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schematic

simulate this circuit – Schematic created using CircuitLab I'm trying to understand if the average of instantaneous power is equal to real power. My reasoning is that the reactive components average out to zero and from there the average of that signal will give you just the power dissipated by the resistor. Is this true or do we always have to use P=IVpf.

Thanks

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Instantaneous power is v x i from the perspective of the power source. So, you average that to produce average power.

Is the average of instanteous power equal to real power?

It sure is - that is what generates the heat in your resistor and that ultimately is what you get billed for.

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  • \$\begingroup\$ great so does reactive power contribute to real power? That is would I still have the same real power loss in that circuit if I got rid of the inductors and capacitors? \$\endgroup\$ – TryingToLearnMath Jun 16 '17 at 12:04
  • \$\begingroup\$ If the inductor and capacitor were resonated i.e. had the same impedances that cancelled out then yes. Unfortunately with your numbers the capacitor has -j2653 ohms whilst the inductor has +j377 ohms. Net reactive impedance is not zero but -j2276 ohms and, if you removed (shorted) that, the whole supply would be across the 300 ohm resistor and you would consume more power. \$\endgroup\$ – Andy aka Jun 16 '17 at 12:10
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I'm trying to understand if the average of instantaneous power is equal to real power.

Yes.

My reasoning is that the reactive components average out to zero ...

This is true only if their impedances match and cancel. In your example $$ Z_C = \frac {1}{2 \pi fC} = \frac {1}{2 \pi \cdot 60 \ 1 \mu} = 2653 \; \Omega $$ and $$ Z_L = 2 \pi f L = 2 \pi \cdot 60 \cdot 1 = 377 \; \Omega $$ so they don't balance. There will be a reactive component in your supply load.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Since the reactive and inductive vectors don't cancel out the result is the kVA vector and it is inductive.

... and from there the average of that signal will give you just the power dissipated by the resistor. Is this true or do we always have to use P=IVpf.

If you are measuring V and I at the supply then you need to use the power factor in your calculations. (If you measure V and I at the resistor then the power factor there will be unity.)

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  • \$\begingroup\$ I agree the reactive vectors don't cancel, however they're sum creates one reactive component vector. If you average that shouldn't it be zero? Hence, that's why if we take the average of the supply that gives us Watts and not VA since the VAR component averages to zero. \$\endgroup\$ – TryingToLearnMath Jun 16 '17 at 11:59
  • \$\begingroup\$ If they don't cancel they can't sum (is that what you mean by average?) to zero. The vector diagram demonstrates that. If the sum of the inductive and capacitive loads do not cancel out then you have some reactive power and the power factor will be < 1 (angle between kVA and kW > 0). You're almost there but missing some little bit in your understanding. Ask again if you can explain where you're getting lost. \$\endgroup\$ – Transistor Jun 16 '17 at 12:42
  • \$\begingroup\$ What I'm asking is if we average the just the total instantaneous reactive power, that should average to zero correct? The instantaneous power through the resistor plus the instantaneous reactive power through the capacitor and inductor should add to the total instantaneous power (S with the units VA) in the whole circuit, correct? And then if we take the average of that that will give us just the average power, also known as IrmsVrmspf. \$\endgroup\$ – TryingToLearnMath Jun 16 '17 at 14:10
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If you measure the instantaneous voltage and current at the power supply, the real power is the average of the instantaneous V X I. The reactive components average to zero. Internally to the load circuit, some or all of the reactive components may cancel out. Any reactive power that does not average to zero inside the load must average to zero between the load and the supply. The real and reactive power can be completely calculated from instantaneous measurements anywhere in the circuit. As long as you are careful about the +/- signs of the instantaneous values, you can calculate the real and reactive values for each circuit component and see that the real power furnished by the supply is equal to the real power dissipated by the resistor and that the reactive VA of the supply, capacitor and inductor sum to zero. The same results would be found using RMS values and power factors.

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  • \$\begingroup\$ Charles, the term "average to zero" is causing the OP some confusion and I've never heard it used regarding reactive loads / power-factor. Can you elaborate? \$\endgroup\$ – Transistor Jun 16 '17 at 13:54
  • \$\begingroup\$ Thanks Charles this answers my question. @Transistor Sorry for the confusion, maybe my response above will clear things up? \$\endgroup\$ – TryingToLearnMath Jun 16 '17 at 14:13
  • \$\begingroup\$ @Transistor We don't usually talk about reactive power as a continuous exchange of energy flowing back and forth among the source, capacitive load elements and inductive load elements. The "average to zero" term that I applied recognizes that there is no net flow of power, only a half-cycle by half-cycle exchange. I believe that explaining it that way is sometimes more effective. \$\endgroup\$ – Charles Cowie Jun 16 '17 at 14:42
  • \$\begingroup\$ Got it. OP is asking about average power so, yes, the average of the reactive power is zero while the RMS current is actually increased. Hence the power utility's need to penalise us for poor power factor since we're taking power from them inefficiently. \$\endgroup\$ – Transistor Jun 16 '17 at 15:35

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