Chapter 12 Solutions to Exercises

Engineering Circuit Analysis, 7th Edition

1.

Chapter Twelve Solutions

10 March 2006

Vbc = Vbe + Vec = 0.7 – 10 = -9.3 V Veb = - Vbe =

-0.7 V

Vcb = Vce + Veb = 10 – 0.7 = 9.3 V

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Engineering Circuit Analysis, 7th Edition

2.

(a) Vgd = Vgs + Vsd = -1 – 5

Chapter Twelve Solutions

10 March 2006

= -6 V

(b) Vsg = Vsd + Vdg = -4 – 2.5

= -6.5 V

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Engineering Circuit Analysis, 7th Edition

3.

Chapter Twelve Solutions

10 March 2006

(a) positive phase sequence Van = |Vp| ∠ 0o Vbn = |Vp| ∠ -60o Vcn = |Vp| ∠ -120o

Vdn = |Vp| ∠ -180o Ven = |Vp| ∠ -240o Vfn = |Vp| ∠ -300o

(b) negative phase sequence Van = |Vp| ∠ 0o Vbn = |Vp| ∠ 60o Vcn = |Vp| ∠ 120o

Vdn = |Vp| ∠ 180o Ven = |Vp| ∠ 240o Vfn = |Vp| ∠ 300o

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Engineering Circuit Analysis, 7th Edition

4.

(a) Vyz = Vyx + Vxz

Chapter Twelve Solutions

10 March 2006

= -110 ∠20o + 160 ∠ -50o = -103.4 – j37.62 + 102.8 – j122.6 = -0.6 – j160.2 = 160.2 ∠ -90.21o V

(b) Vaz = Vay + Vyz

= 80 ∠130o + 160.2 ∠ -90.21o = -51.42 + j61.28 -0.6 – j160.2 = -52.02 – j98.92 = 111.8 ∠ -117.7o V

(c)

Vzx - 160∠ - 50 o 160∠130 o = = = 1.455∠110 o Vxy 110∠20 o 110∠20 o

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Engineering Circuit Analysis, 7th Edition

5.

(a) V25 = V24 + V45

Chapter Twelve Solutions

10 March 2006

= -80 ∠ 120o + 60 ∠ 75o = 40 – j69.28 + 15.53 + j57.96 = 55.53 – j11.32 = 56.67 ∠ -11.52o V

(b) V13 = V12 + V25 + V53

= 100 + 55.53 – j11.32 + j120 = 155.53 + j108.7 = 189.8 ∠ 34.95o V

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Engineering Circuit Analysis, 7th Edition

6.

Chapter Twelve Solutions

10 March 2006

V12 = 9∠87o V = 0.4710 + j8.988 V, V23 = 8∠45o V = 5.657 + j 5.657 V

(

)

(

)

(

)

(a) V21 = – V12 = 9∠ 180o + 87 o V = 9∠ 267o V = 9∠ −93o V

(b) V32 = – V23 = 8∠ (180o + 45o ) V = 8∠ ( 225o ) V = 8∠ ( −135o ) V (c) V12 – V32 = V12 + V23 = 0.4710 + j8.988 + 5.657 + j 5.657 = 6.128 + j14.65 V

= 15.88∠67.29o V

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Engineering Circuit Analysis, 7th Edition

7.

Chapter Twelve Solutions

10 March 2006

Vbn

(a)

135o

75o 45o

Vcn Van

(b) The phase sequence is negative, since sequence is acbacb…. A positive sequence would be abcabc…

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Engineering Circuit Analysis, 7th Edition

8.

Chapter Twelve Solutions

10 March 2006

The temptation is to extend the procedure for voltages, but without the specific circuit topology, we do not have sufficient information to determine I31.

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Engineering Circuit Analysis, 7th Edition

9.

Chapter Twelve Solutions

10 March 2006

The temptation is to extend the procedure for voltages, but without the specific circuit topology, we do not have sufficient information to determine I31.

