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CHAPTER 8 Reactance and Impedance
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product VI the apparent power, because it must be multiplied by the power factor, cos , to get the true power P. This can be expressed by writing eq. (228) in the form true power apparent power cos  that is, Pt Pa cos  VI cos  229 where Pt true power, and Pa apparent power VI. Now recall that in an ac circuit ENERGY is actually expended ONLY in the RESISTIVE component of the circuit impedance. Thus TRUE POWER, in an ac circuit, is always equal to the square of the rms current, times the resistance R ; that is Pt I 2 R hence eq. (229) becomes I 2 R VI cos  230 It is also true that V IZ (eq. (200)), and thus, substituting this value of V in eq. (230), we have the important fact that cos  R=Z 231
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where R is the circuit resistance and Z is the magnitude of the circuit impedance (as can also be seen from inspection of Figs. 130 and 139). Hence another expression for the power factor is R/Z, and thus eq. (229) is extended to the form Pt VI cos  VI R=Z 232 To continue the discussion, let us rst redraw Fig. 130 as Fig. 155, where !L X and " jZ j Z.
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Next, multiply all three sides of the triangle in Fig. 155 by the square of the magnitude of the rms current, I 2 . Doing this preserves the angle , and Fig. 155 becomes Fig. 156 where, from our work above, I 2 R Pt and (since V IZ) VI I 2 Z Pa , as shown in Fig. 157. The meaning of Px is as follows. Since reactance, X, is measured in ohms, it follows that I 2 X is measured in watts; that is, I 2 X represents what is called reactive power. Thus Px represents power that does not represent energy that is transformed into heat, light, or mechanical energy, but only energy that is momentarily stored in electric and/or magnetic elds and then returned to the source. From inspection of Fig. 157, note that P2 P2 P2 a t x 233 To summarize, the true power Pt in an ac circuit is given by eq. (228), in which cos  is the power factor, where cos  R=Z Pt =Pa 234
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CHAPTER 8 Reactance and Impedance
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It should be noted that the above conclusions are valid for any con guration of series, " " parallel, or series-parallel network. Thus, if a generator of V volts delivers a current of I " R jX ohms amperes into a certain network, and if the generator sees an impedance Z looking into the network, then the true power produced by the generator is equal to Pt VI cos , where cos  R=Z. Problem 133 In problem 116, nd (a) the apparent power, (b) the true power, (c) the reactive power. Problem 134 In problem 119 (Fig. 134), nd the true power produced by the generator if V 28 volts, R 12 ohms, and !L 16 ohms. (Answer: 65.333 W, approx.) Problem 135 In Fig. 158, the circuit values are in ohms and henrys, the generator voltage being 32 V rms, as shown. Given that ! 1000 rad/sec, nd the true power produced by the generator. (Answer: 226.98 W, approx.)
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Problem 136 Rework problem 135, this time using the method of loop currents. Remembering that the power delivered to a pure resistance of R ohms is always equal to the square of the magnitude of the rms current, times R P I 2 R , nd (a) power delivered to the 2-ohm resistance, (b) power delivered to the 4-ohm resistance, (c) check to verify that the sum of the answers in (a) and (b) is the same as the answer found in problem 135.
Problem 137 Find (a) the apparent power, (b) the true power, produced by the generator in problem 124 (Fig. 141). Problem 138 Find the true power produced by the generator in Fig. 142, making use of the value " of I1 found in problem 126. Problem 139 In Fig. 142, nd (a) power to the 15-ohm resistance, (b) power to the 10-ohm resistance. Verify that the sum of (a) and (b) is the same as the answer found in problem 138.
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