The power required by the appliance is directly proportional to the current,voltage, power factor. If the power factor is low, more current is required to supply the rated power of the appliance hence the ohmic losses increase. Therefore the efficiency decreases and the voltage regulation increases which is bad for the power company as well as the consumer.
In the study of alternating current [that which supplies our homes and businesses in the United States], 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 it's peak, the current flow is at it's peak as well.
An inductive load [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 it's peak level.
A capacitive load [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.
The power factor of a device is what determines how much useful power is used out of the total amount of power which is supplied to it from the source.
A power factor as close to 1 as possible is desirable because then most of the power transferred from the source to the load is useful power.
1. If a device has a power factor much less than 1, that means more total input current must be supplied for a given output power dissipation and a more powerful source is required to deliver the required output power. This means the device must draw a higher amount of volt-amps (VA) compared to the actual load power it is delivering, which means its conversion of input power VA to output power VA is inefficient.
2. The closer to a power factor of 1 that a device has, the better the total current which has to be supplied will match the output from the device, and the more efficient it will be in its conversion of input power VA to output power VA.
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Not all the loads in an electrical installation operate at their nominal power all the time. A good example are the plug sockets which are rarely used at their maximum output (which in this case would be the limit of the protective fuse of 10 or 16 amps). So, in order to get a realistic image of the power demand, there is also a simultaneity factor applied. But each tipe of consumer can have its own factor depending on the tipe of load, ratio of its usage and also on the interpretation of the engineer. Coming back to the example with sockets, usually a coefficient of 0.2 is taken into consideration but for other loads that are operating at their nominal current/power(like lighting) a higher factor up until 1 would be assumed. By multiplying the nominal power with simultaneity factor you will get the power demand.
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No-one ever aims to reduce the power factor, the ideal power factor is equal to 1, and that is the maximum possible value. A load with a power factor of 0.7 draws 40% more current along the supply wires compared to a equal-power load with a power factor of 1. That means that the power loss in the resistance of the supply wires is doubled in the case of the poor power factor. Since the supply company receives no extra revenue for the lost power, it does not like this situation and sometimes penalises users with poor power factors with extra tariffs. The power factor can often be improved by placing a passive reactor in parallel with the load to draw off the reactive volt-amps (VAR or kVAR) so that the supply wiring sees a load with a good power factor. Normally a bad load like a motor draws inductive VARs and in this case it can be corrected with a parallel capacitor that draws an equal number of capacitive VARs. Looked at another way, the added capacitor 'tunes' the load to resonate at the supply frequency.
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The power in watts equals the VA times the power factor. For a resistive load like a convector heater or an iron the power factor is 1 For other things like motors the power factor might be 0.7. A poor power factor is not a good thing because more current is needed from the supply to produce a given amount of power, so that requires thicker wires (more expensive). For a power factor of 1, 70 kVA = 70 kW For a power factor of 0.7, 70 kVA = 50 kW.
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It's all to do with trigonometry. Apparent power is the vector-sum of true power and reactive power, and you can think of these three quantities forming a right-angled triangle (an oversimplification, but good enough for our purposes). Power factor is defined as the ratio of true power to apparent power, which corresponds to the ratio of adjacent to hypotenuse of this right-angled triangle. This ratio is called a cosine.
In an ac system the kVA is the product of the kV and the amps. But in some loads such as electric motors the current is not in phase with the voltage so some power flows from the load back into the supply for a fraction of the ac cycle. That means that the average power flow in kW is generally less than the kVA by a factor known as the power-factor. The ideal power factor is 1 and power factors of less than about 0.8 are not good because the current supplied, and therefore the power loss in the cable, is more than necessary for the given amount of power. In many cases the power factor of high-power loads can be corrected by adding special components.