chapter 5 · chapter 5 exercises e5.1 (a) we are given . the angular frequency is the coefficient...
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CHAPTER 5
Exercises
E5.1 (a) We are given . The angular frequency is the coefficient of t so we have . Then
)30200cos(150)( o−= ttv π 200 radian/sπω =
E5.4 (a) cos(10
Hz 1002/ == πωf ms 10/1 == fT
(b) 10=I
V 1.1062/1502/ === mrms VV
cos(10 (c)
Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that has units of radians, the positive peak occurs when
(b) (c) A plot of v(t) is shown in Figure 5.4 in the book.
tω
sin(20
ms 8333.0 180
30 maxmax =⇒×= tt πω
W 225/2 == RVP rmsavg
E5.2 We use the trigonometric identity sin( Thus 100 E5.3 T
The period corresponds to 360 therefore 5 ms corresponds to a phase angle of ( . Thus the voltage is
).90cos() o−= zz)30300cos(100)60300sin( oo −=+ tt ππ
radian/s 3772 ≅= fπω ms 67.16/1 ≅= f V 6.1552 ≅= rmsm VVo
o108=o360)67.16/5 ×)108377cos(6.155)( o−= ttv
ooo 4514.14101090100101 −∠≅−=−∠+∠= jV )45cos(14.14)sin(10) o−=+ ttt ωωω
330.45.25660.8605301 jj −++≅−∠+∠ oo
o44.318.11670.016.11 ∠≅+≅ j)44.3cos(18.11)30sin(5)30 ooo +=+++ ttt ωωω
99.125.702060150202 jj −++≅−∠+∠= ooI o28.2541.3099.125.27 −∠≅−≅ j
)28.25cos(41.30)60cos(15)90 ooo −=−++ ttt ωωω
142
E5.5 The phasors are V
v1 lags v by 60 (or we could say v leads v by 60
ooo 4510 and 3010 3010 321 −∠=+∠=−∠= VV
2 2
1
o
o
o
)o
v1 leads v by 15 (or we could say v lags v by 15
3 3
1
)o
v2 leads v3 by 75 (or we could say v lags v by 75
3
2
)o
E5.6 (a) o905050 ∠=== jLjZL ω o0100∠=LV
(b) The phasor diagram is shown in Figure 5.11a in the book.
o90250/100/ −∠=== jZLLL VI
E5.7 (a)
o905050/1 −∠=−== jCjZC ω o0100∠=CVo902)50/(100/ ∠=−== jZCCC VI
(b) The phasor diagram is shown in Figure 5.11b in the book.
E5.8 (a) (b) The phasor diagram is shown in Figure 5.11c in the book.
o05050 ∠=== RZRo0100∠=RV
o02)50/(100/ ∠=== RRR VI
E5.9 (a) The transformed network is:
mA 13528.28
2502509010 o
o
−∠=+−∠
==jZ
sVI
143
mA )135500cos(28.28)( o−= tti
(b) The phasor diagram is shown in Figure 5.17b in the book. (c) i(t) lags vs(t) by
E5.10 The transformed network is:
o13507.7 −∠== IV RRo4507.7 −∠== IV LjL ω
.45o
Ω−∠=++−+
= 31.5647.55)200/(1)50/(1100/1
1 o
jjZ
V 31.564.277 o−∠== IV Z 69.33547.5)50/( o∠=−= jC VI AA 31.146387.1)200/( o−∠== jL VI
A 31.56774.2)100/( o−∠== VIR
E5.11 The transformed network is:
We write KVL equations for each of the meshes:
100)(100100 211 =−+ IIIj
Simplifying, we have 0)(100100200 1222 =−++− IIII jj
100100)100100( 21 =−+ IIj
Solving we find Thus we have
0)100100(100 21 =−+− II jA. 01 and A 45414.1 21
oo ∠=−∠= II1000cos()( and A )451000 2 ).cos(414.1)(1 ttitti =− o=
144
E5.12 (a) For a power factor of 100%, we have which implies that
the current and voltage are in phase and Thus, Also Thus we have
(b) For a power factor of 20% lagging, we have cos( which implies that the current lags the voltage by Thus,
Also, we have Thus we have
,1)cos( =θ.0=θ .0)tan( == θPQ
( ) A.500/[5000]cos/[ 10)]0cos( === θrmsrms VPI.4014.14 and 14.142 o∠== IrmsI=mI
,2.0) =θ.46. o=θ 78)2.0(cos 1 =−
.kVAR 49.24)tan( == θPQ ( )] = A. 0.50cos/[= θrmsrms VPI.46.3871 o−∠.70 and A 71.70 I2 == rmsm II =
(c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a). Wiring costs would be lower for the load of part (a).
E5.13 The first load is a 10 capacitor for which we have
The second load absorbs an apparent power of with a power factor of 80% lagging from which we have Notice that we select a positive angle for because the load has a lagging power factor. Thus we have and .
F µΩ== 90265)/(1 o−∠3.CjZC ω o90−=Cθ A 770.3/ == CrmsCrms ZVI
0)cos( == CCrmsrmsC IVP θ kVAR 770.3)sin( −== CCrmsrmsC IVQ θ
kVA 10=rmsrmsIV.87.36 o=)8.0(cos 1
2 = −θ
2θkW 0.8)cos( 2 =22 = θrmsrmsIVP
kVAR 6)sin(22 == θrmsrms IVQ Now for the source we have:
kW 82 =+= PPP Cs kVAR 23.22 =+= QQQ Cs
kVA 305.822 =+= sssrmsrms QPIV A 305.8/ == rmssrmsrmssrms VIVI
%33.96%100)/(factor power =×= srmsrmss IVP
E5.14 First, we zero the source and combine impedances in series and parallel to determine the Thévenin impedance.
