Chapter 12 Solutions to Exercises

(1)

1. Vbc = Vbe + Vec = 0.7 – 10 = -9.3 V

Veb = - Vbe = -0.7 V

Vcb = Vce + Veb = 10 – 0.7 = 9.3 V

(2)

(3)

3. (a) positive phase sequence

Van = |Vp| ∠ 0o Vdn = |Vp| ∠ -180o

Vbn = |Vp| ∠ -60o Ven = |Vp| ∠ -240o

Vcn = |Vp| ∠ -120o Vfn = |Vp| ∠ -300o

(b) negative phase sequence

Van = |Vp| ∠ 0o Vdn = |Vp| ∠ 180o

Vbn = |Vp| ∠ 60o Ven = |Vp| ∠ 240o

Vcn = |Vp| ∠ 120o Vfn = |Vp| ∠ 300o

(4)

yz yx xz = -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) _o o o o o xy zx 110 1.455 20 110 30 1 160 20 110 50 -160 = ∠ ∠ ∠ = ∠ ∠ = V V

(5)

5. (a) V25 = V24 + V45 = -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

(6)

6. V₁₂ = ∠9 87 V = 0.4710 + 8.988 V, j V₂₃= ∠ 8 45 V = 5.657+ j5.657 V

(a) V21 = – V12 = 9∠

(

180o+87o

)

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+ j5.657= 6.128 + j14.65 V

(7)

7. (a) Vbn 75o Van 45o Vcn 135o

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

(8)

(9)

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

(10)

230 / 460 V rms Z : S 10 40 kVA; Z : 8 10 kVA; 10, 000 40 Z : 4 80 kVA Let V 230 0 V S V I , I 43.48 40 A 230 4000 80 I 43.48 40 A, S V I I 8.696 80 , I 8.696 80 I I I 460 I 43. ∗ ∗ ∗ ∗ = ∠ ° ∠ ° ∠ ° ∠ − ° = ∠ ° ∴ = = = ∠ ° ∠ − ° ∴ = ∠ − ° = ∴ = = ∠ − ° = ∠ °∴ = + ∴ = AN NB AB AN AN AN AN AN AN AB AB AB AB AB aA AN AB aA 48 40 8.696 80 39.85 29.107 I 39.85 A 8000 10 I 34.78 10 , I 34.78 10 A 230 I 34.78 10 8.696 80 35.85 175.96 , I 35.85 A I 43.48 40 34.78 10 21.93 87.52 , I 21.93A − − ∗ + + ∠ ° + ∠ ° = ∠ − ° ∴ = ∠ ° = = ∠ ° = ∠ − ° ∴ = − ∠ − ° − ∠ ° = ∠ − ° ∴ = = − ∠ − ° + ∠ − ° = ∠ ° = aA NB NB bB bB nN nN

(11)

11. (a) InN = 0 since the circuit is balanced. IAN = 12 ∠0 IAB = 12 -36.9o 12 16 0 240 ₌ _∠ + ∠ j 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

(12)

(a) 21 3 10 10 3 10 19 2 8 2 (21 3) (674 167 60 32) 10 3 8 2 36 5 10( 360 50 74 44) (10 3) (80 20 184 77) 5800 1995 6127 18.805 720 10 10 3 720 19 2 8 2 720(614 135 434 94) 7 0 8 2 36 5 + − − − Δ = − + − − = + + − − − − − − + + − − − − − + + + + Δ = + = ∠ ° − − − + − − = + + + = − − + j j j j j j j j j j j j j j j j j j j j j j j ∴ 126.06 1 130.65 1 4.730 10 15.891 17.069 0.224 33.18 kW = × + × + × = + + = ) (b) (c) P_ω_,_tot 2 2 2

(d P_{gen tot}_, =720 126.06 cos 6.479× ° +720 130.65 cos 5.968× − ° =90.18 93.56 183.74 kW+ =

20 1072.7 12.326 720 1072.7 12.326 I 6127 18.805 × 126.06 6.479 A ∠ ° × ∠ ° ∴ = = − ° ∠ ° aA ∠ 21 3 720 10 3 720 (1084 247) 10 720 8 2 720 (1084 247) I 130.65 5.968 A 6127 18.805 10 3 0 36 5 I 130.65 5.968 126.06 6.479 4.730 7.760 A − − + − − + − − − = + ∴ = = ∠ − ° ∠ ° − − + = ∠ − ° − ∠ − ° = ∠ ° Bb nN j j j j j j j ∴

