Electric Circuits ECSE 210
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ECSE 210: Electric Circuits 2 Chapter 9 AC SteadySteady-State Analysis
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KCL/KVL for Phasor Circuits i1 (t )
Æ KCL in time domain:
i5 (t )
i1 (t) + i2 (t) + i3 (t) + i4 (t) + i5 (t) + ... = 0
i4 (t ) i2 (t )
Æ Complex exponential input:
I e 1
jωt
+I e 2
jωt
+I e 3
jωt
+I e 4
jωt
Æ KCL for phasors
+I e 5
jωt
i3 (t )
+ ... = 0 I s1
I s5
I + I + I + I + I + ... = 0 1 2 3 4 5
Æ We can use the same approach to prove KVL for phasor circuits.
I s4
I s2
I s3 2
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Impedance Z and Admittance Y Frequency Domain
Time Domain
V = ZI
v = Ri
I = YV
i = Gv
Æ Phasor terminal relationships are linear and in the same form as in the case of dc circuits Æ Except in this case we have complex numbers. Æ KCL and KVL are valid for phasor circuits. Î We can derive network theorems that are similar to th dc the d case. Î Example: Element series and parallel combinations
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Parallel Combinations
Y3
Y2
Y1
Gpar
G3
G1
G3
Ypar
Ypa par = Y1 + Y2 + Y3 + K
G ppar = G1 + G2 + G3 + K
1 1 1 1 = + + +K Z par Z1 Z 2 Z 3
1 1 1 1 = + + +K R par R1 R2 R3
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Series Combinations
Z1
R1
Z2
R3
Z series
Rseries R3
Z3
Z series = Z1 + Z 2 + Z 3 + K 1 1 1 1 = + + +K Yseries Y1 Y2 Y3
Rseries = R1 + R2 + R3 + K 1 1 1 1 = + + +K Gseries G1 G2 G3
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Example 1 Calculate the equivalent impedance at 60Hz and 400Hz. Æ At 60Hz,
25Ω
ω = 2πf = 120π
Z eq = 25Ω + j (120π × 20 ×10 −3 ) +
1 j (120π × 50 × 10 −6 )
Z eq = 25Ω + j 7.54 − j 53.05 = 25 − j 45.51
Z eq ⇒
20mH
Æ Capacitive Æ At 400Hz 400Hz,
ω = 2πf = 800π
50μF
1 Z eq = 25Ω + j (800π × 20 × 10 ) + j (800π × 50 × 10 −6 ) −3
Z eq = 25Ω + j 50.27 − j 7.96 = 25 + j 42.31 Æ Inductive
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Example 2 At 60Hz, Is = 0.5∠ − 22.98
I
o
Z eq
Calculate vs(t) First determine Vs
Vs
Z R = 20Ω
+ -
20Ω
40mH
Z L = jωL = j (2π × 60) × (40 × 10 −3 )= jj15.08Ω
Z RZ L (20)( j15.08) Z eq = = = 7.25 + j9.61Ω = 12.04∠52.98 Z R + Z L 20 + j15.08 j15 08
(
Vs = Z eq I = 12.04∠52.98
o
)(0.5∠ − 22.98 )= 6.02∠30 )(
vs (t) = 6.02 cos(120π t + 30 o ) volts
o
o
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Example 3: Equivalent Circuits Find an equivalent circuit for the impedance: Æ Option #1:
Z = 10∠30o
Z = 10∠30 o = 8.66 8 66 + j5Ω
8.66 Ω
Æ Positive P iti reactance t Î inductive i d ti
j5 Ω
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Example 3: Equivalent Circuits Find an equivalent circuit for the impedance: Æ Option #2:
Z = 10∠30o
Z = 10∠30o = 8.66 + j 5Ω
1 o Y= = 0.1∠ − 30 = 0.0866 − j0.05 o 10∠30
Negative susceptance Î inductive Positive reactance Î inductive Y circuit Z circuit
Equivalent circuits
?
0.0866S G
-j0.05S 11.55 Ω B R
? j20Ω X 9
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Example 4 1Ω
− j2Ω
Z eq ⇒
4Ω
j2Ω
2Ω j6Ω
j 4Ω
− j2Ω
− j2Ω
Z eq = 1Ω || (−2 jΩ) + [4Ω + j 2Ω + j 4Ω || (− j 2Ω)] || (2Ω + j 6Ω − j 2Ω)
→ Z eq = 3.8 + j 0.6Ω
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AC Steady State Analysis Basis:
KCL, KVL and V=ZI can be used to solve linear, ac steady-state circuit problems.
