Why is power draw measured in WATTS ( which is volts multiplied by amperage ) but power production or power sources are measured in VA ( volt amps ). Are they not identical? What’s the reason for the difference.
In: Engineering
If you have a purely resistive load, the instantaneous current is directly proportional to the voltage, and apparent power (VA) would be equal to real power(watts).
If you have a purely reactive load, real power is zero, the energy stored by the component just gets put back into the grid. The current is still flowing though, so it still puts a burden on the generation and transmission equipment.
For a purely resistive load, like a light bulb or heater, volts times amps equals watts. But many loads are not purely resistive; they can also be capacitive or inductive. By far the most common of the two is inductive: most motors, for example, are inductive loads. These loads consume power but “return” some of it without using it. So there’s a thing called a “power triangle” that shows the relationship between the “real” power (measured in watts; this is the actual amount of power consumed), the “reactive” power (measure in volt amps reactive; this is the power that gets “returned”), and the “apparent” power (measured in volt amps: for a purely resistive load, this will match the real power).
I will try my first ELI5 answer.
Take your little red wagon, some string, and your little brother and go to the bottom of a small hill. Sit your brother on the wagon and tie one end of the string the string to the handle.
Now, grab the string a short distance from the wagon and slowly walk up the hill, pulling the wagon behind you. Be careful not to jerk or pull too hard.
The strength the string needs to be, so it does not break while you are steadily dragging the wagon, is the Watts.
This is slightly like a resistive load like you’d find with a light bulb or a heating element (as others have mentioned)
Now go back to the bottom of the hill. Position the wagon and your little brother in the same place but this time walk back up the hill with the other end of the string.
Pull the wagon up the hill hand over hand.
Each pull brings your brother a bit closer and the wagon may even coast a little bit before your next pull.
This is like an inductive load like in an electric motor or the alternating current in an AC circuit (which ‘tugs’ 60 times a second).
Unlike when you slowly pulled the wagon up with you, in this case you are tugging repeatedly on the string over and over to get the wagon up the hill. This means the string sometimes needs to be a little bit stronger in order to not break when you pull each time. This is like the Volt-Amps.
The amount of effort it takes to get you, your wagon, and your brother to the top of the hill is about the same in both cases.
This is why, in electrical systems, wiring and equipment that needs to handle AC current or feed power to motors, etc, are rated in VA because at certain points in time, they need to handle more. VA is always larger than W in these cases. Situations where there is a steady load VA = W as there is no ‘tugging’.
Edit.. Apologies, I kinda broke rule 4 I think.
If you are familiar with a sine wave, picture in your mind that the current and voltage can each have their own sine wave in the same circuit. So when you have a resistive circuit (space heater, soldering gun, etc.), the voltage and current sine waves are both peaking at the same time, so power in this case is watts = volts x amps since both are sine waves are maximum simultaneously. It can be thought of as the voltage and current being “in phase” with each other, and the only thing impeding current flow is the resistance of the conductor.
In an inductive circuit (motor, coil, or anything with a winding), you have what is called inductive reactance. When current flows and changes (as it is constantly doing in a sine wave) the expanding and contracting magnetic field around the wire is inducing a voltage (called CEMF or counter electro magnetic force) in the wire itself that opposes the current flow creating it. So now you have not only the resistance of the conductor opposing the flow of current, you have the CEMF. The two together would be the total “impedance” of the circuit.
(Quick off-topic analogy. Think of resistance in a water pipe being the constrictive size of the pipe diameter. Now you come to a hill and the pump has to overcome gravity (inductance) as well as the resistance. They both oppose the flow of water, but differently).
If you’ve wrapped your head around that, then think that this CEMF is “slowing” the rise of current in the circuit, so that the current sine wave is lagging the voltage sine wave. In other words, the voltage sine wave peaks and starts decaying before the current sine wave peaks. Now you cannot simply multiply the volts x amps because they are not peaking at the same time, so you have to take into consideration how much gap, or “phase shift” in the sine waves you have.
Read this sentence then forget it if it’s too confusing… if portrayed vectorially, the amps would be lagging the voltage by a certain number of degrees. The numerical “cosine” of that angle is the determining factor in the efficiency of the circuit.
So to calculate power now, you would multiply the volts x the amps x how much out of phase they are (the cosine of the angle) to get power. Since the cosine is always 1 (for a resistive circuit with no shift) or less than one for an inductive circuit, VA will only equal W on a resistive circuit and VA will be greater than watts on any other type of circuit.
So…resistive circuit, volts x amps = watts
Inductive circuit, volts x amps x cosine of the phase shift = watts
The same is true of a capacitive load except that the current leads the voltage instead of lagging it.
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