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UNIT I – BASIC CIRCUIT CONCEPTS – Circuit elements – Kirchhoff’s Law – V-I Relationship of R,L and C – Independent and Dependent sources– Simple Resistive circuits – Networks reduction – Voltage division – current source transformation.- Analysis of circuit using mesh current and nodal voltage methods.
Methods of Analysis1
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Resistance
Methods of Analysis2
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Ohm’s Law
Methods of Analysis3
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Resistors & Passive Sign Convention
Methods of Analysis4
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Example: Ohm’s Law
Methods of Analysis5
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Short Circuit as Zero Resistance
Methods of Analysis6
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Short Circuit as Voltage Source (0V)
Methods of Analysis7
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Open Circuit
Methods of Analysis8
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Open Circuit as Current Source (0 A)
Methods of Analysis9
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Conductance
Methods of Analysis10
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Circuit Building Blocks
Methods of Analysis11
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Branches
Methods of Analysis12
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Nodes
Methods of Analysis13
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Loops
Methods of Analysis14
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Overview of Kirchhoff’s Laws
Methods of Analysis15
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Kirchhoff’s Current Law
Methods of Analysis16
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Kirchhoff’s Current Law for Boundaries
Methods of Analysis17
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KCL - Example
Methods of Analysis18
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Ideal Current Sources: Series
Methods of Analysis19
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Kirchhoff’s Voltage Law - KVL
Methods of Analysis20
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KVL - Example
Methods of Analysis21
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Example – Applying the Basic Laws
Methods of Analysis22
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Example – Applying the Basic Laws
Methods of Analysis23
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Example – Applying the Basic Laws
Methods of Analysis24
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Resistors in Series
Methods of Analysis25
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Resistors in Parallel
Methods of Analysis26
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Resistors in Parallel
Methods of Analysis27
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Voltage Divider
Methods of Analysis28
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Current Divider
Methods of Analysis29
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Resistor Network
Methods of Analysis30
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Resistor Network - Comments
Methods of Analysis31
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Delta Wye Transformations
Methods of Analysis32
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Delta Wye Transformations
Methods of Analysis33
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Example –Delta Wye Transformations
Methods of Analysis34
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Methods of Analysis35
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Methods of Analysis
Methods of Analysis36
• Introduction• Nodal analysis• Nodal analysis with voltage source• Mesh analysis• Mesh analysis with current source• Nodal and mesh analyses by inspection• Nodal versus mesh analysis
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3.2 Nodal Analysis
Steps to Determine Node Voltages:1. Select a node as the reference node. Assign voltage
v1, v2, …vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.
2. Apply KCL to each of the n-1 nonreference nodes. Use Ohm’s law to express the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown node voltages.
Methods of Analysis37
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Figure 3.1
Methods of Analysis38
Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis.
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Figure 3.2
Methods of Analysis39 Typical circuit for nodal analysis
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Methods of Analysis40
322
2121
iiIiiII
Rvv
i lowerhigher
2333
23
21222
212
1111
11
or 0
)(or
or 0
vGiR
vi
vvGiR
vvi
vGiR
vi
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Methods of Analysis41
3
2
2
212
2
21
1
121
Rv
RvvI
Rvv
RvII
232122
2121121
)()(
vGvvGIvvGvGII
2
21
2
1
322
221
III
vv
GGGGGG
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Example 3.1
Calculus the node voltage in the circuit shown in Fig. 3.3(a)
Methods of Analysis42
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Example 3.1
At node 1
Methods of Analysis43
20
45 121
321
vvv
iii
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Example 3.1
At node 2
Methods of Analysis44
60
45 212
5142
vvviiii
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Example 3.1
In matrix form:
Methods of Analysis45
55
41
61
41
41
41
21
2
1
vv
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Practice Problem 3.1
Methods of Analysis46
Fig 3.4
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Example 3.2
Determine the voltage at the nodes in Fig. 3.5(a)
Methods of Analysis47
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Example 3.2
At node 1,
Methods of Analysis48
243
3
2131
1
vvvvii x
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Example 3.2
At node 2
Methods of Analysis49
40
8223221
32
vvvvv
iiix
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Example 3.2
At node 3
Methods of Analysis50
2)(2
84
2
213231
21
vvvvvviii x
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Example 3.2
In matrix form:
Methods of Analysis51
003
83
89
43
81
87
21
41
21
43
3
2
1
vvv
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3.3 Nodal Analysis with Voltage Sources
Case 1: The voltage source is connected between a nonreference node and the reference node: The nonreference node voltage is equal to the magnitude of voltage source and the number of unknown nonreference nodes is reduced by one.
