Sinusoidal 10 Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/- 45 V, I = 20/15 A Therefore 1 P = (100)(20) cos[-45 - (15)] = 500 W, 2 Q = 1000 sin -60 = -866.03 VAR, [b] V = 100/- 45 , I = 20/165 BA AB
AB
BA
P = 1000 cos(-210 ) = -866.03 W, Q = 1000 sin(-210 ) = 500 VAR, [c] V = 100/- 45 , I = 20/- 105 AB
P = 1000 cos(60 ) = 500 W,
Q = 1000 sin(60 ) = 866.03 VAR, [d] V = 100/0 , I = 20/120
AB
P = 1000 cos(-120 ) = -500 W,
BA BA
Q = 1000 sin(-120 ) = -866.03 VAR,
AP 10.2 pf = cos(v - i ) = cos[15 - (75)] = cos(-60 ) = 0.5 leading rf = sin(v - i ) = sin(-60 ) = -0.866
101
102
CHAPTER 10. Sinusoidal Steady State Power Calculations I 0.18 Ieff = = A 3 3 0.0324 (5000) = 54 W 3
AP 10.3 From Ex. 9.4
2 P = Ieff R =
AP 10.4 [a] Z = (39 + j26) (-j52) = 48 - j20 = 52/- 22.62 Therefore I = 250/0 = 4.85/18.08 A(rms) 48 - j20 + 1 + j4
VL = ZI = (52/- 22.62 )(4.85/18.08 ) = 252.20/- 4.54 V(rms) IL = VL = 5.38/- 38.23 A(rms) 39 + j26
[b] SL = VL I = (252.20/- 4.54 )(5.38/+ 38.23 ) = 1357/33.69 L = (1129.09 + j752.73) VA PL = 1129.09 W; QL = 752.73 VAR Q = |I |2 4 = 94.09 VAR
[c] P = |I |2 1 = (4.85)2 1 = 23.52 W;
[d] Sg (delivering) = 250I = (1152.62 - j376.36) VA Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR. [e] Qcap = (252.20)2 |VL |2 = = -1223.18 VAR -52 -52 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 = 250I = (40,000 - j30,000) Therefore I = 160 - j120 = 200/- 36.87 A(rms) I = 200/36.87 A(rms) Z= 250 V = = 1.25/- 36.87 = (1 - j0.75) I 200/36.87 XC = -0.75
Therefore R = 1 ,
Problems Parallel circuit derivation: P = (250)2 ; R (250)2 ; XC therefore R = (250)2 = 1.5625 40,000 (250)2 = -2.083 -30,000
103
Q=
therefore XC =
AP 10.6 S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA S2 = 6000(0.8) + j6000(0.6) = 4800 - j3600 VA ST = S1 + S2 = 13,800 + j8400 VA ST = 200I ; therefore I = 69 + j42 I = 69 - j42 A
Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91 V(rms) AP 10.7 [a] The phasor domain equivalent circuit and the Thvenin equivalent are shown below: Phasor domain equivalent circuit:
Thvenin equivalent:
VTh = 3
-j800 = 48 - j24 = 53.67/- 26.57 V 20 - j40 -j800 = 20 + j10 = 22.36/26.57 20 - j40
ZTh = 4 + j18 +
For maximum power transfer, ZL = (20 - j10)
104
CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I = 53.67/- 26.57 = 1.34/- 26.57 A 40 1.34 2
2
Therefore P =
20 = 18 W
[c] RL = |ZTh | = 22.36 53.67/- 26.57 = 1.23/- 39.85 A [d] I = 42.36 + j10 Therefore P = AP 10.8 1.23 2
2
(22.36) = 17 W
Mesh current equations: 660 = (34 + j50)I1 + j100(I1 - I2 ) + j40I1 + j40(I1 - I2 ) 0 = j100(I2 - I1 ) - j40I1 + 100I2 Solving, I1 = 3.536/- 45 A, I2 = 3.5/0 A; AP 10.9 [a] 1 .. P = (3.5)2 (100) = 612.50 W 2
248 = j400I1 - j500I2 + 375(I1 - I2 ) 0 = 375(I2 - I1 ) + j1000I2 - j500I1 + 400I2 Solving, I1 = 0.80 - j0.62 A; I2 = 0.4 - j0.3 = 0.5/- 36.87 A 1 .. P = (0.25)(400) = 50 W 2
Problems [b] I1 - I2 = 0.4 - j0.32 A 1 P375 = |I1 - I2 |2 (375) = 49.20 W 2 1 [c] Pg = (248)(0.8) = 99.20 W 2 Pabs = 50 + 49.2 = 99.20 W V2 = 1 V1 ; AP 10.10 [a] VTh = 210/0 V; 4 Short circuit equations: 840 = 80I1 - 20I2 + V1 0 = 20(I2 - I1 ) - V2 .. I2 = 14 A; [b] Pmax = 210 30
2
105
(checks) I 1 = 1 I2 4
RTh =
210 = 15 14
15 = 735 W
AP 10.11 [a] VTh = -4(146/0 ) = -584/0 V(rms) = 584/180 V(rms) V2 = 4V1 ; I1 = -4I2
Short circuit equations: 146/0 = 80I1 - 20I2 + V1 0 = 20(I2 - I1 ) + V2 .. I2 = -146/365 = -0.40 A; [b] P = -584 2920
2
RTh =
-584 = 1460 -0.4
1460 = 58.40 W
106
CHAPTER 10. Sinusoidal Steady State Power Calculations
Problems
P 10.1 1 [a] P = (100)(10) cos(50 - 15) = 500 cos 35 = 409.58 W 2 Q = 500 sin 35 = 286.79 VAR (abs) (abs) (abs)
1 [b] P = (40)(20) cos(-15 - 60) = 400 cos(-75 ) = 103.53 W 2 Q = 400 sin(-75 ) = -386.37 VAR (del)
1 [c] P = (400)(10) cos(30 - 150) = 2000 cos(-120 ) = -1000 W 2 Q = 2000 sin(-120 ) = -1732.05 VAR (del) (del) 1 [d] P = (200)(5) cos(160 - 40) = 500 cos(120 ) = -250 W 2 Q = 500 sin(120 ) = 433.01 VAR P 10.2 p = P + P cos 2t - Q sin 2t; dp = 0 when dt (abs)
(del)
dp = -2P sin 2t - 2Q cos 2t dt tan 2t = - Q P
- 2P sin 2t = 2Q cos 2t or
cos 2t =
P2
P ; + Q2
sin 2t = -
P2
Q + Q2
Let = tan-1 (-Q/P ), then p is maximum when 2t = and p is minimum when 2t = ( + ). Therefore pmax = P + P pmin = P - P P Q(-Q) - 2 =P+ P 2 + Q2 P + Q2 P 2 + Q2
and
P2
P Q -Q 2 =P- 2 +Q P + Q2
P 2 + Q2
Problems P 10.3 [a] hair dryer = 600 W sun lamp = 279 W television = 240 W Therefore Ieff = vacuum = 630 W air conditioner = 860 W P = 2609 W
107
2609 = 21.74 A 120 Yes, the breaker will trip. P = 2609 - 909 = 1700 W; Ieff =
[b]
1700 = 14.17 A 120 Yes, the breaker will not trip if the current is reduced to 14.17 A. [b] Ieff = 130/115 1.13 A =
2 vs dt R
P 10.4 P 10.5
[a] Ieff = 40/115 0.35 A; = Wdc = ..
