That is true.
In the case of DC, and in AC when current and voltage are in phase, a volt-ampere is the same as a watt (and therefore, a kilo-volt-ampere is the same as a kW). In the case of AC, when current and voltage are NOT in phase, power = voltage x current x power factor; the power factor is the cosine of the angle between current and voltage, and it is always less than or equal to one. In such a case, a kVA would be less than a kW.
What is the question for this situation
It SHOULD always be less than the divisor... Otherwise your answer is wrong.
A kVA is basically the same as a kW - unless the so-called "power factor" is much less than one. However, it is quite often close to one, so it can be ignored. (Power = volts x amperes x power factor.)
Power Factor = KVA/KW. This has no unit. Its value is always 1 or less.
It can supply 2.5 kW to a load with a power factor of 1. Otherwise the power available must be multiplied by the power factor which is always less than 1. If uncertain assume 0.8 which makes it 2 kW.
Power factor correction is a process that has to do with managing the essential traits of electric loads that are used to create a power factor of less than one.
Power - Watt. Energy - Watt Hour. Power factor - no unit. just number less than or equal to 1.
No relation is there power factor is a unit less quantity.
Always equal to or less than the smaller number, yes.
i know that static capacitors are used to improve the power factor. power factor should be high. Static capacitor supplies lagging reactive power. That means; the current I has 2 components they are magnetising Im (watless or waste current) and useful current Iw. Iw is in phase with voltage and Im is 90 degree away. Phase angle between them is phi 1. power factor is given by cosine of phi 1. phi angle should be less so that cosine of phi is high. To make phi angle less we use capacitor; this is nothing but power factor correction and capacitor used for this is called power factor correction capacitor. now when a capacitor is connected, it induces a current Ic 180 out of phase from Im and less in magnitude from Im. therefore, now the magnetising current is Im1=Im-Ic. due to this the phase angle reduces to phi 2. now the new power factor is cosine of phi 2. it is improved power factor.
With a dc system the kW are always equal to the kV times the amps. It's only with ac that the kW are usually less than the kVA by a factor called the power factor.
i will try my best to answer this quistion, but u must not mind if i make a mistake! >>> the rating of an elctrical machine depend upon the loses in it. if, there are any losses in the machine due to power factor than the machine will b rated in KW and if there is no loss due to power factor than the machine is rated in KVA. so there are no losses in a transformer due to power factor so it is rated in KVA. as the KW= KVA* power factor so, kVA= KW/power factor here, KVA=100 so, KW= 100*power factor u can derive from here that the load on a transformer depends upon the power factor. as the power is always less than unity so the load will be less than 100KW. thankyou!
Is equal to or less than 1
481 amps if the load has a power factor of 1, but if the power factor is less than one (e.g. if it's a motor) you also have to divide by the power factor.
It can supply up to 1.5 kW into a load with a perfect power factor like a convector heater, or less if the power factor is less than 1. For example a motor might have a power factor of 0.75, then only 1125 watts could be used, in other words about 1.5 horse-power.