Æ Consider the network theorems developed for resistive dc circuits for ac steady-state analysis of RLC circuits. q y domain Æ Each time-domain method is valid for frequency analysis. (Derive equivalent frequency domain results based on the phasor form of KCL, KVL and V=IZ.) Æ We will consider a selection of illustrative ac steady-state (phasor) circuit analysis examples. 11
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Example 1: Linearity I1
Vs +-
j4 Ω I3
4Ω I2
-j3Ω
I4
2Ω
I5
-j2 Ω
Use linearity to determine the phasor I4 and the time domain current i4(t) given that: vs (t ) = 12 cos(377t + 30)V Æ The phasor for vs(t) is given by:
Vs = 12∠30oV How do we find the value of L in the above circuit?
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Example 1: Linearity Solve using linearity:
Assume:
I4 = 1∠0
Æ Solve for Vs Æ Using linearity calculate I4 for the given value of Vs
For example:
•Assume that for I4=1A we can solve for Vs=3V. •Then we could immediately state (using linearity) that if Vs=1.5 =1 5 then I4= (1.5/3)1 (1 5/3)1 = 0.5A 0 5A Æ Apply above procedure to the circuit on previous slide. 13
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Example 1: Linearity 4Ω
I1
I2
Vs +-
-j3 Ω
Assume I 4 = 1∠0 A o
I4 = 1∠0 A 2V = j Æ I5 = − jj2Ω o
j4Ω
V1
Æ
Æ
I3
V2 I4
2Ω
I5
-j2Ω
C l l t Vs Calculate
V2 = (1∠0 o A)(2Ω) = 2V
Æ I3 = I4 + I5 = 1 + j A
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Example 1: Linearity I1
Vs +-
4Ω
j4 Ω
V1 I2
-j3Ω
I3
V2 I4
2Ω
I5
-j2 Ω
V1 = I3 ( j 4Ω) + V2 = (1 + j )( j 4) + 2 = −2 + 4 j V
−2 + 4 j −4 2 = −j Æ I2 = − j3 3 3 4 2 1 1 Æ I1 = I 2 + I 3 = − − j + 1 + j = − + j 3 3 3 3 15
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Example 1: Linearity I1
Vs +-
4Ω
j4 Ω
V1 I2
-j3Ω
I3
V2 I4
2Ω
I5
-j2 Ω
1⎞ ⎛ 1 Æ Vs = I1 (4Ω) + V1 = − + j (4Ω) + (−2 + 4 j ) ⎝ 3 3⎠ Æ
10 16 Vs = − + j = 6.29∠122 o V 3 3 16
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Example 1: Linearity I 4 = 1∠0
o
10 16 Vs = − + j = 6.29∠122 o 3 3
Æ
Using linearity:
Vs = 12∠30 o
o o 12∠30 1∠0 ( ) ( ) o = 1.91∠ − 92 Æ I4 = o 6 29∠122 6.29∠122
In the time domain:
i4 (t ) = 1.91 cos(377t − 92 o )
A 17
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Example 2: Nodal Analysis V1
2∠60o A
12∠30o V
V2
+ -
Find V1 and V2.
jj1Ω
2Ω
-j2 j2 Ω
KCL at supernode: V1 V2 V2 + + = 2∠60 o j1 2 − j2
For the supernode: V2 = V1 − 12∠30 o
V1 V1 − 12∠30 V1 − 12∠30 + + = 2∠60 o j1 2 − j2 18
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Example 2: Nodal Analysis V1 V1 − 12∠30 V1 − 12∠30 + + = 2∠60 o j1 2 − j2 ⎛1 1 1 ⎞ 12∠30 o 12∠30 o o V1 ⎜ + + − ⎟ = 2∠60 + ⎝ j 2 −22 j ⎠ 2 j2
How do we combine these?
V1 = 14.69∠117.14 o V Also
V2 = V1 − 12∠30 o = 14.69∠117.14 o − 12∠30 o = −6.70 + j13.07 − (10.4 + j 6 ) From supernode constraint
= −17.1 + j 7.07 = 18.5∠157.54 o V 19
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Example 3: Nodal Analysis 2 V1 2Vx
+ -
V2
6∠0o V
1Ω V3
+ -
+ 1Ω V x o 4∠0 V -
-j1Ω
+
Find Vo.
