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10.19 A circuit with impedance Z 10:0 608
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has the power factor improved by a parallel capacitive reactance of 20
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. What percent reduction in current results
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Since I VY, the current reduction can be obtained from the ratio of the admittances after and before addition of the capacitors. Ybefore 0:100 608 S and Yafter 0:050 908 0:100 608 0:062 36:208 S Iafter 0:062 0:620 Ibefore 0:100 so the reduction is 38 percent.
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10.20 A transformer rated at a maximum of 25 kVA supplies a 12-kW load at power factor 0.60 lagging. What percent of the transformer rating does this load represent How many kW in additional load may be added at unity power factor before the transformer exceeds its rated kVA
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For the 12-kW load, S 12=060 20 kVA. The transformer is at 20=25 100% 80% of full rating. The additional load at unity power factor does not change the reactive power, q Q 20 2 12 2 16 kvar (inductive) Then, at full capacity,  0 sin 1 16=25 39:798 P 0 25 cos 39:798 19:2 kW Padd 19:2 12:0 7:2 kW Note that the full-rated kVA is shown by an arc in Fig. 10-24, of radius 25.
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Fig. 10-24
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10.21 Referring to Problem 10.20, if the additional load has power factor 0.866 leading, how many kVA may be added without exceeding the transformer rating
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The original load is S 12 j16 kVA and the added load is S2 S2 308 S2 0:866 jS2 0:500 The total is St 12 0:866S2 j 16 0:500S2 (kVA). Then, kVA
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2 ST 12 0:866S2 2 16 0:500S2 2 25 2
gives S2 12:8 kVA.
AC POWER
[CHAP. 10
10.22 An induction motor with a shaft power output of 1.56 kW has an e ciency of 85 percent. this load, the power factor is 0.80 lagging. Give complete input power information.
Pout 0:85 Pin Then, from the power triangle, Sin 1:765 2:206 kVA 0:80 Qin q 2:206 2 1:765 2 1:324 kvar (inductive) or Pin 1:5 1:765 kW 0:85
The equivalent circuit of an induction motor contains a variable resistance which is a function of the shaft load. The power factor is therefore variable, ranging from values near 0.30 at starting to 0.85 at full load.
Supplementary Problems
10.23 Given a circuit with an applied voltage v 14:14 cos !t (V) and a resulting current i 17:1 cos !t 14:058 (mA), determine the complete power triangle. Ans: P 117 mW; Q 29:3 mvar (inductive); pf 0:970 lagging Given a circuit with an applied voltage v 340 sin !t 608 (V) and a resulting current i 13:3 sin !t 48:78 (A), determine the complete power triangle. Ans: P 2217 W; Q 443 var (capacitive); pf 0:981 leading A two-element series circuit with R 5:0
and XL 15:0
, has an e ective voltage 31.6 V across the resistance. Find the complex power and the power factor. Ans: 200 j600 Va, 0.316 lagging A circuit with impedance Z 8:0 j6:0
has an applied phasor voltage 70:7 90:08 V. complete power triangle. Ans: P 200 W; Q 150 var (capacitive), pf 0:80 leading Obtain the
Determine the circuit impedance which has a complex power S 5031 26:578 VA for an applied phasor voltage 212:1 08 V. Ans: 4:0 j2:0
Determine the impedance corresponding to apparent power 3500 VA, power factor 0.76 lagging, and e ective current 18.0 A. Ans: 10:8 40:548
A two-branch parallel circuit, with Z1 10 08
and Z2 8:0 30:08
, has a total current i 7:07 cos !t 908 (A). Obtain the complete power triangle. Ans: P 110 W; Q 32:9 var (capacitive), pf 0:958 leading A two-branch parallel circuit has branch impedances Z1 2:0 j5:0
and Z2 1:0 j1:0
. complete power triangle for the circuit if the 2.0-
resistor consumes 20 W. Ans: P 165 W; Q 95 var (inductive), pf 0:867 lagging Obtain the
A two-branch parallel circuit, with impedances Z1 4:0 308
and Z2 5:0 608
, has an applied e ective voltage of 20 V. Obtain the power triangles for the branches and combine them to obtain the total power triangle. Ans: ST 128:1 VA, pf 0:989 lagging Obtain the complex power for the complete circuit of Fig. 10-25 if branch 1 takes 8.0 kvar. Ans: S 8 j12 kVA, pf 0:555 lagging In the circuit of Fig. 10-26, nd Z if ST 3373 Va, pf 0:938 leading, and the 3-
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