Case 2: The voltage source is connected between two nonreferenced nodes: a generalized node (supernode) is formed.
Methods of Analysis52
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3.3 Nodal Analysis with Voltage Sources
56
08
042
32
323121
3241
vv
vvvvvviiii
Methods of Analysis53 Fig. 3.7 A circuit with a supernode.
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A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.
The required two equations for regulating the two nonreference node voltages are obtained by the KCL of the supernode and the relationship of node voltages due to the voltage source. Methods of Analysis54
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Example 3.3 For the circuit shown in Fig. 3.9, find the node
voltages.
2
042
72
02172
21
21
vv
vvii
Methods of Analysis55
i1 i2
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Example 3.4
Methods of Analysis56
Find the node voltages in the circuit of Fig. 3.12.
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Example 3.4
At suopernode 1-2,
Methods of Analysis57
2023
106
21
14123
vv
vvvvv
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Example 3.4
At supernode 3-4,
Methods of Analysis58
)(34163
4143
342341
vvvv
vvvvvv
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3.4 Mesh Analysis
Mesh analysis: another procedure for analyzing circuits, applicable to planar circuit.
A Mesh is a loop which does not contain any other loops within it
Methods of Analysis59
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Fig. 3.15
Methods of Analysis60(a) A Planar circuit with crossing branches,(b) The same circuit redrawn with no crossing branches.
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Fig. 3.16
Methods of Analysis61
A nonplanar circuit.
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Steps to Determine Mesh Currents:1. Assign mesh currents i1, i2, .., in to the n meshes.2. Apply KVL to each of the n meshes. Use Ohm’s
law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.
Methods of Analysis62
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Fig. 3.17
Methods of Analysis63A circuit with two meshes.
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Apply KVL to each mesh. For mesh 1,
For mesh 2,
Methods of Analysis64
123131
213111
)(0)(
ViRiRRiiRiRV
223213
123222
)(0)(
ViRRiRiiRViR
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Solve for the mesh currents.
Use i for a mesh current and I for a branch current. It’s evident from Fig. 3.17 that
Methods of Analysis65
2
1
2
1
323
331
VV
ii
RRRRRR
2132211 , , iiIiIiI
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Example 3.5
Find the branch current I1, I2, and I3 using mesh analysis.
Methods of Analysis66
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Example 3.5
For mesh 1,
For mesh 2,
We can find i1 and i2 by substitution method or Cramer’s rule. Then,
Methods of Analysis67
123010)(10515
21
211
iiiii
12010)(1046
21
1222
iiiiii
2132211 , , iiIiIiI
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Example 3.6
Use mesh analysis to find the current I0 in the circuit of Fig. 3.20.
Methods of Analysis68
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Example 3.6
Apply KVL to each mesh. For mesh 1,
For mesh 2,
Methods of Analysis69
1265110)(12)(1024
321
3121
iiiiiii
021950)(10)(424
321
12322
iiiiiiii
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Example 3.6
For mesh 3,
In matrix from Eqs. (3.6.1) to (3.6.3) become
we can calculus i1, i2 and i3 by Cramer’s rule, and find I0.
Methods of Analysis70
020)(4)(12)(4
, A, nodeAt 0)(4)(124
321
231321
210
23130
iiiiiiiii
iIIiiiiI
00
12
21121956511
3
2
1
iii
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3.5 Mesh Analysis with Current Sources
Methods of Analysis71
Fig. 3.22 A circuit with a current source.
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Case 1– Current source exist only in one mesh
– One mesh variable is reduced Case 2
– Current source exists between two meshes, a super-mesh is obtained.
Methods of Analysis72
A21 i
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Fig. 3.23
a supermesh results when two meshes have a (dependent , independent) current source in common.
Methods of Analysis73
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Properties of a Supermesh
1. The current is not completely ignored– provides the constraint equation necessary to
solve for the mesh current.2. A supermesh has no current of its own.3. Several current sources in adjacency form
a bigger supermesh.