j1Ω
+ 1Ω Vo -
V2 = 6∠0 Vx = Vo − 6∠0 Use KVL on outer loop V1 = Vo + Vx = 2Vo − 6∠0 V3 = Vo − 4∠0
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Example 3: Nodal Analysis 2 V1
Constraints V2 = 6∠0 2Vx
V1 = Vo + Vx = 2Vo − 6∠0
+ -
V2
V3
V3 = Vo − 4∠0
6∠0o V
1Ω
+
Vx = Vo − 6∠0
+ 1Ω V x 4∠0oV -
-j1Ω
j1Ω
+ -
+ 1Ω Vo -
KCL at supernode: Vo − V1 V3 − V1 V3 − V2 V3 Vo + + + + =0 1 −j 1 j 1
Æ Substitute for V1, V2, V3:
Vo = 3.225∠7.125o V
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Example 4: Mesh Analysis j1 Ω
3Ω
Find Vo. I1 = 2∠30 o
+
I1
2Ω
Vo -
I2
+ -
8∠45o
-j2 j Ω
2∠30o A
KVL (I2 − I1 )(2 − 2 j)) + I2 ( j1) 1) + 8∠45 o = 0
(−2∠30 )(2 − 2 j ) + I (2 − 1 j ) + 8∠45 = 0 2∠30 )(2 − 2 j) − 8∠45 ( I = = 3.18∠ − 64.96 o
o
2
o
2
o
2 − 1j
o
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Example 4: Mesh Analysis j1 Ω
3Ω I1 = 2∠30 o
I 2 = 3.18∠ − 64.96 o I1 − I2 = 2∠30 − 3.18∠ − 64.96 o
+ o
I1
= 0.38 + j 3.88 A 2∠30o A
Vo = 2(I1 − I 2 ) = 0.76 + j 7.76
2Ω
Vo -
I2
+ -
8∠45o
-j2 j Ω
V 23
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Example 5: Mesh Analysis 2 Objective: Find I4 then Vo.
-j1Ω
I2
I1
+ - 12∠0o V
1Ω V3 2I x
I3
4∠0o A
1Ω j1Ω
Ix
I4
+ 1Ω Vo -
Constraint Equations I2 = −4∠0 Ix = I4 − I2 = I4 + 4
I3 = 2Ix = 2(I4 + 4 ) 24
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Example 5: Mesh Analysis 2 Constraint Equations I2 = −4∠0
-j1Ω
+ - 12∠0o V
1Ω
Ix = I4 − I2 = I4 + 4
V3
I3 = 2Ix = 2(I4 + 4 )
Mesh Equations
I2
I1
2I x
I3
4∠0o A
1Ω j1Ω
Ix
I4
+ 1Ω Vo -
− jI1 + 1(I1 − I3 ) = −12∠0 j (I 4 − I3 ) + 1(I 4 − I2 ) + 1I 4 = 0 Vo = −4∠ − 36.87 o V
I4 = −4∠ − 36.87 o A 25
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Example 6: Linear Superposition Calculate Vo using superposition:
j1 Ω
3Ω +
Need to assume: 1 Both sources are at 1.Both the same frequency 2.The reactances are given i for f thi this same frequency
2Ω
Vo -
2∠30o A
+ -
8∠45o
-j2 j Ω
•Superposition was shown to hold in the time domain •But also holds in the frequency domain if all frequencies are the same
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Example 6: Linear Superpostion Short circuit voltage source and calculate Vo1.
Io1 2Ω
Current divider: j1Ω Io1 = 2∠30 o ) ( 2 − j2 + j1
j1Ω
3Ω +
Vo1 -
2∠30o A
-j2 j Ω
j1 Vo1 = 2∠30 o )(2) = 1.79∠146.57 o ( 2 − j2 + j1
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Example 6: Linear Superpostion j1Ω
3Ω Open circuit current source and calculate Vo2.
+
2Ω
Vo2 -
2∠30 A o
+ -
8∠45o
-j2 j Ω
Voltage divider:
2 Vo2 = 8∠45 o = 7.14∠71.57 o − j2 + 2 + j1 B superposition: By ii Vo = Vo1 + Vo2 = 1.79∠146.57 o + 7.14∠71.57 o = 0.76 + j 7.76 V
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Example 7: Source Transformation j4 Ω
4Ω
I4
12∠30 V +o
2Ω
-j3 Ω
-j2 Ω
Use sou source ce ttransformation a s o at o to dete determine e tthe ep phasor aso I4. j4 Ω I4
3∠30o A
4Ω
j3Ω -j3
2Ω
-j2 Ω 29
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Example 7: Source Transformation j4Ω I4
3∠30o A
4Ω
2Ω
-j3Ω
-j2Ω
Z1
(3∠30 )Z
j4Ω
o
1
V
+-
(4)(− j 3) − j12 Z1 = = 4 − j3 4 − j3
I4
2Ω
-j2Ω j
30
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Example 7: Source Transformation Z1
(3∠30 )Z
jj4Ω
o
(3∠30 )Z
1
V +-
I4
2Ω
-j2Ω
I4
o
1
A
Z2
2Ω
-j2Ω j2Ω
Z2 12 + j 4 Ω Z 2 = Z1 + j 4 = 4 − j3
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Example 7: Source Transformation
(3∠30 )Z
I4
o
1
Z3
Z2
(3∠30 )Z o
Z2
1
2Ω
(− j 2)( Z 2 ) Z3 = − j2 + Z2 4 − j12 Ω = 3 − j2
= 2.85∠ − 78.43o A
⎛ 4 − j12 ⎞ ⎜ ⎟ ⎝ 3 − j2 ⎠ I4 = 2.85∠ − 78.43o = 1.91∠ − 78.43o A ⎛ 4 − j12 ⎞ 2+⎜ ⎟ ⎝ 3 − j2 ⎠
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Example 8: Thevenin’s Theorem j1Ω
3Ω Calculate Vo using Thevenin’s Theorem.
