13
\$\begingroup\$

Real power makes sense since there is actual consumption, but regarding reactive power; what is consumed / delivered? And how does the circuit change once this happens?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You seem to be quoting some document. Maybe you can provide a link to that quote so it can be taken in context. Also, why would there be circuit changes? \$\endgroup\$
    – Andy aka
    May 11, 2015 at 9:17
  • \$\begingroup\$ This is a question in general. I am encountering many questions of the form: "Load 1 absorbs average power of 10kW and delivers 4kVAR of reactive power". I am confused as to what it means for reactive power to be delivered or absorbed between load and source. what happens when this happens? how can I visualise it. \$\endgroup\$
    – Chris-Al
    May 11, 2015 at 9:33

4 Answers 4

10
\$\begingroup\$

To answer the question: Real power is consumed by a circuit. Reactive power is transferred between the circuit and the source.

Real power in W (P) is useful power. Something we can get out of circuit. Heat, light, mechanical power. Power that is consumed in resistors or motors.

Apparent power in VA (S) is what the source puts into a circuit. The full impact the circuit has on the source.

So the power factor is a kind of efficiency pf = P / S for a circuit. The closer it is to 1, the better.

Reactive power in VAR (Volt Amps Reactive) (Q) is power that circulates between the source and the load. Power that is stored in capacitors or inductors. But it is needed. For example, inductive reactive power in electric motors form the magnetic fields to spin the motor. Without it the motor would not work so it's dangerous to consider it is wasted, but it sort of is.

Capacitors and Inductors are reactive. They store power in their fields (electric and magnetic). For 1/4 of the ac waveform, power is consumed by the reactive device as the field is formed. But the next quarter waveform, the electric or magnetic field collapses and energy is returned to the source. Same for last two quarters, but opposite polarity.

To see it animated, see Waveforms for Series AC Circuits. It shows all 6 series circuits (R, L, C, RL, RC & RLC). Turn on the instantaneous power. When p is positive, source is providing power. When p is negative, power is being sent to source.

For a R, power is consumed. For a L or C, power flows between source and device. For a RL or RC, these two relationships are combined. Resistor consumes and reactive device stores/sends power to source.

The true benefit is when an inductor AND a capacitor are in the circuit. Leading capacitive reactive power is opposite in polarity to lagging inductive reactive power. The capacitor supplies power to the inductor decreasing the reactive power the source has to provide. The basis for power factor correction.

Select RLC in the reference. Notice that the source voltage \$V_S\$ (hypoteneuse) is formed from \$V_R\$ and \$V_L - V_C\$. It is less than if formed from \$V_R\$ and \$V_L\$

If the capacitor supplies all the power of the inductor, the load becomes resistive and P = S and pf = 1. The power triangle disappears. The source current required is less, which means the cabling, circuit protection can be less. Inside the motor, the uncorrected power triangle exists, with additional current coming from the capacitor.

The reference shows series circuits, but any C will supply power to any L in the ac circuit decreasing the apparent power the source must provide.


Edit... ![Power Factor Correction][2]

Let's take an example. P = 1kW motor at 0.707 pf lagging with 120V source.

Before power factor correction: \$Q_L = 1kVAR\$ and \$S_1 = 1.42kVA \$ (dashed line) \$Θ_1 = 45° lagging \$ as in I lags \$V_S\$ by 45°. \$I_1 = 11.8A \$

Increase power factor to 0.95 lagging by adding capacitor in parallel with load.

After factor correction: P and \$Q_L\$ still exist. Capacitor adds \$Q_C = 671VAR\$. This decreases reactive power source has to provide, so net reactive power is \$Q_T = 329VAR\$. \$S_2 = 1.053kVA \$ and \$I_2 = 8.8A \$ A 25.8% saving in current. Everything on power triangle exists except \$S_1 \$.

The capacitor supplies 671VAR of leading reactive power to the lagging reactive power of the motor, decreasing net reactive power to 329VAR. The capacitor acts acts as a source for the inductor (motor coils).

Electric field of capacitor charges up. As the electric field discharges, the magnetic field of coils form. As the magnetic fields collapse, capacitor charges up. Repeat. Power is going back and forth between capacitor and inductor.

Ideal is when \$Q_L = Q_C \$. Power triangle disappears. \$S_2 = P = 1kVA \$ and \$I_2 = 8.33A \$

\$\endgroup\$
3
  • \$\begingroup\$ Alright. So the capacitance provides leading VAR while the inductor provides lagging VAR. The leading VAR ABSORBS any lagging VAR. So we say the Inductance SUPPLIES reactive power, while the Capacitance ABSORBS reactive power. So is VAR or reactive power another name for lagging current (as opposed to lagging OR leading current)? \$\endgroup\$
    – Chris-Al
    May 11, 2015 at 15:32
  • \$\begingroup\$ Better to say Q (than VAR) as in (Q_L and Q_C). It's the other way around. Capacitor supplies leading reactive power to Inductor which needs lagging reactive power. An inductor has a lagging phase angle. I lags V_S. A capacitor has a leading phase angle. I leads V_S. For a circuit with a C & L. The capacitor acts as a supply for the inductor decreasing reactive power source must provide. \$\endgroup\$ May 11, 2015 at 16:43
  • \$\begingroup\$ Makes much more sense now. Reactive power is a measure of the current leading the voltage(source). A capacitor supplies Q, while an inductor absorbs Q (induces lagging current). Zero reactive power when the phases fully cancel each other, resulting in a unity power factor, meaning the source only needs to provide (active) power for resistance. \$\endgroup\$
    – Chris-Al
    May 12, 2015 at 15:15
4
\$\begingroup\$

