ac thevenin's
DESCRIPTION
TRANSCRIPT
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Lecture 16Phasor Circuits, AC Power,
Thevenin
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Kirchhoff’s Laws in Phasor Form
0321 =−+ VVV
We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.
The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
outin II =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuit Analysis Using Phasors and Impedances
1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2. Replace inductances by their compleximpedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC). Resistances have impedances equal to their resistances.
3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
)15500cos(707.0)(15707.0454.141
30100
454.141
10010050150100
50)10)(40)(500(
11150)3.0)(500(
30100)30500cos(100
6
ooo
o
o
oo
−=→−∠=∠∠
==
∠=
+=−+=++=
Ω−=−=−=
Ω===∠=→+=
−
ttiZ
jjjZZRZ
jjC
jZ
jjLjZtV
S
CLeq
C
L
SS
VI
V
ω
ω
Find i(t):
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
oooo
oooo
o
o
o
1054.35)15707.0)(9050(9011751.106)15707.0)(90()90()(
)15707.0)(100(
15707.0
30100
−∠=−∠−∠=⎟⎠⎞
⎜⎝⎛ −∠=⎟
⎠⎞
⎜⎝⎛−=
∠=−∠∠=∠==
−∠==
−∠=
∠=
IIV
IIV
IV
I
V
CCj
LLjwL
R
C
L
R
S
S
ωω
ωω
Find the phasorvoltage across each element:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
50504571.704501414.0
01
01.001.01
1001
1001
111
1
100)1010)(1000(
11100)1.0)(1000(
9010)901000cos(10)1000sin(10)(
6
j
jjZR
Z
jx
jC
jZ
jjjwLZtttv
C
RC
C
L
ss
−=−∠=∠
∠=
+=
−+
=+
=
Ω−=−=−=
Ω===−∠=→−==
−ω
V
Find the voltage vc(t) in steady state:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
)1000cos(10)1801000cos(10)(
18010)9010)(901()9010(4571.704571.70
)9010(5050
4571.70)9010(100)5050(
4571.70
tttv
jjjZZZ
V
C
SLRC
RCC
−=−=
−∠=−∠−∠=−∠⎟⎟⎠
⎞⎜⎜⎝
⎛∠−∠
=
−∠⎟⎟⎠
⎞⎜⎜⎝
⎛+−∠
=−∠⎟⎟⎠
⎞⎜⎜⎝
⎛+−−∠
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
o
oooo
o
o
oo
oo
V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
ooo
oo
o
ooo
901.090100
1801010018010
1801.0100
18010
1351414.04571.709010
50509010
)5050(1009010
−∠=−∠
−∠=
−−∠
==
−∠=−∠
==
−∠=∠−∠
=+−∠
=−+
−∠=
+=
jZ
R
jjjZZ
C
CC
CR
RCL
S
VI
VI
VI
Find the phasor current through each element:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.5
5.11.02.005.1510
:2
22.0)2.01.0(902510
:1
21122
21211
=+−→∠=−−
+
−=−+→−∠=−−
+
VVVVV
VVVVV
jjjj
nodeatKCL
jjjj
nodeatKCL
o
o
Use the node voltage technique to find v1(t):
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.5
)7.29100cos(1.16)(
74.2912.168142.01.0
2323)2.01.0(
:32.04.0
22.0)2.01.0(
5.11.02.022.0)2.01.0(
1
11
21
21
21
21
o
o
+=
∠=+=−−
=→−=−
=+−−=−+
=+−−=−+
ttv
jj
jjj
Addingjj
jjj
jjjjj
VV
VVVV
VVVV
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.9
)135500cos(104.28)(
135104.28456.353
9010250250
:250)5.0)(500(
2509010)90500cos(10)500sin(10)(
3
3
o
oo
o
o
−=
−∠=∠−∠
=+
=+
=→=+
===Ω=
−∠=→−==
−
−
txti
xjZZ
ZZ
KVLjjLjZ
Ztttv
S
LR
SSLR
L
R
SS
VVIVII
V
ω
Find i(t):
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.9
( )( )
9010
4507.7
135)250)(103.28(90)5.0)(500(
13507.7135)250)(103.28(
3
3
−∠=
−∠=
−∠∠==
−∠=−∠==
−
−
S
L
R
xLj
xR
V
IV
IV
o
o
oo
ω
Construct the a phasor diagram showing all three voltages and the current:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Find the phasor voltage and phasor current through each element:
o31.5647.5515.4677.3032
200
20032
1001
200501001
2001
5011
200)1)(200(
50)10100)(200(
11
1111
6
−∠=−=+
=
+=+−=++
−=
===
−=−=−=
++=
−
jj
Z
jjjjjZ
jjLjZ
jx
jC
jZ
ZZZZ
eff
eff
L
C
RLCeff
ωω
Exercise 5.10
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.10
o
o
o
o
o
o
o
o
o
ooo
31.5677.20100
31.563.277
3.14639.190200
31.563.277
69.3355.59050
31.563.277
31.563.277)31.5647.55)(05(
−∠=∠−∠
==
−∠=∠−∠
==
∠=−∠−∠
==
−∠=−∠∠====
R
RR
L
LL
C
CC
effRLC
Z
Z
Z
Z
VI
VI
VI
IVVV
Find the phasor voltage and current through each element:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0)100100(100100)100100(
0)(
)(
21
21
1222
211
2
1
=−+−=−+
=−++
=−+
IjIVIIj
ZIIZIZI
VZIIZI
S
RLC
SRL
Exercise 5.11
Solve for the mesh currents:
100j
100j
-200j
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.11
)451000cos(41.1)(
452009.1009.19881
1097.91097.90)100(0)141(
4514100
)0100)(0100()45141)(45141()0100)(0()45141)(0100(