(13)

13. V_AN =220 Vrms, 60 Hz (a) PF 1= ∴I 220 0 40.85 21.80 A; I 377C 440 5 2 I 40.85 cos 21.80 (377C440 40.85sin 21.80 ) 40.85sin 21.80 C 377 440 + ∠ ° = = ∠ − ° = × + (b) I_AB =377 91.47 10× × −6×440 15.172 A= ∴VA=440 15.172× =6.676 kVA 91.47 F ∴ = ° + − ° ° = = × AN AB aA j j j μ ∴

(14)

14. (a) IaA = IAN + IAB = AB R 3 12+ j + = 15.69 – j3.922 + R_AB

Since we know that |IaA| = 30 A rms = 42.43 A,

2 2 AB 3.922 R 400 69 . 15 43 . 42 _⎟⎟ + ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = or RAB = 15.06 Ω (b) IaA = IAN + IAB = AB AB X 400 922 . 3 -15.69 X -0 400 3 12 0 200 j j j j = + ∠ + + ∠

In order for the angle of IaA to be zero,

AB

X 400

(15)

15. + seq. V_BC =120 60 V rms, R∠ ° _w =0.6Ω P_load =5 kVA, 0.6 lag (a) 2 120 5000 5000 V = ∠150 V S 0.8 0.6 3 3 3 120 S 150 I I 24.06 113.13 A 3 I 24.06 113.13 P 3 24.06 0.6 AN AN AN aA aA aA wire j ∗ ∗ ° ∴ = × + ∴ = ∠ ° ∴ = ∠ − ° 1041.7 W (b) ∴ = ∠ ° ∴ = × × = V =0.6 24.06 113.13× 14.434 113.13 V 120 V V V 14.434 113.13 158 3 aA an aA AN ∠ ° = ∠ ° 81.29 143.88 V ∴ = + = ∠ ° + ∠ ° = ∠ °

(16)

16. ↑V_an =2300 0 V , R∠ ° _rms _w= Ω +2 , seq., S_tot =100+ j30 kVA (a) 1(100, 000 30, 000) 2300 I I 15.131 16.699 A 3 j aA aA ∗ + = ∴ = ∠ − ° (b) VAN =2300 2 15.131− × ∠ −16.699° =2271 0.2194 V∠ ° (c) Z V / I 2271 0.2194 143.60 43.67 15.131 16.699 p AN aA j ∠ ° = = = + ∠ − ° Ω (d) trans. eff. 143.60 0.9863, or98.63% 145.60 = =

(17)

17. ↑Z_p =12+ Ωj5 , I_bB=20 0 A rms, +seq., PF∠ ° =0.935

(a) cos 0.9351 20.77 5 tan 20.77 , R 1.1821

12 R_w w θ ₌ − ₌ _°∴ ₌ _° ₌ + Ω (b) (c) VAB = 3 VBN /∠VBN +150° =450.3 172.62 V∠ ° (d) V_BN =I Z_bB _p =20 (12+ j5)=240+ j100 V ∴V_bn =20 (13.1821+ j5)=281.97 20.77 V∠ ° S_source =3 V I∗ = ×3 281.97∠ −20.77 (20)° 15.819 6.000 kVA = − Bn bB j

(18)

55 μF → -j/(2π)(60)(55×10-6) = -j48.23 Ω

The per-phase current magnitude |I| is then I =

) 23 . 48 12 . 47 ( 75 125 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.

The power factor of the load is 1.000

75 23 . 48 12 . 47 cos ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ −

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.

