Download Sinusoidal Steady State Power Calculations: Assessment Problems and more Exercises Electrical Circuit Analysis in PDF only on Docsity!
Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/− 45 ◦^ V, I = 20/15◦^ A Therefore P =
(100)(20) cos[− 45 − (15)] = 500 W, A → B
Q = 1000 sin − 60 ◦^ = − 866 .03 VAR, B → A
[b] V = 100/− 45 ◦, I = 20/165◦
P = 1000 cos(− 210 ◦) = − 866. 03 W, B → A
Q = 1000 sin(− 210 ◦) = 500 VAR, A → B
[c] V = 100/− 45 ◦, I = 20/− 105 ◦
P = 1000 cos(60◦) = 500 W, A → B
Q = 1000 sin(60◦) = 866.03 VAR, A → B
[d] V = 100/0◦, I = 20/120◦ P = 1000 cos(− 120 ◦) = − 500 W, B → A
Q = 1000 sin(− 120 ◦) = − 866 .03 VAR, B → A
AP 10.2 pf = cos(θv − θi) = cos[15 − (75)] = cos(− 60 ◦) = 0. 5 leading
rf = sin(θv − θi) = sin(− 60 ◦) = − 0. 866
10–2 CHAPTER 10. Sinusoidal Steady State Power Calculations
AP 10.3 From Ex. 9.4 Ieff = √Iρ 3
A
P = Ieff^2 R =
) (5000) = 54 W
AP 10.4 [a] Z = (39 + j26)‖(−j52) = 48 − j20 = 52/− 22. 62 ◦^ Ω
Therefore I =
48 − j20 + 1 + j 4 = 4.85/18. 08 ◦^ A(rms)
V L = Z I = (52/− 22. 62 ◦)(4.85/18. 08 ◦) = 252.20/− 4. 54 ◦^ V(rms)
I L =
V L
39 + j 26 = 5.38/− 38. 23 ◦^ A(rms)
[b] SL = V L I ∗ L = (252.20/− 4. 54 ◦)(5.38/+ 38. 23 ◦) = 1357/33. 69 ◦ = (1129.09 + j 752 .73) VA
PL = 1129. 09 W; QL = 752.73 VAR [c] P = | I |^2 1 = (4.85)^2 · 1 = 23. 52 W; Q = | I |^2 4 = 94.09 VAR [d] Sg(delivering) = 250 I ∗ = (1152. 62 − j 376 .36) VA Therefore the source is delivering 1152. 62 W and absorbing 376. magnetizing VAR.
[e] Qcap =
| V L|^2
(252.20)^2
= − 1223 .18 VAR
Therefore the capacitor is delivering 1223.18 magnetizing VAR.
Check: 94 .09 + 752.73 + 376.36 = 1223.18 VAR and 1129 .09 + 23.52 = 1152. 62 W
AP 10.5 Series circuit derivation:
S = 250 I ∗^ = (40, 000 − j 30 ,000)
Therefore I ∗^ = 160 − j120 = 200/− 36. 87 ◦^ A(rms)
I = 200/36. 87 ◦^ A(rms)
Z =
V
I
200/36. 87 ◦^
= 1.25/− 36. 87 ◦^ = (1 − j 0 .75) Ω
Therefore R = 1 Ω, XC = − 0 .75 Ω
10–4 CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =
= 1.34/− 26. 57 ◦^ A
Therefore P =
(
- 34 √ 2
) 2 20 = 18 W
[c] RL = |ZTh| = 22.36 Ω [d] I =
42 .36 + j 10
= 1.23/− 39. 85 ◦^ A
Therefore P =
( (^1) √. 23 2
) 2 (22.36) = 17 W
AP 10.
