This is not really the reason. For most circuits, frequency remains constant throughout anyway.
The real reason is that for circuit analysis, you want to convert sinusoidal functions to phasors. A phasor is essentially a complex number (it has a magnitude and angle), and makes the maths of...
Having thought about this some more, defining reactive power in the first way is best because:
Both Q and P are conservative at every node, whereas VI/2 isn't conservative. This allows a conservation of power approach to circuit analysis (much like a conservation of energy approach in physics)...
OK I understand what you're saying here. I wasn't aware that the term "Watts" was reserved only for time rate change of work and not energy.
Isn't this reasoning circular? Reactive volt-amps are by definition VARS?
I know it's S. If S is capable of quantifying the circulating volt-amps on top...
Reactive power is defined as (VI/2).sinφ. This is the amplitude of the second component in the equation:
p(t) = v(t).i(t) = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt)
But the above equation can also be written as:
(VI/2)cosφ + (VI/2).cos(2ωt -φ)
Why not define reactive power as the...
Leakage inductance is a result of leakage flux. This is the flux which does not perfectly couple both windings in a transformer and goes to waste.
https://commons.wikimedia.org/wiki/File:Flux_leakage.png
The more leakage inductance (leakage flux) you have, the less power you will see on the...
Agree. Just emphasising that reactive power is measurable in Watts, and has real practical effects on the output of a generator (i.e. it serves some useful purpose). The more reactive power you supply, the less real power you can supply (as per the capability curve you attached). It also costs a...
Over the cycle it doesn't waste energy. The power station (or voltage source) always gets back what it puts out. The problem is that with a power station, this still amounts to the generator running even if you are capable of recouping all the energy back. This wears your plant and costs money...
Read a reliable source?
They draw instantaneous power from the supply given by:
p(t) = v(t).i(t) = Vsinφcos(wt + 90 - φ).Icos(wt - φ) = (VI/2).sinφ.sin(2φ - 2wt)
Reactive power is defined as the maximum of this value Q = (VI/2).sinφ.
If you worked in a power station, your plant would need...
Imagine you have an electrical device that is largely inductive that you need to connect to the grid. Let's say the average real power it consumes is 1MW. Guess what is going to happen if the power station only burns 1MW worth of coal? Your device is not going to run because it's not getting...
Not everything is about calculating supply losses. The reactive power Q is related to the rate of change of energy stored in your capacitors (1/2⋅CV^2) or inductors (1/2⋅LI^2). One utility of Q is that it allows you to calculate the maximum energy stored in your reactive elements during the...
For a purely reactive component p(t) = (VI/2).sinφ.sin(2φ - 2wt). Q is therefore the peak value of the AC power waveform.
When you have both resistive and reactive components, p(t) = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt). Q is the peak of the sine component, but there is no nice...
You might want to know the reactive power if you ever want to control the voltage on a power grid. Or size your capacitors for power factor correction. Or size your power cables efficiently. Also, peak power is an important design spec for almost all electrical devices...