Useless oscillation of energy

1. Aug 9, 2010

neduet

Hi friends

""Active power is the share of the apparent power which transmits energy from the source (generator) to the user. Reactive power is the share of the apparent power which represents a useless oscillation of energy from the source to the user and back again""

This Texts from a website but i cant understand this lines "useless oscillation of energy from the source to the user and back again

what is this energy and where is it stored and which form it is back again.?

2. Aug 9, 2010

harroxelas

Well, I think this was poorly worded. Reactive power is not useless, it's just not able to do work (in the physical sense).

Answering your question, this "useless oscillation of energy", A.K.A Reactive Power is stord on inductances and capacitances of circuits. You simply cannot have a REAL circuit without capacitances and inductances.

3. Aug 9, 2010

Averagesupernova

I would say it is poorly worded because of the word oscillation. Reactive power is indeed useless in the sense of transmitting power. When power factor correction capacitors are installed at an inductive load such as a motor the current drawn from the transmission line will drop with no change in power output from the motor. So, reactive power in that scenario is indeed useless. To better understand reactive power I like to think that the reactance of a circuit is changing its value constantly through the AC cycle. When the sine wave hits peak the reactance decides it will offer a high resistance to current. When the sine wave hits zero the reactance decides it will offer a low resistance to current. In both cases, the current and voltage are never peaking at the same instant so maximum power is not achieved even though it appears that on the average there is I amps flowing in a E voltage circuit and of course P=IE. Of course this is very much oversimplified and is simply a perception. I've often wondered if it is called reactance since it is a perception that the resistance (and I'm using the word resistance very loosely) offered is reacting and changing its value based on the voltage. Please don't confuse this with what is really happening in reactive components. It is just a way that has made it easier for me to envision what is happening in an AC circuit when parts of the AC cycle are freeze framed.

4. Aug 15, 2010

m.s.j

Active power is average of rate of electrical energy production and/or consumption but reactive power -which I think the reactive energy is better term for it -is a bit of electrical energy for make of necessary electromagnetic body which used for transmission of active power.
For similar conceptual discussion you can refer to http://electrical-riddles.com/topic.php?lang=en&cat=23&topic=389"

Last edited by a moderator: Apr 25, 2017
5. Aug 15, 2010

Bob S

Not true. Reactive power, oscillating between electrical energy stored as high voltage (½cV2) or high currents (½LI2) with little input power, is the basis for efficiently coupling circuits to antennas and for RF-frequency wireless electromagnetic radiation.

Bob S

6. Aug 15, 2010

skeptic2

Reactive power is responsible for the Q of a resonant circuit. Without reactive power the circuit would not be resonant. One might even argue that without reactive power the circuit couldn't oscillate.

7. Aug 15, 2010

Studiot

Anybody here heard of the transformer?

8. Aug 15, 2010

Dickfore

Let's say the voltage difference between two points oscillates with the law:

$$V(t) = V_{0} \, \cos{(\omega \, t)}$$

while the current has a phase shift $\varphi$:

$$I(t) = I_{0} \, \cos{(\omega \, t - \varphi)}$$

The instantaneous power delivered to this circuit is:

$$P(t) = V(t) \, I(t) = V_{0} \, I_{0} \, cos{(\omega \, t)} \, \cos{(\omega \, t - \varphi)}$$

Using the trigonometric identity:

$$\cos{\alpha} \, \cos{\beta} = \frac{1}{2} \, \left[\cos{(\alpha - \beta)} + \cos{(\alpha + \beta)}\right]$$

and the relation between the effective and maximum value of a oscillating quantity:

$$X_{\mathrm{eff}} = \frac{X_{0}}{\sqrt{2}}$$

we can write:

$$P(t) = V_{\mathrm{eff}} \, I_{\mathrm{eff}} \, \left[\cos{\varphi} - \cos{(2 \, \omega \, t - \varphi)} \right]$$

The first term represent a constant average power output:

$$P_{\mathrm{av}} = V_{\mathrm{eff}} \, I_{\mathrm{eff}} \, \cos{\varphi}$$

while the second term is an oscillating power output with twice the frequency of the alternating voltage (current). Negative power output actually means that the circuit delivers power to the source.

The instants where $P(t) = 0$ indicate the times when this power transmission reversal takes place. From the above formula, we see that this happens when:

$$cos{\varphi} = \cos{(2 \, \omega \, t - \varphi)}$$

$$\left\{\begin{array}{lcl} 2 \, \omega \, t - \varphi & = & \varphi + 2 \, k \, \pi \\ 2 \, \omega \, t - \varphi & = & -\varphi + 2 \, k \, \pi \end{array}\right.$$

$$\left\{\begin{array}{lcl} t & = & -\frac{\varphi}{\omega} + k \, \frac{T}{2} \\ t & = & k \, \frac{T}{2} \end{array}\right.$$

We see that in the intervals:

$$t \in (-\frac{\varphi}{\omega} + k \, \frac{T}{2}, k \, \frac{T}{2}), \; k \in \mathbb{Z}$$

the power output is negative. These intervals shrink to zero only in the case when $\varphi = 0$ and that means there is no phase shift between the voltage and the current (and $\cos{\varphi} = 1$ then).

Last edited: Aug 15, 2010
9. Aug 15, 2010

zomgineer

About the reactive power being useless. That's not quite right. The reactive power has an important use. Probably, the most common cause of reactive power is going to be the magnetic fields in motors as they charge and discharge. The fields give the rotors in the motors something to push against so that the motor can do physical work. If we have a really big motor, or maybe and industrial park that's full of motors, it would be better to put a capacitor bank at that location so that the energy bounces between the caps and induction coils in the motors. That way the reactive current doesn't have to travel all the way across the resistance of the regional power lines.

10. Aug 15, 2010

Dickfore

The load on the motor causes it to have more active impedance than one would calculate using the geometry and material of the coils. This is because some of the energy of the magnetic field is transformed into mechanical work.