# Thread: Does volt amps = watts?

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## Does volt amps = watts?

Pardon my ignorance. I am a mechanic and a master of digging ditches, running conduit, pulling wire and troubleshooting but get easily confused on simple technical issues.

Is VA basically the same as watts?

2. If you have unity power factor than VA=Watts. This means that the phase angle between the voltage and current is zero. If that is not the case than VA will be larger than Watts.

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Originally Posted by 72.5kv
If you have unity power factor than VA=Watts. This means that the phase angle between the voltage and current is zero. If that is not the case than VA will be larger than Watts.

Thanks???

If the phase angle (whatever that is) is not zero, is the difference between VA and watts significant?

I am installing some 120v plug mold in a commercial building and code says 90 volt amps per outlet. In this case is 90 VA equal to 90 watts? Does the voltage/phase figure into the unity power factor?

4. The plug mold is rated in VA because the power factor the load it may supply is unknown. power factor = watts/VA

EX. 1
A 90 Watt incandescent bulb has a power factor of 1. In this case, 90VA = 90Watts

Ex2
An electric motor 90VA with a power factor of .8: VA= 90 watts = 72. conductors suppling the load is size to carry the VA not only the Watts

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VA = Volts X Amps.That is all that is necessary to consider in the case of the plug mold.

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Where do you get the power factor from? is it on specs of motor,device, etc.

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VA = Apparent Power
Watts = True Power

Ex. Line side of xfmr = apparent power (VA) expressed in kVA. The load side of xfmr would be true power (watts), which is the power actually being used, and not lost through the xfmr for various reasons. In the case of plug mold I wouldn't worry about it.

8. Here's a plain old English explanation.

Simply put, a wattmeter is an electric motor whose speed depends on both the voltage between line terminals and the current through the lines. The wattmeter actually responds to volts and amps (volt-amps), not watts.

Voltage peaks twice per cycle, and current peaks twice per cycle. A wattmeter measures power by simultaneously measuring the voltage and the current, and then 'calculates' the power consumed as what we call watts.

If the voltage and current peaks occur simultaneously, the volt-amps and watts are genuinely the same, but if there is a time difference between the two peaks, the wattmeter turns slower for a given amount of volt-amps.

However, in spite of a deceptively-low wattmeter reading, the power system must be sized to safely carry both peaks, whether they occur simultaneously or not. Since the voltage is a given, the current is really the variable.

The circuit must be designed for the voltage and the current. The insulation doesn't care what the current is (of properly-sized conductors, of course), and the (properly-insulated) conductor doesn't care what the voltage is.

Power companies and electricians alike must design the system components to safely carry the voltage and the current, not the power. Poor power-factor results in a system that must carry current not useable by the load.

To answer the OP directly, yes, for the sake of Plugmold, you can consider VA to equal wattage. Since voltage is considered to be a constant, amperage is the variable that must be accomodated. I hope this made some sense.

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A slightly different way to think about it. As best I can determine, the below is consistent with what Larry wrote above, just from a different perspective:

instantaneous volts * instantaneous amps = instantaneous watts

In an AC circuit, volts, amps and watts are constantly changing.

To describe these constantly changing values, we use 'RMS' measurements. RMS measurements are a type of average suited for measuring electrical quantities. (Side note: the most common meaning of the term "average" describes a specific mathematical operation, and RMS is _not_ that operation, so if you prefer, think of RMS as being analogous to "average", but for electrical measurements.)

RMS voltage is a useful measurement, because when you apply AC of X volts RMS to a resistor, you dissipate the same amount of power as applying DC of X volts to that resistor.

But like any averaging process, you are throwing away a bunch of information.

For 'pure resistive loads', RMS amps * RMS volts = average watts

For many loads, the above equation is not true.

So we use the term 'power factor' to describe the difference.

average watts = RMS amps * RMS voltage * PF

There are many reasons that a load could have a power factor, but what it boils down to is that voltage and current are not changing in lock step. Both a time difference between peak voltage and peak current, or a difference in _shape_ between voltage waveform and current waveform, can result in a power factor less than 1.

-Jon

10. Originally Posted by LarryFine
Here's a plain old English explanation.

Simply put, a wattmeter is an electric motor whose speed depends on both the voltage between line terminals and the current through the lines. The wattmeter actually responds to volts and amps (volt-amps), not watts.

Voltage peaks twice per cycle, and current peaks twice per cycle. A wattmeter measures power by simultaneously measuring the voltage and the current, and then 'calculates' the power consumed as what we call watts.

If the voltage and current peaks occur simultaneously, the volt-amps and watts are genuinely the same, but if there is a time difference between the two peaks, the wattmeter turns slower for a given amount of volt-amps.

However, in spite of a deceptively-low wattmeter reading, the power system must be sized to safely carry both peaks, whether they occur simultaneously or not. Since the voltage is a given, the current is really the variable.

The circuit must be designed for the voltage and the current. The insulation doesn't care what the current is (of properly-sized conductors, of course), and the (properly-insulated) conductor doesn't care what the voltage is.

Power companies and electricians alike must design the system components to safely carry the voltage and the current, not the power. Poor power-factor results in a system that must carry current not useable by the load.

To answer the OP directly, yes, for the sake of Plugmold, you can consider VA to equal wattage. Since voltage is considered to be a constant, amperage is the variable that must be accomodated. I hope this made some sense.
The simplest way to think of it is two people trying to pull a block of concrete, one guy is named Mr. Amperes and the other one is simply known as Volts. If they line up along a single line to pull, all their strength will be fully additive. when they are trying to pull in different directions, say Mr. Amperes is pulling toward 12 o'clock, but Volts is heading to 10 o'clock they each will need to exert more power than before and eventually will produce no movement regardless of the strength excerted when Mr Amperes tries to head toward 9 o'clock while Volts is pulling toward 3 o'clock.

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