2 Vdc T; R to +T to
Ws =
2 vs dt R
to +T to
2 Vdc T = R
2 Vdc =
1 T 1 T
to +T to
2 vs dt
Vdc = P 10.6
to +T to
2 vs dt = Vrms = Veff
2 [a] Area under one cycle of vg :
A = (52 )(2)(30 10-6 ) + 22 (2)(37.5 10-6 ) = 1800 10-6
2 Mean value of vg :
1800 10-6 A = =9 200 10-6 200 10-6 .. Vrms = 9 = 3 V(rms) M.V. = [b] P = P 10.7
2 Vrms 32 = 4W = R 2.25
i(t) = 200t
0 t 75 ms 75 ms t 100 ms
0.075 0
i(t) = 60 - 600t Irms = 1 0.1 =
(200)2 t2 dt +
0.1 0.075
(60 - 600t)2 dt 75 = 8.66 A(rms)
10(5.625) + 10(1.875) =
108
CHAPTER 10. Sinusoidal Steady State Power Calculations
2 P = Irms R
P 10.8 P 10.9
.. R =
3 103 = 40 75
Ig = 40/0 mA jL = j10,000 ; 1 = -j10,000 jC
Io =
j10,000 (40/0 ) = 80/90 mA 5000
1 1 P = |Io |2 (5000) = (0.08)2 (5000) = 16 W 2 2 1 Q = |Io |2 (-10,000) = -32 VAR 2 S = P + jQ = 16 - j32 VA |S| = 35.78 VA P 10.10 Ig = 4/0 mA; 1 = -j1250 ; jC jL = j500
Zeq = 500 + [-j1250 (1000 + j500)] = 1500 - j500 1 1 Pg = - |I|2 Re{Zeq } = - (0.004)2 (1500) = -12 mW 2 2 The source delivers 12 mW of power to the circuit.
Problems P 10.11 jL = j105 (0.5 10-3 ) = j50 ;
109
1 1 = -j30 = 5 [(1/3) 10-6 ] jC j10
-4 +
Vo Vo - 50I =0 + j50 40 - j30 Vo j50
I =
Place the equations in standard form: Vo 1 -50 1 + I + j50 40 - j30 40 - j30 1 + I (-1) = 0 j50 =4
Vo
Solving, Vo = 200 - j400 V; I = -8 - j4 A
Io = 4 - (-8 - j4) = 12 + j4 A 1 1 P40 = |Io |2 (40) = (160)(40) = 3200 W 2 2 P 10.12 [a] line loss = 7500 - 2500 = 5 kW line loss = |Ig |2 20 .. |Ig |2 = 250
|Ig | =
250 A
1010
CHAPTER 10. Sinusoidal Steady State Power Calculations |Ig |2 RL = 2500 |Ig |2 XL = -5000 Thus, .. RL = 10 .. XL = -20
|Z| =
(30)2 + (X - 20)2
|Ig | =
500 900 + (X - 20)2
.. 900 + (X - 20)2 = Solving, Thus,
25 104 = 1000 250 (X - 20) = 10. or X = 30
X = 10
[b] If X = 30 : 500 = 15 - j5 A Ig = 30 + j10 Sg = -500I = -7500 - j2500 VA g Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars. Qj30 = |Ig |2 X = 250(30) = 7500 VAR Therefore the line reactance is absorbing 7500 magnetizing vars. Q-j20 = |Ig |2 XL = 250(-20) = -5000 VAR Therefore the load reactance is generating 5000 magnetizing vars. Qgen = 7500 VAR = If X = 10 : 500 = 15 + j5 A Ig = 30 - j10 Sg = -500I = -7500 + j2500 VA g Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizing vars. Qj10 = |Ig |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 Qabs
Problems P 10.13 Zf = -j10,000 20,000 = 4000 - j8000 Zi = 2000 - j2000 .. Zf 4000 - j8000 = 3 - j1 = Zi 2000 - j2000 Zf Vg ; Zi Vg = 1/0 V
1011
Vo = -
Vo = (3 - j1)(1) = 3 - j1 = 3.16/- 18.43 V P =
2 1 (10) 1 Vm = = 5 10-3 = 5 mW 2 R 2 1000
P 10.14 [a] P = -
1 (240)2 = 60 W 2 480
1 -9 106 = = -360 C (5000)(5) 1 (240)2 = -80 VAR 2 (-360)
Q=
pmax = P + P 2 + Q2 = 60 + (60)2 + (80)2 = 160 W(del) [b] pmin = 60 - 602 + 802 = -40 W(abs) [c] P = 60 W from (a) [d] Q = -80 VAR from (a) [e] generate, because Q < 0 [f] pf = cos(v - i ) I= 240 240 + = 0.5 + j0.67 = 0.83/53.13 A 480 -j360
.. pf = cos(0 - 53.13 ) = 0.6 leading [g] rf = sin(-53.13 ) = -0.8
1012 P 10.15 [a]
CHAPTER 10. Sinusoidal Steady State Power Calculations
The mesh equations are: (10 - j20)I1 + (j20)I2 = 170 (j20)I1 + (12 - j4)I2 = 0 Solving, I1 = 4 + j1 A; I2 = 3.5 - j5.5 A
S = -Vg I = -(170)(4 - j1) = -680 + j170 VA 1 [b] Source is delivering 680 W. [c] Source is absorbing 170 magnetizing VAR. [d] P10 = ( 17)2 (10) = 170 W P12 = ( 42.5)2 (12) = 510 W (I1 - I2 ) = 0.5 + j6.5 A Q-j20 = ( 42.5)2 (20) = -850 VAR |I1 - I2 | = 42.5 Qj16 = ( 42.5)2 (16) = 680 VAR [e] Pdel = 680 W Pdiss = 170 + 510 = 680 W .. [f] Pdel = Pdiss = 680 W
Qabs = 170 + 680 = 850 VAR Qdev = 850 VAR .. mag VAR dev = mag VAR abs = 850
Problems P 10.16 [a] 1 = -j40 ; jC jL = j80
1013
Zeq = 40 - j40 + j80 + 60 = 80 + j60 Ig = 40/0 = 0.32 - j0.24 A 80 + j60
1 1 Sg = - Vg I = - 40(0.32 + j0.24) = -6.4 - j4.8 VA g 2 2 P = 6.4 W(del); |S| = |Sg | = 8 VA [b] I1 = -j40 Ig = 0.04 - j0.28 A 40 - j40 Q = 4.8 VAR(del)
