# Watts vs. volts amperes

What I understand:

1. In simple DC circuits, this is a product of the current and voltage, such that 1 watt = 1 ampere x 1 volt
2. I understand that a watt is a unit of power (change in energy per unit time) that describes the rate at which physical work can be done
3. Reactive power is power for which current flow occurs, and that isn't actually used by the source. Even though it's not used by the destination, the intermediate current flow causes transmission losses, which is why it's undesirable. Hence why power factor correction exists, and why utility companies bill non-residential customers for a low power factor.

What I don't understand:

1. Is reactive power truly unusable? Suppose I had a magnet coil in a fictitious AC circuit, that is purely capacitive reactive, with no resistive component at all. It would exhibit purely reactive power flow. Wouldn't the coil produce a changing magnetic field, that I can use to deflect a nearby magnet (converting electric energy into kinetic energy)?
2. Are watt and volt amperes dimensionally equivalent, just e.g. like km/s and miles/hour?
1. If so, would using watts to denote apparent power be "technically correct" (and merely "wrong by convention")? If that's the case, then my understanding that VA is used instead of W, because it has a "documenting" purpose, by commenting exactly how these watts can be used. But this sounds strange, because then it seems like "not all watts are created equal".

(I know this question has been asked before, but none of the Q&As I found answered the precise points of confusion I had.)

• Regarding the latter, they are not just dimensionally equivalent but dimensionally identical in SI units. (A volt is defined as the potential difference that does one joule of work on a test charge of one ampere-second.) Regarding the first I lack experience with this but of industry jargon, but you might edit out the second question since it is easily answered, to better fit the site’s “one question per question” conventions. – CR Drost Jun 3 at 15:19
• @CRDrost My question 2 is a precursor to 2.1., which tries to distinguish "technically correct" vs "more conventionally correct". Splitting into its own question wouldn't be useful IMO, because the answer would be so heavily coupled with question 1. – Alexander Jun 3 at 15:30
• As a practicality comment, reactive power is deemed unusable because, practically speaking, when you have an inductive/capacitive load big enough for us to be concerned with the reactive power, it's rare that we find a capacitive/inductive use for that reactive power that happens to have the opposite reactive power. So from a practical perspective, it tends to be one of those things that's just a waste. The best you end up doing is balancing it with a corresponding reactance. – Cort Ammon Jun 3 at 23:59
• Just as a remark: It is actually a common convention to have "different" units for differnet quantities hat are actually the same unit: E.g. frequency is given in Hertz, radioactivity is given in Becquerel, but both are just $1/\text{s}$. However, you wouldn't express your CPU speed as 4 GBq, although that would be correct in a strict sense. Likewise, torque is given in Newton-metres, but not in Joules, although that is again the same unit. – Toffomat Jun 4 at 8:50
• @Toffomat Interesting, I didn't know those examples, so this instance stood out in my head as a strange outlier. The definition section outlines a pretty good justification for why Becquerel was a better choice than to use s^-1. I'm not a fan of this though: "Whereas 1 Hz is 1 cycle per second, 1 Bq is 1 aperiodic radioactivity event per second." The distinction seems to subtle/pedantic there, but it's similar to the reasoning behind the Baud, so I digress. – Alexander Jun 4 at 15:58

What you understand is basically correct. Regarding DC circuits it is important to point out that 1 watt = 1 amp x 1 volt under steady state (long time) conditions when transients are gone. Under those conditions an ideal capacitor looks like an open circuit (no current flow) and an ideal inductor looks like a short-circuit (no voltage across the inductor). The only power is that dissipated in resistance.

Is reactive power truly unusable?

No. Reactive power results in energy being stored in the reactive circuit components. At any instant in time the energy stored in the electric field of a capacitor is

$$E_{C}(t)=\frac{CV(t)^2}{2}$$

And energy stored in the magnetic field of an inductor is

$$E_{L}(t)=\frac{LI(t)^2}{2}$$

That energy is available to either do work or generate heat. The energy of the magnetic field, for example, is essential to motor operation. You yourself pointed out that you could use a changing magnetic field to deflect a magnet (do work).

Are Watt and Volt Amps dimensionally equivalent, just e.g. like km/s and miles/hour

Yes. Both are units of Joules/sec. A volt is a Joule/Coulomb and an ampere is a Coulomb per second. The units of watts, whether they be electrical watts or watts of mechanical work, are Joules/sec.

If so, would using watts to denote apparent power be "technically correct" (and merely "wrong by convention")?

No it would not be technically correct. They both have units of Joules/sec but Volt-Amperes does not take into account any phase difference between voltage and current that would occur with reactive components (inductors, capacitors) in the circuit. On the other hand, instantaneous power in watts, is given by V(t)I(t)cosθ where θ is the angle between voltage and current. The angle is zero for purely resistive circuits so that the cos is 1.