(19)

19. Bal.,+ seq. Z 8 6 , Z 12 16 , Z 5 0, V 120 0 V rms 120 0 120 120 120 120 R =0.5Ω ( ) I 6.803 83.86 A 8.5 6 12.5 16 5.5 I 6.803 96.14 A rms AN BN CN AN w nN nN j j j a j j + Ω = − Ω = + = ∠ ° ∠ ° ∠ − ° ∠ ° − = + + = ∠ ° + − ↑ = = ∠ − ° ∴

(20)

(

)

2 2 w 5 10 R 40 + + = I (a) RW = 0 Then 3.578A 10 25 40 2 = + =

I , and the power delivered to each phase of the load is

(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 3.123A 10 64 40 2 = + =

I , and the power delivered to each phase of the load is

(21)

21. ↑Z_p = ∠ ° Ω75 25 25 F, Vμ _an =240 0 V rms, 60 Hz, R∠ ° _w = Ω2 w = × load = = source = ° (a) 6 10 75 25 ( 106.10) Z = − 106.10 Z 75.34 23.63 377 25 75 25 106.10 240 Z 77.34 23.63 I 77.34 23.63 + ∠ ° − = − Ω ∴ = = − Ω × ∠ ° − ∴ = − ∴ = =2.968 16.989 A∠ ° − cap p p w aA j j j j j j j (b) P 3(2.968)2 2=52.84 W (c) P 3(2.968) 75.34 1990.6 W2 (d) PF cos16.989 =0.9564 lead

(22)

impedance,

-j6366 || 100 ∠ 28o

= 89.59 + j46.04 Ω so that the current flowing through the combined

load is rms A 362 . 2 04 . 46 90.59 240 2 2 ₊ = = I

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.

(23)

23. ↑Bal., R_w =0, Z_p =10+ j5 ,Ω f =60 Hz (a) 10+ j5 11.180 26.57= ∠ ° ∴PF=cos 26.57° =0.8944 (b) _P_F₌_{0.93 lag,} ₌_{21.57 , Y} 1 _0.08 _0.04S 11.180 26.57 377C 0.04 Y 0.08 (377C 0.04) tan 21.57 0.3952 0.08 377C 0.04 0.08 0.3952 0.00838 C p j p j θ ° = = − ∠ ° − ′ = + − ∴ = − ° = − 22.23 Fμ ∴ = − × = ∴ = (c) 6 , 10 440 / 3 440 V rms, Z 119.30 , I 2.129 A 120 22.23 119.30 440 VAR 2.129 3 − = = = − Ω = = V 540.9 VAR ( .) ∴ = × L load c c j j π = cap

(24)

Van rms = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∠ 2 0 115 3 1 o _{= 46.9 ∠0}o V rms 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

(25)

25. R_w =0, V_an =200 60 V rms. S∠ ° _p = −2 j1 kVA seq.+ (a) V_bc =220 3∠ − ° =30 346.4∠ − °30 V (b) (c) S =2000− 1000 V I 346.4 30 I I 6.455 3.435 , I 6.455 3.435 200 3 30 Z 53.67 26.57 6.455 3.435 BC BC BC BC BC BC p j ∗ ∗ ∗ − − − = = ∠ − ° 48 j24 ∴ = ∠ ° = ∠ − ° ∠ − ° ∴ = = ∠ − ° = − Ω ∠ − ° IaA = IAB ICA 6.455 120 3.43 6.455 120 3.43 − − − = ∠ ° − ° − ∠ − ° − ° =11.180 86.57 A rms∠ °

(26)

26. ↑15kVA, 0.8lag, +seq., V_BC =180 30 V rms, R∠ ° _w =0.75Ω (a) 1 (b) V 180 30 V 180 150 V, S 5000 cos 0.8 5000 36.87 180 30 I I 27.78 6.87 and I 27.78 113.13 A I I I I 27.78(1 6.87 1 113.13 ) 48.11 36.87 A V 0.75( I I ) V 0.75 48.11(1 − ∗ ° ∴ = ∠ ° = ∠ = ∠ ° = ∠ ° = ∠ ∴ = ∠ − ° = ∠ ° ∴ = − ∴ = ∠ − ° − ∠ ° = ∠ − ° ∴ = − ∴ = × ∠ − BC AB p BC BC AB bB BC AB bB bC bB cC bC 36.87° − ∠ −1 156.87 ) 180 30° + ∠ ° 233.0 20.74 V= ∠ ° 2 P 3 48.11 0.75 5208 W S 5208 15, 000 36.87 wire gen = × × = = + ∠ ° =17.208+ j9.000 kVA

(27)

27. ↑Bal., S_L = +3 j1.8 kVA, S_gen =3.45+ j1.8 kVA, R_w =5Ω (a) P 450 W 1 450 I2_A 5 I_aA=5.477 A rms 3 = ∴ × = × ∴ w a (b) I 1 5.477=3.162 A rms 3 AB = ×