451410100010045141
45141001000100
0)45141()0100(0100)0100()45141(
0)100100(100100)100100(
1
33
22
1
21
21
21
21
o−=
−∠=−=−
=∠−∠
−∠=
∠−∠−−−∠∠∠−−−∠∠
=
−∠∠−∠−∠−∠∠−∠
=
=−∠+∠−∠=∠−∠
=−+−=−+
tti
jxjx
I
IIII
IjIVIIj S
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
ZV
IwhereIZ
VZ
mmm
m =−∠=∠∠
== θθ
o0VI
For θ>0 the load is called “inductive” since Z=jωL for an inductor
For θ<0 the load is “capacitive” since Z=-j/ωC for a capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for a Resistive Load
)(cos)(
)cos()()cos()(
0
2 tIVtp
tItitVtv
RZ
mm
m
m
ω
ωω
=
==
∠=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for an Inductive Load
)2sin(2
)sin()cos()(
)sin()90cos()(
9090900
0)cos()(
90
tIV
ttIVtp
tItIti
IL
VLV
VtVtv
LZ
mmmm
mm
mmm
mm
ωωω
ωωωω
ω
ω
==
=−=
−∠=−∠=∠
∠=∠=→=
∠=
o
ooo
o
IV
“Reactive” power
using cos(x)sin(x) = (1/2)sin(2x)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits• For a pure inductance half of the
time the power is positive and power flows to the inductor where it is stored as energy in the magnetic field
• Half of the time the power is negative and power flows from the inductor to the source
• Even though the average power is zero for a pure inductor, current flows between the source and the inductor and power is dissipated in the lines connecting the source to the inductor
Reactive Power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for a Capacitive Load
)2sin(2
)sin()cos()(
)sin()90cos()(
90901
00)cos()(
901
tIV
ttIVtp
tItIti
I
C
VVtVtv
CZ
mmmm
mm
mm
mm
ωωω
ωωω
ω
ω
−=−=
−=+=
∠=−∠
∠=→∠==
−∠=
o
o
o
oo
o
IV
Reactive power for a capacitor opposite the sign for an inductor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Inductor
CapacitorIf a load contains both inductance and capacitance with reactive powers of equal magnitude, the reactive powers cancel.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power for a General Load
[ ]
)cos()cos(
)cos(22
)cos(2
)2sin()sin(2
)2cos(1)cos(2
)cos()cos()()cos()(
)cos()(
iirmsrms
imm
imm
av
imm
imm
imm
im
m
PFIV
IVIVP
tIV
tIV
ttIVtptItitVtv
θθ
θθ
ωθωθ
θωωθω
ω
==
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛==
++=
−=−=
=
For a general RLC circuit in which the phase can be any value between -90° to +90°
PF is the “Power Factor”
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power for a General Load
)cos()cos(
θθ
θθθ
==
−=
PFIVP rmsrms
currentvoltage
Power angle:
If the phase angle for the voltage is not zero, we define the power angle θ:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
( ) [ ]WIVP θcosrmsrms=
( ) [ ]VARIVQ θsinrmsrms=
Average Power:
Reactive Power:
Apparent Power: [ ]VAIV RMSRMS
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( )2rmsrms22 IVQP =+
Power Triangles
Average power
Reactive power
Average power
Apparent power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Load Impedance in the Complex Plane
ZXZR
jXRZZ
=
=
+=∠=
)sin(
)cos(
θ
θ
θ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Additional Power Relations
RIP 2rms=
XIQ 2rms=
RVP R
2rms=
XVQ X
2rms=
Average power
Reactive power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
Compute the power and reactive power from the source in this circuit:
AI
VV
rms
rms
s
ss
iv
1.02
1414.02
071.72
102
45)135(90
===
===
=−−−=−=
I
V
oooθθθ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
VARx
IVQWx
IVP
rmss
rmss
rms
rms
5.0)45sin(1.0071.7
)sin(5.0)45cos(1.0071.7
)cos(
==
===
=
o
o
θ
θ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
VARVARVARQQQ
VARXIQ
VARXIQ
CL
CCC
LrmsL
rms
5.05.00.1
5.0)100(21.0
0.1)100()1.0(2
2
22
=−=+=
−=−⎟⎟⎠
⎞⎜⎜⎝
⎛==
===
The reactive power delivered to the inductor and capacitor:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
The power delivered to the resistor:
PWRRIP RRR rms
==⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛== 5.0100
21.0
222
2
I
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
Find the power, reactive power and power factor for the source. Also find the phasor current I:
Load A: 10kVA apparent power, PF=0.5 leading (capacitive)
Load B: 5kW power, PF=0.7 lagging (inductive)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