Mesh current equations: 660 = (34 + j50) I 1 + j100( I 1 − I 2 ) + j 40 I 1 + j40( I 1 − I 2 )
0 = j100( I 2 − I 1 ) − j 40 I 1 + 100 I 2
Solving, I 1 = 3.536/− 45 ◦^ A,
I 2 = 3.5/0◦^ A;. .· P =
(3.5)^2 (100) = 612. 50 W
AP 10.9 [a]
248 = j 400 I 1 − j 500 I 2 + 375( I 1 − I 2 ) 0 = 375( I 2 − I 1 ) + j 1000 I 2 − j 500 I 1 + 400 I 2 Solving, I 1 = 0. 80 − j 0. 62 A; I 2 = 0. 4 − j 0 .3 = 0.5/− 36. 87 ◦^ A
..^ · P =^1 2
(0.25)(400) = 50 W
Problems 10–
[b] I 1 − I 2 = 0. 4 − j 0. 32 A
P 375 =
| I 1 − I 2 |^2 (375) = 49. 20 W
[c] Pg =
(248)(0.8) = 99. 20 W
∑ Pabs = 50 + 49.2 = 99. 20 W (checks)
AP 10.10 [a] VTh = 210/0◦^ V; V 2 = 14 V 1 ; I 1 = 14 I 2 Short circuit equations: 840 = 80 I 1 − 20 I 2 + V 1 0 = 20( I 2 − I 1 ) − V 2
..^ · I 2 = 14 A; RTh =^210 14
[b] Pmax =
) 2 15 = 735 W
AP 10.11 [a] V Th = −4(146/0◦) = −584/0◦^ V(rms) = 584/180◦^ V(rms)
V 2 = 4 V 1 ; I 1 = − 4 I 2 Short circuit equations: 146/0◦^ = 80 I 1 − 20 I 2 + V 1
0 = 20( I 2 − I 1 ) + V 2
..^ · I 2 = − 146 /365 = − 0. 40 A; RTh = −^584 − 0. 4
[b] P =
) 2 1460 = 58. 40 W
Problems 10–
P 10.3 [a] hair dryer = 600 W vacuum = 630 W
sun lamp = 279 W air conditioner = 860 W television = 240 W ∑ P = 2609 W
Therefore Ieff =
= 21. 74 A
Yes, the breaker will trip. [b]
∑ P = 2609 − 909 = 1700 W; Ieff =
= 14. 17 A
Yes, the breaker will not trip if the current is reduced to 14. 17 A.
P 10.4 [a] Ieff = 40/ 115 ∼= 0. 35 A; [b] Ieff = 130/ 115 ∼= 1. 13 A
P 10.5 Wdc = V (^) dc^2 R
T ; Ws =
∫ (^) to+T to
v s^2 R
dt
..^ · V^
dc^2 R
T =
∫ (^) to+T to
v^2 s R dt
V (^) dc^2 =
T
∫ (^) to+T to v^2 s dt
Vdc =
√ 1 T
∫ (^) to+T to v^2 s dt = Vrms = Veff
P 10.6 [a] Area under one cycle of v g^2 :
A = (5^2 )(2)(30 × 10 −^6 ) + 2^2 (2)(37. 5 × 10 −^6 ) = 1800 × 10 −^6 Mean value of v g^2 :
M.V. =
A
200 × 10 −^6
1800 × 10 −^6
200 × 10 −^6
..^ · Vrms = √9 = 3 V(rms)
[b] P = V (^) rms^2 R
= 4 W
P 10.7 i(t) = 200t 0 ≤ t ≤ 75 ms
i(t) = 60 − 600 t 75 ms ≤ t ≤ 100 ms
Irms =
√ 1
- 1
{∫ (^0). 075 0 (200)^2 t^2 dt +
∫ (^0). 1
- 075 (60 − 600 t)^2 dt
}
√ 10(5.625) + 10(1.875) =
75 = 8. 66 A(rms)
10–8 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.8 P = Irms^2 R. .· R =
3 × 103
P 10.9 I g = 40/0◦^ mA
jωL = j 10 ,000 Ω;
jωC = −j 10 ,000 Ω
I o = j 10 , 000 5000 (40/0◦) = 80/90◦^ mA
P =
| I o|^2 (5000) =
(0.08)^2 (5000) = 16 W
Q =
| I o|^2 (− 10 ,000) = −32 VAR
S = P + jQ = 16 − j 32 VA
|S| = 35. 78 VA
P 10.10 I g = 4/0◦^ mA;
jωC
= −j1250 Ω; jωL = j500 Ω
Zeq = 500 + [−j 1250 ‖(1000 + j500)] = 1500 − j500 Ω
Pg = −
|I|^2 Re{Zeq} = −
(0.004)^2 (1500) = − 12 mW
The source delivers 12 mW of power to the circuit.