1 P40 = |I1 |2 (40) = 1.6 W 2 1 P60 = |Ig |2 (60) = 4.8 W 2 Pdiss = 1.6 + 4.8 = 6.4 W = [c] I-j40 = Ig - I1 = 0.28 + j0.04 A 1 Q-j40 = |I-j40 |2 (-40) = -1.6 VAR(del) 2 1 Qj80 = |Ig |2 (80) = 6.4 VAR(abs) 2 Qabs = 6.4 - 1.6 = 4.8 VAR = P 10.17 [a] Z1 = 240 + j70 = 250/16.26 pf = cos(16.26 ) = 0.96 lagging rf = sin(16.26 ) = 0.28 Qdev Pdev
1014
CHAPTER 10. Sinusoidal Steady State Power Calculations Z2 = 160 - j120 = 200/- 36.87 pf = cos(-36.87 ) = 0.80 leading rf = sin(-36.87 ) = -0.60 Z3 = 30 - j40 = 50/- 53.13 pf = cos(-53.13 ) = 0.6 leading rf = sin(-53.13 ) = -0.8
[b] Y = Y1 + Y2 + Y3 Y1 = 1 ; 250/16.26 Y2 = 1 ; 200/- 36.87 Y3 = 1 50/- 53.13
Y = 19.84 + j17.88 mS Z= 1 = 37.44/- 42.03 Y
pf = cos(-42.03 ) = 0.74 leading rf = sin(-42.03 ) = -0.67 P 10.18 [a] S1 = 16 + j18 kVA; S2 = 6 - j8 kVA; S3 = 8 + j0 kVA
ST = S1 + S2 + S3 = 30 + j10 kVA 250I = (30 + j10) 103 ; Z= .. I = 120 - j40 A
250 = 1.875 + j0.625 = 1.98/18.43 120 - j40
[b] pf = cos(18.43 ) = 0.9487 lagging P 10.19 [a] From the solution to Problem 10.18 we have IL = 120 - j40 A(rms) .. Vs = 250/0 + (120 - j40)(0.01 + j0.08) = 254.4 + j9.2 = 254.57/2.07 V(rms) [b] |IL | = 16,000 Q = (16,000)(0.08) = 1280 VAR Qs = 10,000 + 1280 = 11.28 kVAR
P = (16,000)(0.01) = 160 W [c] Ps = 30,000 + 160 = 30.16 kW 30 (100) = 99.47% [d] = 30.16
Problems P 10.20 ST = 4500 - j S1 = 4500 (0.28) = 4500 - j1312.5 VA 0.96
1015
2700 (0.8 + j0.6) = 2700 + j2025 VA 0.8
S2 = ST - S1 = 1800 - j3337.5 = 3791.95/- 61.66 VA pf = cos(-61.66 ) = 0.4747 leading P 10.21
2400I = 60,000 + j40,000 1 I = 25 + j16.67; 1 .. I1 = 25 - j16.67 A(rms)
2400I = 20,000 - j10,000 2 I = 8.33 - j4, 167; 2 I3 = .. I2 = 8.33 + j4.167 A(rms) I4 = 2400/0 = 0 - j25 A j96
2400/0 = 16.67 + j0 A; 144
Ig = I1 + I2 + I3 + I4 = 50 - j37.5 A Vg = 2400 + (j4)(50 - j37.5) = 2550 + j200 = 2557.83/4.48 V(rms) P 10.22 [a] S1 = 60,000 - j70,000 VA S2 = |VL |2 (2500)2 = 240,000 + j70,000 VA = Z2 24 - j7 .. IL = 120/0 A(rms)
S1 + S2 = 300,000 VA 2500I = 300,000; L
Vg = VL + IL (0.1 + j1) = 2500 + (120)(0.1 + j1) = 2512 + j120 = 2514.86/2.735 V(rms)
1016
CHAPTER 10. Sinusoidal Steady State Power Calculations 1 1 = = 16.67 ms f 60 .. t = 126.62 s
[b] T =
t 2.735 = ; 360 16.67 ms
[c] VL lags Vg by 2.735 or 126.62 s
P 10.23 [a] From the solution to Problem 9.56 we have:
Vo = j80 = 80/90 V 1 1 Sg = - Vo I = - (j80)(10 - j10) = -400 - j400 VA g 2 2 Therefore, the independent current source is delivering 400 W and 400 magnetizing vars. I1 = P5 Vo = j16 A 5 1 = (16)2 (5) = 640 W 2 Vo = -10 A -j8
Therefore, the 8 resistor is absorbing 640 W. I =
1 Qcap = (10)2 (-8) = -400 VAR 2 Therefore, the -j8 capacitor is developing 400 magnetizing vars. 2.4I = -24 V I2 = j80 + 24 Vo - 2.4I = j4 j4
= 20 - j6 A = 20.88/- 16.7 A
Problems 1 Qj4 = |I2 |2 (4) = 872 VAR 2 Therefore, the j4 inductor is absorbing 872 magnetizing vars. Sd.s. = 1 (2.4I )I = 1 (-24)(20 + j6) 2 2 2 = -240 - j72 VA
1017
Thus the dependent source is delivering 240 W and 72 magnetizing vars. [b] [c] Pgen = 400 + 240 = 640 W = Pabs Qabs Qgen = 400 + 400 + 72 = 872 VAR =
P 10.24 [a] From the solution to Problem 9.58 we have
Ia = -j10 A;
Ib = -20 + j10 A;
Io = 20 - j20 A
1 S100V = - (100)I = -50(j10) = -j500 VA a 2 Thus, the 100 V source is developing 500 magnetizing vars. Sj100V = - 1 (j100)I = -j50(-20 - j10) b 2 = -500 + j1000 VA Thus, the j100 V source is developing 500 W and absorbing 1000 magnetizing vars. 1 P10 = |Ia |2 (10) = 500 W 2 Thus the 10 resistor is absorbing 500 W. 1 Q-j10 = |Ib |2 (-10) = -2500 VAR 2 Thus the -j10 capacitor is developing 2500 magnetizing vars. 1 Qj5 = |Io |2 (5) = 2000 VAR 2 Thus the j5 inductor is absorbing 2000 magnetizing vars. [b] Pdev = 500 W = Pabs
1018 [c]
CHAPTER 10. Sinusoidal Steady State Power Calculations Qdev = 500 + 2500 = 3000 VAR Qabs = 1000 + 2000 = 3000 VAR = Qdev
465/0 P 10.25 [a] I = = 2.4 - j1.8 = 3/- 36.87 A(rms) 124 + j93 P = (3)2 (4) = 36 W [b] YL = 1 = 5.33 - j4 mS 120 + j90 1 = -250 -4 10-3
.. XC = [c] ZL =
1 = 187.5 5.33 10-3 465/0 [d] I = = 2.43/- 0.9 A 191.5 + j3 P = (2.43)2 (4) = 23.58 W [e] % = 23.58 (100) = 65.5% 36 Thus the power loss after the capacitor is added is 65.6% of the power loss before the capacitor is added.