Load A: 10kVA apparent power, PF=0.5 leading (capacitive)
• Reactive power QA and the power angle θA are negative
Load B: 5kW power, PF=0.7 lagging (inductive)
• Reactive power QB and the power angle θB are positive
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
( ) ( ) ( )
kVARPQ
kVAR
PIVQ
kWIVP
BBB
B
AArmsA
AArmsA
rms
rms
101.5)57.45tan(5000)tan(
57.45)7.0arccos(
660.8
500010
5)5.0(10)cos(
22422
4
===
==
−=
−=−=
===
o
o
θ
θ
θ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
( ) ( ) kVAxQPIV
PQ
kVARkVARkVARQQQkWkWkWPPP
rmsrms
BA
BA
61.1010559.310
59.1910
559.3arctanarctan
559.3101.5660.81055
232322 =−+=+=
−=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛=
−=+−=+==+=+=
oθ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
o
ooo
59.4915
59.49)59.19(30
15)61.10)(414.1(2
61.10161.10
1414.1
14142
∠=∠=
=−−=−=
===
===
===
i
vi
rms
rms
rmsrmsrms
rms
I
AAI
AkV
kVAV
IVI
kVV
θ
θθθ
I
I
V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
o
o
59.4915
301414
∠=
∠=
I
V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.12o401.707 ∠=VVoltage source:
Delivers 5 kW to a load with a power factor of 100 percent. Find the reactive power and the phasor current:
o
o
4014.14
14.142
10500105
50041.11.707
50)sin(PowerReactive
01)cos(%100
3
∠=
==
==
==
===
==→=
I
AII
AVWxI
VVV
kWVIIVQ
PF
rms
rms
rms
rmsrms
rnsrms θθθ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.12If the PF=20% lagging, cos(θ) = 0.20 → θ = arcos(0.20) = 78.46°
θ
Q = 24.49kVAR
P = 5kW
o
ooo
o
46.3871.70
46.3846.7840
71.70)50)(41.1(2
50)cos(
49.24)46.78tan()5()tan()tan(
−∠=
−=−=−=
===
==
===→=
I
θθθ
θ
θθ
vi
rms
rmsrms
AAII
AV
PI
kVARkWPQPQ
VrmsIrms
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Thevenin Equivalent Circuits
The Thevenin equivalent for an ac circuit consists of a phasor voltage source Vt in series with a complex impedance Zt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
ocVV =t
We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.
Thevenin Equivalent Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
scsc
oc
IV
IV t
tZ ==
Thevenin Equivalent Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Norton Equivalent Circuit
The Norton equivalent for an ac circuit consists of a phasor current source In in parallel with a complex impedance Zt
scII =n
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
50504571.704501414.0
0101.001.0
1)100/(1100/1
1 jjj
Zt −=−∠=∠
∠=
+=
−+= o
o
o
Circuit with the voltage and current source zeroed to find the Thevenin impedance
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
Circuit with the output shorted to find the short circuit current.
ooo
o
oo
45414.1190101
901
01100
0100100
−∠=−=∠−∠=
∠=
∠=∠
==
−=
j
V
SC
S
SR
SRSC
I
I
I
III
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
o
ooo
45414.1
90100)4571.70)(45414.1(
−∠==
−∠=−∠−∠==
SCn
tSCt Z
II
IV
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The Thevenin equivalent of a two-terminal circuit delivering power to a load impedance.
Maximum Average Power Transfer
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.
If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.
Maximum Average Power Transfer
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Maximum Average Power TransferFind the load for the maximum power transfer if the load can have any complex value and if the load must be a pure resistance:
Ω=+=
−=
+=
−=
71.705050
5050
5050
5050:
22
*
t
t
t
t
Z
jZloadReal
jZ
jZloadComplex
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
50502
10010011
1100
1100
1001
1001
100111
jjjj
jjZ
jjRLjZ
eff
eff
+=+
=⎟⎟⎠
⎞⎜⎜⎝
⎛++
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=−
=
−=+
−=+=
ω
Find the Thevenin impedance, the Thevenin voltage and the Norton current.
25100255050502550 jjjjZZ efft +=−++=−+=
Zero the sourceZeff
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
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o
o
o
o
o
o
4571.704520100
4521000100
4521000100
100100100 2
−∠=∠∠
=∠∠
=∠∠
=+
= SSOC jVVV
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
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04.59686.004.141.1034571.70
251004571.70
−∠=∠
−∠=
+−∠
===jZT
TSCN
VII