+
2Ω
Vo -
2∠30o A
+ -
8∠45o
-j2Ω j
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Example 8: Thevenin’s Theorem j1Ω
3Ω Calculate Vo using Thevenin’s Theorem.
+
Voc -
2∠30o A
Voc = (2∠30 o )( j1) + 8∠45 o = 8.73∠57.75 o V Z th = − j 2 + j1 = − jΩ
Vth = Voc = 8.73∠57.75 o V
+ -
8∠45o
-j2Ω j
No currentt N through this capacitor
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Example 8: Thevenin’s Theorem Z th = − jΩ
8.73∠57.75
o
V +-
+
2Ω
Vo
-
2 o V0 = 88.73∠57.75 73∠57 75 = 0.76 0 76 + j7.76 j7 76 2− j
V
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Summary 1. All the methods derived for solving dc steady-state circuits are also valid for ac steady-state circuit analysis i the in th frequency f domain. d i These Th methods th d iinclude: l d Æ Æ Æ Æ Æ
Nodal analysis. Mesh and loop analysis. Superposition. Source transformation. Thevenin and Norton’s theorems
2 The principle of linearity is valid for the analysis of ac 2. steady-state circuit analysis in the frequency domain under certain conditions . 36
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Consider Linear Superposition Again •See Section 9.4 in J & J •Superposition only holds true in general for linear circuits in the time domain •See Example 9.9 for how to apply this concept to the case where the circuit has sources at different frequencies •We cannot superpose phasors at different frequencies!
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Sources With Different Frequencies Find
12cos(2t) V
- v(t)
+ -
+
2Ω
0.5F 1Ω
1H
+ - 5cos(1t ( + 45 o ) Æ Two sources at different frequencies. Æ We must use superposition.
(Only meaningful in the time domain!) 38
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Sources With Different Frequencies () - v(t)
12 cos(2t ) V
+ -
+
2Ω
0 5F 0.5F 1Ω
1H
+ - 5 cos(t + 45o )V - v2(t) +
+ -
() + - v1(t)
2Ω
-j2Ω
-j1Ω
12∠0o V
2Ω
1Ω
j2Ω
1Ω + - 5∠45o V
j1Ω 39
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Sources With Different Frequencies In parallel to ground & replaced by
- v1(t) +
VC
2Ω
-j2Ω 1Ω
+
v1(t)
VC
5∠45o V
1Ω + - 5∠45o V
-
j1Ω
+ -
voltage divider (− j2Ω) || (2Ω + j1Ω) VC = (5∠45 o ) 1Ω + (− j2Ω) || (2Ω + j1Ω)
VC = (0.7∠ (0 7∠ − 12.09 12 09o )(5∠45o ) = 3.5∠32.9 3 5∠32 9 o
v1 (t ) = 3.5 cos(t + 32.9o )
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Sources With Different Frequencies - v2(t) + (2Ω + j2Ω) || (1Ω) VC = (12∠0 o ) (2Ω + j2Ω) || (1Ω) − j1Ω
+ -
2Ω
-j1Ω 1Ω
12∠0o V
j2Ω
VC
VC = (0.686∠59.04 o )(12∠0 o ) = 8.23∠59.04 o
V
V2 = (8.23∠59.04 o ) − (12∠0 o ) = 10.5∠138 o
V
v2 (t ) = 10.5 cos((2t + 138o )V 41
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Sources With Different Frequencies v1 (t ) = 3.5 cos(t + 32.9 o )
V
v2 (t ) = 10.5 cos(2t + 138o )
V
v(t ) = v1 (t ) + v2 (t ) = 3.5 cos((t + 32.9 o ) + 10.5 cos((2t + 138o ) Æ Note we can only add waveforms at different frequencies in the time domain. We CANNOT add their phasors. Each phasor is defined at a specific frequency and phasors with diff different t frequencies f i cannott be b added. dd d
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Next Topic: Power Recall: Power as a function of time
i (t )
Passive Sign Convention: pp to Æ Power supplied circuit/circuit element:
p (t ) = i (t )v(t )
+
v(t ) -
Electric El t i Circuit/ Ci it/ Circuit Element
Power as a function of time when the currents and voltages are sinusoids 43
ECSE 210: Electric Circuits 2 Chapter 9 AC SteadySteady-State Analysis
1
12/22/08
KCL/KVL for Phasor Circuits i1 (t )
Æ KCL in time domain:
i5 (t )
i1 (t) + i2 (t) + i3 (t) + i4 (t) + i5 (t) + ... = 0
i4 (t ) i2 (t )
Æ Complex exponential input:
I e 1
jωt
+I e 2
jωt
+I e 3
jωt
+I e 4
jωt
Æ KCL for phasors
+I e 5
jωt
i3 (t )
+ ... = 0 I s1
I s5
I + I + I + I + I + ... = 0 1 2 3 4 5
Æ We can use the same approach to prove KVL for phasor circuits.