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

10. 230 / 460 V rms Z AN : S = 10∠40° kVA; Z NB: 8∠10° kVA; Z AB : 4∠ − 80° kVA

∗ ∗ Let VAN = 230∠0° V ∴ SAN = VAN IAN , IAN =

10, 000∠40° = 43.48∠40° A 230

4000∠ − 80° = 8.696∠ − 80°, IAB = 8.696∠80°∴ IaA = IAN + IAB 460 ∴ IaA = 43.48∠40° + 8.696∠80° = 39.85− ∠ − 29.107° ∴ I aA = 39.85− A ∗ ∗ ∴ IAN = 43.48∠ − 40° A, SAB = VAB IAB ∴ IAB =

8000∠10° = 34.78∠10°, INB = 34.78∠ − 10° A 230 ∴ IbB = −34.78∠ − 10° − 8.696∠80° = 35.85+ ∠ − 175.96°, ∴ IbB = 35.85+ A ∗ INB =

InN = −43.48∠ − 40° + 34.78∠ − 10° = 21.93∠87.52°, I nN = 21.93A

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Engineering Circuit Analysis, 7th Edition

11.

Chapter Twelve Solutions

10 March 2006

(a) InN = 0 since the circuit is balanced. 240∠0 IAN = 12 ∠0 IAB = = 12 ∠ - 36.9 o 16 + j12 IaA = IAN + IAB = 12 + 9.596 – j7.205 = 22.77 ∠ -18.45o A (b) IAN = 24 ∠ 0o A IBN = -12 ∠ 0o A InN = -12 ∠ 0o A

The voltage across the 16-Ω resistor and j12-Ω impedance has not changed, so IAB has not changed from above. IaA = IAN + IAB = 24 ∠ 0o + 12 ∠ -36.9o = 34.36 ∠ -12.10o A IbB = IBN - IAB = -12 ∠ 0o - 12 ∠ -36.9o = 7.595 ∠ -108.5o A InN = IBN – IAN = -12 – 24 = 36 ∠180o A

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

12.

(a)

−10 −10 − j 3 21 + j 3 19 + j 2 −8 − j 2 = (21 + j 3) (674 + j167 − 60 − j 32) Δ = −10 −10 − j 3 −8 − j 2 36 + j5 + 10(−360 − j 50 − 74 − j 44) − (10 + j 3) (80 + j 20 + 184 + j 77) ∴ Δ = 5800 + j1995 = 6127 ∠18.805° 720 −10 −10 − j 3 720 19 + j 2 −8 − j 2 = 720(614 + j135 + 434 + j 94) = 720 ×1072.7∠12.326° 0 −8 − j 2 36 + j 5

∴ IaA =

(b)

720 × 1072.7∠12.326° = 126.06∠ − 6.479° A 6127∠18.805°

21 + j 3 720 −10 − j 3 720 (1084 + j 247) −10 720 −8 − j 2 = 720 (1084 + j 247) ∴ IBb = = 130.65− ∠ − 5.968° A 6127∠18.805° −10 − j 3 0 36 + j 5 ∴ I nN = 130.65− ∠ − 5.968° − 126.06∠ − 6.479° = 4.730∠7.760° A

(c)

Pω ,tot = 126.062 × 1 + 130.652 × 1 + 4.7302 × 10 = 15.891 + 17.069 + 0.224 = 33.18 kW

(d)

Pgen ,tot = 720 × 126.06 cos 6.479° + 720 × 130.65− cos 5.968° = 90.18 + 93.56 = 183.74 kW

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Engineering Circuit Analysis, 7th Edition

13.

VAN = 220 Vrms, 60 Hz

(a)

PF = 1 ∴ IAN =

Chapter Twelve Solutions

10 March 2006

220∠0° = 40.85+ ∠ − 21.80° A; IAB = j 377C × 440 5 + j2

∴ IaA = 40.85cos 21.80° + j (377C440 − 40.85sin 21.80°) ∴C =

(b)

40.85sin 21.80° = 91.47 μ F 377 × 440

IAB = 377 × 91.47 ×10−6 × 440 = 15.172 A ∴ VA = 440 × 15.172 = 6.676 kVA

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Engineering Circuit Analysis, 7th Edition

14.