P 10.26 [a]
250I = 7500 + j2500; 1 250I = 2800 - j9600; 2 I3 =
.. I1 = 30 - j10 A(rms) .. I2 = 11.2 + j38.4 A(rms)
500 500 + = 40 - j10 A(rms) 12.5 j50
Ig1 = I1 + I3 = 70 - j20 A Sg1 = 250(70 + j20) = 17,500 + j5000 VA
Problems Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars. Ig2 = I2 + I3 = 51.2 + j28.4 A(rms) Sg2 = 250(51.2 - j28.4) = 12,800 - j7100 VA
1019
Thus the Vg2 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 = 30.3kW = 12.5 Pgen
Qdel = 9600 + 5000 = 14.6 kVAR Qabs (500)2 = 14.6 kVAR = = 2500 + 7100 + 50 Qdel
P 10.27 S1 = 1200 + 1196 = 2396 + j0 VA .. I1 = 2396 = 19.97 A 120 1700 = 14.167 A 120 16,674 = 69.48 A 240
S2 = 860 + 600 + 240 = 1700 + j0 VA .. I2 =
S3 = 4474 + 12,200 = 16,674 + j0 VA .. I3 =
Ig1 = I1 + I3 = 89.44 A Ig2 = I2 + I3 = 83.64 A Breakers will not trip since both feeder currents are less than 100 A. P 10.28 [a]
I1 =
4000 - j1000 = 32 - j8 A (rms) 125
1020
CHAPTER 10. Sinusoidal Steady State Power Calculations I2 = I3 = 5000 - j2000 = 40 - j16 A (rms) 125 10,000 + j0 = 40 + j0 A (rms) 250
.. Ig1 = I1 + I3 = 72 - j8 A (rms) In = I1 - I2 = -8 + j8 A (rms) Ig2 = I2 + I3 = 80 - j16 A(rms) Vg1 = 0.05Ig1 + 125 + 0.15In = 127.4 + j0.8 V(rms) Vg2 = -0.15In + 125 + 0.05Ig2 = 130.2 - j2 V(rms) Sg1 = [(127.4 + j0.8)(72 + j8)] = [9166.4 + j1076.8] VA Sg2 = [(130.2 - j2)(80 + j16)] = [10,448 + j1923.2] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P0.05 = |Ig1 |2 (0.05) = 262.4 W P0.15 = |In |2 (0.15) = 19.2 W P0.05 = |Ig2 |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 = Qabs = 1000 + 2000 = 3000 VAR Qdel = 1076.8 + 1923.2 = 3000 VAR = P 10.29 [a] Let VL = Vm /0 : Qabs Pdis
SL = 600(0.8 + j0.6) = 480 + j360 VA I = 480 360 +j ; Vm Vm I = 480 360 -j Vm Vm
Problems 120/ = Vm + 480 360 -j (1 + j2) Vm Vm
1021
2 2 120Vm / = Vm + (480 - j360)(1 + j2) = Vm + 1200 + j600 2 120Vm cos = Vm + 1200;
120Vm sin = 600
2 2 (120)2 Vm = (Vm + 1200)2 + 6002 2 4 2 14,400Vm = Vm + 2400Vm + 18 105
or
4 2 Vm - 12,000Vm + 18 105 = 0
Solving, Vm = 108.85 V and Vm = 12.326 V If Vm = 108.85 V: sin = 600 = 0.045935; (108.85)(120) 600 = 0.405647; (12.326)(120) .. = 2.63
If Vm = 12.326 V: sin = [b] .. = 23.93
P 10.30 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA 125I = (17,000 + j10,535.65); L .. IL = 136 - j84.29 A(rms) Vs = 125 + (136 - j84.29)(0.01 + j0.08) = 133.10 + j10.04 = 133.48/4.31 V(rms) |Vs | = 133.48 V(rms) [b] P = |I |2 (0.01) = (160)2 (0.01) = 256 W I = 136 + j84.29 A(rms) L
1022
CHAPTER 10. Sinusoidal Steady State Power Calculations (125)2 = -10,535.65; XC - 1 = -1.48; C C= XC = -1.483 1 = 1788.59 F (1.48)(120)
[c]
[d] I = 136 + j0 A(rms) Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88 = 126.83/4.92 V(rms) |Vs | = 126.83 V(rms) [e] P = (136)2 (0.01) = 184.96 W P 10.31
IL = IC =
153,600 - j115,200 = 32 - j24 A(rms) 4800 4800 4800 =j = jIC -jXC XC
I = 32 - j24 + jIC = 32 + j(IC - 24) Vs = 4800 + (2 + j10)[32 + j(IC - 24)] = (5104 - 10IC ) + j(272 + 2IC ) |Vs |2 = (5104 - 10IC )2 + (272 + 2IC )2 = (4800)2
2 .. 104IC - 100,992IC + 3,084,800 = 0
Solving,
IC = 31.57 A(rms);
IC = 939.51 A(rms)
*Select the smaller value of IC to minimize the magnitude of I . .. XC = - .. C = 4800 = -152.04 31.57
1 = 17.45 F (152.04)(120)
Problems P 10.32 ZL = |ZL |/ = |ZL | cos + j|ZL | sin Thus |I| = |VTh | (RTh + |ZL | cos )2 + (XTh + |ZL | sin )2 0.5|VTh |2 |ZL | cos (RTh + |ZL | cos )2 + (XTh + |ZL | sin )2
1023
Therefore P =
Let D = demoninator in the expression for P, then (0.5|VTh |2 cos )(D 1 - |ZL |dD/d|ZL |) dP = d|ZL | D2 dD = 2(RTh + |ZL | cos ) cos + 2(XTh + |ZL | sin ) sin d|ZL | dP = 0 when d|ZL | D = |ZL | dD d|ZL |
Substituting the expressions for D and (dD/d|ZL |) into this equation gives us the 2 2 relationship RTh + XTh = |ZL |2 or |ZTh | = |ZL |. P 10.33 [a] ZTh = j40 40 - j40 = 20 - j20
.. ZL = ZTh = 20 + j20
[b] VTh =
40 (120) = 60 - j60 V 40 + j40
I=
60 - j60 = 1.5 - j1.5 A 40 1 Pload = |I|2 (20) = 45 W 2
P 10.34 [a]
115.2 + j33.6 - 240 115.2 + j33.6 =0 + ZTh 80 - j60 .. ZTh = 40 - j100 .. ZL = 40 + j100
1024
CHAPTER 10. Sinusoidal Steady State Power Calculations 240 = 3 A(rms) 80
[b] I =
P = (3)2 (40) = 360 W P 10.35 [a] ZTh = [(3 + j4) - j8] + 7.32 - j17.24 = 15 - j15 .. R = |ZTh | = 21.21 [b] VTh = -j8 (112.5) = 144 - j108 V(rms) 3 - j4
I=
144 - j108 = 4.45 - j1.14 35.21 - j15
P = |I|2 (21.21) = 447.35 W P 10.36 [a] Open circuit voltage:
V1 = 5I = 5
100 - 5I 25 + j10
(25 + j10)I = 100 - 5I I = 100 = 3 - j1 A 30 + j10 j3 (5I ) = 15/0 V 1 + j3
VTh =
Problems Short circuit current:
1025
V2 = 5I =
100 5I - 25 + j10
I = 3 - j1 A Isc = ZTh 5I = 15 - j5 A 1 15 = 0.9 + j0.3 = 15 - j5
ZL = ZTh = 0.9 - j0.3
IL =
15 = 8.33 A(rms) 1.8
P = |IL |2 (0.9) = 62.5 W [b] VL = (0.9 - j0.3)(8.33) = 7.5 - j2.5 V(rms)
I1 =
VL = -0.833 - j2.5 A(rms) j3
1026
CHAPTER 10. Sinusoidal Steady State Power Calculations I2 = I1 + IL = 7.5 - j2.5 A(rms) 5I = I2 + VL .. I = 3 - j1 A
Id.s. = I - I2 = -4.5 + j1.5 A Sg = -100(3 + j1) = -300 - j100 VA Sd.s. = 5(3 - j1)(-4.5 - j1.5) = -75 + j0 VA Pdev = 300 + 75 = 375 W % developed = Checks: P25 = (10)(25) = 250 W P1 = (62.5)(1) = 62.5 W P0.9 = 62.5 W Pabs = 250 + 62.5 + 62.5 = 375 W = Qj10 = (10)(10) = 100 VAR Qj3 = (6.94)(3) = 20.83 VAR Q-j0.3 = (69.4)(-0.3) = -20.83 VAR Qsource = -100 VAR Q = 100 + 20.83 - 20.83 - 100 = 0 P 10.37 [a] Open circuit voltage: Pdev 62.5 (100) = 16.67% 375