I s4
I s2
I s3 2
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Impedance Z and Admittance Y Frequency Domain
Time Domain
V = ZI
v = Ri
I = YV
i = Gv
Æ Phasor terminal relationships are linear and in the same form as in the case of dc circuits Æ Except in this case we have complex numbers. Æ KCL and KVL are valid for phasor circuits. Î We can derive network theorems that are similar to th dc the d case. Î Example: Element series and parallel combinations
3
12/22/08
Parallel Combinations
Y3
Y2
Y1
Gpar
G3
G1
G3
Ypar
Ypa par = Y1 + Y2 + Y3 + K
G ppar = G1 + G2 + G3 + K
1 1 1 1 = + + +K Z par Z1 Z 2 Z 3
1 1 1 1 = + + +K R par R1 R2 R3
4
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Series Combinations
Z1
R1
Z2
R3
Z series
Rseries R3
Z3
Z series = Z1 + Z 2 + Z 3 + K 1 1 1 1 = + + +K Yseries Y1 Y2 Y3
Rseries = R1 + R2 + R3 + K 1 1 1 1 = + + +K Gseries G1 G2 G3
5
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Example 1 Calculate the equivalent impedance at 60Hz and 400Hz. Æ At 60Hz,
25Ω
ω = 2πf = 120π
Z eq = 25Ω + j (120π × 20 ×10 −3 ) +
1 j (120π × 50 × 10 −6 )
Z eq = 25Ω + j 7.54 − j 53.05 = 25 − j 45.51
Z eq ⇒
20mH
Æ Capacitive Æ At 400Hz 400Hz,
ω = 2πf = 800π
50μF
1 Z eq = 25Ω + j (800π × 20 × 10 ) + j (800π × 50 × 10 −6 ) −3
Z eq = 25Ω + j 50.27 − j 7.96 = 25 + j 42.31 Æ Inductive
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Example 2 At 60Hz, Is = 0.5∠ − 22.98
I
o
Z eq
Calculate vs(t) First determine Vs
Vs
Z R = 20Ω
+ -
20Ω
40mH
Z L = jωL = j (2π × 60) × (40 × 10 −3 )= jj15.08Ω
Z RZ L (20)( j15.08) Z eq = = = 7.25 + j9.61Ω = 12.04∠52.98 Z R + Z L 20 + j15.08 j15 08
(
Vs = Z eq I = 12.04∠52.98
o
)(0.5∠ − 22.98 )= 6.02∠30 )(
vs (t) = 6.02 cos(120π t + 30 o ) volts
o
o
7
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Example 3: Equivalent Circuits Find an equivalent circuit for the impedance: Æ Option #1:
Z = 10∠30o
Z = 10∠30 o = 8.66 8 66 + j5Ω
8.66 Ω
Æ Positive P iti reactance t Î inductive i d ti
j5 Ω
8
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Example 3: Equivalent Circuits Find an equivalent circuit for the impedance: Æ Option #2:
Z = 10∠30o
Z = 10∠30o = 8.66 + j 5Ω
1 o Y= = 0.1∠ − 30 = 0.0866 − j0.05 o 10∠30
Negative susceptance Î inductive Positive reactance Î inductive Y circuit Z circuit
Equivalent circuits
?
0.0866S G
-j0.05S 11.55 Ω B R
? j20Ω X 9
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Example 4 1Ω
− j2Ω
Z eq ⇒
4Ω
j2Ω
2Ω j6Ω
j 4Ω
− j2Ω
− j2Ω
Z eq = 1Ω || (−2 jΩ) + [4Ω + j 2Ω + j 4Ω || (− j 2Ω)] || (2Ω + j 6Ω − j 2Ω)
→ Z eq = 3.8 + j 0.6Ω
10
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AC Steady State Analysis Basis:
KCL, KVL and V=ZI can be used to solve linear, ac steady-state circuit problems.
