of phase, so no actual energy is converted into heat.

Now any wires are basically resistive and the voltage across them (rather than between them) is in synch with the current through them. So they incur just the same amount of power loss at the given current as if the power factor were 1. At the same point of time, you only pay for the power loss inside of your home: the losses outside are not metered.

One can actually use large synchronous motors as capacitors or inductors depending on how large you make the electric field of the rotor. The energy is stored here in the rotating mass, so you want a bit of mass on the axle, and you want the machine to be able to deal with the resulting torques and angular vibration.

Answer 8

Do not consider it as a voltage source, consider it as a power source.

The fact that a load is taking it's peak current at a different time from the peak voltage means the generator needs to supply a higher current to send the same amount of real power.

The generator does not know it has to deliver such and such a power, the load pulls whatever real power it needs. That drawn power tends to brake the generator, and the generator then consumes more power from the prime mover to work harder to keep up to speed.