V - 100 V + - 0.1V = 0 5 j5 .. V = 40 + j80 V(rms)
Problems VTh = V + 0.1V (-j5) = V (1 - j0.5) = 80 + j60 V(rms) Short circuit current:
1027
Isc = 0.1V +
V = (0.1 + j0.2)V -j5
V V - 100 V + =0 + j5 -j5 5 .. V = 100 V(rms) Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms) ZTh = VTh 80 + j60 = 4 - j2 = Isc 10 + j20
.. Ro = |ZTh | = 4.47 [b]
I=
80 + j60 = 7.36 + j8.82 A(rms) 4 + 20 - j2 P = (11.49)2 ( 20) = 590.17 W
1028 [c]
CHAPTER 10. Sinusoidal Steady State Power Calculations
I=
80 + j60 = 10 + j7.5 A(rms) 8
P = (102 + 7.52 )(4) = 625 W [d]
V - 100 V V - (25 + j50) + + =0 5 j5 -j5 V = 50 + j25 V(rms) 0.1V = 5 + j2.5 5 + j2.5 + IC = 10 + j7.5 IC = 5 + j5 A(rms) IL = V = 5 - j10 A(rms) j5
IR = IC + IL = 10 - j5 A(rms) Ig = IR + 0.1V = 15 - j2.5 A(rms) Sg = -100I = -1500 - j250 VA g 100 = 5(5 + j2.5) + Vcs + 25 + j50 .. Vcs = 50 - j62.5 V(rms) Scs = (50 - j62.5)(5 - j2.5) = 93.75 - j437.5 VA Thus, Pdev = 1500 % delivered to Ro = 625 (100) = 41.67% 1500
Problems P 10.38 [a] First find the Thvenin equivalent: jL = j3000 ZTh = 6000 12,000 + j3000 = 4000 + j3000 VTh = 12,000 (180) = 120/0 V 6000 + 12,000
1029
-j = -j1000 C
I=
120 = 18 - j6 mA 6000 + j2000
1 P = |I|2 (2000) = 360 mW 2 [b] Set Co = 0.1 F so -j/C = -j2000 Set Ro as close as possible to Ro = 40002 + 10002 = 4123.1 .. Ro = 4000 [c] I = 120 = 14.77 - j1.85 mA 8000 + j1000 j3000 - j2000 = j1000
1 P = |I|2 (4000) = 443.1 mW 2 Yes; 443.1 mW > 360 mW 120 = 15 mA [d] I = 8000 1 P = (0.015)2 (4000) = 450 mW 2 [e] Ro = 4000 ; [f] Yes; Co = 66.67 nF 450 mW > 443.1 mW
1030
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.39 [a] Set Co = 0.1 F, so -j/C = -j2000 ; also set Ro = 4123.1 I= 120 = 14.55 - j1.79 mA 8123.1 + j1000
1 P = |I|2 (4123.1) = 443.18 mW 2 [b] Yes; [c] Yes; P 10.40 [a] 443.18 mW > 360 mW 443.18 mW < 450 mW
1 1 = 100 ; C= = 26.53 F C (100)(120) 13,800 13,800 [b] Iwo = + = 46 - j138 A(rms) 300 j100 Vswo = 13,800 + (46 - j138)(1.5 + j12) = 15,525 + j345 = 15,528.83/1.27 V(rms) Iw = 13,800 = 46 A(rms) 300 15,528.82 - 1 (100) = 11.88% 13,879.98
Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28 V(rms) % increase =
[c] P wo = |46 - j138|2 1.5 = 31.74 kW P w = 462 (1.5) = 3174 W % increase = 31,740 - 1 (100) = 900% 3174 1600 (0.6) = 1600 + j1200 kVA 0.8
P 10.41 [a] So = original load = 1600 + j Sf = final load = 1920 + j
1920 (0.28) = 1920 + j560 kVA 0.96
.. Qadded = 560 - 1200 = -640 kVAR [b] deliver [c] Sa = added load = 320 - j640 = 715.54/- 63.43 kVA pf = cos(-63.43) = 0.4472 leading
Problems [d] I = L (1600 + j1200) 103 = 666.67 + j500 A 2400
1031
IL = 666.67 - j500 = 833.33/- 36.87 A(rms) |IL | = 833.33 A(rms) [e] I = L (1920 + j560) 103 = 800 + j233.33 2400
IL = 800 - j233.33 = 833.33/- 16.26 A(rms) |IL | = 833.33 A(rms) P 10.42 [a] Pbefore = Pafter = (833.33)2 (0.05) = 34,722.22 W [b] Vs (before) = 2400 + (666.67 - j500)(0.05 + j0.4) = 2633.33 + j241.67 = 2644.4/5.24 V(rms) |Vs (before)| = 2644.4 V(rms) Vs (after) = 2400 + (800 + j233.33)(0.05 + j0.4) = 2346.67 + j331.67 = 2369.99/8.04 V(rms) |Vs (after)| = 2369.99 V(rms) P 10.43 [a]
180 = 3I1 + j4I1 + j3(I2 - I1 ) + j9(I1 - I2 ) - j3I1 0 = 9I2 + j9(I2 - I1 ) + j3I1 Solving, I1 = 18 - j18 A(rms); I2 = 12/0 A(rms) .. Vo = (12)(9) = 108/0 V(rms) [b] P = (12)2 (9) = 1296 W [c] Sg = -(180)(18 + j18) = -3240 - j3240 VA % delivered = 1296 (100) = 40% 3240 .. Pg = -3240 W
1032
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.44 [a] Open circuit voltage:
180 = 3I1 + j4I1 - j3I1 + j9I1 - j3I1 .. I1 = 180 = 9.31 - j21.72 A(rms) 3 + j7
VTh = j9I1 - j3I1 = j6I1 = 130.34 + j55.86 V = 141.81/23.20 V(rms) Short circuit current:
180 = 3I1 + j4I1 + j3(Isc - I1 ) + j9(I1 - Isc ) - j3I1 0 = j9(Isc - I1 ) + j3I1 Solving, Isc = 20 - j20 A ZTh = I1 = 30 - j20 A VTh 130.34 + j55.86 = 1.86 + j4.66 = Isc 20 - j20
IL =
130.34 + j55.86 = 35 + j15 = 38.08/23.20 A 3.72
Problems PL = (38.12)2 (1.86) = 2700 W [b] I1 = Zo + j9 1.86 - j4.66 + j9 I2 = (35 + j15) = 30/0 A(rms) j6 j6
1033
Pdev = (180)(30) = 5400 W P 10.45 [a]
54 = I1 + j2(I1 - I2 ) + j3I2 0 = 7I2 + j2(I2 - I1 ) - j3I2 + j8I2 + j3(I1 - I2 ) Solving, I1 = 12 - j21 A(rms); I2 = -3 A(rms)
Vo = 7I2 = -21/180 V(rms) [b] P = |I2 |2 (7) = 63 W [c] Pg = (54)(12) = 648 W % delivered = P 10.46 [a] 63 (100) = 9.72% 648
Open circuit: VTh = -j3I1 + j2 I1 = -jI1 I1 = 54 = 10.8 - j21.6 A 1 + j2
VTh = -21.6 - j10.8 V
1034
CHAPTER 10. Sinusoidal Steady State Power Calculations Short circuit: 54 = I1 + j2(I1 - Isc ) + j3Isc 0 = j2(Isc - I1 ) - j3Isc + j8Isc + j3(I1 - Isc ) Solving, Isc = -3.32 + j5.82 ZTh = VTh -21.6 - j10.8 = 0.2 + 3.6j = 3.6/86.82 = Isc -3.32 + j5.82
.. RL = |ZTh | = 3.606 [b]
I=
-21.6 - j10.8 = 4.610/163.2 A 3.806 + j3.6 which is greater than when RL = 7
P = |I|2 (3.6) = 76.6 W, P 10.47 [a]
54 = I1 + j2(I1 - I2 ) + j4kI2 0 = 7I2 + j2(I2 - I1 ) - j4kI2 + j8I2 + j4k(I1 - I2 ) Place the equations in standard form: 54 = (1 + j2)I1 + j(4k - 2)I2 0 = j(4k - 2)I1 + [7 + j(10 - 8k)]I2 I1 = 54 - I2 j(4k - 2) (1 + j2) j54(4k - 2) [7 + j(10 - 8k)](1 + j2) + (4k - 2)2
Substituting, I2 = -
For Vo = 0, I2 = 0, so if 4k - 2 = 0, then k = 0.5.
