【正文】 that is, the area above the axis will equal the area below.3. Then switch to DC (to permit both the dc and the ac ponents of the waveform to enter the oscilloscope), and note the shift in the chosen level of part 2, as shown in Fig. (b). Equation() can then be used to determine the dc or average value of the waveform. For the waveform of Fig. (b), the average value is aboutThe procedure outlined above can be applied to any alternating waveform such as the one in Fig. . In some cases the average value may require moving the starting position of the waveform under the AC option to a different region of the screen or choosing a higher voltage scale. DMMs can read the average or dc level of any waveform by simply choosing the appropriate scale. EFFECTIVE (rms) VALUESThis section will begin to relate dc and ac quantities with respect to the power delivered to a load. It will help us determine the amplitude of a sinusoidal ac current required to deliver the same power as a particular dc current. The question frequently arises, How is it possible for a sinusoidal ac quantity to deliver a power if, over a full cycle, the current in any one direction is zero (average value
0)? It would almost appear that the power delivered during the positive portion of the sinusoidal waveform is withdrawn during the negative portion, and since the two are equal in magnitude, the power delivered is zero. However, understand that irrespective of direction, current of any magnitude through a resistor will deliver power to that resistor. In other words, during the positive or negative portions of a sinusoidal ac current, power is being delivered at eachinstant of time to the resistor. The power delivered at each instant will, of course, vary with the magnitude of the sinusoidal ac current, but there will be a flow during either the positive or the negative pulses with a flow over the full cycle. The power flow will equal twice that delivered by either the positive or the negative regions of sinusoidal quantity. A fixed relationship between ac and dc voltages and currents can be derived from the experimental setup shown in Fig. . A resistor in a water bath is connected by switches to a dc and an ac supply. If switch 1 is closed, a dc current I, determined by the resistance R and battery voltage E, will be established through the resistor R. The temperature reached by the 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 b