145
50502550
100/1100/112550 jj
jjZt ++−=
++−=
o04.141.103j25100 ∠=+= Then we analyze the circuit to determine the open-circuit voltage.
o4571.70100100
100100 −∠=+
×==joct VV
o04.596858.0/ −∠== tt ZVnI E5.15 (a) For a complex load, maximum power is transferred for
. The Thévenin equivalent with the load attached is:
LLtL jXRjZZ +=−== 25100*
The current is given by
oo
453536.02510025100
4571.70−∠=
−++−∠
=jj
I
The load power is W 25.6)2/3536.0(100 22 === rmsLL IRP
146
(b) For a purely resistive load, maximum power is transferred for The Thévenin equivalent with the load attached is:
. 1.10325100 22 Ω=+== tL ZR
The current is given by
oo
98.373456.0251001.103
4571.70−∠=
−+−∠
=j
I
The load power is
E5.16 The line-to-neutral voltage is 1000 No phase angle was
specified in the problem statement, so we will assume that the phase of Van is zero. Then we have
The circuit for the a phase is shown below. (We can consider a neutral connection to exist in a balanced Y-Y connection even if one is not physically present.)
W 157.6)2/3456.0(1.103 22 === rmsLL IRP
V. 4.5773/ =
ooo 1204.577 1204.577 04.577 ∠=−∠=∠= cnbnan VVV
The a-phase line current is
The currents for phases b and c are the same except for phase.
oo
02.37610.440.75100
04.577−∠=
+∠
==jZL
anaA
VI
oo 98.82610.4 02.157610.4 ∠=−∠= cCbB II
kW 188.3)02.37cos(2
610.44.5773)cos(2
3 =×
== oθLY IVP
147
kVAR 404.2)02.37sin(2
610.44.5773)sin(2
3 =×
== oθLY IVQ
E5.17 The a-phase line-to-neutral voltage is anV
The phase impedance of the equivalent Y is
oo 04.57703/1000 ∠=∠=
. 67.163/503/ Ω=== ∆ZZY
Thus the line current is
A 063.3467.16
04.577 oo
∠=∠
==Y
an
ZVIaA
Similarly, and Finally, the power is
A 12063.34 o−∠=bBI A. 12063.34 o∠=cCI
kW 00.30)2/(3 2 == yaA RIP
Problems
P5.1 (a) Increasing the peak amplitude Vm? 1. Stretches the sinusoidal curve vertically.
(b) Increasing the frequency f? 4. Compresses the sinusoidal curve horizontally. (c) Decreasing θ? 5. Translates the sinusoidal curve to the right. (d) Decreasing the angular frequency ω? 3. Stretches the sinusoidal curve horizontally. (e) Increasing the period? 3. Stretches the sinusoidal curve horizontally.
P5.2 The units of angular frequency ω are radians per second. The units of frequency f are hertz, which are equivalent to inverse seconds. The relationship between them is ω .2 fπ=
148
P5.3*
( ) ( ) ( )oo 601000cos10301000sin10 −=+= ππttv
( ) W 1V 071.72102
ms 21radians 360 angle phase
Hz 500rad/s 1000
2 ==
===
==
−=−==
=
=
RVPVVfT
f
rms
mrms
πθ
πω
o
The first positive peak occurs for
ms 3333.0 031000
=
=−
peak
peak
tt ππ
P5.4
( ) ( ) ( )oo 30500cos50120500sin50 +=+= tttv ππ
( ) W 25V 36.352502
ms 41radians 630 angle phase
Hz 250rad/s 500
2 ==
===
==
===
=
=
RVPVVfT
f
rms
mrms
πθ
πω
o
Positive peak occurs for
=t
ms 3333.0
06500
1
1
−
=+t ππ
The first positive peak after 0 is at ms 667.31 =+= Tt
=t tpeak
149
P5.5* Sinusoidal voltages can be expressed in the form
The peak voltage is V. The frequency is 10 kHz and the angular frequency is ω
radians/s. The phase corresponding to a time interval of 20 µs is . Thus we have V.
).cos()( θω += tVtv m
28.282022 rms =×== VVm
== Tf /1 41022 ππ == f=t∆
o72o360) =×/(= Tt∆θ )72102cos(28. 4 o−28)( = ttv π
P5.6 The angular frequency is radians/s. We have ms. The sinusoid reaches a positive peak one quarter of a cycle after crossing zero with a positive slope. Thus, the first positive peak occurs at t = 3 ms. The phase corresponding to a time interval of 3 ms is
. Thus, we have V. P5.7
ππω 2502 == f 8/1 == fT
=t∆oo 135360)/( =×∆= Ttθ )135o−250cos(15)( = ttv π
( ) ( A 1657cos12 )tti π=( ) ( ) ( ) ( )[ ]
( ) ( ) W 9500212132
W 4000cos195002000cos101922
232
===
+=×==
rmsavg IRP
tttRitp ππ
150
P5.8
P5.9 A Hz
rad/s
Pavg
( ) ( )( ) ( ) ( ) ( )[ ]
( ) W 6240W 1192sin16240596sin12480
V 596sin2503
2
22
==
−===
=
RVttRtvtp
ttv
rms
ππ
π
071.7252 === rmsm II 1001==
Tf
3.6282 == fπω o108360 max −=−=T
tθ
P5.10 A sequence of MATLAB instructions to accomplish the desired plot for
part (a) is
108200cos(071.7)( o−= tti π )
Wx = 2*pi; Wy = 2*pi; Theta = 90*pi/180; t = 0:0.01:20; x = cos(Wx*t); y = cos(Wy*t + Theta); plot(x,y)
By changing the parameters, we can obtain the plots for parts b, c, and d. The resulting plots are
151
(a) (b)
(c) (d)
P5.11*
( ) V 536.32521 1
00
2 === ∫∫ dtdttvTiV
T
rms
P5.12*
( ) [ ] [ ]∫∫∫ +===5.0
0
5.0
0
2
0
2 )4sin(5.125.12)2sin(51 dttdttdttvT
VT
rms ππ
V 5.225.6)4cos(4
5.125.125.0
0==
−=
=
=
t
trms ttV π
π
P5.13* ( ) A 581.114411 2
0
4
20
2 =
+== ∫ ∫∫ dtdtdtti
TI
T
rms
152
P5.14
( ) [ ]∫∫ +==1
0
2
0
2 )2sin()2cos(1 dttBtAdttvT
VT
rms ππ
[ ]
+=
++=
++=
=
=
∫
2
20
2
)2(sin)2sin()2cos(2)2(cos
22
1
0
22
5.0
0
2222
BA
tBtA
dttBttABtAV
t
t
rms ππππ
P5.15
( )dttvTiV
T
rms ∫=0
2
[ ]
V 44.13
09.2802.72.0
1
)]20cos(5.1445.144)10cos(204362.0
1
)]10(cos289)10cos(204362.0
1
)]10(cos289)10cos(20436[2.0
1
)]10cos(176[2.0
1
2.0
0
2.0
0
2.0
0
2.0
0
2.0
0
2.0
0
22.0
0
2.0
0
2
2.0
0
2
=
+++=
+++=
++=
++=
+=
∫ ∫∫∫
∫ ∫∫
∫
∫
dttdtdttdt
dttdttdt
dttt
dttVrms
ππ
ππ
ππ
π
P5.16
( ) [ ] [ ]∫∫∫ −=−==1
0
1
0
2
0
2 )2exp(9)exp(31 dttdttdttvT
VT
rms
[ ] V 973.1)]2exp(1[5.4)2exp(5.4 10 =−−=−−= =
=ttt
P5.17
rmsV ( ) ( ) V 887.2
325.6
215.2
211
2
0
32
0
2
0
2 =
===
=
=∫∫
t
t
T tdttdttvT
153
P5.18 The rms values of all periodic waveforms are not equal to their peak values divided by the square root of two. However, they are for sinusoids, which are important special cases.
