Reactive power is basically defined as that part of apprant power that shuttles between source and loads.Then why we call inductive ( absorbers of reactive power) nd capacitive (dissipator of reactive power) loads??
I think your misunderstanding begins with descriptions (not by you, since your description of reactive power is OK) of generators as providers of "reactive power" with an ASSUMPTION this is UNDERSTOOD to mean capability to provide lagging current unavoidably drawn (absorbed) by motors and transformers using magnetic principles. Both capacitors and inductors have reactance. Inductive and capacitative REACTIVE loads do not absorb much energy (watts) - in theory none -but they do add to the amps (current) load - so transformers and generators have to have bigger volt-amp ratings to supply them. Look up "absorb" and "dissipate" in an English dictionary, I think you have misunderstood their meaning. I have seen "dissipate" applied to resistors many times - they really do "waste/throw away/spread about/use up energy", but not to capacitors, except to describe their "losses" (which have to be dissipated as heat). A capacitive current can cancel out or compensate an inductive current (power factor compensatION), because there is 180 degrees phase difference, but "dissipate" is the wrong word for this.
You understand OK when my "free" time wastes your "valuable" time - also your writing improves! "Load" can mean many things. In electricity, "load" is what takes current from a generator, not just what uses energy. It is not just the load you want to power, like a light bulb. A motor (due to its inductive current), with 0.8 power factor takes 1/0.8 = 1.25 times the current essential to turn it. Cables are sized by current. Transformers and generators are rated and cost money by size in VA. Note VA = rated volts x rated amps. Consequently, the extra current taken by the motor takes real money to supply. The primary current essential to magnetise transformers also increases the load on generators.
I am impressed that even after the arrogance expressed by the "guest" OP, you still continued to offer information on the subject in question. Well done...
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Thanx for your reply. I may be wrong with selection of word like dissipate,,but then wat i exactly like to ask is capacitor supplies reactive power while inductor absorbs. then we aslo say recative power bounces back load and source.
If reactive power shuttles between load and source,,how come capacitor and inductor action come into picture.
I await some good response from u that can remove my doubt..
Capacitor and inductors are the cause of reactive power, they cannot be other than "in the picture"! In any electrical circuit there are the following fundamental actions:-
1.
"Burning-up" or "dissipation" of energy (which commonly finishes up as heat, but could, for example, be the power radiated from a radio antenna or mechanical shaft power output) - represented by resistance.
2.
Storage of energy by electric fields - represented by capacitors.
3.
Storage of energy by magnetic fields - represented by inductors.
Reactive "power" is all to do with stored energy. Take an inductor connected to a sinusoidal voltage source (under steady-state conditions) :-
1.
Over a quarter cycle of current increase from zero, the source provides energy that is stored in the inductor's magnetic field. At the end of the quarter cycle, the current is maximum and the magnetic field strength/stored energy is maximum.
2.
During the next quarter cycle, the current falls to zero, at which the stored energy is zero.
3.
Over the next half cycle, 1/ and 2/ repeat, but the current has the the opposite polarity.
4.
Go back to 1/ and the whole cycle repeats.
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One can take a generator driven by an engine or motor and apply an inductive load to it, which takes a large current (which would have a very noticeable effect on the fuel flow/ drive torque if it were a resistive load like lamps for a football stadium).
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But for this inductive load, the increase in fuel/torque is much less than that.
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In fact, the most careful analysis (e.g. using a torquemeter and tricks like putting the inductor in a heat insulated tank of oil and measuring temperature rise over a time interval) would only show that the extra fuel/torque is due to the inductor losses and not in proportion to the amps x volts at the inductor terminals.
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The conclusion must be that energy taken from the source over a quarter cycle (starting at zero current) is stored and returned to it over the following quarter cycle.
One can follow similar arguments, for a capacitor connected to a sinusoidal voltage source.
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