Click here to purchase the entire book in PDF format. RMS Look at a light bulb. Not directly - you'll hurt your eyes - actually let's just think of a lightbulb. I turn on the switch on my wall and that closes a connection which sends electricity to the bulb. That electricity flows through the bulb which is slightly resistive. The result of the resistance in the bulb is that it has to burn off power which it does by heating up - so much that it starts to glow. But remember, the electricity which I'm sending to the bulb is not constant - it's fluctuating up and down between -170 and 170 volts. Since it takes a little while for the bulb to heat up and cool down, its always lagging behing the voltage change - actually, it's so slow that it stays virtually constant in temperature and therefore brightness. The light bulb does not respond to instantaneous voltage values - instead, it burns off an average amount of power over time. That average is essentially an equivalent DC voltage that would result in the same power dissipation. The question is, how do we calculate it? First we'll begin by looking at the average voltage delivered to your lightbulb by the hydro company. If we average the voltage for the full 360$^{\circ }$ of the sine wave that they provide to the outlets in your house, you'd wind up with 0 V - because the voltage is negative as much as it's positive in the full wave - it's symmetrical around 0 V. This is not a good way for the hydro company to decide on how much to bill you, because your monthly cost would be 0 dollars. (Sounds good to me - but bad to the hydro company...) Figure 2.3: A full cycle of a sinusoidal AC waveform with a level of $170 V_{p}$. Note that the total average for this waveform would be 0 V because the negative voltage is identically opposite to the positive voltage. What if we only consider one half of a cycle of the 60 Hz waveform? Therefore, the voltage curve looks like the first half of a sine wave. There are 180$^{\circ }$ in this section of the wave. If we were to measure the voltage at each degree of the wave, add the results together and divide by 180 (in other words, find the average voltage) we would come up with a number which is 63.6$\%$ of the peak value of the wave. For example, the hydro company gives me a 170 volt peak sine wave. Therefore, the average voltage which I receive for the positive half of each wave is 170 V * 0.636 or 108.1 V as is shown in Figure 2.4. Figure 2.4: The positive half of one cycle of a sinusoidal AC waveform with a level of $170 V_{p}$. Note that the total average for this waveform would be 63.6$\%$ of the peak voltage as is shown by the red horizontal line at 108.1 V (63.6$\%$ of $170 V_{p}$). This does not, however give me the equivalent DC voltage level which would match my AC power usage, because our calculation did not bring power into account. In order to find this level, we have to complicate matters a little. We know from Watt's law and Ohm's law that $P=V^{2}/R$. Therefore, if we have an AC wave of $170 V_{peak}$ in a circuit containing a $1 \Omega$ resistor, the peak power consumption is $$\frac{170 \textrm{ Volts} ^{2}}{1 \Omega} = 28900 \textrm{ Watts}$$ (3.7) But this is the power consumption for one point in time, when the voltage level is actually at 170 V. The rest of the time, the voltage is either swinging on its way up to 170 V or on its way down from 170 V. The power consumption curve would no longer be a sine wave, but a $\sin^{2}$ wave. Think of it as taking all of those 180 voltage measurements and squaring each one. From this list of 180 numbers (the instantaneous power consumption for each of the 180$^{\circ }$) we can find the average power consumed for a half of a waveform. This number turns out to be 0.5 of the peak power, or, in the above case, 0.5*28900 Watts, or 14450 W as is shown in Figure 2.5. Figure 2.5: The power consumed during the positive half of one cycle of a sinusoidal AC waveform with a level of $170 V_{p}$ and a resistance of 1 $\Omega$. Note that the total average for this waveform would be 50$\%$ of the peak power as is shown by the red horizontal line at 14450 W (50$\%$ of 28900 W). This gives us the average power consumption of the resistor, but what is the equivalent DC voltage which would result in this consumption? We find this by using Watt's law in reverse as follows: $$P = \frac{V^{2}}{R}$$ (3.8) $$14450 \textrm{ Watts} = \frac{V^{2}}{1 \Omega}$$ (3.9) $$\sqrt{14450 \textrm{ Watts}} = V$$ (3.10) $$V = 120 \textrm{ Volts}$$ (3.11) Therefore, 120 VDC would result in the same power consumption over a period of time as a 170 VAC wave. This equivalent is called the Root Mean Square or RMS of the AC voltage. We call it this because it's the square root of the mean (or average) of the square of the original voltage. In other words, a lightbulb in a lamp plugged into the wall (remember, it's being fed $170 V_{peak}$ AC sine wave) will be exactly as bright if it's fed 120 VDC. Just for a point of reference, the RMS value of a sine wave is always 0.707 of the peak value and the RMS value of a square wave (with a 50$\%$ duty cycle) is always the peak value. If you use other waveforms, the relationship between the peak value and the RMS value changes. This relationship between the RMS and the peak value of a waveform is called the crest factor. This is a number that describes the ratio of the peak to the RMS of the signal, therefore $$\textrm{Crest factor} = \frac{V_{peak}}{V_{RMS}}$$ (3.12) So, the crest factor of a sine wave is 1.41 (or $\sqrt{2}$). The crest factor of a square wave is 1. This causes a small problem when you're using a digital volt meter. The reading on these devices ostensibly show you the RMS value of the AC waveform you're measuring, but they don't really measure the RMS value. They measure the peak value of the wave, and then multiply that value by 0.707 - therefore they're assuming that you're measuring a sine wave. If the waveform is anything other than a sine, then the measurement will be incorrect (unless you've thrown out a ton of money on a True RMS multimeter...)