P5.19 To add sinusoids we: 1. Convert each of the sinusoids to be added into a phasor. 2. Add the phasors and convert the sum to polar form using complex arithimetic. 3. Convert the resulting phasor to a sinusoid. All of the sinusoids to be added must have the same frequency.
P5.20 Two methods to determine the phase relationship between two sinusoids
are: 1. Examine the phasor diagram and consider it to rotate counterclockwise. If phasor A points in a given direction before phasor B by an angle θ, we say that A leads B by the angle θ or that B lags A by the angle θ. 2. Examine plots of the sinusoidal waveforms versus time. If A reaches a point (such as a positive peak or a zero crossing with positive slope) by an interval ∆t before B reaches the corresponding point, we say that A leads B by the angle or that B lags A by the angle θ. °×∆= 360)/( Ttθ
P5.21*
( ) ( )( ) ( ) ( )
( ) ( )oo
o
o
o
45cos4.141454.141100100
100901001000100
90cos100sin100cos100
21s
2
1
2
1
−=
−∠=−=+=
−=−∠=
=∠=
−==
=
ttvjj
tttvttv
s ω
ωωω
VVVVV
o
o
o
45 by leads 45 by lags 90 by lags
2s
1s
12
VVVVVV
154
P5.22*
5.23* )
( ) ( )( )( ) ( )
( ) ( )( ) ( )( ) ( ) o
o
o
o
o
o
60 by leads 60 by lags 120 by lags
90400cos10)150400cos(5
30400cos10
4002
32
31
21
3
2
1
tvtvtvtvtvtv
ttvttvttv
f
+π=
+π=
+π=
π=π=ω
P We are given the expression 5 o
Conver 5
sin(4)75cos(3)75cos( ttt ωωω +−−+ o ting to phasors we obtain
=−∠+−∠−∠ ooo 90475375 +2941.1
)09.82cos(763.375cos(5
o+
+
tt
ωω
The magnitudes of the phasors for the two voltages are and 2 The phase angles are not known. If the ph angles are the same, thephasor sum would have its maximum magnitude which is 27 . On the other hand, if the phase angles differ by 180
=−−− 4)8978.27765.0(8296.4 jjj o09.82763.37274.35176.0 ∠=+ j Thus, we have
)sin(4)75cos(3) oo =+−− tt ωω
P5.24 26
o , the phasor sum would
V. ase
ha
(v
ve its minimum magnitude which is 25of the sum is 7 V and the minimum is 5 V.
) ( )( ) ( ) ( )
o
o
oo
o
759.1293015071.7071.7045100
30cos15060sin15045cos100
1
2
1
−=−∠=
+=∠=
−=+=
+=
jj
tttvtt
ωωω
VV
. Thus, the maximum rms value
P5.25
( ) ( )oo
23.1cos6.20023.16.20029.46.20021s
2
−=
−∠=−=+=
tt
vj
s ωVVV
155
P5.26 V s
o
o
o
77.28 by leads 23.46 by lags
75 by lags
2s
1s
12
VVVVVV
3=mV 5.0=T
21==
Tf π=π=ω 42 f Hz rad/s
V
oo 45360 max −=−=θT
t
)454cos(3)( o−π= ttvo453 −∠=V V
V
121.223
==rmsV
)
P5.27
V
( ) ( )
( ) ( )oo
o
o
50cos071.750071.7
3010071.72
30cos10
1
1
1
1rms1
1
+=
∠=
∠=
=×=
+=
tti
IIttv
m
ω
ω
I
P5.28 We are given the expression sin(5 Converting to phasors we obtain
150cos(5)30cos(5) oo ++++ ttt ωωω
5 5− j Thus, we have
=∠+∠+−∠ ooo 1505305905.23301.4500.23301.4 =+−++ jj 0
0)150cos(5)30cos(5)sin(5 =++++ oo ttt ωωω
156
P5.29 A sequence of MATLAB commands to generate the desired plot is: t = 0:0.01:2; v = cos(19*pi*t) + cos(21*pi*t); plot(v,t)
The resulting plot is
Notice that the first term cos(19πt) has a frequency of 9.5 Hz while the second term cos(21πt) has a frequency of 10.5 Hz. At t = 0, the rotating vectors are in phase and add constructively. At t = 0.5, the vector for the second term has rotated one half turn more than the vector for the first term and the vectors cancel. At t = 1, the first vector has rotated 9.5 turns and the second vector has rotated 10.5 turns so they both point in the same direction and add constructively.