Æ Consider the network theorems developed for resistive dc circuits for ac steady-state analysis of RLC circuits. q y domain Æ Each time-domain method is valid for frequency analysis. (Derive equivalent frequency domain results based on the phasor form of KCL, KVL and V=IZ.) Æ We will consider a selection of illustrative ac steady-state (phasor) circuit analysis examples. 11
12/22/08
Example 1: Linearity I1
Vs +-
j4 Ω I3
4Ω I2
-j3Ω
I4
2Ω
I5
-j2 Ω
Use linearity to determine the phasor I4 and the time domain current i4(t) given that: vs (t ) = 12 cos(377t + 30)V Æ The phasor for vs(t) is given by:
Vs = 12∠30oV How do we find the value of L in the above circuit?
12
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Example 1: Linearity Solve using linearity:
Assume:
I4 = 1∠0
Æ Solve for Vs Æ Using linearity calculate I4 for the given value of Vs
For example:
•Assume that for I4=1A we can solve for Vs=3V. •Then we could immediately state (using linearity) that if Vs=1.5 =1 5 then I4= (1.5/3)1 (1 5/3)1 = 0.5A 0 5A Æ Apply above procedure to the circuit on previous slide. 13
12/22/08
Example 1: Linearity 4Ω
I1
I2
Vs +-
-j3 Ω
Assume I 4 = 1∠0 A o
I4 = 1∠0 A 2V = j Æ I5 = − jj2Ω o
j4Ω
V1
Æ
Æ
I3
V2 I4
2Ω
I5
-j2Ω
C l l t Vs Calculate
V2 = (1∠0 o A)(2Ω) = 2V
Æ I3 = I4 + I5 = 1 + j A
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Example 1: Linearity I1
Vs +-
4Ω
j4 Ω
V1 I2
-j3Ω
I3
V2 I4
2Ω
I5
-j2 Ω
V1 = I3 ( j 4Ω) + V2 = (1 + j )( j 4) + 2 = −2 + 4 j V
−2 + 4 j −4 2 = −j Æ I2 = − j3 3 3 4 2 1 1 Æ I1 = I 2 + I 3 = − − j + 1 + j = − + j 3 3 3 3 15
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Example 1: Linearity I1
Vs +-
4Ω
j4 Ω
V1 I2
-j3Ω
I3
V2 I4
2Ω
I5
-j2 Ω
1⎞ ⎛ 1 Æ Vs = I1 (4Ω) + V1 = − + j (4Ω) + (−2 + 4 j ) ⎝ 3 3⎠ Æ
10 16 Vs = − + j = 6.29∠122 o V 3 3 16
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Example 1: Linearity I 4 = 1∠0
o
10 16 Vs = − + j = 6.29∠122 o 3 3
Æ
Using linearity:
Vs = 12∠30 o
o o 12∠30 1∠0 ( ) ( ) o = 1.91∠ − 92 Æ I4 = o 6 29∠122 6.29∠122
In the time domain:
i4 (t ) = 1.91 cos(377t − 92 o )
A 17
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Example 2: Nodal Analysis V1
2∠60o A
12∠30o V
V2
+ -
Find V1 and V2.
jj1Ω
2Ω
-j2 j2 Ω
KCL at supernode: V1 V2 V2 + + = 2∠60 o j1 2 − j2
For the supernode: V2 = V1 − 12∠30 o
V1 V1 − 12∠30 V1 − 12∠30 + + = 2∠60 o j1 2 − j2 18
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Example 2: Nodal Analysis V1 V1 − 12∠30 V1 − 12∠30 + + = 2∠60 o j1 2 − j2 ⎛1 1 1 ⎞ 12∠30 o 12∠30 o o V1 ⎜ + + − ⎟ = 2∠60 + ⎝ j 2 −22 j ⎠ 2 j2
How do we combine these?
V1 = 14.69∠117.14 o V Also
V2 = V1 − 12∠30 o = 14.69∠117.14 o − 12∠30 o = −6.70 + j13.07 − (10.4 + j 6 ) From supernode constraint
= −17.1 + j 7.07 = 18.5∠157.54 o V 19
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Example 3: Nodal Analysis 2 V1 2Vx
+ -
V2
6∠0o V
1Ω V3
+ -
+ 1Ω V x o 4∠0 V -
-j1Ω
+
Find Vo.
