The 10 Percent Rule: This rule is a standard method for selecting R 1 and R 2 that takes into account
2.21 AC and Resistors, RMS Voltage, and Current
Another special case occurs in Fig. 2.85d, where waveforms A and B are 180° out of phase. Here it doesn’t matter which waveform is considered the leading or lagging waveform. Waveform B is always positive when waveform A is negative, and vice versa. If you combine these two equal voltage or current waveforms together within the same circuit, they completely cancel each other out.
load isn’t purely resistive (e.g., has capacitance and inductance), it’s a whole different story—more on that later.
To find the power dissipated by the resistor under sinusoidal conditions, we can simply plug the sinusoidal voltage expression into Ohm’s power law to get an instan- taneous power expression:
P t V t R
V
RP ft ( ) ( )
sin (2 )
2 2
= = 2 π (2.26)
Expressing voltage, current, and power in an instantaneous fashion is fine, math- ematically speaking; however, in order to get a useful result, you need to plug in a specific time—say, t = 1.3 s. But how often do you need to know that the voltage, current, or power are at exactly t = 1.3 s? Better yet, when do you start counting?
These instantaneous values are typically inconvenient for any practical use. Instead, it is more important to come up with a kind of averaging scheme that can be used to calculate effective power dissipation without dealing with sinusoidal functions.
Now, you might be clever and consider averaging the sinusoidal voltage or cur- rent over a complete cycle to get some meaningful value. However, the average turns out to be zero—positive and negative sides of waveforms cancel. This may be a bit confusing, in terms of power, since the positive- going part of the wave still delivers energy, as does the negative- going part. If you’ve ever received a shock from 120- V line voltage, you’ll be able to attest to that.
The measurement that is used instead of the average value is the RMS or root mean square value, which is found by squaring the instantaneous values of the ac voltage or current, then calculating their mean (i.e., their average), and finally taking the square root of this—which gives the effective value of the ac voltage or current.
These effective, or RMS, values don’t average out to zero and are essentially the ac equivalents of dc voltages and currents. The RMS values of ac voltage and current are based upon equating the values of ac and dc power required to heat a resistive element to exactly the same degree. The peak ac power required for this condition
FIGURE 2.86
is twice the dc power needed. Therefore, the average ac power equivalent to a cor- responding average dc power is half the peak ac power.
ave 2 P Ppeak
= (Average dc power equivalent of ac waveform) (2.27) Mathematically, we can determine the RMS voltage and current values for sinu- soidal waveforms V(t) = VP sin (2π × f × t), and I(t) = IP sin (2π × f × t):
1 ( ) 1
2 0.707
RMS 0
V 2
T V t dt V V
T
P P
∫
= = × = × (2.28) RMS
voltage
1 ( ) 1
2 0.707
RMS 0
I 2
T I t dt I I
T
P P
∫
= = × = × (2.29) RMS current
Notice that the RMS voltage and current depend only on the peak voltage or current—
they are independent of time or frequency. Here are the important relations, without the scary calculus:
2 1.414 0.707 1.414
2 1.414 0.707 1.414
RMS RMS
RMS RMS
V V V
V V V
I I I
I I I
P P
P P
P P
P P
= = = × = ×
= = = × = ×
For example, a U.S. electric utility provides your home with 60 Hz, 120 VAC (in Europe and many other countries it’s 50 Hz, 240V AC). The “VAC” unit tells you that the supplied voltage is given in RMS. If you were to attach an oscilloscope to the outlet, the displayed waveform on the screen would resemble the following function:
V(t) = 170 V sin (2π × 60 Hz × t), where 170 V is the peak voltage.
With our new RMS values for voltage and current, we can now substitute them into Ohm’s law to get what’s called ac Ohm’s law:
VRMS = IRMS × R (2.30) ac Ohm’s law Likewise, we can use the RMS voltage and current and substitute them into Ohm’s power law to get what’s called the ac power law, which gives the effective power dis- sipated (energy lost per second):
= × = =
P I V V
R I R
RMS RMS
2RMS 2
RMS (2.31) ac power law Again, these equations apply only to circuits that are purely resistive, meaning there is virtually no capacitance and/or inductance. Doing power calculation on cir- cuits with inductance and capacitance is a bit more complicated, as we’ll see a bit later.