But this sounds strange, because then it seems like "not all watts are created equal"

I think it's more accurate to say "not all volt-amperes are considered equal"

The product of voltage and current, without any knowledge of the circuit elements, is called volt-amperes and not watts. If the circuit is purely resistive, than volt-amps is the same as watts.

I guess you could say that all watts are volt-amperes but not all volt-amperes are watts.

Most of this makes sense. But what's troubling me is the idea that you use a different unit just because of context. "I guess you could say that all watts are volt-amperes but not all volt-amperes are watts." this makes sense, but it doesn't explain why we don't just universally use "VA", or universally use "W", and add on the stipulation of what we mean "Watts apparent" or "VA resistive".

I understand why this can be troublesome. First of all, the units are not different, but the same, i.e., Joules/sec. You don't have to think of the labels "watts" and "volt-amperes" as being electrical units. It is just that "watts" connotes in phase current and voltage and has traditionally been associated with resistive loads while "volt-amperes" leaves it open. I think you will find the term "volt-amperes" used when no detail is give on just what the voltage across and the current to particular circuit elements are. When in doubt, I always use the term volt-amperes since this is the most all inclusive term.

Hope this helps.

• Most of this makes sense. But what's troubling me is the idea that you use a different unit just because of context. "I guess you could say that all watts are volt-amperes but not all volt-amperes are watts." this makes sense, but it doesn't explain why we don't just universally use "VA", or universally use "W", and add on the stipulation of what we mean "Watts apparent" or "VA resistive". – Alexander Jun 3 at 17:12
• Consider an analogy, we use m/s regardless of what direction we're going in. Imagine if I invent a unit, call it the Alex, which is a unit of 1 m/s northward. When expressing motion, without any knowledge of the motion's direction, we use m/s. If the motion is purely Northward, than we use Alexs, which are the same a m/s. It just doesn't make sense to me, it makes much more sense to say "1 m/s North", than to define a new unit, with equivalent dimensionality, whose only purpose is to document context. – Alexander Jun 3 at 17:14
• @Alexander Regarding the following statement from your second comment " When expressing motion, without any knowledge of the motion's direction, we use m/s." Yes, that is true, but without knowledge of the motions direction we call the m/s the speed of the object, not its velocity. When we add the direction of the motion, we call it velocity. The units of the magnitude of each is the same (m/s), but we give them different "labels", speed vs velocity. Hope this helps. – Bob D Jun 3 at 17:29
• @IllidanS4 When expressed using the complex power triangle, real power, reactive power, and apparent power are vectors. Real power (watts) is a vector on the real axis. Reactive power (volt-amperes reactive) is a vector on the imaginary axis, and apparent power (volt-amperes) is the vector sum of the real and reactive power. – Bob D Jun 4 at 12:06
• An ideal coil with purely reactive characteristics is infinitely small and/or has a core with infinite permeability, so that its magnetic field never leaves the coil. The moment you use that field to move a motor, or simply let it escape as EM waves, your coil stops being ideal and starts consuming active power. – Dmitry Grigoryev Jun 4 at 14:02

I understand that a watt is a unit of power (change in energy per unit time) that describes the rate at which physical work can be done

Right.

The key thing to observe is that energy can move in both directions. It can move from the "supply" to the "load", but it can also move from the "load" to the "supply".

The power at any given instant is found by multiplying the current at that instant by the voltage at that instant. In a DC system our life is simple, voltage is a constant, current is a constant, so power is also a constant.

In an AC system however, voltage current and power all vary over time. So we attempt to use RMS voltage, RMS current and mean power to describe our AC systems.

There is a problem though, If voltage and current are sinusiodal and in-phase with each other then mean power is equal to RMS current times RMS voltage. However if they are 90 degrees out of phase then there is no net transfer or power. For two of the four quarter cycles energy flows from source to load. For the other two quarter cycles energy flows from load to source. Net energy transfer over the cycle is zero.

Is reactive power truly unusable? Suppose I had a magnet coil in a fictitious AC circuit, that is purely capacitive reactive, with no resistive component at all. It would exhibit purely reactive power flow. Wouldn't the coil produce a changing magnetic field,

Yes

that I can use to deflect a nearby magnet (converting electric energy into kinetic energy)?

What you would find is that when you introduced the magnet and placed a load on it that the current in the coil would no longer be purely reactive.

Are Watt and Volt Amps dimensionally equivalent, just e.g. like km/s and miles/hour?

Yes they are dimensionally equivilent, but by convention we use watts for "real" (mean) power and for instantationus power but not for "reactive" or "apparent" power.