P5.30 i1 leads i2 by the angle
Therefore, the angle for i2 is which is equivalent to
ms 049.4247/1/1 === fT.7.355360)049.4/4(360)/( °=°×=°×∆=∆ Ttθ
°−=∆−= 7.30812 θθθ.3.51 °+
P5.31 For an inductance, we have V .
For a capacitance, we have V .
LL Lj Iω=
CjC
C ωI
=
157
P5.32 For a pure resistance, current and voltage are in phase. For a pure inductance, current lags voltage by 90 .o
For a pure capacitance, current leads voltage by
.90o
P5.33*
( ) ( )
( )( ) ( ) ( ) ( ) ( )ttti
Z
jLjZ
ttv
L
LLL
L
L
L
ππ=−ππ=
−∠π==
∠=
∠π=π=ω=
π=ω
π=
2000sin201902000cos20190201
01090200200
20002000cos10
o
o
o
o
VIV
( ) ( ) o90 by lags tvti LL
P5.34*
I ( )iC
( ) ( )
( ) ( ) ( )tttiZ
jCjZ
ttv
C
C
C
C
ππ
ω
πωπ
2000sin6283.0902000cos6283.0906283.0
010
9092.1592.15
20002000cos10
CC
−=+=
∠==
∠
−∠=−=−
=
=
=
o
o
o
o
V
Ω
C =V
( ) o90 by leads tvt C
158
P5.35 (a) Notice that the voltage is a sine rather than a cosine.
1009010030160100 jZ −=−∠==∠=−∠= ooo
IVIV
Because Z is pure imaginary and negative, the element is a capacitance.
200=ω F501µ=
ω=
ZC
(b) 500=V
Because Z is pure real, the element is a resistance of 250 Ω.
0250025050250 jZ +=∠==∠=∠ ooo
IVI
(c) Notice that the current is a sine rather than a cosine.
Because Z is pure imaginary and positive, the element is an inductance.
1009010060130100 jZ =∠==−∠=∠= ooo
IVIV
400=ω H25.0=ω
=ZL
P5.36 A Matlab m-file that produces the desired plots is:
f=0:1:1000; ZL=2*pi*f*0.01; ZC=1./(2*pi*f*10e-6); R=50 plot(f,ZL) axis([0 1000 0 100]) hold plot(f,ZC) plot(f,R)
The resulting plot is:
159
P5.37 (a) Z = 209020 j−=−∠= o
IV
Because Z is pure imaginary and negative, the element is a capacitance.
(b) 4Z ==IV
Because Z is pure imaginary and positive, the element is an inductance.
500=ω F1001 µω
==Z
C
490 j=∠ o
500=ω mH8==ωZL
(c) 10==VZ
Because Z is pure real, the element is a resistance of 10 Ω.
5.38 (a) From the plot, we see that ms, so we have Hz and Also, we see that the current lags the voltage by 1 ms or 90°, so we have an inductance. Finally, from which we find that 3.18 mH.
100 =∠ o
I
4=T 250/1 == Tf.500πω =
, 5 Ω/ == mm IVLω=L
(b) From the plot, we see that ms, so we have Hz and Also, we see that the current leads the voltage by 2 ms or
8=T 125/1 == Tf.250πω =
160
90°, so we have a capacitance. Finally, 1 from which we find that 0.5093 µF.
P5.39 The steps in analysis of steady-state ac circuits are: 1. Replace sources with their phasors. 2. Replace inductances and capacitances with their complex impedances. 3. Use series/parallel, node voltages, or mesh currents to solve for the quantities of interest.
, 2500// Ω== mm IVCω=C
jω
−
= o41.416.5450.88∠
= jZ
o05.8046.312 ∠==Z
o32.8013.321 ∠==Z
o45
V 45V 45.7
mA 71.70100
010
o
o
o
∠==
−==
∠=
+∠
=
+=
IVIV
VI
LjR
j
LjR
L
R
s
ω
ω
071.7071
45100
o
∠
−
All of the sources must have the same frequency.
P5.40* C
RLjZ ω 1+=
=ω
4.1654 5466.92 :565
=Ω+
−+=
jjω
8.30754 :2027 Ω+ j =ω
55.31654 :2077 Ω+ j
P5.41*
I by lags sV
161
P5.42
P5.43
V 44.63472.4V 56.26944.8
mA 56.2644.8950100
010
o
o
o
o
∠==
−∠==
−∠=
+∠
=
+=
IVIV
VI
LjR
j
LjR
L
R
s
ω
ω
o56.26 by lags sVI
CjLjZZZ
cL ωω +=
+=
11
111
=ω
Ω+= 3.134 :529 jZΩ−== 10.40 :1758 jZωΩ−== 41.69 :1255 jZω
P5.44
CjLjRZZR
ZcL ω+ω+=
++=
1/11
11/11
.capacitive is impedance the negative, is part imaginary the Because26.3058.3 :571 Ω−== jZω
.capacitive is impedance the negative, is part imaginary the Because163.40669.0 :1066 Ω−== jZω
.capacitive is impedance the negative, is part imaginary the Because339.10069.0 :2648 Ω−== jZω
P5.45
I
V 45071.7mA 45071.7
10001000010
o
o
o
∠==
∠=
−∠
=
−=
IV
VI
R
j
CjR
R
s
ω
( ) V 45071.7 o−∠=−= IV CjC ω
o45 by leads sV
162
P5.46*
I
( ) V57.26944.8V 43.63472.4
mA 43.63472.420001000
010
o
o
o
o
−∠=−=
∠==
∠=
−∠
=
−=
IVIV
VI
CjR
j
CjR
C
R
s
ω
ω
o43.63 by leads sV
P5.47
V
o44.63 by leads sI
b lags sIV
P5.48*
o
o
44.6372.4410012001
105.0
∠=
+=
∠=
js
s
IV
I
o
o44.632236.0 ∠== RR VI56.264472.0 −∠== LjL ωVI
( )200110011
mA 0100 o
−+=
∠=
js
s
IV
I
( ) mA 44.6372.44mA 56.2644.89
V 56.26944.8
o
o
o
∠==
−∠==
−∠=
CjR
C
R
ωVIVI
o56.26 y
163
P5.49*
V
( ) mA 9050mA 9050
mA 010V 010
005.0005.010001110
111
mA 010
2
o
o
o
o
o
∠==
−∠==
∠==
∠=
+−=
++=
∠=
−
CjLj
R
jj
CjLjR
C
L
R
s
s
ω
ω
ωω
VIVIVI
I
I
The peak value of is five times larger than the source current! This
is possible because current in the capacitance balances the current in the inductance (i.e., ).
