| RMS Voltage |
RMS Voltage
San Luis Obispo, Calif. In the last article in this series, we
discovered that the average voltage of a full cycle of a sine waveform is 0
volts (since the positive side "cancels out" the negative side), while the
average of a half cycle of a sine waveform is the peak voltage times
2/pi. This was determined first by taking the average of several instantaneous
voltages through the half-cycle, then by increasing the number of samples
towards infinity by applying a little calculus. This time, we'll look at the RMS
voltage of a sine wave.
Root Mean Square
We've heard that RMS stands for "Root Mean Square", but that expression may
not be the model of clarity. If we recall that the arithmetic mean of two
numbers is just the average of those two numbers, the expression starts
to make a little sense. What we are doing is taking the square root of the
average of the squares of the instantaneous voltages.
But he means well...
As a side note, I found this summer while tutoring algebra that the mean
is a number that is part of a sequence of numbers that is between two other
numbers. If the sequence of numbers is an arithemtic sequence (where each
successive number is the previous number plus some constant), each of the
elements of the sequence between two other elements of the sequence are means
of the other two. For example, in the below sequence of numbers, each successive
number is determined by adding three to the previous number.
1, 4, 7, 10, 13, 16
In this sequence, the means of 4 and 13 are 7 and 10. How about if
there is only a single mean? That single mean is then the arithmetic
mean. If we try a couple examples, we find that 7 is the arithmetic mean of
4 and 10. Further, 7 is the average of 4 and 10 (since (4+10)/2=7). So,
to find the arithmetic mean of a couple numbers, just take the average.
Another kind of sequence is a geometric sequence. In a geometric
sequence, each successive number is determined by multiplying the previous
number by a constant (as opposed to adding a constant, which we did
before). The below geometric sequence is formed using a constant of 3.
1, 3, 9, 27, 81, 243
9 and 27 are geometric means of 3 and 81 in this sequence. 9 is the
geometric mean of 3 and 27 in this sequence. We normally find the geometric
mean by taking the squre root of the product of the two numbers. Here, 3*27=81,
and the square root of 81 is indeed 9. How about 27 and 243? 27*43=6,561 and the
square root of 6,561 is 81.
Back on RMS, we'll take the square root of the arithmetic mean of the
squares of the instantaneous voltages. Why? It has something to do with power.
Let's see what the average power delivered by a 1 volt peak sine wave is into a
1 ohm resistor. Starting with the formula for power,
P=IV
and substituting Ohm's Law's I=V/R for I, we get
P=V2/R
Further, the instantaneous voltage of the sine wave is
V(a)=VP*sin(a)
where VP is the peak voltage and a is how far we are in to
the waveform in degrees or radians (depending upon which sine function we're
using). Table 1 shows the voltage at various points through a single cycle of a
1 volt peak sine wave.
| Radians |
Degrees |
Volts |
Power (W) |
| 0 |
0 |
0 |
0 |
| 0.392699 |
22.5 |
0.382683 |
0.1464463 |
| 0.785398 |
45 |
0.707107 |
0.5 |
| 1.178097 |
67.5 |
0.92388 |
0.8535543 |
| 1.570796 |
90 |
1 |
1 |
| 1.963495 |
112.5 |
0.92388 |
0.8535543 |
| 2.356194 |
135 |
0.707107 |
0.5 |
| 2.748894 |
157.5 |
0.382683 |
0.1464463 |
| 3.141593 |
180 |
0 |
0 |
| 3.534292 |
202.5 |
-0.382683 |
0.1464463 |
| 3.926991 |
225 |
-0.707107 |
0.5 |
| 4.31969 |
247.5 |
-0.92388 |
0.8535543 |
| 4.712389 |
270 |
-1 |
1 |
| 5.105088 |
292.5 |
-0.92388 |
0.8535543 |
| 5.497787 |
315 |
-0.707107 |
0.5 |
| 5.890486 |
337.5 |
-0.382683 |
0.1464463 |
Table 1 - Instantaneous voltage and power.
To find the average of these instantaneous powers, we can merely add them up
and divide by the number of samples. My calculator shows the sum of the powers
to be 8.0000. Dividing by the number of samples (16), we find the average power
is 1/2 watt (or 500 mW). This continuously varying voltage dissipates 500 mW in
a 1 ohm resistor. What DC voltage would dissipate that same power?
Above, we found that P=V2/R. Solving for V, we get V=sqrt(P*R).
Since, in this case, R=1 ohm, we can find the voltage by taking the square root
of the power. So... A DC voltage source of sqrt(500mW) would deliver the same
power to a 1 ohm load as a 1 volt peak sine wave does. Doing the square root
(more popular than the Macarena?), we find the equivalent DC voltage is 707 mV.
What did we do? We took the square root of the mean of the squares of the
instantaneous voltages, hence RMS.
If we have a higher peak voltage, all the voltages in the volt column of
table 1 would be multiplied by a constant (the peak voltage). All the powers in
the power column would be multiplied by the peak voltage squared, which would
result in an average of VP2/2. Taking the square root of
the average, we get VP/sqrt(2) as the RMS value of any sine
wave. Messing around with the equation, we find that VP=VRMS*sqrt(2).
Hence, a 117 VAC power line has a peak voltage of 165.463 volts. The
instantaneous voltage varies between 165.463 and -165.463 volts.
Calculus
We were lucky that our average turned out as well as it did. Choosing 16
samples in a cycle gave us the exact relationship between RMS and peak voltage.
Recall from our discussion of average voltage last month that we can keep
increasing the number of samples towards infinity and find the true average of a
continuoously varying waveform. Figure 1 shows how calculus can demonstrate the
relationship between peak voltage and RMS voltage.
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