(c) Assume I 3.162 0 and +seq. 1(3000 1800) (3.162 0 )

3 V 368.8 30.96 V V V V V V V 5 I 5 5.477 30 27.39 30 , V 27.39 150 V 27.39 30 27.39 150 368.8 30.96 V (1 ∗ = ∠ ° ∴ + = = ∠ ° ∴ = ∠ ° ∴ = + − + = = × ∠ − ° = ∠ − ° = ∠ − ° ∴ = ∠ − ° − ∠ − ° + ∠ ° + ∠ − AB AB AB AB AB an aA AB bB bn aA aA bB an an j V I V 120 ) 27.39 30 27.39 150 368.8 30.96 V 236.8 2.447 1 1 120 ° ∠ − ° − ∠ − ° + ∠ ° 236.8 V rms = = ° ∴ = − ∠ − ° an ∠ − Van ∴

(28)

w

segment. Thus, the line current is 5.898A rms 3

.

280 = . Since this is a D-connected load,

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

Ptotal = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⋅ ⋅ 2 2 (5.898) 3 PF

3V_line I_line V_line = 1800

where |Vline| =

(

)( )

249.2V (5.898) 3 2 2 1800 ₌

This is the voltage at the load, so we need to add the voltage lost across the wire, which

(taking the load voltage as the reference phase) is 1

( )

_W

R 2 1 cos 898 . 5 _⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ∠ − = 13.57 ∠-45o

V. Thus, the line voltage magnitude of the source is |249.2 ∠ 0o_{+ 13.57 ∠ -45}o

(29)

29. Bal., +seq. 120 3 30 V 120 0 V 120 3 30 , ., I 20.78 30 A 10 120 3 90 120 3 150 I 41.57 ; I 20.78 120 A 5 10 I I I 20.78(1 30 1 120 ) 40.15 45 A rms ∠ ° = ∠ ° ∴ = ∠ ° = = ∠ ° ∠ − ° ∠ ° = = − = = ∠ − − = − = ∠ ° − ∠ − ° = ∠ ° an ab AB BC CA aA AB CA etc A j j (a) ° (b) I_bB = −41.57 20.78 30− ∠ ° =60.47∠ −170.10 A rms° (c) I_cC =20.78∠ −120° +41.57=36.00∠ − °30 A rms (d) _S _{= V I} _{V I} _{V I} _{120 3} ₃₀ _20.78 ₃₀ _{120 3} _{90 ( 41.57)} 120 3 150 20.78 120 4320 0 0 8640 0 4320 tot AB AB BC BC CA CA j j j ∗ ₊ ∗ ₊ ∗ ₌ _{∠ °×} _{∠ − ° +} _{∠ − ° −} ₊ ∠ °× ∠ ° = + + + + − =4320+ j4320 VA

(30)

30. IAB = 21.1 -18.4 A 4 . 18 49 . 9 30 || 10 = ∠ o = ∠ j |IA| = 3IAB = 36.5A

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. V 11.6 7.3 ) 11.6 (0.2)(36.5 o o RW = ∠ = ∠ V With VAN = 30o 3

200 ∠ , and noting that Van = VAN + VR_W = 122 ∠ 28.9

o

, we may now compute the power factor of the source as

(31)

31. ↑Bal., V_an =140 0 V∠ ° _rms, seq., R+ _w =0, S_L =15+ j9 kVA

(a) V_ab =V_AB = 3 140 30∠ ° =242.5−∠ °30 V

(b) V I_{AB AB}∗ =5000+ j3000=242.5−∠ °30 I_AB∗ ∴I_AB =24.05−∠ −0.9638 A rms°

(c) I_aA = I_AB−I_CA =24.05−∠ −0.9638° −24.05−∠119.03° =41.65−∠ −30.96 A rms°

(32)

VAB = 120 3∠30o V

VBC = 120 3∠−90o V

VCA = 120 3∠−210o 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 [1]

VBC = -10 I1 + (10 – j10.6) I2 + j10.6 I3 [2]

VCA = - j5.65 I1 + j10.6 I2 + (j5.65 – j10.6) I3 [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, so VAN = 300.7 V

(b) VBN = 10(I2 – I1) = 245.7 ∠ 127.4o V, so VBN = 245.7 V

(c) VCN = -j10.6 (-I2) = 429.8 ∠ 110.3o 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