( )tiL
0=+ CL II
P5.50
VC
The peak value of is five times larger than the source voltage! This is possible because the impedance of the capacitor cancels the impedance of the inductance.
P5.51
( )tvL
o
o
o
o
0101009050500
A 01.0500100500
10
010
∠==
∠=×=
∠=
−+=
−+=
∠=
IVIV
VI
V
j
jj
CjRLj
L
s
s
ωω
o9050500 −∠=−= IjR
164
( )
o
o
o
oo
6.5645.77506.1469.154100
6.146549.150100
15060.8650100
5060.8610030100 10090100
21
21
−∠==
−∠==
−∠=
+−−
=+
+−−=
+−
=
∠=−=−∠=
IVIV
VVI
VV
j
jj
jjj
LjR
j
L
R
ω
o
o
90 by lags 6.56 by lags 1
LVIVI
P5.52
P5.53* First we write the KVL equation: −V Then we enclose nodes 1 and 2 in a closed surface to form a supernode
and write a KCL equation:
( )
( )o
o
o
o
oo
o
01100100
10045414.1
901100100
10045414.1
45414.101004571.70
50505050100
01.001.01100
11
∠=
−×−∠=
+=
−∠=
−−
×−∠=+
=
−∠=∠
=
∠=
+=
−+=
++=
++=
jZRR
jj
ZRZ
Z
jjj
jj
CjRLjZ
CC
C
CR
total
total
II
II
I
ωω
o01021 ∠=V
165
+V 0
51520102211 =
−++
jjVVV
The solution to these equations is:
=V
P5.54 The KCL equation is Solving, we
find V V.
o58.29402.91 ∠=Vo45.111986.42 ∠
.01625021
1548330 111 =
−°∠−
++°∠−
jjVVV
o98.1614898.0 ∠=1
P5.55 The KCL equation is Solving, we find
V. P5.56 Writing KCL equations at nodes 1 and 2 we obtain
.0417128
214 11 =°∠++
+°∠−
jVV
°∠= 30.10941.31V
o0115510
211 ∠=+−
+jVVV
Solving these equations, we obtain
o30115510
122 ∠=+−
+− jj
VVV
o54.38735.61 −∠=Vo52.5525.162 −∠=V
P5.57* Writing KVL equations around the meshes, we obtain 4 °∠=−+ 8330)(15 211 III j − 16j Solving, we obtain: =I
°∠−=−+ 25021)(15 122 III j
o60.159191.11 ∠
=I P5.58 The current through the current source is
o26.162140.32 ∠
°∠=− 4121 II Writing KVL around the perimeter of the circuit, we have
Solving, we obtain:
°∠=++ 214)712(8 1 2II j
=I °−∠=−∠ 95.192837.0 and 23.1264.1 21 Io
166
P5.59 Writing KVL equations around the meshes, we obtain 10 )(20 211 =−+ III j 0 20j 5− j Solving, we obtain: =I
10)(15)( 3212 −=−+− IIII)(15 233 =−+ III 0
o4.1509402.01 −∠
=I =I
o0.177051.12 −∠o6.1589972.03 −∠
P5.60 A Matlab program that produces the desired plots is: w=0:1:2000;
L=20e-3; C=50e-6; Zmaga=abs(w*L-1./(w*C)); Zmagb=abs(1./((1./(i*w*L))+i*w*C)); plot(w,Zmaga) hold on plot(w,Zmagb) axis([0 2000 0 100]) hold off
The result is:
167
P5.61 A Matlab program that produces the desired plots is: w=0:1:5000;
L=20e-3; R=50; Zmaga=abs(R+i*w*L); Zmagb=abs(1./((1./(i*w*L))+1/R)); plot(w,Zmaga) hold on plot(w,Zmagb) hold off
The result is:
P5.62 The units for real power are watts (W). For reactive power, the units are
volt-amperes reactive (VAR). For apparent power, the units are volt-amperes (VA).
P5.63 Power factor is the cosine of the power angle. It is often expressed as a
percentage. PF
%100)cos( ×= θ
168
P5.64 (a) For a pure resistance, the power is positive and the reactive power is zero.
(b) For a pure inductance, the power is zero and the reactive power is positive.
(c) For a pure capacitance, the power is zero, and the reactive power is negative.
P5.65 A load with a leading power factor is capacitive and has negative reactive
power. A load with a lagging power factor is inductive and has positive reactive power.
P5.66 See Figure 5.23 in the book. P5.67 Real power represents a net flow, over time, of energy from the source
to the load. This energy must be supplied to the system from hydro, fossil-fuel, or nuclear sources.
Reactive power represents energy that flows back and forth from the source to the load. Aside from losses in the transmission system (lines and transformers), no net energy must be supplied to the system to create the reactive power. Reactive power is important mainly because of the increased system losses associated with it.
P5.68 Usually, power factor correction refers to adding capacitances in parallel
with an inductive load to reduce the reactive power flowing from the power plant through the distribution system to the load. To minimize power system losses, we need 100% power factor for the combined load. Ultimately, an economic analysis is needed to balance the costs of power factor correction against the costs of losses and additional distribution capacity needed because of reactive power.
P5.69* This is a capacitive load because the reactance is negative.
kW 5.22100)15( 22 === RIP rms
power factor cos( ==
kVAR 25.11)50()15( 22 −=−== XIQ rms
o57.26)5.0(tantan 11 =−=
= −−
PQθ
%44.89)θ apparent power 2 += QP KVA 16.252 =
169
P5.70 This is an inductive load because the reactance is positive.