Chapter Twelve Solutions

10 March 2006

200∠0 400∠0 400 + = 15.69 – j3.922 + 12 + j 3 R AB R AB Since we know that |IaA| = 30 A rms = 42.43 A,

(a) IaA = IAN + IAB =

2

⎛ 400 ⎞ ⎟⎟ + 3.922 2 42.43 = ⎜⎜15.69 + R AB ⎠ ⎝

or RAB = 15.06 Ω 200∠0 400∠0 j 400 + = 15.69 - j 3.922 + 12 + j 3 - jX AB X AB 400 In order for the angle of IaA to be zero, = 3.922, so that XAB = 102 Ω capacitive. X AB (b) IaA = IAN + IAB =

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

15.

+ seq. VBC = 120∠60° V rms, R w = 0.6 Ω

(a)

120 5000 5000 ∠150° V ∴ SAN = 0.6 × 0.8 + j 3 3 3 120 ∴ SAN = ∠150° IaA∗ ∴ IaA∗ = 24.06 ∠ − 113.13° A 3 ∴ IaA = 24.06∠113.13° ∴ Pwire = 3 × 24.062 × 0.6 = 1041.7 W

(b)

VaA = 0.6 × 24.06∠113.13° = 14.434∠113.13° V

10 March 2006

Pload = 5 kVA, 0.6 lag

VAN =

∴ Van = VaA + VAN = 14.434∠113.13° +

120 ∠158° = 81.29∠143.88° V 3

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

16.

↑ Van = 2300∠0° Vrms , R w = 2 Ω, + seq., Stot = 100 + j 30 kVA

(a)

1 (100, 000 + j 30, 000) = 2300 I∗aA ∴ IaA = 15.131∠ − 16.699° A 3

(b)

VAN = 2300 − 2 × 15.131∠ − 16.699° = 2271∠0.2194° V

(c)

Z p = VAN / IaA =

(d)

trans. eff. =

10 March 2006

2271∠0.2194° = 143.60 + j 43.67 Ω 15.131∠ − 16.699°

143.60 = 0.9863, or 98.63% 145.60

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

17.

↑ Z p = 12 + j 5Ω, IbB = 20∠0° A rms, +seq., PF = 0.935

(a)

θ = cos −1 0.935 = 20.77°∴

(b)

VBN = IbB Z p = 20 (12 + j5) = 240 + j100 V ∴Vbn = 20(13.1821 + j 5) = 281.97∠20.77° V

(c)

VAB = 3 VBN / ∠VBN + 150° = 450.3∠172.62° V

(d)

Ssource = 3 VBn IbB∗ = 3 × 281.97 ∠ − 20.77° (20)

5 = tan 20.77°, R w = 1.1821Ω 12 + R w

= 15.819 − j 6.000 kVA

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Engineering Circuit Analysis, 7th Edition

18.

Chapter Twelve Solutions

125 mH → j(2π)(60)(0.125) = j47.12 Ω 55 μF → -j/(2π)(60)(55×10-6) = -j48.23 Ω The per-phase current magnitude |I| is then I =

10 March 2006

75 Ω → 75 Ω 125 75 + (47.12 − 48.23) 2 2

= 1.667 A.

The power in each phase = (1.667)2 (75) = 208.4 W, so that the total power taken by the load is 3(208.4) = 625.2 W. ⎛ 47.12 − 48.23 ⎞ The power factor of the load is cos⎜ ⎟ = 1.000 75 ⎠ ⎝ This isn’t surprising, as the impedance of the inductor and the impedance of the capacitor essentially cancel each other out as they have approximately the same magnitude but opposite sign and are connected in series.

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

19.

