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Work, Power and Apparent Power

If we have electricity, what can we do with it? Well, we can do work, which in physics is defined as transferring energy from one object or another through the application of force. Essentially, any time you have a circuit with electricity flowing in it and you are doing something with it, you are accomplishing work. The basic unit of work (or energy) is a joule ("J").

Now, finally, we get to where we wanted to go in our discussion of electricity: the definition of power. Power is simply the rate at which work is done. The more power you have in a system, the more work you can get done in the same period of time. In terms of electricity, increasing power means the ability to do more electrical work (for example, running more appliances, or spinning a motor faster, or running a faster CPU, etc.) in the same number of seconds. Power is measured in watts ("W"). Since power is the rate at which work is done, one watt equals one joule of energy expended in one second:

Power (W) = Work (J) / Time (seconds)

Conversely, the amount of energy used by a device can be computed as the amount of power it uses multiplied by the length of time over which that power is applied:

Work (J) = Power (W) * Time (seconds)

Computing electrical power can be very simple or very complicated, depending on the type of electricity you are looking at. Let's start with direct current. Here, power (in watts) is just the product of the voltage (in volts) and the current (in amps) of the circuit:

P (W) = V (V) * I (A)

Fairly simple stuff, and it makes sense: you do more work when you have electrons pushing with more force (higher voltage) and also when you have more of them per period of time (higher current). Since P = V*I, and I = V/R, another way to express power is:

P = V² / R

For example, if you have a simple 5 V circuit running through a 20 ohm simple resistance, you will have 250 mA of current, and the total power is 5*0.250=1.25 W. Double the voltage to 10 V, and the power doesn't double; it increases by a factor of four, because doubling the voltage while leaving the resistance the same will also double the current. The new power is 5 W.

Unsurprisingly, with alternating current the answer is more complicated. To understand it, it is necessary to introduce the concept of phase, which I will try to do without overcomplicating things (not an easy task! :^) ). As illustrated on this page, alternating current is a wave of voltage that swings between a large positive and negative value. The current also makes this sinusoidal trip the same number of times per second. However, sometimes the current and voltage don't peak at the exact same time. The timing relationship between current and voltage of a flow is called its phase, and is expressed in degrees. Why degrees? Well, a cycle of a sine wave is analogous to a circle. 360 degrees is a full cycle, 180 degree half a cycle, and so on.

Now, what determines the phasing between the current and voltage? Primarily, it depends on the kinds of loads being powered. Simple loads, such as light bulbs, heater elements and the like, are said to be primarily resistive. These loads will cause the phase between the current and voltage to be close to zero. When the phase angle is zero, the voltage and current applied to the load is equal to the voltage and current used by the load.

 115 VAC current and voltage driving a purely resistive load. The phase angle between voltage and current is about 0 degrees. Note that the voltage and current peak together.

Other loads, particularly items such as motors, are said to be reactive. Reactive loads are caused by more complex opposition to the flow of alternating current such as that produced by capacitors and inductors. They can cause the current and voltage to be out of phase, in theory by as much as 90 degrees.

 115 VAC current and voltage driving a (theoretical) purely reactive load. The current is lagging behind the voltage by about 90 degrees. (It is possible for the current to be leading by 90 degrees also.) Note that whenever one of voltage or current hits a peak, the other one is at zero!

If the phase angle between current and voltage is 90 degrees, then whenever voltage is at its peak (either positive or negative), current is zero, and vice-versa. This is a "worst-case" situation that doesn't normally arise in the real world because real loads aren't purely reactive. A more typical situation is where the phase angle is about 45 degrees.

 115 VAC current and voltage driving a partly resistive and partly reactive load. The current is lagging behind the voltage by about 45 degrees, making this an inductive load. If the load were capacitive, the current would be leading the voltage. See here for more on inductors and capacitors.

"Alright, alright," you are saying. "Why do I care about all of this?" Well, here's one important reason: PC power supplies are partly reactive loads, and often exhibit a phase difference between voltage and current of about 45 degrees. This means that the voltage and current applied to the load do not equal the voltage and current used by the load, and you cannot compute the power used by the supply by simply multiplying the current and voltage. OK, now here's where it gets interesting. :^) The voltage and current applied to the load can be multiplied together to yield what is called apparent power, measured in Volt-Amps (VA):

Apparent Power (VA) = V (V) * I (A)

Apparent power represents the voltage and current being sent to the device, and is used to measure draw from the utility, for determining heat generation by equipment under use, and for sizing wires and circuit breakers. The actual power used by the load is called "true" power, or just power, and is measured in Watts. (Even though Watts = Volts * Amps, apparent power is measured in VA to differentiate it from true power.) The relationship between power and apparent power is expressed using this formula:

P (W)  = cosine(phase) * Apparent Power (VA)

where "cosine" is the trigonometric function. "cosine(phase)" is also called the power factor of the load. Let's try an example. Let's suppose we are trying to run a power supply and the power supplied is 115V voltage and 2A of current. The apparent power is 115 * 2 = 230 VA. If the nature of the power supply is that its voltage and current are out of phase by 50 degrees, then the power factor is cosine(50º) = 0.642 (sometimes expressed as 64.2%) and the power used by the load is 148 W.

There's a particular place where all of this comes into play, and that is in the capacity and sizing of uninterruptible power supplies. UPSes are normally specified in terms of apparent power (VA), whereas PC power supplies are specified in terms of true power (W). Many people use the numbers interchangeably, when they most definitely are not the same! Now that you understand the difference between the two, and you know what a power factor is, you are light years ahead of 95% of the population when it comes to figuring out how to purchase a properly-sized UPS or similar device.

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