If so, would using watts to denote apparent power be "technically correct" (and merely "wrong by convention")? If that's the case, then my understanding that VA is used instead of W, because it has a "documenting" purpose, by commenting exactly how these watts can be used. But this sounds strange, because then it seems like "not all watts are created equal"

Dimension is not a full specification of what a quantity means. For example both joules and newton-meters work out to kg⋅m2⋅s−2 but noone would argue that torque and energy are the same thing.

• This is more correct and more useful than the accepted answer, especially the point about the nearby magnet absorbing power from the varying magnetic field and so making the coil no longer pure reactive. It's interesting that when you take a torque, and turn it through an angle, that is, multiply it by a dimensionless number radians, it becomes work. – Neil_UK Jun 4 at 9:59

Suppose I had a magnet coil in a fictitious AC circuit, that is purely capacitive, with no resistive component at all. It would exhibit purely reactive power flow. Wouldn't the coil produce a changing magnetic field, that I can use to deflect a nearby magnet

The coil should have a purely inductive impedance (not capacitive), only it it were ideal, like an infinite solenoid with no resistance. A real solenoid could have a negligible resistance as well, but it would radiate, thus the impedance would deviate from the purely inductive. If you put a magnet that can move, the resistive part of the impedance will be even more. Basically, the energy transfer takes place only due to the resistive part of the impedance.

Are Watt and Volt Amps dimensionally equivalent, just e.g. like km/s and miles/hour?

Yes! The "mechanical" unit is the W (defined only in terms of kg, m and s). To get V and A you need the electrical charge, the C. But when you multiply V*A the C simplyfies.

If so, would using watts to denote apparent power be "technically correct"

Rigorously, it is correct. In principle, you can say that you have an ideal capacitor, and drive it with an AC current with 1 W of product between current and voltage, although no energy is transferred. It's just a unit of measurement. It is also necessary to clarify what are the reported currents and voltages: the peak values, the peak-to-peak, or the "power-average" values? (the last is the value which gives the average power). In any case, it is better to write what you are reporting and not to rely on the unit of measurement.

• "If you put a magnet that can move, the resistive part of the impedance will be even more." Woah, what, really? "Basically, the energy transfer takes place only due to the resistive part of the impedance." that makes sense! – Alexander Jun 3 at 16:13
• Now I'm curious why VA "stuck", and not something subscript on W, like W_a, W_r, etc. – Alexander Jun 3 at 16:13
• Really! You can use impedance measurements to measure the dissipation in a dielectric (of a capacitor). The real part (resistive part) represents the dissipation. If you put a moveable magnet, that can really move due to the AC (at 1MHz it does not!), then the movements of the magnet will be such that they generate a voltage in the coil, which appears as a resistive impedance. About "VA", actually, I do not know but I'm curious too! – Doriano Brogioli Jun 3 at 16:17
• "Then the movements of the magnet will be such that they generate a voltage in the coil" Ah yes, this makes sense! "they generate a voltage in the coil, which appears as a resistive impedance" But if it's an ideal coil (super conducting: 0 residence), doesn't that still mean 0 loss? – Alexander Jun 3 at 17:07
• @Alexander no, still have the back EMF even in superconducting motors, e.g. nari.arc.nasa.gov/sites/default/files/… – Pete Kirkham Jun 4 at 7:47

Everything is already explained in depth, and I want just attempt to say a few words about different units in everyday usage:

1. Watt (W) is used to characterize [useful] power at consumer side; kWh measures actual energy used
2. Volt x ampere (VA, VAR) is used to characterize transmission effects; kVARh measures energy transferred over power lines (in both directions). The ampere part of it (current) accounts for losses.
• Got it, thanks. And welcome to stack exchange :) – Alexander Jun 4 at 16:00
1. There will always be some power loss in transmission, however the higher the voltage, the less loss to resistance. That is why power companies use high voltage transmission lines (7,200 volts or 14,400 volts) with transformers at delivery points to reduce to required usable voltage, 110 V, 220 V, etc. This power loss from resistance could theoretically be used, and it can not be destroyed, merely converted to heat and magnetism.

2. Since volts times amperes equals watts, volt amperes equals watts, it is two ways of saying the same thing.

• The higher is the voltage, the lower is the loss, at same resistance. – Doriano Brogioli Jun 3 at 16:13
• Hi Adrian, I appreciate your answer but it kind of misses the essence of my question. For one, I explicitly specify that my circuit is "ideal", i.e. no loss in transmission. Imagine it was an infinitely long super conducting coil, purely inductive with no resistive losses. – Alexander Jun 3 at 16:14
• no resistive loss, no magnetic field, the energy must come from somewhere. – Adrian Howard Jun 3 at 19:51
• infinite coil would be the same as an open circuit – Adrian Howard Jun 3 at 19:51
• "the higher the voltage, the less resistance". No, the (ohmic) resistance stays the same. Perhaps you meant to say something else? – Peter Mortensen Jun 4 at 8:43