10–10 CHAPTER 10. Sinusoidal Steady State Power Calculations
| I g|^2 RL = 2500. .· RL = 10 Ω | I g|^2 XL = − 5000. .· XL = −20 Ω Thus,
|Z| =
√ (30)^2 + (X − 20)^2 | I g| =
√^500
900 + (X − 20)^2
..^ · 900 + (X − 20)^2 =^25 ×^10
4 250
Solving, (X − 20) = ± 10. Thus, X = 10 Ω or X = 30 Ω [b] If X = 30 Ω:
I g =
30 + j 10 = 15 − j 5 A
Sg = − 500 I ∗ g = − 7500 − j 2500 VA Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars. Qj 30 = | I g|^2 X = 250(30) = 7500 VAR Therefore the line reactance is absorbing 7500 magnetizing vars. Q−j 20 = | I g|^2 XL = 250(−20) = −5000 VAR Therefore the load reactance is generating 5000 magnetizing vars. ∑ Qgen = 7500 VAR =
∑ Qabs If X = 10 Ω: I g =
30 − j 10
= 15 + j 5 A
Sg = − 500 I ∗ g = −7500 + j 2500 VA Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizing vars. Qj 10 = | I g|^2 (10) = 250(10) = 2500 VAR Therefore the line reactance is absorbing 2500 magnetizing vars. The load continues to generate 5000 magnetizing vars. ∑ Qgen = 5000 VAR =
∑ Qabs
Problems 10–
P 10.13 Zf = −j 10 , 000 ‖ 20 ,000 = 4000 − j8000 Ω
Zi = 2000 − j2000 Ω
..^ · Zf Zi
4000 − j 8000 2000 − j 2000 = 3 − j 1
V o = − Zf Zi
V g; V g = 1/0◦^ V
V o = (3 − j1)(1) = 3 − j1 = 3.16/− 18. 43 ◦^ V
P =
V (^) m^2 R
= 5 × 10 −^3 = 5 mW
P 10.14 [a] P =
(240)^2
= 60 W
ωC
− 9 × 106
Q =
(240)^2
= −80 VAR
pmax = P +
√ P 2 + Q^2 = 60 +
√ (60)^2 + (80)^2 = 160 W(del)
[b] pmin = 60 −
602 + 80^2 = − 40 W(abs) [c] P = 60 W from (a) [d] Q = −80 VAR from (a) [e] generate, because Q < 0 [f] pf = cos(θv − θi)
I =
−j 360 = 0.5 + j 0 .67 = 0.83/53. 13 ◦^ A
..^ · pf = cos(0 − 53. 13 ◦) = 0. 6 leading
[g] rf = sin(− 53. 13 ◦) = − 0. 8
Problems 10–
P 10.16 [a]
jωC = −j40 Ω; jωL = j80 Ω
Zeq = 40‖ − j40 + j80 + 60 = 80 + j60 Ω
I g =
80 + j 60 = 0. 32 − j 0. 24 A
Sg = −
V g I ∗ g = −
40(0.32 + j 0 .24) = − 6. 4 − j 4. 8 VA
P = 6. 4 W(del); Q = 4.8 VAR(del) |S| = |Sg| = 8 VA
[b] I 1 = −j 40 40 − j 40
I g = 0. 04 − j 0. 28 A
P40Ω =
| I 1 |^2 (40) = 1. 6 W
P60Ω =
| I g|^2 (60) = 4. 8 W ∑ Pdiss = 1.6 + 4.8 = 6. 4 W =
∑ Pdev
[c] I −j40Ω = I g − I 1 = 0.28 + j 0. 04 A
Q−j40Ω =
| I −j40Ω|^2 (−40) = − 1 .6 VAR(del)
Qj80Ω =
| I g|^2 (80) = 6.4 VAR(abs) ∑ Qabs = 6. 4 − 1 .6 = 4.8 VAR =
∑ Qdev
P 10.17 [a] Z 1 = 240 + j70 = 250/16. 26 ◦^ Ω
pf = cos(16. 26 ◦) = 0. 96 lagging rf = sin(16. 26 ◦) = 0. 28