°∠=+= 54.6162.463827 jZ
°−∠=°∠°∠
== 61.28292.3161.5462.46
2621488Z
I V
kW 51.2727)92.31( 22 === RIP rms
kVAR 72.38)38()92.31( 22 === XIQ rms
power factor )cos( == θ apparent power = rmsrmsIV KVA
o61.54=θ%91.57
50.4792.311488 =×=
P5.71 power factor cos( == leading
ooo 306030 −=−=−= iv θθθ%60.86)θ
apparent power ==V KVA
kW 99.12)cos( == θrmsrms IVPkVAR 5.7)sin( −== θrmsrms IVQ
15151000 =×rmsrmsI
P5.72 kV A
o
o
o
3067.6660215
3021000−∠=
∠∠
==IVZ
90=rmsV 20=rmsI ooo 38)23(15 =−−=−= iv θθθlagging %80.78%100cos factor power =×= θ
This is an inductive load.
kW 1418)cos( == θrmsrmsIVP kVAR 1108)sin( == θrmsrmsIVQKVA 1800Power Apparent == rmsrmsIV
P5.73 (a) For a resistance in series with an inductance, real power is positive
and reactive power is positive. (b) For a resistance in series with a capacitance, real power is positive and reactive power is negative.
P5.74 If the inductive reactance is greater than the capacitive reactance, the
total impedance is inductive, real power is zero, and reactive power is positive. If the inductive reactance equals the capacitive reactance, the total impedance is zero, and the current is infinite. Real power and reactive
170
power are indeterminate. This case is rarely, if ever, of practical importance. If the inductive reactance is less than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative.
P5.75 If the inductive reactance is greater than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative. If the inductive reactance equals the capacitive reactance, the total impedance is infinite, and the current is zero. Real power and reactive power are zero. If the inductive reactance is less than the capacitive reactance, the total impedance is inductive, real power is zero, and reactive power is positive.
P5.76
A Delivered by Source A:
kW AQ kVAR
ooo
55.4522025
272186452268∠=
+∠−∠
=j
I
20=rmsI
272.5)45.5545cos(268 =−= rmsA IP972.0)45.5545sin(268 −=−= rmsI
Absorbed by Source B: kW 3.271)45.5527cos(186 =−= rmsB IP
AQ kVAR Absorbed by resistor:
kW
1.772)45.5527sin(186 −=−= rmsI
2 == RIP rmsR 2Absorbed by inductor: kVAR
P5.77
800.02 == XIQ rmsL
ooo 67.8822.206162280161213)1316( −∠=−∠+∠+= jAVrmsV 2.206rms =AV
Delivered by Source A: kW
AQ kVAR 931.0)16167.88cos()13(2.206 −=−−=AP
513.2)16167.88sin()13(3.206 =−−=
Absorbed by Source B: kW 635.3)16116cos()13(280 −=−−=BP
171
BQ kVAR 191.0)16116sin()13(280 −=−−=
Absorbed by resistor: kW Absorbed by inductor:
704.22 == RIP rmsR
=QL kVAR
P5.78 Apparent power ⇒ ⇒ I
704.22 =XIrms
rmsrmsIV= rmsI2202500 = A11.37=rms
36.19==rms
rms
IVZ
25001500)cos( ==
rmsrmsIVPθ °= 53.13θ ⇒ (We know that θ is positive
because the impedance is inductive.)
Ω L H
LjRjZ ω+=+=°∠= 49.1562.1113.5336.19
62.11=R 0411.060249.15
==π
P5.79
( ) lagging %30.828230.063.34cos factor PowerkVA 55.26 power Apparent
kVAR 09.15sinkW 85.21cos
63.22275.16 5.1661221585
1151221585
===
==
==
==
−∠=
∠+
∠=
o
o
oo
rmsrms
rmsrms
rmsrms
IVIVQIVP
j
θθ
I
P5.80*
kVAR 374.8sinVQ
kW 85.21cos974.8276.14
3001221585
1151221585
rms −==
==
−∠=
−∠
+∠
=
o
oo
rms
rmsrms
IIVP
j
θθ
I
( ) leading %4.93934.097.20cos factor Power
kVA 39.23 power Apparent===
==o
rmsrmsIV
172
P5.81* Load A:
Load B:
( )kVAR 843.4tan
84.259.0coskW 10
1A
==
==
=−
AAA
A
PQ
P
θθ o
Source:
P5.82 oad A:
( )( ) kW 12cos
kVAR 9sin=
=
BBrms
BBrms
II
θθ
kVAR 84.13kW 22
=+=
=+=
BAs
BAs
QQQPPP
( ) 87.368.0cos 1
=
=
== −
rmsB
rmsB
B
Brmsrms
VPVQ
θ o
kVA 15=IV
Pow
( )
Apparent factor er
power Apparent 2
=
+=
s
s
PP ( )
lagging %62.848462.0power
kVA 262
==
=sQ
L ( )
kVAR 421.2tan84.259.0cos
kW 51
A
==
==
=−
AAA
A
PQ
P
θθ o
Load B:
( )( )
kVAR 5.7tanfactor. power leading a for negative is
86.368.0coskW 10
1
−==
−==
=−
BBB
B
B
PQ
P
θθ
θ o
Source:
( ) ( )
leading 94.72%0.9472power Apparent
factor Power
kVA 84.15 power Apparent
kVAR 079.5kW 15
22
===
=+=
−=+=
=+=
s
ss
BAs
BAs
PQP
QQQPPP
173
P5
.83
.84
( )
( ) ( )
leading %48.99 power Apparent
factor Power
VA 3651 power Apparent
VAR 5.370)3.192(
447280447
W 363355
447
22
22
22
22
==
=+=
−=−
+=
+=
+=
===
PQP
XV
XV
QQQR
VP
C
rms
L
rms
CL
rms
P5
( )( ) ( )
PowerVA 1930 power Apparent
VAR 1635
W 102591.572318.4
3.192280550244702447
22
2
==
−=
+=+=
==
−∠=
−+∠
=−+∠
=
rmsrms
rmsCrmsLCL
rms
IV
IXIXQQQIRP
jjCjLjRo
oo
ωωI
( ) leading %13.535313.091.57cos factor ===
( )
.85* (a)
(b)
( ) ( )o52.752400
25.0kV 1cos−∠=IθrmsV
( )
1582.2
103.387
0QkVAR 3.387sin
23
total
θ
−=−=
=×−=
+==
==
X
XVQ
QQIVQ
C
rmsC
Cload
rmsrmsload
F1027 µω
=CCC
o
A 400kW 100cos
52.7525.0cos
===
=
=
=
θθθ
rms
rmsrms
PI
IVP
P5
174
The capacitor must be rated for at least 387.3 kVAR. With the apacitor in place, we have:
(c) ine current is smaller by a factor of 4 with the capacitor in
e, reducing
c
o0100A 100
kW 100
∠=
=
==
Irms
rmsrms
IIVP
The l plac
The ac steady state Thévenin equivalent circuit for a two-terminal circconsists of
RI 2 losses in the line by a factor of 16. P5.86 uit
a phasor voltage source Vt in series with a complex impedance Zt.
