# Watts VS Volt Amps is stumping me

Discussion in 'General Electronics Chat' started by yourownfree, Sep 11, 2015.

1. ### yourownfree Thread Starter Active Member

Jul 16, 2008
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0
I am a little stumped as I have never thought about this before. I read on the Internet somewhere that apparent power (Volt Amps) can draw current but consume no watts. The reason it doesn't make sense to me is that I am old. I have stamped in my brain that when something draws current it consumes power. Now I see this and It needs a little more explaining before I can see it. I can see they are talking about power factor cause it says so. I understand or at least I think I do that watts is the power consumed. So is Volt Amp, apparent power, power being consumed but not being used, or is it some instant power charge, like charging a capacitor or apparently charging a capacitor as an example. Either way to me if the current is drawn it is consumed. If current was water and you draw some from the reservoir its consumed unless you put it back but even that doesn't make sense. As far as I know unless Ohms law has been changed no current can be drawn unless there is resistance or reactance so how is it you can say you can draw current but is is never drawn basically, it just appears it is being drawn. That's like someone else driving my car. Oh he's not really driving it, it just appears he is. I need help on this one. Below is what they said...... and may I add this if it was a MOSFET with a zero resistance would it be the same? Are they just talking about parts that have no resistance such as a perfect MOSFET with no resistance when it turns on? I was born that no current can be drawn unless you have a load and that load consumes energy. I'm lost and obviously maybe just confused and should be put in a mental home.

Power factor is always a number between zero and one because the watts drawn by a device are always less than or equal to the volt-amperes. Note that it is possible for a circuit to have a large voltage across it and to draw substantial current, but consume no energy (dissipate zero watts).

While this seems counterintuitive, it is true if the circuit is purely reactive (a pure capacitor or pure inductor). The circuit will do no work and produce no heat, so it is drawing (and dissipating) zero watts. Yet it can draw substantial current, resulting in substantial VA.

In this case, the power factor is zero. This is possible because the phase relationship between the voltage and current waveforms is such that the circuit is alternately absorbing real power and giving that real power back, so the net real power consumption is zero.

2. ### crutschow Expert

Mar 14, 2008
13,475
3,362
I'm a little confused here myself.
First you say you don't understand how a circuit can draw current and not dissipate power and then, in the last paragraphs, you explain how it can.
If the load is a pure reactance then the power factor is zero and no power is drawn.
If the load is part resistance and part reactance then the current through the resistance will dissipate power but the current through the reactance won't, so the power factor will greater than 0 but less than 1.
A resistance dissipates energy from the current through it but a reactance stores the energy from the current through it (and then returns it to the source in an AC circuit).
So where does your confusion arise?

3. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,009
2,352
The first fallacy is to look at current as power or analogous to water outside of the realm of a simple static DC circuit. Current is not consumed, current is just a movement of charge that drifts slowly in a loop in response to electric fields giving rise to magnetic fields that the energy of the circuit propagates in. First, even if the current is reactive (because the fields are reactive) it still must be moved by the power source by it's limited electric field generation capacity and because it circulates in wires and in the conductors of components with resistance the reactive component of the total current will generate voltage drops (with real electric and magnetic field values) across 'real' resistance and have 'real' power losses due to reactive current flow.