Problems [b] When I2 = 0 I1 = 54 = 10.8 - j21.6 A(rms) 1 + j2
1035
Pg = (54)(10.8) = 583.2 W Check: Ploss = |I1 |2 (1) = 583.2 W P 10.48 [a] From Problem 9.67, ZTh = 85 + j85 and VTh = 850 + j850 V. Thus, for maximum power transfer, ZL = ZTh = 85 - j85 :
I2 =
850 + j850 = 5 + j5 A 170
425/0 = (5 + j5)I1 - j20(5 + j5) .. I1 = 325 + j100 = 42.5 - j22.5 A 5 + j5
Sg (del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA Pg = 18,062.5 W [b] Ploss = |I1 |2 (5) + |I2 |2 (45) = 11,562.5 + 2250 = 13,812.5 W % loss in transformer = 18,062.5 - 13,812.5 (100) = 23.53% 18,062.5
1036
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.49 [a] From Problem 9.70, Zab = 100 + j136.26 I1 = I2 = so 50 50 = = 160 - j120 mA 100 + j13.74 + 100 + 136.26 200 + j150 jM j270 (0.16 - j0.12) = 51.84 + j15.12 mA I1 = Z22 800 + j600
VL = (300 + j100)(51.84 + j15.12)103 = 14.04 + j9.72 V |VL | = 17.08 V [b] Pg (ideal) = 50(0.16) = 8 W Pg (practical) = 8 - |I1 |2 (100) = 4 W PL = |I2 |2 (300) = 874.8 mW % delivered = P 10.50 [a] 0.8748 (100) = 21.87% 4
Open circuit: 120 VTh = (j10) = 36 + j48 V 16 + j12 Short circuit: (16 + j12)I1 - j10Isc = 120 -j10I1 + (11 + j23)Isc = 0 Solving, Isc = 2.4/0 A ZTh = 36 + j48 = 15 + j20 2.4
.. ZL = ZTh = 15 - j20
IL =
VTh 36 + j48 = 1.2 + j1.6 A(rms) = 2.0/53.13 A(rms) = ZTh + ZL 30
PL = |IL |2 (15) = 60 W
Problems [b] I1 = 26 + j3 Z22 I2 = (1.2 + j1.6) = 5.23/- 30.29 A)rms) jM j10 60 (100) = 13.86% 432.8
1037
Ptransformer = (120)(5.23) cos(-30.29 ) - (5.23)2 (4) = 432.8 W % delivered =
P 10.51 [a] jL1 = j(10,000)(1 10-3 ) = j10 jL2 = j(10,000)(1 10-3 ) = j10 jM = j10
200 = (5 + j10)Ig + j5IL 0 = j5Ig + (15 + j10)IL Solving, Ig = 10 - j15 A; Thus, ig = 18.03 cos(10,000t - 56.31 ) A iL = 5 cos(10,000t - 180 ) A [b] k = M 0.5 = = 0.5 L1 L2 1 [c] When t = 50 s: 10,000t = (10,000)(50) 10-6 = 0.5 rad = 90 ig (50 s) = 18.03 cos(90 - 56.31 ) = 15 A iL (50 s) = 5 cos(90 - 180 ) = 0 A 1 1 1 w = L1 i2 + L2 i2 + M i1 i2 = (10-3 )(15)2 + 0 + 0 = 112.5 mJ 1 2 2 2 2 When t = 100 s: 10,000t = (104 )(100) 10-6 = = 180 ig (100 s) = 18.03 cos(180 - 56.31 ) = -10 A iL (100 s) = 5 cos(180 - 180 ) = 5 A 1 1 w = (10-3 )(10)2 + (10-3 )(5)2 + 0.5 10-3 (-10)(5) = 37.5 mJ 2 2 IL = -5 A
1038
CHAPTER 10. Sinusoidal Steady State Power Calculations
[d] From (a), Im = 5 A, 1 .. P = (5)2 (15) = 187.5 W 2 [e] Open circuit: VTh = 200 (-j5) = -80 - j40 V 5 + j10
Short circuit: 200 = (5 + j10)I1 + j5Isc 0 = j10Isc + j5I1 Solving, Isc = -11.094/123.69 A; ZTh = I1 = 22.188/- 56.31 A
VTh -80 - j40 = = 1 + j8 Isc 11.094/123.69
.. RL = 8.962 [f]
I=
-80 - j40 = 7.399/165.13 A 9.062 + j8
1 P = (7.399)2 (8.062) = 220.70 W 2
[g] ZL = ZTh = 1 - j8 -80 - j40 = 44.72/- 153.43 A [h] I = 2
1 P = (44.72)2 (1) = 1000 W 2
Problems P 10.52 [a]
1039
10 = j1(I1 - I2 ) + j1(I3 - I2 ) - j1(I1 - I3 ) 0 = 1I2 + j2(I2 - I3 ) + j1(I2 - I1 ) + j1(I2 - I1 ) + j1(I2 - I3 ) 0 = 1I3 - j1(I3 - I1 ) + j2(I3 - I2 ) + j1(I1 - I2 ) Solving, I1 = 6.25 + j7.5 A(rms); Ia = I1 = 6.25 + j7.5 A Ic = I2 = 5 + j2.5 A Ie = I1 - I3 = 1.25 + j10 A [b] I2 = 5 + j2.5 A(rms); I3 = 5 - j2.5 A(rms)
Ib = I1 - I2 = 1.25 + j5 A Id = I3 - I2 = -j5 A If = I3 = 5 - j2.5 A
Va = 10 V Vc = 1Ic = 5 + j2.5 V Ve = -j1Ie = 10 - j1.25 V Sa = -10I = -62.5 + j75 VA a Sb = Vb I = 6.25 + j1.5625 VA b
Vb = j1Ib + j1Id = j1.25 V Vd = j2Id + j1Ib = 5 + j1.25 V Vf = 1If = 5 - j2.5 V
1040
CHAPTER 10. Sinusoidal Steady State Power Calculations Sc = Vc I = 31.25 + j0 VA c Sd = Vd I = -6.25 + j25 VA d Se = Ve I = 0 - j101.5625 VA e Sf = Vf I = 31.25 VA f
[c]
Pdev = 62.5 W Pabs = 6.25 + 31.25 - 6.25 + 31.25 = 62.5 W Note that the total power absorbed by the coupled coils is zero: 6.25 - 6.25 = 0 = Pb + Pd
[d]
Qdev = 101.5625 VAR The capacitor is developing magnetizing vars. Qabs = 75 + 1.5625 + 25 = 101.5625 VAR Q absorbed by the coupled coils is Qb + Qd = 26.5625 VAR
P 10.53 Open circuit voltage:
I1 =
10/0 = 2 - j4 A 1 + j2
VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57 V Short circuit current:
10/0 = (1 + j2)I1 - j3.2Isc
Problems 0 = -j3.2I1 + j5.4Isc Solving, Isc = 5.89/- 5.92 A ZTh = 14.31/26.57 = 2.43/32.49 = 2.048 + j1.304 5.89/- 5.92 14.31/26.57 = 3.49/26.57 A 4.096
1041
.. I2 =
10/0 = (1 + j2)I1 - j3.2I2 .. Zg = P 10.54 [a] I1 = 10 + j3.2(3.49/26.57 ) 10 + j3.2I2 = = 5/0 A 1 + j2 1 + j2
10/0 = 2 + j0 = 2/0 5/0
272/0 = 2Ig + j10Ig + j14(Ig - I2 ) - j6I2 +j14Ig - j8I2 + j20(Ig - I2 ) 0 = j20(I2 - Ig ) - j14Ig + j8I2 + j4I2 +j8(I2 - Ig ) - j6Ig + 8I2
1042
CHAPTER 10. Sinusoidal Steady State Power Calculations Solving, Ig = 20 - j4 A(rms); P8 = (24)2 (8) = 4608 W I2 = 24/0 A(rms)
[b] Pg (developed) = (272)(20) = 5440 W Vg 272 - 2 = 11.08 + j2.62 = 11.38/13.28 [c] Zab = -2= Ig 20 - j4 [d] P2 = |Ig |2 (2) = 832 W Pdiss = 832 + 4608 = 5440 W = P 10.55 [a] Pdev
300 = 60I1 + V1 + 20(I1 - I2 ) 0 = 20(I2 - I1 ) + V2 + 40I2 1 V2 = V1 ; 4 Solving, V1 = 260 V(rms); I1 = 0.25 A(rms); V2 = 65 V(rms) I2 = -1.0 A(rms) I2 = -4I1
V5A = V1 + 20(I1 - I2 ) = 285 V(rms) .. P = -(285)(5) = -1425 W Thus 1425 W is delivered by the current source to the circuit. [b] I20 = I1 - I2 = 1.25 A(rms) .. P20 = (1.25)2 (20) = 31.25 W
Problems P 10.56
1043
30Vo = Va ; -Va Vb = ; 1 20
Io = Ia ; 30 Ib = -20Ia ;
Vo = 10Io therefore
therefore
Va = 9 k Ia
Vb 9000 = 22.5 = Ib 400
Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers are lossless, P10 = P22.5 , and the power delivered to the 22.5 resistor is 22 (22.5) or 90 W. P 10.57 [a] Vb a2 10 = = 2.5 ; therefore a2 = 100, Ib 400 50 = 10 A; P = (100)(2.5) = 250 W [b] Ib = 5 200 50
2
a = 10
P 10.58 [a] ZTh = 720 + j1500 + .. Zab = 1700 Zab = ZL (1 + N1 /N2 )2
(40 - j30) = 1360 + j1020 = 1700/36.87
(1 + N1 /N2 )2 = 6800/1700 = 4 .. N1 /N2 = 1 [b] VTh or N2 = N1 = 1000 turns 255/0 (j200) = 1020/53.13 V = 40 + j30
IL =
1020/53.13 = 0.316/34.7 A(rms) 3060 + j1020
Since the transformer is ideal, P6800 = P1700 . P = |IL |2 (1700) = 170 W
1044 [c]
CHAPTER 10. Sinusoidal Steady State Power Calculations
255/0 = (40 + j30)I1 - j200(0.26 + j0.18) .. I1 = 4.13 - j1.80 A(rms) Pgen = (255)(4.13) = 1053 W Ptrans = 1053 - 170 = 883 W % transmitted = P 10.59 [a] 883 (100) = 83.85% 1053
For maximum power transfer, Zab = 90 k Zab = 1 + .. 1+ N1 N2
2 2
ZL = 90,000 = 225 400 N1 = 15 - 1 = 14 N2 180 180,000 180 180,000
2
N1 1+ N2
N1 = 15; N2
2
[b] P = |Ii | (90,000) = [c] V1 = Ri Ii = (90,000) [d]
(90,000) = 90 mW = 90 V
Vg = (2.25 10-3 )(100,000 80,000) = 100 V
Problems Pg (del) = (2.25 10-3 )(100) = 225 mW % delivered = P 10.60 [a] Zab = 1 + .. I1 = I2 = N1 N2 90 (100) = 40% 225
2
1045
(1 - j2) = 25 - j50
100/0 = 2.5/0 A 15 + j50 + 25 - j50
N1 I1 = 10/0 A N2
.. IL = I1 + I2 = 12.5/0 A(rms) P1 = (12.5)2 (1) = 156.25 W P15 = (2.5)2 (15) = 93.75 W [b] Pg = -100(2.5/0 ) = -250 W Pabs = 156.25 + 93.75 = 250 W = P 10.61 [a] 25a2 + 4a2 = 500 1 2 I25 = a1 I; I4 = a2 I; P4 = 4P25 ; .. P25 = a2 I2 (25) 1 P4 = a2 I2 (4) 2 a2 I2 4 = 100a2 I2 2 1 Pdev
100a2 = 4a2 1 2 a1 = 2 a2 = 10
25a2 + 100a2 = 500; 1 1 25(4) + 4a2 = 500; 2 [b] I =
2000/0 = 2/0 A(rms) 500 + 500
I25 = a1 I = 4 A P25 = (16)(25) = 400 W [c] I4 = a2 I = 10(2) = 20 A(rms) V4 = (20)(4) = 80/0 V(rms)
1046
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.62 [a] Open circuit voltage:
500 = 100I1 + V1 V2 = 400I2 V2 V1 = 1 2 I1 = 2I2 Substitute and solve: 2V1 = 400I1 /2 = 200I1 500 = 100I1 + 100I1 .. .. .. V1 = 100I1 I1 = 500/200 = 2.5 A .. V2 = 2V1
1 I2 = I1 = 1.25 A 2 V2 = 2V1 = 500 V
V1 = 100(2.5) = 250 V;
VTh = 20I1 + V1 - V2 + 40I2 = -150 V(rms) Short circuit current:
500 = 80(Isc + I1 ) + 360(Isc + 0.5I1 ) 2V1 = 40 I1 + 360(Isc + 0.5I1 ) 2
500 = 80(I1 + Isc ) + 20I1 + V1
Problems Solving, Isc = -1.47 A; RTh = I1 = 4.41 A; V1 = 176.47 V VTh -150 = 102 = Isc -1.47
1047
P = [b]
752 = 55.15 W 102
500 = 80[I1 - (75/102)] - 75 + 360[I2 - (75/102)] 575 + .. 6000 27,000 + = 80I1 + 180I1 102 102 I1 = 3.456 A 55.15 (100) = 4.05% 1360.29 75 102
2