ady state Norton equivalent circuit for a two-terminal circuit
zeroing the independent sources and finding the impedance looking into he terminals of the original network. Also, we have
The ac ste
consists of a phasor current source In in parallel with a complex impedance Zt.
Vt is the open-circuit voltage of the original circuit. In is the short
circuit current of the original network. The impedance can be found by
t
To attain maximum power, the load must equal (a ) the complex coof the Théve in
ttt Z IV = .
P5.87 njugate impedance if the load can have any complex value; (b) the
of the Thévenin impedance if the load must be a pure
P5.88
nmagnituderesistance.
The load voltage is given by the voltage divider principle. LZ
LttL ZZ +
= VV
The load voltage LV is larger than the Thévenin voltage in magnitude if the magnitude of Lt ZZ + is less than the magnitude of LZ which can happen when the imaginary parts of Z and Z have opposite signsdoes not hap
t L . This pen in purely resistive circuits.
175
P5.89* (a) Zeroing the current source, we have:
in
Thus, the Thévenin impedance is Ω 57.268.11150100 o∠=+= jZt Under open circuit conditions, there is zero voltage across the
ductance, the current flows through the resistance, and the Thévenin voltage is
o
o
57.26789.1 −∠== ttn ZVIThus, the Thévenin and Norton equivalent circuits are:
0200∠== oct VV
(b) For maximum power transfer, the load impedance is
15010050100
20050100
=−++
=+
=
−=
loadt
tload
load
jjZZ
jZVI
( ) W 50)2/1(100 22 === −loadrmsloadload IRP
(c
) In the case for which the load must be pure resistance, the load for
maximum power transfer is
W 21.47
28.139190.02008.111
−∠===
==
t
tload
I
ZZoV
( )8.11150100
2 ==
+++
−loadrmsloadload
loadtload
IRPjZZ
176
P5.90 Zeroing sources, we have:
Thus, the Thévenin impedance is
4243.63472.451101
1 jj
Zt +=∠=+
= o
Writing a current equation for the node at the upper end of the current source under open circuit conditions, we have
Thus, the Thévenin and Norton equivalent circuits are:
o
o
o
36.3099.1380.9356.62
5510
45100
∠==
∠==
=+∠−
ttn
oct
ococ
Z
j
VIVV
VV
For the maximum power transfer, the load impedance is
loadI
( ) W 6.244
80.9364.154242
80.9356.6242
2 ==
∠=−++
∠=
+=
−=
−loadrmsloadload
loadt
t
load
IRPjjZZ
jZo
oV
In the case for which the load must be pure resistance, the load for maximum power transfer is
loadI
( ) W 2.151
08.62223.8472.44280.9356.62
472.4
2 ==
−∠=++∠
=+
=
==
−loadrmsloadload
loadt
t
tload
IRPjZZ
ZZ
ooV
177
P5.91 At the lower left-hand node under open-circuit conditions, KCL yields
from which we have Then, the voltages across the 5-Ω resistance and the j5-Ω inductance are zero, and KVL yields
xx II 5.0= .0=xI
°∠== − 303ocbat VV
With short circuit conditions, we have
°∠=+
°∠−= 1654243.0
55303jxI
°−∠=−== − 152121.05.0sc xxban IIIIThe Thévenin impedance is given by
Finally, the equivalent circuits are:
°∠=°−∠
°∠== 4514.14
152121.0303
t
ttZ
IV
P5.92 Under open-circuit conditions, we have
V °∠=°∠+== − 87.361002)34(oc jabt VVWith the source zeroed, we look back into the terminals and see Ω Next, the Norton current is
433 =++−= jjZt 4
°∠== 87.365.2t
tn Z
VI
178
P5.93 For maximum power transfer, the impedance of the load should be the
complex conjugate of the Thévenin impedance:
loadZ Setting real parts equal:
( )loadload CjRj ω−=−= 822
Setting imaginary parts equal:
Ω= 22loadR
( )( ) F382.6 81
18µω
ω==
−=−
load
load
CC
P5.94* For maximum power transfer, the impedance of the load should be the
complex conjugate of the Thévenin impedance:
Setting real parts equal:
0146.00401.010146.00401.01
822
jCjRYjZY
jZ
loadloadload
loadload
load
+=+=
+==
−=
ω
Ω=
=
93.240401.01
load
load
RR
Setting imaginary parts equal:
F 7.44
0146.0µ
ω=
=
load
load
CC
P5.95 We are given
( ) ( )o66302cos182 += ttvan
179
(a) By inspection, Hz 06.48
3022=
==
ffπω
(b) (c)
( ) ( )( ) ( )o
o
174302cos18254302cos182
−=
−=
ttvttv
cn
bn
( ) ( )( ) ( )o
o
54302cos182174302cos182
−=
−=
ttvttv
cn
bn
P5.96 We are given:
( )van
( ) ( )
( )( ) ( )ttv
ttttv
cn
bn
ωωω
cos10060cos10060cos100
−=
+=
−=o
o
The phasor diagram is:
For counterclockwise rotation, the sequence of phasors is acb. Thus, this is a negative sequence source. From the phasor diagram, we can determine that
( ) ( )( ) ( )( ) ( )o
o
o
o
o
o
150cos310030cos310090cos3100
1503100
303100
903100
+=
+=
−=
+∠=−=
+∠=−=
−∠=−=
ttvttvttv
ca
bc
ab
anca
cnbnbc
bnanab
ω
ω
ω
VVV
VVV
VVV
cn
180
P5.97*
.98*
P5.99
he power flow pulsates. Reduced vibration in generators nd motors is a potential advantage for the three-phase system. In
ositive seque urce. The phasor diagram is shown in Figure book. Thus, ve
( ) ( )kW 36.19
0cos67.144403cos3V
rms A 67.1430440
rmsV 1.76244033
Y
=
×××==
===
=×=×=
θL
YL
YL
IPR
VI
VV
P5
−∠=
−=
×+=
+=
−
05.6243.2370.2098.10
103775011
11
4
o
Y
jj
CjRZ
ω
Total power flow in a balanced system is constant with time. For a single phase system taaddition, less wire is needed for the same power flow in a balanced three-phase system.