(33)

33. 1↑R_line= Ω (a) 1 207.8[1 30 (70 40) 1( 10 45)] 21.690 34.86 I 33.87 45.20 I 630 115 630 115 aA j j j j j ∠ ° + + − − ∠ ° ∴ = = = ∠ − − ° = 2 12 1 30 10 1 1 5 207.8 10 0 10 5 207.8[ 1 30 ( 10 45) 1(20 60)] I 630 115 630 115 16,136 162.01 25.20 172.36 A 630 115 ∠ ° − − − − − − − ∠ ° − − − − ∴ = = − − ∠ ° = = ∠ ° − j j j j j j j j j (b) ∴I_cC =25.20∠ −7.641 A° (c) ∴I_bB = −I_aA−I_CC = −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 j5190.4 ° ∠ − ° + ∠ ° ∠ ° = − − + =6096+ j3344 VA 1 207.8 30 1 10 1 30 1 10 207.8 90 2 5 5 207.8 1 2 5 5 0 5 10 5 0 5 10 5 120 3 207.8 I 12 1 10 12(70 40) ( 10 45) 10(20 55) 1 2 5 5 10 5 10 5 j j j j j j j j j j j j j j j j j j ∠ ° − − ∠ ° − − ∠ − ° + − − + − − − − − = = = − − + + − − − + − + − − − −

(34)

34. |Vline| = 240 V. Set Vab = 240∠0o V. Then Van = 30 3 ∠− . IA2 = 23.8 61.0 A 3 5 30 3 240 o o − ∠ = + − ∠ j IA1B1 =

(

)

20.0 4.76 mA 10 2 1 0 3 240 o 3 o − ∠ = × + ∠ j

Iphase leads Iline by 30o, so

IA1 = 20 3∠−34.8o mA = 34.6∠−34.8o 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.

(35)

35. Define I flowing from the ‘+’ terminal of the source. Then, I = _o 12.41 29.74o 29.74 16.12 0 200 ) 20 || 10 ( 10 0 200 − ∠ = ∠ ∠ = + ∠ j (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.

(36)

I1 =

[

(

)

]

1.86 21 A 106 100 377 50 0 440 = ∠ o + ∠ -j || j IC = I 2.43 41.3 A 106 100 377 377 = ∠ o − + j j j V2 = (106∠-90o)(2.43∠-41.3o) = 257∠-48.7o V Pmeasured = (257)(1.86) cos (21o + 48.7o) = 166 W.

No reversal of meter leads is needed. PSpice verification:

FREQ VM($N_0002,0) VP($N_0002,0) 6.000E+01 2.581E+02 -4.871E+01 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E+01 1.863E+00 2.103E+01

(37)

37. 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 so that I1 = 4.023 74.78 A 40 -20 ) 90 -20)(1.77 (-20 0 141 o o ∠ = ∠ + + ∠ j j and IC = I1 – I2 = 2.361 ∠ 63.43o A. Imeter = -I1 = 4.023∠-105.2o Vmeter = 20 IC = 47.23 ∠63.43o V

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.

(38)

hand meshes, respectively. Then we may write:

100 ∠0 = (10 – j10) I1 - (10 – j10) I3

50 ∠90o = (8 + j6) I2 – (8 + j6) I3

0 = -(10 – j10) I1 – (8 + j6) I2 + (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) IP(V_PRINT1) 6.280E+00 1.014E+01 6.144E-02

(39)

39. This circuit is equivalent to a Y-connected load in parallel with a Δ-connected load.

For the Y-connected load, Iline = 4.62 60 A

30 25 30 3 200 o o o − ∠ = ∠ − ∠ PY =

(

4.62

)

cos30 1.386kW 3 200 ) 3 ( ⎟ o = ⎠ ⎞ ⎜ ⎝ ⎛

For the Δ-connected load, Iline = 4 60 A

60 50 0 200 o o ∠ = − ∠ ∠ PΔ = (3)(200)(4 cos 60o) = 1.2 kW Ptotal = PY + PΔ = 2.586 kW Pwattmeter = Ptotal / 3 = 862 W

(40)

from the source connection: (a)

(41)

41. We assume that the wire resistance cannot be separated from the load, so we measure from the source connection:

(a)

(b)