j1Ω
+ 1Ω Vo -
V2 = 6∠0 Vx = Vo − 6∠0 Use KVL on outer loop V1 = Vo + Vx = 2Vo − 6∠0 V3 = Vo − 4∠0
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Example 3: Nodal Analysis 2 V1
Constraints V2 = 6∠0 2Vx
V1 = Vo + Vx = 2Vo − 6∠0
+ -
V2
V3
V3 = Vo − 4∠0
6∠0o V
1Ω
+
Vx = Vo − 6∠0
+ 1Ω V x 4∠0oV -
-j1Ω
j1Ω
+ -
+ 1Ω Vo -
KCL at supernode: Vo − V1 V3 − V1 V3 − V2 V3 Vo + + + + =0 1 −j 1 j 1
Æ Substitute for V1, V2, V3:
Vo = 3.225∠7.125o V
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Example 4: Mesh Analysis j1 Ω
3Ω
Find Vo. I1 = 2∠30 o
+
I1
2Ω
Vo -
I2
+ -
8∠45o
-j2 j Ω
2∠30o A
KVL (I2 − I1 )(2 − 2 j)) + I2 ( j1) 1) + 8∠45 o = 0
(−2∠30 )(2 − 2 j ) + I (2 − 1 j ) + 8∠45 = 0 2∠30 )(2 − 2 j) − 8∠45 ( I = = 3.18∠ − 64.96 o
o
2
o
2
o
2 − 1j
o
22
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Example 4: Mesh Analysis j1 Ω
3Ω I1 = 2∠30 o
I 2 = 3.18∠ − 64.96 o I1 − I2 = 2∠30 − 3.18∠ − 64.96 o
+ o
I1
= 0.38 + j 3.88 A 2∠30o A
Vo = 2(I1 − I 2 ) = 0.76 + j 7.76
2Ω
Vo -
I2
+ -
8∠45o
-j2 j Ω
V 23
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Example 5: Mesh Analysis 2 Objective: Find I4 then Vo.
-j1Ω
I2
I1
+ - 12∠0o V
1Ω V3 2I x
I3
4∠0o A
1Ω j1Ω
Ix
I4
+ 1Ω Vo -
Constraint Equations I2 = −4∠0 Ix = I4 − I2 = I4 + 4
I3 = 2Ix = 2(I4 + 4 ) 24
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Example 5: Mesh Analysis 2 Constraint Equations I2 = −4∠0
-j1Ω
+ - 12∠0o V
1Ω
Ix = I4 − I2 = I4 + 4
V3
I3 = 2Ix = 2(I4 + 4 )
Mesh Equations
I2
I1
2I x
I3
4∠0o A
1Ω j1Ω
Ix
I4
+ 1Ω Vo -
− jI1 + 1(I1 − I3 ) = −12∠0 j (I 4 − I3 ) + 1(I 4 − I2 ) + 1I 4 = 0 Vo = −4∠ − 36.87 o V
I4 = −4∠ − 36.87 o A 25
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Example 6: Linear Superposition Calculate Vo using superposition:
j1 Ω
3Ω +
Need to assume: 1 Both sources are at 1.Both the same frequency 2.The reactances are given i for f thi this same frequency
2Ω
Vo -
2∠30o A
+ -
8∠45o
-j2 j Ω
•Superposition was shown to hold in the time domain •But also holds in the frequency domain if all frequencies are the same
26
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Example 6: Linear Superpostion Short circuit voltage source and calculate Vo1.
Io1 2Ω
Current divider: j1Ω Io1 = 2∠30 o ) ( 2 − j2 + j1
j1Ω
3Ω +
Vo1 -
2∠30o A
-j2 j Ω
j1 Vo1 = 2∠30 o )(2) = 1.79∠146.57 o ( 2 − j2 + j1
27
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Example 6: Linear Superpostion j1Ω
3Ω Open circuit current source and calculate Vo2.
+
2Ω
Vo2 -
2∠30 A o
+ -
8∠45o
-j2 j Ω
Voltage divider:
2 Vo2 = 8∠45 o = 7.14∠71.57 o − j2 + 2 + j1 B superposition: By ii Vo = Vo1 + Vo2 = 1.79∠146.57 o + 7.14∠71.57 o = 0.76 + j 7.76 V
28
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Example 7: Source Transformation j4 Ω
4Ω
I4
12∠30 V +o
2Ω
-j3 Ω
-j2 Ω
Use sou source ce ttransformation a s o at o to dete determine e tthe ep phasor aso I4. j4 Ω I4
3∠30o A
4Ω
j3Ω -j3
2Ω
-j2 Ω 29
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Example 7: Source Transformation j4Ω I4
3∠30o A
4Ω
2Ω
-j3Ω
-j2Ω
Z1
(3∠30 )Z
j4Ω
o
1
V
+-
(4)(− j 3) − j12 Z1 = = 4 − j3 4 − j3
I4
2Ω
-j2Ω j
30
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Example 7: Source Transformation Z1
(3∠30 )Z
jj4Ω
o
(3∠30 )Z
1
V +-
I4
2Ω
-j2Ω
I4
o
1
A
Z2
2Ω
-j2Ω j2Ω
Z2 12 + j 4 Ω Z 2 = Z1 + j 4 = 4 − j3
31
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Example 7: Source Transformation
(3∠30 )Z
I4
o
1
Z3
Z2
(3∠30 )Z o
Z2
1
2Ω
(− j 2)( Z 2 ) Z3 = − j2 + Z2 4 − j12 Ω = 3 − j2
= 2.85∠ − 78.43o A
⎛ 4 − j12 ⎞ ⎜ ⎟ ⎝ 3 − j2 ⎠ I4 = 2.85∠ − 78.43o = 1.91∠ − 78.43o A ⎛ 4 − j12 ⎞ 2+⎜ ⎟ ⎝ 3 − j2 ⎠
32
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Example 8: Thevenin’s Theorem j1Ω
3Ω Calculate Vo using Thevenin’s Theorem.
