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How to Figure out Watts, Power, Current, Voltage

I was trying to figure out the relative wattages of various Apple power adapters I have–for the MacBook, MacBook Air, and MacBook Pro–to see which is compatible with other MacBooks. While doing this it occurred to me that it might be helpful to provide a quick little factoid/tutorial for non-EEs who get confused by the concepts of voltage, current, and power.

The electrons flowing through a conductor (wire) are a current, represented by i or I, and measured in amperes, or amps. It’s analogous to water flowing through a hose. (Note: the imaginary number i is thus referred to as j by EEs to distinguish it from current, i.) The voltage, v or V, represents the amount of “pressure” pushing the current through the conductor. These concepts are often confused by laymen, used interchangeably, etc. For example, you can be killed if a large enough current passes through your body–amps. Your body has a certain resistance R, measured in ohms, ? (or impedance, if you take into account complex (imaginary, frequency) aspects; incidentally, the analogous concept for magnetic flux is called reluctance–I’ve always loved those three terms: resistance, impedance, and reluctance; if we need another, I suppose stubbornness might do). The bigger the resistance, the more pressure or voltage V is needed to push through a given current I. There is a simple relation: V=IR, or I = V/R.  For example, if you have 10V and a 10 ? resistor, there is 1A of current. If you get electrocuted, it’s the current passing through you–the amps–that kills you.

Anyway, electrical power is measured in terms of Watts, and power, P, is determined by a simple equation: P=VI. That is, it’s directly proportional to the voltage and the current. When you see a power adapter that is rated to provide, say, 18.5V at 4.6A, as with Apple’s 85W MagSafe Power Adapter, you’ll notice that you get the 85W figure by multiplying the voltage times the current: 18.5 x 4.6 = 85.1. The point is, this is a very handy relationship to know, and it’s easy to remember: P=VI. If you know the V and I, you can figure out the wattage (power). In addition, because of the V=IR relationship, you can figure out, say, current, if you know resistance and power. Or say you have a 100? lightbulb, and it’s powered by a 110V outlet. If you want to figure out the current, it’s I=V/R = 110/100 or about 1 amp. This helps to explain why some yards use “low voltage” lighting: if there is low voltage, it’s hard for a human (say) to get shocked badly. If you touch the terminals of a 1.5V battery, the resistance of your body combined with this voltage means the current is very small. To have a decent light from low voltage, you need much lower resistance bulbs, so that you get the wattage (light) output that you want. Likewise, although homes use 110V power, this is stepped down from the higher-voltage power lines. The voltage on power lines is very high so that there is less current for a given power. P = VI, so for the same power P, if you increase V, you can decrease I. The reason you want to do this is the more current, the more heat is generated and energy wasted.

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  • Paul Mollon May 7, 2010, 3:33 pm

    Good stuff Stephan. Hey wait a minute; how come a lawyer knows Ohm’s Law? Lawyers I know wouldn’t know an amp from a lobster.

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