'Reactive Power', which is expressed in reactive volt amperes, describes the rate at which energy is alternately stored (in a circuit's electric or magnetic field) and returned to the a.c. supply when the field collapses. It differs from true power, expressed in watts, because true power describes the rate at which energy is permanently lost by heat transfer due to the resistive component of the circuit.
Reactive power doesn't 'have an use', per se, it's merely a way of quantifying the movement of energy in the reactive component of an a.c. circuit.
The vector sum of a circuit's reactive power and its true power is called the apparent power of the circuit, expressed in volt amperes.
Use a wattmeter, as it only reads 'real power' of your load. Use an ammeter and a voltmeter, and the product of the two readings will give you 'apparent power' of your load. Since apparent power is the vector sum of real power and reactive power, use the following equation to find the reactive power of your load: (reactive power)2 = (apparent power)2 - (real power)2
current that does't use in power measurement is called reactive current.
Reactive power is useless part of the energy supplied. A minimum value of reactive power is always required to keep the voltage constant and supply the useful active power. In order to understand the concept completely, do refer PV and QV curves which explains the interdependency of active power, reactive power and voltage.
Real power is power that is used to do work. This is the power a light bulb uses to glow. Reactive power is power that is stored in one part of the AC voltage wave, and released in another. This causes the power to appear as a phase shift, and is generally caused by inductive or capactive loads. With regard to the power system, you can also get reactive power flowing due to unequal source voltages. Under these conditions, the apparent impedance of one source to the other will appear reactive/inductive, resulting in reactive power flow.
There are various ways in which you can determine the reactive power (in reactive volt amperes) of a load. From the practical point of view, you can use a voltmeter and an ammeter and use the product of their readings to determine the apparent power (in volt amperes) of the load, and a wattmeter to determine the true power (in watts) of the load, then find the vectorial difference: (reactive power)2=(apparent power)2-(true power)2
Apparent power is the vectorial sum of the true power and reactive power. In this case, the total reactive power is the difference between 7200 var and 3600 var -i.e. 3600 var.So you can now use the equation,(apparent power)2 = (true power)2 + (total reactive power)2,to determine your answer.
only the appliance power
You use power factor when the load is not resistive, i.e. when it is reactive, and the phase angle between voltage and current is not zero.
volts
Imaginary power; i.e., the power that flows into a reactive circuit. "Kilo" means thousands; "var" is short for volt-ampere reactive. Since it measures power, you could just as well use kilowatts.
An ammeter and a wattmeter are the instruments needed.
hydroelectric