10–14 CHAPTER 10. Sinusoidal Steady State Power Calculations
Z 2 = 160 − j120 = 200/− 36. 87 ◦^ Ω
pf = cos(− 36. 87 ◦) = 0. 80 leading
rf = sin(− 36. 87 ◦) = − 0. 60
Z 3 = 30 − j40 = 50/− 53. 13 ◦^ Ω pf = cos(− 53. 13 ◦) = 0. 6 leading
rf = sin(− 53. 13 ◦) = − 0. 8
[b] Y = Y 1 + Y 2 + Y 3
Y 1 =
250/16. 26 ◦^
; Y 2 =
200/− 36. 87 ◦^
; Y 3 =
Y = 19.84 + j 17. 88 mS
Z =
Y
= 37.44/− 42. 03 ◦^ Ω
pf = cos(− 42. 03 ◦) = 0. 74 leading
rf = sin(− 42. 03 ◦) = − 0. 67
P 10.18 [a] S 1 = 16 + j 18 kVA; S 2 = 6 − j 8 kVA; S 3 = 8 + j 0 kVA
ST = S 1 + S 2 + S 3 = 30 + j 10 kVA 250 I ∗^ = (30 + j10) × 103 ;. .· I = 120 − j 40 A
Z =
120 − j 40 = 1.875 + j 0 .625 Ω = 1.98/18. 43 ◦^ Ω
[b] pf = cos(18. 43 ◦) = 0. 9487 lagging
P 10.19 [a] From the solution to Problem 10.18 we have
I L = 120 − j 40 A(rms)
..^ · V s = 250/0◦^ + (120 − j40)(0.01 + j 0 .08) = 254.4 + j 9. 2 = 254.57/2. 07 ◦^ V(rms)
[b] | I L| =
√ 16 , 000
P = (16,000)(0.01) = 160 W Q = (16,000)(0.08) = 1280 VAR
[c] Ps = 30,000 + 160 = 30. 16 kW Qs = 10,000 + 1280 = 11. 28 kVAR [d] η =
10–16 CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] T =
f
= 16. 67 ms
- 735 ◦ 360 ◦^
t
- 67 ms ;. .· t = 126. 62 μs
[c] V L lags V g by 2. 735 ◦^ or 126. 62 μs
P 10.23 [a] From the solution to Problem 9.56 we have:
V o = j80 = 80/90◦^ V
Sg = −
V o I ∗ g = −
(j80)(10 − j10) = − 400 − j 400 VA
Therefore, the independent current source is delivering 400 W and 400 magnetizing vars.
I 1 = V o 5 = j 16 A
P5Ω =
(16)^2 (5) = 640 W
Therefore, the 8 Ω resistor is absorbing 640 W.
I ∆ =
V o −j 8
= − 10 A
Qcap =
(10)^2 (−8) = −400 VAR
Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars.
- 4 I ∆ = − 24 V
I 2 = V o − 2. 4 I ∆ j 4
j80 + 24 j 4
= 20 − j 6 A = 20.88/− 16. 7 ◦^ A
Problems 10–
Qj 4 =
| I 2 |^2 (4) = 872 VAR
Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars.
Sd.s. = 12 (2. 4 I ∆) I ∗ 2 = 12 (−24)(20 + j6) = − 240 − j 72 VA Thus the dependent source is delivering 240 W and 72 magnetizing vars. [b]
∑ Pgen = 400 + 240 = 640 W =
∑ Pabs [c]
∑ Qgen = 400 + 400 + 72 = 872 VAR =
∑ Qabs
P 10.24 [a] From the solution to Problem 9.58 we have
I a = −j 10 A; I b = −20 + j 10 A; I o = 20 − j20 A
S 100 V = −
(100) I ∗ a = −50(j10) = −j 500 VA
Thus, the 100 V source is developing 500 magnetizing vars.