Figure 2.87 shows the relationships between RMS, peak, peak- to- peak and half- wave average values of voltage and current. Being able to convert from one type to another is important, especially when dealing with component maximum voltage and current ratings—sometimes you’ll be given the peak value, other times the RMS value. Understanding the differences also becomes crucial when making test
measurements—see the note on pages 91–92 about making RMS test measurements.
Most of the time, when dealing with ac voltage you can assume that voltage is expressed as an RMS value unless otherwise stated.
Example 1: How much current would flow through a 100- Ω resistor connected across the hot and neutral sockets of a 120- VAC outlet? How much power would be dis- sipated? What would the results be using 1000- Ω, 10,000- Ω, and 100,000- Ω resistors?
Answer:
First, don’t try this with any ordinary resistor; you’d need a power resistor or spe- cial heating element with a power rating of greater than 144 W! (Also, simply don’t attempt attaching a resistor with the outlets powered.) A 1000- Ω resistor plugged into the same outlet would dissipate 14.4 W; a 10,000- Ω resistor would dissipate 1.44 W; a 100,000- Ω resistor would dissipate 0.14 W.
Example 2: What is the peak voltage on a capacitor if the RMS voltage of a sinusoidal waveform signal across it is 10.00 VAC?
Answer: VAC means RMS, so VP = 2 × VRMS = 1.414 × 10 V = 14.14 V.
Example 3: A sinusoidal voltage displayed on an oscilloscope has a peak amplitude of 3.15 V. What is the RMS value of the waveform?
Conversion Factors for AC Voltage and Current
FROM TO MULTIPLY BY
Peak Peak- to- peak 2 Peak- to- peak Peak 0.5
Peak RMS 1/ 2 or 0.7071
RMS Peak 2 or 1.4142
Peak- to- peak RMS 1/( 2 2) or 0.35355 RMS Peak- to- peak (2 × 2) or 2.828
Peak Average* 2/o or 0.6366
Average* Peak o/2 or 1.5708
RMS Average* (2 × 2)/o or 0.9003 Average* RMS o/( 2 × 2) or 1.1107
* Represents average over half a cycle.
FIGURE 2.87
FIGURE 2.88
Answer:
V VP
2
3.15 V
1.414 2.23 VAC
RMS = = =
Example 4: A 200- W resistive element in a heater is connected to a 120- VAC outlet.
How much current is flowing through the resistive element? What’s the resistance of the element, assuming it’s an ideal resistor?
Answer: IRMS = PAVE/VRMS = 200 W/120 VAC = 1.7 A. R = VRMS/IRMS = 120 V/1.7 A = 72 Ω.
Example 5: A sinusoidal voltage supplied by a function generator is specified as 20 V peak to peak at 1000 Hz. What is the minimum resistance value of a 1⁄8-W resistor you can place across the generator’s output before exceeding the resistor’s power rating?
Answer: VP = 1/2 VP− P = 10 V; VRMS = 0.707 × VP = 7.1 VAC; R = VRMS2/PAVE = (7.1)2/ (1/8 W) = 400 Ω.
Example 6: The output of an oscillator circuit is specified as 680 mVAC. If you feed this into an op amp with an input resistance of 10 MΩ, how much current enters the IC?
Answer: IRMS = VRMS/R = 0.68 V/(10,000,000 Ω) = 0.000000068 = 68 nA.
MEASURING RMS VOLTAGES AND CURRENTS
There are many digital multimeters that do not measure the RMS value of an ac voltage directly. Often the meter will simply measure the peak value and calculate the equivalent RMS value—assuming the measured waveform is sinusoidal—and then display this value. Analog meters usually measure the half- wave average value, but are made to indicate the equivalent RMS.
True RMS multimeters, on the other hand, measure the true RMS value of voltages and current, and are especially handy for nonsinusoidal voltage and currents. Though relatively expensive, these meters are worth the price. It’s important to note that true RMS meters will also include the contribution of any dc voltages or current components present along with the ac.
Fortunately, you can still get a fairly accurate idea of the RMS value of a sine waveform, knowing one of the other measurements such as the half- wave average, peak, or peak- to- peak value. This can be done by calculation, or using the table in Fig. 2.89. As you can see, it’s also possible to work out the RMS value of a few other symmetrical and regular waveforms, such as square and triangular waves, knowing their peak, average, or peak- to- peak values.
An important thing to note when using the table is that you need to know the exact basis on which your meter’s measurement is made. For example, if your meter measures the peak value, and then calculates and displays the equivalent sine wave RMS figure, this means you’ll need to use the table dif- ferently when compared to the situation where the meter really measures the