【正文】 water is determined by the dc power dissipated in the form of heat by the resistor.If switch 2 is closed and switch 1 left open, the ac current through the resistor will have a peak value of Im. The temperature reached by the water is now determined by the ac power dissipated in the form of heat by the resistor. The ac input is varied until the temperature is the same as that reached with the dc input. When this is acplished, the average electrical power delivered to the resistor R by the ac source is the same as that delivered by the dc source. The power delivered by the ac supply at any instant of time isThe average power delivered by the ac source is just the first term, since the average value of a cosine wave is zero even though the wave may have twice the frequency of the original input current waveform. Equating the average power delivered by the ac generator to that delivered by the dc source,which, in words, states thatthe equivalent dc value of a sinusoidal current or voltage is 1/2
or of its maximum value.The equivalent dc value is called the effective value of the sinusoidal quantity.In summary,As a simple numerical example, it would require an ac current with a peak value of 2
(10)
A to deliver the same power to the resistor in Fig. as a dc current of 10 A. The effective value of any quantity plotted as a function of time can be found by using the following equation derived from the experiment just described: which, in words, states that to find the effective value, the function i(t) must first be squared. After i(t) is squared, the area under the curve isfound by integration. It is then divided by T, the length of the cycle or the period of the waveform, to obtain the average or mean value of thesquared waveform. The final step is to take the square root of the meanvalue. This procedure gives us another designation for the effectivevalue, the rootmeansquare (rms) value. In fact, since the rms term isthe most monly used in the educational and industrial munities,it will used throughout this text.EXAMPLE Find the rms values of the sinusoidal waveform in each part of Fig. .Solution: For part (a), Irms
(12 10 3 A)
part (b), again Irms
mA. Note that frequency did notchange the effective value in (b) above pared to (a). For part (c),Vrms
( V)
120 V, the same as available from a home outlet.EXAMPLE The 120V dc source of Fig. (a) delivers W to the load. Determine the peak value of the applied voltage (Em) and the current (Im) if the ac source [Fig. (b)] is to deliver the same power to the load.Solution:EXAMPLE Find the effective or rms value of the waveform of Fig. .Solution:EXAMPLE Calculate the rms value of the voltage of Fig. .Solution:EXAMPLE Determine the average and rms values of the square wave of Fig. .Solution: By inspection, the average value is zero.The waveforms appearing in these examples are the same as thoseused in the examples on the average value. It might prove interesting topare the rms and average values of these rms values of sinusoidal quantities such as voltage or currentwill be represented by E and I. These symbols are the same as thoseused for dc voltages and currents. To avoid confusion, the peak valueof a waveform will always have a subscript m associated with it: Imsin qt. Caution: When finding the rms value of the positive pulse of asine wave, note that the squared area is not simply (2Am)2
4A2m。 it is not necessary to be proficient in its use to continue with this text. It is a useful mathematical tool, however,and should be learned. Finding the area under the positive pulse of a sine wave using integration, we havewhere ∫?is the sign of integration, 0 and p are the limits of integration, Am sin a is the function to be integrated, and da indicates that we are integrating with respect to a. Integrating, we obtainSince we know the area under the positive (or negative) pulse, we can easily determine the average value of the positive (or negative) region of a sine wave pulse by applying Eq. ():
For the waveform of Fig. ,
EXAMPLE Determine the average value of the sinusoidal waveform of Fig. .Solution: By inspection it is fairly obvious thatthe average value of a pure sinusoidal waveform over one full cycle iszero.EXAMPLE Determine the average value of the waveform of Fig. .Solution: The peaktopeak value of the sinusoidal function is16 mV +2 mV
= 18 mV. The peak amplitude of the sinusoidal waveform is, therefore, 18 mV/2
= 9 mV. Counting down 9 mV from 2 mV(or 9 mV up from 16 mV) results in an average or dc level of 7 mV,as noted by the dashed line of Fig. .EXAMPLE Determine the average value of the waveform of Fig. .Solution:EXAMPLE For the waveform of Fig. , determine whether the average value is positive or negative, and determine its approximate value.Solution: From the appearance of the waveform, the average value is positive and in the vicinity of 2 mV. Occasionally, judgments of this type will have to be made.InstrumentationThe dc level or average value of any waveform can be found using a digital multimeter (DMM) or an oscilloscope. For purely dc circuits,simply set the DMM on dc, and read the voltage or current are limited to voltage levels using the sequence of steps listed below:1. First choose GND from the DCGNDAC option list associated with each vertical channel. The GND option blocks any signal to which the oscilloscope probe may be connected from entering the oscilloscope and responds with just a horizontal line. Set the resulting line in the middle of the vertical axis on the horizontal axis, as shown in Fig. (a).2. Apply the oscilloscope probe to the voltage to be measured (if not already connected), and switch to the DC option. If a dc voltage is present, the horizontal line will shift up or down, as demonstrated in Fig. (b)