Ω−∠=
=∆
05.6229.703
o
YZZ
P5.100 This is a p nce so
5.41 in the we ha
o03
2430∠=anV
The impedance of an equivalent wye-connected load is
Ω−== ∆ 1667.93
jZZY
The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:
181
Thus, the line current is
( )
( )( ) W 192 1.03
kW 58.1829326.3567.20
36.295.60264.09.347
1.01.026.580.351.01.0
2
2
=×=
=×=
∠=
∠=
−∠=
+−=
∠=
++=
aArmsline
ABrmsload
AB
aAanAn
Y
anaA
IPIP
j
Zj
o
o
o
o
V
IVV
VI
=∆
ABAB Z
VI
P5.101* is shown in Figure
.41 in the book. Thus, we have: This is a positive sequence source. The phasor diagram5
3
o02430∠=anV
The impedance of a equivalent wye-connected load is
Ω−== ∆ 6667.0667.13
jZZY
The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:
182
Thus, the line current is:
( )
( )( ) kW 37.51.03
kW 53.895378.473.109
98.254.58802.47.339
1.01.078.172.189
1.01.0
2
2
aA
=×=
=×=
∠=
=
∠=
−∠=
+−=
∠=
++=
∆
aArmsline
ABrmsload
ABAB
AB
aAanAn
Y
an
IPIP
Z
j
Zj
o
o
o
o
VI
V
IVV
VI
183
P5.102
o
o
o
1503
903
303
−∠×=−=
∠×=−=
−∠×=−=
Yancnca
Ybc
Ybnanab
VVV
VVV
VVV
VVV
cnbn
184
P5.103
rmsV 0.2543
4403
=== LY
VV
o
o
o
1202254
120225402254
∠=
−∠=
∠=
cn
bn
an
V
V
V
o
o
o
1502440902440
302440
∠=
−∠=
∠=
ca
bc
ab
V
V
V
( )
o
o
o
85.532054.215.1862054.2
15.662054.23.037750
∠=
−∠=
−∠=+
=
c
b
ana j
I
I
VI
P5.104 The line-to-line voltage is 240
( ) ( )
( ) ( )VAR 1431
15.66sin054.22543sin3W 9.632
15.66cos054.22543cos3
=
×××==
=
×××==
o
o
θ
θ
LrsmYrms
LrmsYrms
IVQ
IVP
rms.V 7.4153 =The impedance of each arm of the delta is
=∆Z
The equivalent wye has impedances of
o63.444.472)5/(110/1
1∠=
+ j
ZY
Then working with one phase of the wye-wye, we have
lineI A rms
o63.44491.13
∠== ∆Z
0.161240==
YZ power =P
71.44%100)cos(63.44 factor =×= o %kW 83.51)4471.0)(0.161)(240(3 =
185
P5.105 As suggested in the hint given in the book, the impedances of the circuits between terminals a and b with c open must be identical.
Equating the impedances, we obtain:
(1) CBA
CBCA
BACba ZZZ
ZZZZZZZ
ZZ++
+=
++=+
)/(1/11
Similarly for the other pairs of terminals, we obtain
(2)
CBA
CBBA
CABca ZZZ
ZZZZZZZ
ZZ++
+=
++=+
)/(1/11
CBA
BACA
BCAcb ZZZ
ZZZZZZZ
ZZ++
+=
++=+
)/(1/11 (3)
Then adding the respective sides of Equations 1 and 2, subtacting the corresponding sides of Equation 3, and dividing boths sides of the result by 2, we have:
Similarly we obtain: CBA
CBa ZZZ
ZZZ++
=
and
CBA
CAb ZZZ
ZZZ++
=CBA
BAc ZZZ
ZZZ++
=
186
P5.106 As suggested in the hint, consider the circuits shown below. The admittances of the circuits between terminals must be identical.
First, we will solve for the admittances of the delta in terms of the impedances of the wye. Then we will invert the results to obtain relationships between the impedances.
(1) cacbba
cb
cba
BC ZZZZZZZZ
ZZZ
YY++
+=
++
=+
/1/11
1
Similarly working with the other terminials, we obtain
(2)
(3)
Then adding the respective sides of Equations 2 and 3, subtacting the corresponding sides of Equation 1, and dividing boths sides of the result by 2, we have:
cacbba
baBA ZZZZZZ
ZZYY++
+=+
cacbba
caCA ZZZZZZ
ZZYY++
+=+
Inverting both sides of this result yields:
cacbba
aA ZZZZZZ
ZY++
=
Similarly, we obtain:
and
a
cacbbaA Z
ZZZZZZZ ++=
b
cacbbaB Z
ZZZZZZZ ++=
c
cacbbaC Z
ZZZZZZZ ++=
187
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