↑ Bal.,+ seq. Z AN = 8 + j 6 Ω, Z BN = 12 − j16 Ω, ZCN = 5 + j 0, VAN = 120∠0° V rms R w = 0.5 Ω (a) − InN =

120∠0° 120∠ − 120° 120∠120° + + = 6.803∠83.86° A 8.5 + j 6 12.5 − j16 5.5

∴ InN = 6.803∠ − 96.14° A rms

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Engineering Circuit Analysis, 7th Edition

20.

Chapter Twelve Solutions

10 March 2006

Working on a per-phase basis, the line current magnitude is simply

I =

40

(R w + 5)2 + 10 2

(a) RW = 0 Then I =

40

= 3.578 A , and the power delivered to each phase of the load is 25 + 10 2 (3.578)2(5) = 64.01 W. The total power of the load is therefore 3(64.01) = 192.0 W.

(b) RW = 3 Ω Then I =

40

= 3.123 A , and the power delivered to each phase of the load is 64 + 10 2 (3.123)2(5) = 48.77 W. The total power of the load is therefore 3(48.77) = 146.3 W.

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Engineering Circuit Analysis, 7th Edition

21. (a)

Chapter Twelve Solutions

10 March 2006

↑ Z p = 75∠25° Ω 25 μ F, Van = 240∠0° V rms, 60 Hz, R w = 2 Ω 106 75∠25°(− j106.10) = − j 106.10 Ω ∴ Z p = = 75.34 − j 23.63 Ω Zcap = − j 377 × 25 75∠25° − j106.10 240 ∴ Z p + w = 77.34 − j 23.63 ∴ IaA = = 2.968∠16.989° A 77.34 − j 23.63

(b)

Pw = 3(2.968) 2 × 2 = 52.84 W

(c)

Pload = 3(2.968) 2 75.34 = 1990.6 W

(d)

PFsource = cos16.989° = 0.9564 lead

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Engineering Circuit Analysis, 7th Edition

22.

Chapter Twelve Solutions

10 March 2006

Working on a per-phase basis and noting that the capacitor corresponds to a –j6366-Ω impedance,

-j6366 || 100 ∠ 28o = 89.59 + j46.04 Ω so that the current flowing through the combined load is 240 I = = 2.362 A rms 90.59 2 + 46.04 2 The power in each phase is (2.362)2 (90.59) = 505.4 W, so that the power deliverd to the total load is 3(505.4) = 1.516 kW. The power lost in the wiring is (3)(2.362)2 (1) = 16.74 W.

Simulation Result: FREQ IM(V_PRINT1) 5.000E+01 1.181E+00

IP(V_PRINT1) -2.694E+01

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

23.

↑ Bal., R w = 0, Z p = 10 + j 5 Ω, f = 60 Hz

(a)

10 + j 5 = 11.180∠26.57° ∴ PF = cos 26.57° = 0.8944

(b)

1 = 0.08 − j 0.04S 11.180∠26.57° 377C − 0.04 = − tan 21.57° = −0.3952 Yp′ = 0.08 + j (377C − 0.04) ∴ 0.08 ∴ 377C = 0.04 − 0.08 × 0.3952 = 0.00838∴ C = 22.23 μ F

(c)

VL ,load = 440 V rms, Zc =

10 March 2006

PF = 0.93 lag, θ = 21.57°, Yp =

∴ VAR = 2.129 ×

440 / 3 − j106 = − j119.30 Ω, Ic = = 2.129 A 120π 22.23 119.30

440 = 540.9 VAR (cap.) 3

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Engineering Circuit Analysis, 7th Edition

24.

Chapter Twelve Solutions

10 March 2006

Working from the single-phase equivalent, 1 ⎛ 115∠0 o ⎞ ⎟⎟ = 46.9 ∠0o V rms ⎜⎜ Van rms = 3⎝ 2 ⎠ 1.5 H → j565 Ω, 100 μF → -j26.5 Ω and 1 kΩ → 1 kΩ. These three impedances appear in parallel, with a combined value of 27.8 ∠ -88.4o Ω. Thus, |Irms| = 46.9/ 27.8 = 1.69 A rms Zload = 27.8 ∠88.4o = 0.776 – j 27.8 Ω, so Pload = (3)(1.69)2 (0.776) = 2.22 W.

