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Hence, for the same amount of real power, the current is reduced by 400 250 =400 0:375 or 37.5 percent. EXAMPLE 10.15 A fourth load Q4 is added in parallel to the three parallel loads of Example 10.12 such that the total power factor becomes 0.8 lagging while the total power remains the same. Find Q4 and the resulting S. discuss the e ect on the current. In Example 10.12 we found total real and reactive powers to be P P1 P2 P3 45 kW and Q Q1 Q2 Q3 54:6 kvar, respectively. For compensation, we now add load Q4 (with P4 0 such that the new power factor is pf cos  0:8 lagging,  36:878. Then, tan 36:878 Q Q4 =P 54:6 Q4 =45 0:75 Q4 20:85 kvar
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The results are summarized in Table 10-2. Addition of the compensating load Q4 reduces the reactive power from 54.6 kvar to 33.75 kvar and improves the power factor. This reduces the apparent power S from 70.75 kVA to 56.25 kVA. The current is proportionally reduced. Table 10-2 Load #1 #2 #3 # 1 2 3 #4 Total P, kW 10 20 15 45 0 45 1 0.5 lagging 0.6 lagging 0.64 lagging 0 leading 0.8 lagging pf Q, kvar 0 34:6 20 54:6 20:85 33:75 S, kVA 10 40 25 70.75 20.85 56.25
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The average power delivered to a load Z1 from a sinusoidal signal generator with open circuit voltage Vg and internal impedance Zg R jX is maximum when Z1 is equal to the complex conjugate 2 of Zg so that Z1 R jX. The maximum average power delivered to Z1 is Pmax Vg =4R.
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EXAMPLE 10.16 A generator, with Vg 100 V(rms) and Zg 1 j, feeds a load Z1 2 (Fig. 10-12). (a) Find the average power PZ1 (absorbed by Z1 ), the power Pg (dissipated in Zg ) and PT (provided by the generator). (b) Compute the value of a second load Z2 such that, when in parallel with Z1 , the equivalent impedance is Z Z1 kZ2 Z g . (c) Connect in parallel Z2 found in (b) with Z1 and then nd the powers PZ ; PZ1 ; PZ2 (absorbed by Z, Z1 , and Z2 , respectively), Pg (dissipated in Zg ) and PT (provided by the generator). p p (a) jZ1 Zg j j2 1 jj 10. Thus I Vg = Z1 Zg 100= 2 1 j and jIj 10 10 A. The required powers are p PZ1 Re Z1 jIj2 2 10 10 2 2000 W p Pg Re Zg jIj2 1 10 10 2 1000 W PT PZ1 Pg 2000 1000 3000 W
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(b) Let Z2 a jb.
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To nd a and b, we set Z1 kZ2 Z 1 j. g
Then,
Z1 Z2 2 a jb 1 j Z1 Z2 2 a jb from which a b 2 0 and a b 2 0. Solving these simultaneous equations, a 0 and b 2; substituting into the equation above, Z2 j2: (b) Z Z1 kZ2 1 j and Z Zg 1 j 1 j 2. Then, I Vg = Z Zg 100= 1 j 1 j 100=2 50 A, and so PZ Re Z I2 1 502 2500 W Pg Re Zg I2 1 502 2500 W
To p PZ1 and PZ2 , we rst nd VZ across Z: VZ IZ 50 1 j . Then IZ1 VZ =Z1 50 1 j =2 nd 25 2 458, and p PZ2 0 W PT Pg PZ1 5000 W PZ1 Re Z1 jIZ1 j2 2 25 2 2 2500 W Alternatively, we can state that PZ2 0 and PZ1 PZ 2500 W
SUPERPOSITION OF AVERAGE POWERS
Consider a network containing two AC sources with two di erent frequencies, !1 and !2 . If a common period T may be found for the sources (i.e., !1 m!, !2 n!, where ! 2=T and m 6 n), then superposition of individual powers applies (i.e., P P1 P2 ), where P1 and P2 are average powers due to application of each source. The preceding result may be generalized to the case of any n number of sinusoidal sources operating simultaneously on a network. If the n sinusoids form harmonics of a fundamental frequency, then superposition of powers applies. P