If you applied an AC voltage supply to a load that comprised only capacitance or inductance the phase angle of the current relative to the voltage is shifted by 90 degrees. When voltage and current are displaced by 90 degrees there is no real power delivered to that load. What is delivered to the load is called reactive power.

If the load were a resistor, the current and voltage would be exactly in-phase (as per ohms law) and there would be no reactive power delivered - the power delivered will be real power and it will heat the resistor.

In between these two limits, both reactive and real power can be delivered. The cosine of the phase angle of the current relative to voltage is called power-factor - you may have heard of this; when the phase is zero (resistive load) cos(zero) is 1. When the phase is 90 (reactive impedance load) cos(90) is zero.

enter image description here
(source: oru.com)

The diagonal (red) line in the drawing above is VA i.e. the volt-amps applied to the load - basically it's RMS voltage x RMS current. VA is called "apparent power" and would equal the real/true power (green) should the load be totally resistive.

If the load were purely reactive, "apparent power" = "reactive power" (blue)

Note that in the diagram above, the angle between real and reactive power is 90 degrees always. Following on from further comments, the diagram below should help clarify a few things about reactive power: -

enter image description here

There are four scenarios, resistive, inductive, capacitive and mixed loads. The black curve on all four is "power" i.e. \$v\cdot i\$. Note that for the inductor and capacitor, the power has an average value of zero.

\$\endgroup\$
12
  • \$\begingroup\$ thank you, although i am familiar with the above. What confuses me is exactly what happens with this reactive power. Correct me if I am wrong... So the inductors and capacitors resisting change in current and voltage cause the phase shift, and I can view reactive power as being consumed as a result for this. AC source --> reactive load --> phase shift --> (temporary) reactive power consumption \$\endgroup\$
    – Chris-Al
    May 11, 2015 at 10:10
  • \$\begingroup\$ I don't personally like the term "reactive power" because power implies watts which implies heat. I much prefer to see it is energy put into something that can be liberated without cost. In a purely reactive load there is a net energy transfer in one half cycle and that energy is released back to the AC supply in the 2nd half cycle. It's no different for inductors and capacitors in various electronic circuits - mathematically there is no difference but the "power" guys like to use terms they feel happy with. \$\endgroup\$
    – Andy aka
    May 11, 2015 at 10:26
  • \$\begingroup\$ See this diagram for instance: physics.sjsu.edu/becker/physics51/images/… - it shows voltage, current and power waveforms for resistors, inductors and capacitors. You should be able to see the cyclic nature of energy going in and then presenting itself back out. \$\endgroup\$
    – Andy aka
    May 11, 2015 at 10:30
  • \$\begingroup\$ @Andy aka I fully agree with your comment on the use of the term 'reactive power'. I was taught that the correct term is 'reactive VA'. \$\endgroup\$
    – Chu
    May 11, 2015 at 10:36
  • 1
    \$\begingroup\$ @ChrisAl It's the word 'power' in this context that I have a problem with. Referring to reactive VA as 'power' implies that you get energy if you multiply it by time. \$\endgroup\$
    – Chu
    May 11, 2015 at 11:33
3
\$\begingroup\$

Reactive power is not consumed. Reactive power is the consequence of the electrical reactance of the circuit, that means, the difference of phase between the source and the load. All the power will be delivered to the active load, but since the circuit is not 100% active, there will be a reactive power needed to "move" the active energy through a reactive circuit. That means that you will need bigger cables to move all this power (active + reactive).

\$\endgroup\$
-1
\$\begingroup\$

Take this humorous explanation. Active power is like the cash you spend on food which you eat. All of it goes directly to perform the required function, which is to satisfy your hunger. Reactive power is like the cash you spend on a stove, you cant eat it but you need it to prepare your food. You can keep on using the stove, it is not used up but, you still cant eat it.

In devices like a transformer or motor, reactive power is needed to set up the magnetic field that is used for power conversion from secondary to primary or energy conversion from electrical to mechanical energy. You cant directly perform work with it, but it is necessary for work to be done. You can also think of it like fuel and oil in a car. The oil doesn't make the car run but without it the engine can not function. This is a loose analogy.

The problem in an electrical system is that reactive power and active power are produced by the generator from the same energy input. (Like in our stove and food analogy, all cash comes out of your pocket.) Thus, we want to have only the minimum reactive power which our system absolutely needs and then have all the remaining source power produced as active power. Although, there are some cases where reactive power is preferred

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.