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what is the disadvantage of leading power factor...why can't
we use it (pf=0) for a long period of time?
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Answer Posted / shareef
P= SQRT(3) * V * I * P.F
P.F= Cosin(AN)
AN= angle between V&I
If you are talking about P.F that means you are talking
about loads.
1- If P.F=1 that means the load is resistive load (its the
ideal case, the consumed power is active power only: MW)
2- If P.F=0 that means the load is inductive or capacitive
-(If the power is inductive that means the load will
consume power: +MVAR)
-(If the power is capacitive that means the load will
deliver power to the system: -MVAR)
In electrical systems the loads are resistive and reactive
that means the P.F value will be between 0 & 1
Examples
* If P.F = 0.9, its good for electrical system because it
is nearset the ideal case (the Voltage drop is less & the
voltage at end bus is within the limits)
* If P.F = 0.6, its bad for electrical system (the Voltage
drop is high & the voltage at end bus is out oflimits), in
this case we need to install capacitor bank to improve the
P.F and the voltage at end bus
Note. capacitor bank will deliver/give MVAR to the
electrical system
| Is This Answer Correct ? | 3 Yes | 6 No |
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Which are resistive loads & which one are inductive loads : Refrigerator, Split air conditioner, ceiling fans, well pump, fluoresecent bulbs, T.V, CRT Monitors, Adsl Routers, C.P.U, Printers, Deep Freezers, Speakers, DVD Players, Microwave oven etc. ? Output Power for these Appliances Would be same like this : P = V * A * P.F Or would be changed ? thanks
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