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

25. R w = 0, Van = 200∠60° V rms. S p = 2 − j1 kVA + seq.

(a)

Vbc = 220 3∠ − 30° = 346.4∠ − 30° V

(b)

∗ ∗ SBC = 2000 − j1000 = VBC IBC = 346.4∠ − 30° IBC ∗ ∴ IBC = 6.455− ∠3.435°, IBC = 6.455− ∠ − 3.435°

∴Zp =

(c)

200 3∠ − 30° = 53.67∠ − 26.57° = 48 − j 24 Ω 6.455− ∠ − 3.435°

IaA = IAB − ICA = 6.455− ∠120° − 3.43° − 6.455− ∠ − 120° − 3.43° = 11.180∠86.57° A rms

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

26.

↑ 15kVA, 0.8lag, +seq., VBC = 180∠30° V rms, R w = 0.75 Ω

(a)

∗ VBC = 180∠30° ∴ VAB = 180∠150° V, Sp = 5000∠ cos −1 0.8 = 5000∠36.87° = 180∠30° IBC

∴ IBC = 27.78∠ − 6.87° and IAB = 27.78 ∠113.13° A ∴ IbB = IBC − IAB ∴ IbB = 27.78(1∠ − 6.87° − 1∠113.13°) = 48.11∠ − 36.87° A ∴ VbC = 0.75( IbB − IcC ) ∴ VbC = 0.75 × 48.11(1∠ − 36.87° − 1∠ − 156.87°) + 180∠30° = 233.0∠20.74° V (b)

Pwire = 3 × 48.112 × 0.75 = 5208 W Sgen = 5208 + 15, 000 ∠36.87° = 17.208 + j 9.000 kVA

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

27.

↑ Bal., SL = 3 + j1.8 kVA, Sgen = 3.45 + j1.8 kVA, R w = 5 Ω

(a)

1 Pw = 450 W ∴ × 450 = I 2aA × 5 ∴ I aA = 5.477 A rms 3

(b)

I AB =

(c)

10 March 2006

1 × 5.477 = 3.162 A rms 3

1 ∗ = VAB (3.162∠0°) Assume IAB = 3.162∠0° and +seq. ∴ (3000 + j1800) = VAB I AB 3 ∴ VAB = 368.8∠30.96° V ∴ Van = VaA + VAB − VbB + Vbn VaA = 5 IaA = 5 × 5.477∠ − 30° = 27.39∠ − 30°, VbB = 27.39∠ − 150° ∴ Van = 27.39∠ − 30° − 27.39∠ − 150° + 368.8∠30.96° + Van (1∠ − 120°) ∴ Van =

27.39∠ − 30° − 27.39∠ − 150° + 368.8∠30.96° = 236.8∠ − 2.447° ∴Van = 236.8 V rms 1 − 1∠ − 120°

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Engineering Circuit Analysis, 7th Edition

28.

Chapter Twelve Solutions

10 March 2006

If a total of 240 W is lost in the three wires marked Rw, then 80 W is lost in each 2.3-Ω 80 = 5.898 A rms . Since this is a D-connected load, segment. Thus, the line current is 2.3 the phase current is 1/ 3 times the line current, or 3.405 A rms. In order to determine the phase voltage of the source, we note that ⎛ 2⎞ ⎟ = 1800 Ptotal = 3 Vline ⋅ I line ⋅ PF = 3 Vline (5.898) ⎜⎜ ⎟ ⎝ 2 ⎠ (1800)(2) = 249.2 V where |Vline| = 2 3 (5.898) This is the voltage at the load, so we need to add the voltage lost across the wire, which ⎡ ⎛ 1 ⎞⎤ (taking the load voltage as the reference phase) is ⎢5.898∠ − cos −1 ⎜ ⎟⎥ (R W ) 2 ⎝ ⎠⎦ ⎣ o = 13.57 ∠-45 V. Thus, the line voltage magnitude of the source is |249.2 ∠ 0o + 13.57 ∠ -45o| = 259.0 V rms.

