As a child I considered the reindeer Rudolph, with his nose so bright, to be a marvel of engineering. Now an adult, I remain perplexed as to the mystery behind the selfgenerating power source behind his nose. Did it ever overheat? I wondered. Perhaps today’s discussion can shed some light on the matter. During the course of our discussion of electricity certain terms have been tossed about, like voltage and current. For some the distinction between the two may be unclear, and that is what we’ll be addressing today. Electricity is a rather abstract phenomenon, but you may consider the flow of electrical current through a wire to be much like water flowing through a garden hose. The water won’t flow unless there’s sufficient pressure behind it, and that pressure is supplied by pumps, either at your city water works or your personal well. Take away the pressure, and the water stops flowing through the hose. Electricity flows in much the same manner. It requires a pushing pressure to get it on its journey from power plant to home, and that pressure is voltage. Take away voltage, and the current stops flowing through the wire. Voltage is, of course, produced by an electrical generator at the power plant. Last time we saw how an electrical transformer can reduce high voltage to low voltage and how this process also works in reverse. But how can that be? How can low voltage be turned into high? Is it really possible to get “something from nothing”? Let’s take a closer look. When a light bulb burns out in your home, you routinely look at the bulb to see how many watts it is so you can replace it with the same type. But what exactly is a “watt”? It’s a unit of power, and the markings on the bulb tell you how much electrical power it consumes when you use it. Generally speaking, this electrical power is related to voltage and current by this formula: Power = Volts × Electrical Current Knowing this, if I have a 60 watt bulb in a table lamp, and I plug it into a 120 volt wall outlet, then how much electrical current is the lamp going to draw from the outlet? Using the formula above and a little algebra, we get: Electrical Current = Power ÷ Volts Electrical Current = 60 watts ÷ 120 volts = 0.5 amperes And believe it or not, this same formula that’s used to assess power of a light bulb also applies to electrical transformers. Basically, the power going into the transformer is equal to the power coming out. To see how this works, consider the example stepup transformer shown in Figure 1, which converts a low voltage to a higher one. By the way, “step up” transformers have all sorts of applications. For example, they are used by electric utilities to raise the voltage produced by a power plant to make it more economical to transmit to far away customers. We’ll get into that in another article. Figure 1 – A StepUp Transformer In this example the input voltage on the primary coil is stepped up from 120 volts to 480 volts on the secondary coil, and this works according to the formula we learned about in last week’s blog: N_{P} ÷ N_{S} = V_{P} ÷ V_{S} where N_{P} and N_{S} are the number of turns of wire in the primary and secondary coils respectively, and V_{P} and V_{S} are the voltages of the primary and secondary coils respectively. Plugging in the numbers we get: 50 turns ÷ 200 turns = 120 volts ÷ V_{S} [(200 turns ÷ 50 turns) × 120 volts] = V_{S} = 480 volts Okay, for the sake of our example let’s say that an electric motor is connected to the 480 volt secondary coil. We have an electric meter hooked up to the primary coil and we measure a 2 ampere (a.k.a. “amps”) electrical current flowing through it. Without having the benefit of another electric meter positioned at the secondary coil, how can we measure how much electrical current is flowing through it? The current flowing through the secondary coil is found by equalizing the power in the primary and secondary coils: Power_{P} = Power_{S} Another way of stating this is to say that electrical power is equal to volts times current, so the equation becomes: V_{P} × I_{P} = V_{S} × I_{S} where I_{P} and I_{S} are the primary coil and secondary coil currents, respectively. Plugging in the numbers and working a little algebra we get the electrical current in the secondary coil: 120 volts × 2 amps = 480 volts × I_{S} I_{S} = (120 volts × 2 amps) ÷ 480 volts = 0.5 amps This shows us that the current flowing in the secondary coil is lower than that of the primary coil. It is therefore obvious that the voltage increase in the secondary coil comes at the expense of electrical current that can flow through the secondary coil. Squeeze down on current, voltage goes up. Squeeze down on voltage, current goes up. The power flowing through the transformer stays the same. Conversely, stepdown transformers reduce the voltage coming in, and thereby produce the reverse effect. There is an actual increase in current that can flow through the secondary coil. This principle exemplifies the tradeoff process which is often present in science and engineering. Next time we’ll explore how both stepup and stepdown transformers are used by electric utilities to transmit power from power plants to its customers tied into the utility grid. As for Rudolph and his power source, that’s still under investigation. _____________________________________________
