That's the ratio between true power, and apparent power. It applies to AC. Apparent power is voltage x current, and in the case of DC, this is also the true power. In the case of AC, it is possible that voltage and current are not in phase; in the extreme case that they are 90 degrees out of phase, they true power is zero (instantaneous power will be positive for part of the cycle, and negative for another part of the cycle).
how to calculate the powerfactor in capacitor
its nothing but a lagging powerfactor...its due to inductive loads......
armature , electrostatic , electromaganetic , commutator,dampe winding,dashpart,voltage ,powerfactor etc..
the powerfactor actuator are available in market. Use them with ADC to switch required capacitors. I hope you can work out with this
It depends on the application the desgn is for, but 0.8 is the most commonly used powerfactor for general design especially home use.
Volts = current (In amps) x Resistance (In ohms) Watts = Volts x Current x PowerFactor Power Factor = 1 in a pure resistive circuit
It varies based on the ballast. Most ballasts nowadays are "power factor corrected" to try and compensate for the bad powerfactor in switching power supplies. The actual power factor of a individual fixture is entirely dependent on the circuit of the ballast, which changes from design to design.
An ampere-hour is one ampere for one hour.A ampere is one coulomb per second, so an ampere-hour is a transfer of charge of 3600 coulombs.See the discussion page.
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.
Depends on the supply voltage. For UK 230 Volts, 1800 Watts would mean the current drawn was 7.83 Amps.Watts=Volts x Amps, so Watts divided by Volts gives the current in amps.
since at no load only excitation current(responsible for core loss ie iron loss) flow on the primary side so core loss current will be 1A and core loss = v1*i1*powerfactor. core loss = 1*11000*0.24= 2640watt.
These are not convertible quantities. The same way that you cannot convert seconds into pounds. Or pounds into miles per hour. If you reconstituted your question into the form of "I have a device that draws 2 amps at 125 volts, how many Watts is it consuming?" we'd be off to the races. Hope that helps.