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

29.

Bal., +seq.

(a)

Van = 120∠0° ∴ Vab = 120 3 ∠30°, etc., IAB = IBC =

10 March 2006

120 3 ∠30° = 20.78∠30° A 10

120 3 ∠ − 90° 120 3 ∠150° = −41.57 A; ICA = = 20.78∠ − 120° A j5 − j10

IaA = IAB − ICA = 20.78(1∠30° − 1∠ − 120°) = 40.15∠45° A rms

(b)

IbB = −41.57 − 20.78∠30° = 60.47∠ − 170.10° A rms

(c)

IcC = 20.78∠ − 120° + 41.57 = 36.00∠ − 30° A rms

(d)

∗ ∗ ∗ Stot = VAB IAB + VBC IBC + VCA ICA = 120 3 ∠30°× 20.78∠ − 30° + 120 3 ∠ − 90°(−41.57) +

120 3 ∠150°× 20.78∠120° = 4320 + j 0 + 0 + j8640 + 0 − j 4320 = 4320 + j 4320 VA

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Engineering Circuit Analysis, 7th Edition

30.

IAB =

Chapter Twelve Solutions

10 March 2006

200∠0 200∠0 = 21.1 ∠ - 18.4 o A = o 10 || j 30 9.49∠18.4

|IA| = 3 I AB = 36.5 A The power supplied by the source = (3) |IA|2 (0.2) + (3) (200)2 / 10 = 12.8 kW Define transmission efficiency as η = 100 × Pload/ Psource. Then η = 93.8%. IA leads IAB by 30o, so that IA = 36.5 ∠ 11.6o. VR W = (0.2)(36.5 ∠11.6 o ) = 7.3 ∠11.6 o V

With VAN =

200

∠30 o , and noting that Van = VAN + VR W = 122 ∠ 28.9o, we may now

3 compute the power factor of the source as PF = cos (ang(Van) – ang(IA)) = cos (28.9o – 11.6o) = 0.955.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

31.

↑ Bal., Van = 140∠0° Vrms , + seq., R w = 0, SL = 15 + j 9 kVA

(a)

Vab = VAB = 3 140∠30° = 242.5− ∠30° V

(b)

∗ ∗ VAB IAB = 5000 + j 3000 = 242.5− ∠30° IAB ∴ IAB = 24.05− ∠ − 0.9638° A rms

(c)

IaA = IAB − ICA = 24.05− ∠ − 0.9638° − 24.05− ∠119.03° = 41.65− ∠ − 30.96° A rms

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Engineering Circuit Analysis, 7th Edition

32.

Chapter Twelve Solutions

10 March 2006

15 mH → j5.65 Ω, 0.25 mF → -j10.6 Ω VAB = 120 3 ∠30 o V VBC = 120 3 ∠ − 90 o V VCA = 120 3 ∠ − 210 o V

Defining three clockwise mesh currents I1, I2 and I3 corresponding to sources VAB, VBC and VCA, respectively, we may write: VAB = (10 + j5.65) I1 – 10 I2 + j5.65 I3 VBC = -10 I1 + (10 – j10.6) I2 + j10.6 I3 VCA = - j5.65 I1 + j10.6 I2 + (j5.65 – j10.6) I3

[1] [2] [3]

Solving using MATLAB or a scientific calculator, we find that I1 = 53.23 ∠ -5.873o A, I2 = 40.55 ∠ 20.31o A, and I3 = 0 (a) VAN = j5.65(I1 – I3) = 300.7 ∠ 84.13o V, (b) VBN = 10(I2 – I1) = 245.7 ∠ 127.4o V, (c) VCN = -j10.6 (-I2) = 429.8 ∠ 110.3o V,

so VAN = 300.7 V so VBN = 245.7 V so VCN = 429.8 V

PSpice Simulation Results (agree with hand calculations) FREQ VM(A,N) VP(A,N) 6.000E+01 3.007E+02 8.410E+01 FREQ VM(B,N) VP(B,N) 6.000E+01 2.456E+02 1.274E+02 FREQ VM(C,N) VP(C,N) 6.000E+01 4.297E+02 1.103E+02

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Engineering Circuit Analysis, 7th Edition

Chapter Twelve Solutions

10 March 2006

33.

