 0

# How can the power factor be calculated without using a power factor meter?

Copy

Use a voltmeter and an ammeter to measure the supply voltage and load current; the product of these two readings will give you the apparent power in volt amperes. Use a wattmeter to measure the true power of the load, in watts. Divide the true power by the apparent power, and this will give you the power factor.

-----

Try using an osciloscope, connect the voltage to one channel and the other to the current (CAUTION: provide proper shunting) You'll see the two wave forms, the distance between them would give you the angle between the voltage phasor and the current phasor, the cosine of this angle in degrees is the power factor.

----

You can calculate the power factor if you have a meter that measures voltage in volts and current in amperes and have access to a wattmeter that measures power. Power factor can then be calculated by this formula:

Pf=P/S , S=V*I where S is apparent Power.

Power Factor, simply put is the relationship between real power (Watts) and reactive power (VARS). It isn't related to efficiency, at least not in terms of the ratio of output power/energy to input power. A motor might have a power factor of 0.87 (30 degree phase angle), but an electrical efficiency either *more* than 87%, or *less* than 87%.

Since the input power to a motor should properly be measured as *real* power, the power factor is not considered in calculations. One reason is that the motor's power factor (which is most likely inductive) can readily be corrected back to unity (1.0) by either adding a parallel capacitor (as is done for the inductive ballast coils in fluorescent lights), or by installing a synchronous motor in the circuit and adjusting the amount of excitation.

Strangely enough, a synchronous motor can be made to appear either inductive (like other motors) or *capacitive* according to its excitation power. Such motors are often used for constant-service applications such as airconditioning.

-----

THIS DOESN'T MATTER IN HOUSEHOLD ELECTRICITY. If a factory has a Power Factor of 95 %, then it will draw 105 % of the current it would draw if it were at 100%, or a Power factor of 1 (also called unity).

The electric company charges a customer more for ineffecient systems, i.e. Power Factor lower than 1 (100 % efficient).

The "Power Factor" is the ratio of volt-amps to watts. To get volt-amps, you also multiply volts times amps. With a resistive load, such as an incandescent lamp, volts times amps equals watts. All of the power gets dissipated heating up the lamp filament to make it glow. In this case, volt-amps is equal to watts, giving a ratio of 1:1, or 100 %. With inductive loads like Transformers, electric motors, fluorescent lamps, etc., there is very little resistance. Something called "reactance" limits current flow. Larger currents flow with little power being dissipated. With a power factor of 50 %, double the current would flow. For example, a 40 watt incandescent lamp draws 0.33 amps. (40 watts / 120 volts = 0.33 amps) This bulb, being a resistive load, has a power factor of 100 %. A single tube fluorescent lamp rated at 40 watts may draw double the current of the 40 watt incandescent, but still only use 40 watts of power. This fixture has a power factor of 50 %.

-----

Home meters

Electrical meters for homes measure only resistive (real, apparent, or actual) power. They do not measure reactive power.

In the study of alternating current, it will be observed that there are alternating waves of both voltage and current. In a circuit with purely resistance load, the waves of current and voltage are in exact phase relationship to each other. This means that when the voltage is at its peak, the current flow is at its peak as well. An inductive load (that is, a coil) causes the current wave to lag or fall behind the voltage wave, so that the peak current flow is some time after the voltage wave is at its peak level. A capacitive load (that is, a capacitor) causes the current wave to lead or advance ahead of the voltage wave, so that the peak current flow is some time in advance of the peak of the voltage wave.

The consequence of this is that the AVAILABLE REAL POWER is the relationship between the current and voltage waves.

Resistive circuits have a power factor of 1.0, or unity, because the waves are in phase.

The more out of phase the relationship between voltage and current, the less efficient the use of available power, the more "waste" energy.

The less efficient the use of energy, the larger the size of transmission and generating equipment required to provide for energy needs and the more costly the operation of utilization equipment.

Scroll down to related links and look at "What is reactive power?"

No improvement to these good answers. I would just add that single phase power will rarely veer from unity. If you are dealing with household electrical service, you are likely to be at or near unity.

"... likely to be at or near unity..."

Umm... no.

Motors use magnetic windings, and are therefore inductive. Inductive components reduce the power factor below the ideal value of 1.0. Old style fluorescent lights use ballast chokes, which are also inductive. A correction capacitor in the housing will correct the PF back to 1.0, but if the capacitor is faulty/has been removed, the fluoro will also give an inductive power factor. In fact, *most* loads, other than heating elements, are inductive.

So you start up the washing machine, dishwasher, benchtop mixer and electric drill. It is unlikely that any of these appliances is power-factor-corrected. Your PF will drop to less than 1.0. Our electricity supplier does all of their house load calculations based on a PF of 0.8.

"...waste energy..." Yes and no. The extra current does cause extra resistive loss in transformers, cables and switches, but the real problem is back at the generator, where more current must be generated than would be needed by a system with a perfect PF of 1.0.

🙏
0
🤨
0
😮
0
😂
0