+
2Ω
Vo -
2∠30o A
+ -
8∠45o
-j2Ω j
33
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Example 8: Thevenin’s Theorem j1Ω
3Ω Calculate Vo using Thevenin’s Theorem.
+
Voc -
2∠30o A
Voc = (2∠30 o )( j1) + 8∠45 o = 8.73∠57.75 o V Z th = − j 2 + j1 = − jΩ
Vth = Voc = 8.73∠57.75 o V
+ -
8∠45o
-j2Ω j
No currentt N through this capacitor
34
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Example 8: Thevenin’s Theorem Z th = − jΩ
8.73∠57.75
o
V +-
+
2Ω
Vo
-
2 o V0 = 88.73∠57.75 73∠57 75 = 0.76 0 76 + j7.76 j7 76 2− j
V
35
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Summary 1. All the methods derived for solving dc steady-state circuits are also valid for ac steady-state circuit analysis i the in th frequency f domain. d i These Th methods th d iinclude: l d Æ Æ Æ Æ Æ
Nodal analysis. Mesh and loop analysis. Superposition. Source transformation. Thevenin and Norton’s theorems
2 The principle of linearity is valid for the analysis of ac 2. steady-state circuit analysis in the frequency domain under certain conditions . 36
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Consider Linear Superposition Again •See Section 9.4 in J & J •Superposition only holds true in general for linear circuits in the time domain •See Example 9.9 for how to apply this concept to the case where the circuit has sources at different frequencies •We cannot superpose phasors at different frequencies!
37
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Sources With Different Frequencies Find
12cos(2t) V
- v(t)
+ -
+
2Ω
0.5F 1Ω
1H
+ - 5cos(1t ( + 45 o ) Æ Two sources at different frequencies. Æ We must use superposition.
(Only meaningful in the time domain!) 38
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Sources With Different Frequencies () - v(t)
12 cos(2t ) V
+ -
+
2Ω
0 5F 0.5F 1Ω
1H
+ - 5 cos(t + 45o )V - v2(t) +
+ -
() + - v1(t)
2Ω
-j2Ω
-j1Ω
12∠0o V
2Ω
1Ω
j2Ω
1Ω + - 5∠45o V
j1Ω 39
12/22/08
Sources With Different Frequencies In parallel to ground & replaced by
- v1(t) +
VC
2Ω
-j2Ω 1Ω
+
v1(t)
VC
5∠45o V
1Ω + - 5∠45o V
-
j1Ω
+ -
voltage divider (− j2Ω) || (2Ω + j1Ω) VC = (5∠45 o ) 1Ω + (− j2Ω) || (2Ω + j1Ω)
VC = (0.7∠ (0 7∠ − 12.09 12 09o )(5∠45o ) = 3.5∠32.9 3 5∠32 9 o
v1 (t ) = 3.5 cos(t + 32.9o )
40
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Sources With Different Frequencies - v2(t) + (2Ω + j2Ω) || (1Ω) VC = (12∠0 o ) (2Ω + j2Ω) || (1Ω) − j1Ω
+ -
2Ω
-j1Ω 1Ω
12∠0o V
j2Ω
VC
VC = (0.686∠59.04 o )(12∠0 o ) = 8.23∠59.04 o
V
V2 = (8.23∠59.04 o ) − (12∠0 o ) = 10.5∠138 o
V
v2 (t ) = 10.5 cos((2t + 138o )V 41
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Sources With Different Frequencies v1 (t ) = 3.5 cos(t + 32.9 o )
V
v2 (t ) = 10.5 cos(2t + 138o )
V
v(t ) = v1 (t ) + v2 (t ) = 3.5 cos((t + 32.9 o ) + 10.5 cos((2t + 138o ) Æ Note we can only add waveforms at different frequencies in the time domain. We CANNOT add their phasors. Each phasor is defined at a specific frequency and phasors with diff different t frequencies f i cannott be b added. dd d
42
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Next Topic: Power Recall: Power as a function of time
i (t )
Passive Sign Convention: pp to Æ Power supplied circuit/circuit element:
p (t ) = i (t )v(t )
+
v(t ) -
Electric El t i Circuit/ Ci it/ Circuit Element
Power as a function of time when the currents and voltages are sinusoids 43
