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The method of loop currents, introduced in section 4.4, applies to ac networks as well as dc networks. All we need do is properly label the ac network, then apply the rules laid down in that section. To illustrate the procedure, let us now work through the solution of an ac network problem using the loop current method. In doing this, we ll go into much more detail than a person normally would in working such a problem; we do this, of course, to emphasize the fundamental ideas involved. Thus, for purposes of basic explanation, let us begin with our example network expressed in terms of the fundamental variable time, t, as shown in Fig. 104, in which resistance values are in ohms. The voltages and currents are sinusoidal, as always, with the voltage and current arrows having the meanings described in the discussion following eq. (108). In the gure, note that only the (peak) generator voltages, including their polarity arrows and their phase angles, and the resistance values are given. The unknowns are thus the peak values, Ip1 and Ip2 , of the current waves and their corresponding phase angles a and b, and the peak value Vp of the output voltage wave vo and its phase angle c.{ All phase angles are to be stated with reference to the voltage wave, 43 sin !t, which, since it passes through the origin of the horizontal time axis, will be said to have zero phase
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* See problem 50 and also the footnote in section 4.7. { Angle c, in this case, will be equal to angle b c b because, as pointed out following Fig. 98, there is zero phase shift between the voltage across, and the current through, a resistance of R ohms.
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CHAPTER 5 Sinusoidal Waves. rms Value
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Fig. 104
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shift. (Since the angular amount !t is in radians, the phase angles a, b, c, as they appear in the gure, must also be expressed in radians.) In the gure, let the voltage appearing across the 12-ohm resistor be the output voltage of the network. Thus vo is the instantaneous value of the output voltage. As time continues to increase, vo continually swings back and forth, sinusoidally, between the positive and negative peak values Vp and Vp . Now let the basic PROBLEM be to FIND THE MAGNITUDE OF THE RMS VOLTAGE that would appear across the 12-ohm load resistor (this being the value of voltage that would be read by an ac voltmeter connected across the 12-ohm resistance). To solve the above problem by the method of loop currents, let us begin by returning to the three-step procedure outlined in section 4.4. The three-step rule, as given in Chap. 4, was applied to dc circuits; it should be emphasized, however, that the rules apply AT ANY AND ALL INSTANTS OF TIME in any network. This is true because Kirchho s current and voltage laws (sections 4.2 and 4.3) are, and must be, satis ed at any and all instants of time in any network. Now, in regard to our problem here and the above-mentioned three steps, notice that we automatically satis ed step I when we drew and labeled the current arrows in Fig. 104. Next, step II de nes the rules to be used in writing the VOLTAGE EQUATIONS around each of the n loops in a network (n 2 in Fig. 104). The same rules, given in the original statement of step II in section 4.4, will also apply to the ac circuit of Fig. 104 except that now the plus and minus battery polarity marks will be replaced by voltage arrows associated with each ac generator (see Fig. 89). Thus the original statement of step II would now, for the alternating-current (ac) case, be stated as follows. All generator voltages will be written on the right-hand sides of the equations, and will be considered positive if we go through a generator with the voltage arrow, but negative if we go through the generator against the voltage arrow. The original statement about voltage drops across resistors will remain unchanged. Thus, applying these step II rules to the ac network of Fig. 104, we have the following two voltage equations, which are valid at all instants of time, rst around the left-hand loop (eq. (122)), then around the right-hand loop (eq. (123)). 18Ip1 sin !t a 8Ip2 sin !t b 43 sin !t 36 sin !t 5=6 8Ip1 sin !t a 20Ip2 sin !t b 36 sin !t 5=6 64 sin !t =3 122 123
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