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As far as the load is concerned, they don't. But, as far as the supply lines are concerned, they have to cater for an increase in load current and, therefore, increased line losses, if the reactive power of the load increases.
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∙ 12y ago
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Difference between real and reactive power?
real =VIcosO reactive = VI sino
Why the power factor of a transformer is poor at no load?
The inductance of the transformer is much higher than the resistance of the transformer, resulting in very low real power losses (in watts), but some reactive power (vars).
Is current reduces with increase in powerfactor?
If power factor is increased, current will be reduced for a specific real power (kWs) relative to before. Total power is real power plus the vector of reactive power (you have to do polar math). So total current = (current from real power) + (current from reactive power). By changing the power factor, you decrease the second quantity. Go to Wikipedia.com and search for Power Factor if you need a more extensive description.
Is reactive power greater than true power?
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
Why VAR meter does not measure real and apparent power?
A VAr meter only measures the reactive (imaginary) power. Apparent power is a combination of real and reactive power; thus having a VAr meter will not suffice to measure apparent power. Likewise, because the VAr meter only measures reactive power, it does not provide any information on real power.
What is real power in electricity?
real power is to do do work and it is supplied to the load ,where as the reactive power is used for magnetising and the apparent power is the total power
What is commercial power factor?
Electrical inductive motors, transformers and magnetic ballasts bring real power that are measured in kilowats and reactive power (measured in kilovolt-amperes reactive, kvar). Real power genrates "productive" work. Reactive power does the the magnetic field required for inductive devices to operate.
What is happen if real and reactive power is absent?
There is no load connected!
What is reactive power and why you call like this and what are disadvantage of this power?
Reactive power is a result of capacitance or inductance, which are reactive elements. This is power that is not useful for doing work - these reactive elements hold power, then release it during each half cycle of the AC waveform. This causes a phase shift from the real power. The main disadvantage is it's not useful for doing work, and you have to size equipment for it anyway. This is why much electrical equipment is rated in KVA, which is the vector sum of real power and reactive power, instead of kW.
Power factor formula?
Power factor is defined as the ratio of real power over total power. Total power is the vector sum of real and reactive power.
What is the Reactive power in a capacitive modeand inducitive mode?
Generators can be required to generate real and reactive power. When operating in a leading mode, the generator is generating real and leading reactive power (inductive power). This means the generator is "sucking in VARS", which will pull down the terminal voltage similar to an inductor. It can also be operated in a lagging mode, which means it is generating real and lagging reactive power (capacitive power). The generator, then, is "pushing out VARS" like a capacitor, which will cause the terminal voltage to increase. Generators can only create so many leading and lagging VARs; in general lagging VARs are limited by the automatic voltage regulator output capabilities; leading vars are limitted by how much heat the stator can dissipate.
What does 193 kva set mean?
KVA is an acronym for killo volt - amperes. It is a measure of power (both real and reactive) defined as P = Vrms * Irms. If there is a power factor associated with this, it can be used to separate the real and reactive power:P(real) = Vrms * Irms * pfP(reactive) = Vrms * Irms * (1 - pf)