↑ Rline = 1Ω

(a)

207.8∠30° 1∠30° −1 − j10 −1 −10 207.8∠ − 90° 2 + j 5 207.8 − j1 2 + j 5 − j5 − j5 0 0 − j 5 10 − j 5 − j 5 10 − j5 120 3 = 207.8 I1 = = 12 −1 − j10 12(70 + j 40) + (−10 − j 45) − 10(20 + j 55) −1 2 + j 5 − j5 −10 − j 5 10 − j 5 ∴ I1 =

207.8[1∠30°(70 + j 40) + j1(−10 − j 45)] 21.690∠34.86° = = 33.87∠45.20° = IaA 630 − j115 630 − j115 12

1∠30°

−10

−1 −10

− j1 − j 5 207.8 0 10 − j 5 207.8[−1∠30°(−10 − j 45) − j1(20 − j 60)] ∴ I2 = = 630 − j115 630 − j115 16,136∠162.01° = = 25.20∠172.36° A 630 − j115

(b)

∴ IcC = 25.20∠ − 7.641° A

(c)

∴ IbB = − IaA − ICC = −33.87∠45.20° − 25.20∠ − 7.641° = 53.03∠ − 157.05° A rms

(d)

S = 120 3 ∠30°(33.87∠ − 45.20°) + 120 3 ∠90°(25.20∠7.641°) = 6793 − j1846.1 − 696.3 + j 5190.4 = 6096 + j 3344 VA

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Engineering Circuit Analysis, 7th Edition

34.

Chapter Twelve Solutions

|Vline| = 240 V. Set Vab = 240∠0o V. Then Van = 240

240 3

10 March 2006

∠ − 30 o .

∠ − 30 o

3 = 23.8∠ − 61.0 o A 5 + j3 240 ∠0 o 3 IA1B1 = = 20.0∠ − 4.76 o mA 3 (12 + j ) × 10

IA2 =

Iphase leads Iline by 30o, so IA1 = 20 3∠ − 34.8 o mA = 34.6∠ − 34.8 o mA Ia = IA1 + IA2 = 11.5 – j20.8 + 28.4 – j19.7 mA = 56.9∠-45.4o mA The power factor at the source = cos (45.4o – 30o) = 0.964 lagging. The power taken by the load = (3)(20×10-3)2 (12×103) + (3)(23.8×10-3)2 (5000) = 22.9 W.

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Engineering Circuit Analysis, 7th Edition

35.

Chapter Twelve Solutions

10 March 2006

Define I flowing from the ‘+’ terminal of the source. Then,

I =

200∠0 200∠0 = 12.41∠ − 29.74 o = 10 + ( j10 || 20) 16.12∠29.74 o

(a) Vxy = 10 I = 124.1 ∠-29.74o V. Thus, Pxy = (12.41)(124.1) = 1.54 kW (b) Pxz = (200)(12.41) cos (29.74o) = 2.155 kW (c) Vyz = 200 ∠0 – 124.1 ∠-29.74o = 110.9 ∠ 33.72o V Thus, Pyz = (110.9)(12.41) cos (33.72o + 29.74o)

= 614.9 W

No reversal of meter leads is required for any of the above measurements.

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Engineering Circuit Analysis, 7th Edition

36.