Sj 100 V = −^12 (j100) I ∗ b = −j50(− 20 − j10) = −500 + j 1000 VA Thus, the j 100 V source is developing 500 W and absorbing 1000 magnetizing vars. P10Ω =
| I a|^2 (10) = 500 W
Thus the 10 Ω resistor is absorbing 500 W.
Q−j10Ω =
| I b|^2 (−10) = −2500 VAR
Thus the −j10 Ω capacitor is developing 2500 magnetizing vars.
Qj5Ω =
| I o|^2 (5) = 2000 VAR
Thus the j5 Ω inductor is absorbing 2000 magnetizing vars. [b]
∑ Pdev = 500 W =
∑ Pabs
Problems 10–
Thus the V g 1 source is delivering 17. 5 kW and 5000 magnetizing vars. I g 2 = I 2 + I 3 = 51.2 + j 28. 4 A(rms) Sg 2 = 250(51. 2 − j 28 .4) = 12, 800 − j 7100 VA Thus the V g 2 source is delivering 12. 8 kW and absorbing 7100 magnetizing vars. [b]
∑ Pgen = 17.5 + 12.8 = 30. 3 kW ∑ Pabs = 7500 + 2800 +
(500)^2
= 30. 3 kW =
∑ Pgen ∑ Qdel = 9600 + 5000 = 14. 6 kVAR ∑ Qabs = 2500 + 7100 +
(500)^2
= 14. 6 kVAR =
∑ Qdel
P 10.27 S 1 = 1200 + 1196 = 2396 + j 0 VA
..^ · I 1 =^2396 120
= 19. 97 A
S 2 = 860 + 600 + 240 = 1700 + j 0 VA
..^ · I 2 =^1700 120
= 14. 167 A
S 3 = 4474 + 12,200 = 16,674 + j 0 VA
..^ · I 3 =^16 ,^674 240
= 69. 48 A
I g 1 = I 1 + I 3 = 89. 44 A
I g 2 = I 2 + I 3 = 83. 64 A
Breakers will not trip since both feeder currents are less than 100 A.
P 10.28 [a]
I 1 =
4000 − j 1000 125 = 32 − j 8 A (rms)
10–20 CHAPTER 10. Sinusoidal Steady State Power Calculations
I 2 =
5000 − j 2000 125 = 40 − j 16 A (rms)
I 3 =
10 ,000 + j 0 250
= 40 + j 0 A (rms)
..^ · I g 1 = I 1 + I 3 = 72 − j 8 A (rms)
I n = I 1 − I 2 = −8 + j 8 A (rms)
I g 2 = I 2 + I 3 = 80 − j 16 A(rms)
V g 1 = 0. 05 I g 1 + 125 + 0. 15 I n = 127.4 + j 0. 8 V(rms)
V g 2 = − 0. 15 I n + 125 + 0. 05 I g 2 = 130. 2 − j 2 V(rms)
Sg 1 = [(127.4 + j 0 .8)(72 + j8)] = [9166.4 + j 1076 .8] VA
Sg 2 = [(130. 2 − j2)(80 + j16)] = [10,448 + j 1923 .2] VA
Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P 0. 05 = | I g 1 |^2 (0.05) = 262. 4 W
P 0. 15 = | I n|^2 (0.15) = 19. 2 W
P 0. 05 = | I g 2 |^2 (0.05) = 332. 8 W ∑ Pdis = 262.4 + 19.2 + 332.8 + 4000 + 5000 + 10,000 = 19, 614. 4 W ∑ Pdev = 9166.4 + 10,448 = 19, 614. 4 W =
∑ Pdis ∑ Qabs = 1000 + 2000 = 3000 VAR ∑ Qdel = 1076.8 + 1923.2 = 3000 VAR =
∑ Qabs
P 10.29 [a] Let V L = Vm/0◦:
SL = 600(0.8 + j 0 .6) = 480 + j 360 VA
I ∗ =
Vm
Vm
; I =
Vm − j
Vm