Psource = (500)[3.456 - (75/102)] = 1360.29 W % delivered =
[c] P80 = 80 I1 -
= 592.13 W
P20 = 20I2 = 238.86 W 1 P40 = 40I2 = 119.43 W 2 P102 = 752 = 55.15 W 102 75 102
2
P360 = 360 I2 -
= 354.73 W Pdev
Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W =
1048
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.63 [a] Open circuit voltage:
40/0 = 4(I1 + I3 ) + 12I3 + VTh I1 = -I3 ; 4 Solving, VTh = 40/0 V Short circuit current: I1 = -4I3
40/0 = 4I1 + 4I3 + I1 + V1 4V1 = 16(I1 /4) = 4I1 ; .. 40/0 = 6I1 + 4I3 Also, 40/0 = 4(I1 + I3 ) + 12I3 Solving, I1 = 6 A; RTh = I3 = 1 A; Isc = I1 /4 + I3 = 2.5 A VTh 40 = 16 = Isc 2.5 .. V1 = I1
Problems
1049
40/0 I= = 1.25/0 A(rms) 32 P = (1.25)2 (16) = 25 W [b]
40 = 4(I1 + I3 ) + 12I3 + 20 4V1 = 4I1 + 16(I1 /4 + I3 ); 40 = 4I1 + 4I3 + I1 + V1 .. I1 = 6 A; I3 = -0.25 A; I1 + I3 = 5.75/0 A; V1 = 11/0 V .. V1 = 2I1 + 4I3
P40V (developed) = 40(5.75) = 230 W .. % delivered = [c] PRL = 25 W; 25 (100) = 10.87% 230
P16 = (1.5)2 (16) = 36 W P1 = (6)2 (1) = 36 W
P4 = (5.75)2 (4) = 132.25 W; P12 = (-0.25)2 (12) = 0.75 W
Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W =
Pdev
P 10.64 [a] Replace the circuit to the left of the primary winding with a Thvenin equivalent: VTh = (15)(20 j10) = 60 + j120 V ZTh = 2 + 20 j10 = 6 + j8
1050
CHAPTER 10. Sinusoidal Steady State Power Calculations Transfer the secondary impedance to the primary side: Zp = 1 XC (100 + jXC ) = 4 + j 25 25
Now maximize I by setting (XC /25) = -8 : .. C = [b] I = 1 = 0.25 F 200(20 103 )
60 + j120 = 6 + j12 A 10
P = |I|2 (4) = 720 W Ro = 6 ; .. Ro = 150 25 60 + j120 = 5 + j10 A [d] I = 12 [c] P = |I|2 (6) = 750 W P 10.65 [a] Zab = 50 - j400 = 1 - N1 N2
2
ZL = 1 -
2800 700
2
ZL = 9ZL
1 .. ZL = (50 - j400) = 5.556 - j44.444 9 [b]
I1 =
24 = 240/0 mA 100
Problems
1051
N1 I1 = -N2 I2 I2 = -4I1 = 960/180 mA IL = I1 + I2 = 720/180 mA(rms) VL = (5.556 - j44.444)IL = -4 + j32 = 32.25/97.13 V(rms) P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves only the resistor R2 . Then, Pmed = 500 W = Thus, 1202 = 28.8 500 [b] Now move to the LOW setting, as shown in Fig. 10.30, which involves the resistors R1 and R2 connected in series: R2 = Plow V2 V2 = 250 W = = R1 + R2 R1 + 28.8 V2 1202 = R2 R2
Thus, 1202 - 28.8 = 28.8 250 [c] Note that the HIGH setting has R1 and R2 in parallel: R1 = Phigh V2 1202 = = = 1000 W R1 R2 28.8 28.8
If the HIGH setting has required power other than 1000 W, this problem sould not have been solved. In other words, the HIGH power setting was chosen in such a way that it would be satisfied once the two resistor values were calculated to satisfy the LOW and MEDIUM power settings.
1052
CHAPTER 10. Sinusoidal Steady State Power Calculations V2 ; R1 + R2 V2 ; R2 R1 + R2 = R2 = V2 PM V2 PL
P 10.67 [a] PL =
PM = PH =
V 2 (R1 + R2 ) R1 R2 V2 ; PL
V PM
2
R1 + R2 = PH = PH = [b] PH =
R1 = =
V2 V2 - PL PM PM PL PM PL (PM - PL )
V 2 V 2 /PL
V PL
2
-
V PM
2
2 PM PM - PL
(750)2 = 1125 W (750 - 250)
P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH . Thus P M - PL =
2 PM PH
or
2 PM - PM + PL = 0 PH
2 P M - PM P H + PL P H = 0 2
.. PM = =
PH 2
PH 2
- PL PH
1 PL PH PH - 2 4 PH
For the specified values of PL and PH PM = 500 1000 0.25 - 0.24 = 500 100 .. PM 1 = 600 W; PM 2 = 400 W
Note in this case we design for two medium power ratings If PM 1 = 600 W R2 = (120)2 = 24 600
Problems R1 + R2 = (120)2 = 60 240
1053
R1 = 60 - 24 = 36 CHECK: PH = (120)2 (60) = 1000 W (36)(24)
If PM 2 = 400 W R2 = (120)2 = 36 400
R1 + R2 = 60 (as before) R1 = 24 CHECK: PH = 1000 W P 10.69 R1 + R2 + R3 = R2 + R3 = (120)2 = 24 600
(120)2 = 16 900
.. R1 = 24 - 16 = 8 R3 + R1 R2 = .. 16 - R2 + R2 - (120)2 = 12 1200 8R2 = 12 8 + R2
8R2 =4 8 + R2
2 8R2 + R2 - 8R2 = 32 + 4R2 2 R2 - 4R2 - 32 = 0
R2 = 2
4 + 32 = 2 6 .. R3 = 8
.. R2 = 8 ;
1054
CHAPTER 10. Sinusoidal Steady State Power Calculations (220)2 = 96.8 500 (220)2 = 193.6 250
P 10.70 R2 =
R1 + R2 =
.. R1 = 96.8 CHECK: R1 R2 = 48.4 PH = (220)2 = 1000 W 48.4