Chapter Twelve Solutions

10 March 2006

1 H → j377 Ω, 25 μF → -j106 Ω

I1 =

440∠0 = 1.86∠21o A 50 + [ j 377||(100-j106 )]

j 377 = 2.43∠41.3o A j 377 + 100 − j106 V2 = (106∠-90o)(2.43∠-41.3o) = 257∠-48.7o V

IC = I

Pmeasured = (257)(1.86) cos (21o + 48.7o) = 166 W. No reversal of meter leads is needed. PSpice verification:

FREQ VM($N_0002,0) 6.000E+01 2.581E+02

VP($N_0002,0) -4.871E+01

FREQ IM(V_PRINT1) 6.000E+01 1.863E+00

IP(V_PRINT1) 2.103E+01

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Engineering Circuit Analysis, 7th Edition

37.

Chapter Twelve Solutions

10 March 2006

2.5 A peak = 1.77 A rms. 200 V peak = 141 V rms. 100 μF → -j20 Ω. Define the clockwise mesh current I1 in the bottom mesh, and the clockwise mesh current I2 in the top mesh. IC = I1 – I2. Since I2 = -177∠-90o, we need write only one mesh equation: 141∠0o = (20 - j40o) I1 + (-20 + j20) I2 141∠0 + (-20 + j 20)(1.77∠ - 90 o ) = 4.023∠74.78 o A 20 - j 40 and IC = I1 – I2 = 2.361 ∠ 63.43o A. Imeter = -I1 = 4.023∠-105.2o Vmeter = 20 IC = 47.23 ∠63.43o V

so that I1 =

Thus, Pmeter = (47.23)(4.023)cos(63.43o + 105.2o) = -186.3 W. Since this would result in pegging the meter, we would need to swap the potential leads.

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Engineering Circuit Analysis, 7th Edition

38.

Chapter Twelve Solutions

10 March 2006

(a) Define three clockwise mesh currents I1, I2 and I3 in the top left, bottom left and righthand meshes, respectively. Then we may write: 100 ∠0 = (10 – j10) I1 50 ∠90o = 0 = -(10 – j10) I1

(8 + j6) I2 – (8 + j6) I2

- (10 – j10) I3 – (8 + j6) I3 + (48 + j6) I3

Solving, we find that I1 = 10.12∠ 32.91o A, I2 = 7.906 ∠ 34.7o and I3 = 3.536 ∠ 8.13o A. Thus, PA = (100)(10.12) cos (-32.91o) = 849.6 W and PB = (5)(7.906) cos (90o – 34.7o) = 225.0 W (b) Yes, the total power absorbed by the combined load (1.075 kW) is the sum of the wattmeter readings. PSpice verification:

FREQ IM(V_PRINT1) 6.280E+00 1.014E+01

IP(V_PRINT1) 6.144E-02

FREQ IM(V_PRINT2) 6.280E+00 4.268E-01

IP(V_PRINT2) 1.465E+02

FREQ VM($N_0002,$N_0006) 6.280E+00 1.000E+02

VP($N_0002,$N_0006) 0.000E+00

FREQ VM($N_0004,$N_0006) 6.280E+00 5.000E+01 -

VP($N_0004,$N_0006) 9.000E+01

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Engineering Circuit Analysis, 7th Edition

39.

Chapter Twelve Solutions

10 March 2006

This circuit is equivalent to a Y-connected load in parallel with a Δ-connected load. 200 ∠ − 30o 3 = 4.62∠ − 60o A For the Y-connected load, Iline = 25∠30o ⎛ 200 ⎞ PY = (3)⎜ ⎟(4.62)cos 30o = 1.386 kW ⎝ 3⎠ 200∠0 = 4∠60o A 50∠ − 60o PΔ = (3)(200)(4 cos 60o) = 1.2 kW

For the Δ-connected load, Iline =

Ptotal = PY + PΔ = 2.586 kW Pwattmeter = Ptotal / 3 = 862 W

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Engineering Circuit Analysis, 7th Edition

40.

Chapter Twelve Solutions

10 March 2006

We assume that the wire resistance cannot be separated from the load, so we measure from the source connection: (a)

(b)

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Engineering Circuit Analysis, 7th Edition

41.

Chapter Twelve Solutions

10 March 2006

We assume that the wire resistance cannot be separated from the load, so we measure from the source connection: (a)

(b)

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