Methods of Analysis74
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Example 3.7
For the circuit in Fig. 3.24, find i1 to i4 using mesh analysis.
Methods of Analysis75
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If a supermesh consists of two meshes, two equations are needed; one is obtained using KVL and Ohm’s law to the supermesh and the other is obtained by relation regulated due to the current source.
620146
21
21
iiii
Methods of Analysis76
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Similarly, a supermesh formed from three meshes needs three equations: one is from the supermesh and the other two equations are obtained from the two current sources.
Methods of Analysis77
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0102)(8
506)(842
443
432
21
24331
iiiiii
iiiiiii
Methods of Analysis78
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3.6 Nodal and Mesh Analysis by Inspection
Methods of Analysis79
(a) For circuits with only resistors and independent current sources
(b) For planar circuits with only resistors and independent voltage sources
The analysis equations can be obtained by direct inspection
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In the Fig. 3.26 (a), the circuit has two nonreference nodes and the node equations
Methods of Analysis80
2
21
2
1
322
221
232122
2121121
)8.3()()7.3()(
III
vv
GGGGGG
MATRIXvGvvGI
vvGvGII
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In general, the node voltage equations in terms of the conductance is
Methods of Analysis81
NNNNNN
N
N
i
ii
v
vv
GGG
GGGGGG
2
1
2
1
21
22221
11211or simply
Gv = i
where G : the conductance matrix, v : the output vector, i : the input vector
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The circuit has two nonreference nodes and the node equations were derived as
Methods of Analysis82
2
1
2
1
323
331 vv
ii
RRRRRR
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In general, if the circuit has N meshes, the mesh-current equations as the resistances term is
Methods of Analysis83
NNNNNN
N
N
v
vv
i
ii
RRR
RRRRRR
2
1
2
1
21
22221
11211
or simply
Rv = i
where R : the resistance matrix, i : the output vector, v : the input vector
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Example 3.8
Write the node voltage matrix equations in Fig.3.27.
Methods of Analysis84
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Example 3.8
The circuit has 4 nonreference nodes, so
The off-diagonal terms are
Methods of Analysis85
625.111
21
81 ,5.0
41
81
81
325.111
81
51 ,3.0
101
51
4433
2211
GG
GG
125.0 ,1 ,0125.0 ,125.0 ,0
111 ,125.0
81 ,2.0
0 ,2.051
434241
343231
242321
141312
GGGGGG
GGG
GGG
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Example 3.8
The input current vector i in amperes
The node-voltage equations are
Methods of Analysis86
642 ,0 ,321 ,3 4321 iiii
6 0 33
.6251 0.125 1 0
0.125 .50 0.125 0 1 0.125 .3251 0.20 0 0.2 .30
4
3
2
1
vvvv
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Example 3.9
Write the mesh current equations in Fig.3.27.
Methods of Analysis87
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Example 3.9
The input voltage vector v in volts
The mesh-current equations are
Methods of Analysis88
6 ,0 ,6612 ,6410 ,4
543
21
vvvvv
60
664
4 3 0 1 0 3 8 0 1 0 0 0 9 4 2
1 1 4 01 20 0 2 2 9
5
4
3
2
1
iiiii
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3.7 Nodal Versus Mesh Analysis
Both nodal and mesh analyses provide a systematic way of analyzing a complex network.
The choice of the better method dictated by two factors.– First factor : nature of the particular network. The
key is to select the method that results in the smaller number of equations.
– Second factor : information required.Methods of Analysis89
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BJT Circuit Models
Methods of Analysis90(a) dc equivalent model.(b) An npn transistor,
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Example 3.13
For the BJT circuit in Fig.3.43, =150 and VBE = 0.7 V. Find v0.
Methods of Analysis91
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Example 3.13
Use mesh analysis or nodal analysis
Methods of Analysis92
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Example 3.13
Methods of Analysis93
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3.10 Summary
1. Nodal analysis: the application of KCL at the nonreference nodes
– A circuit has fewer node equations2. A supernode: two nonreference nodes3. Mesh analysis: the application of KVL
– A circuit has fewer mesh equations4. A supermesh: two meshes
Methods of Analysis94
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UNIT II – SINUSOIDAL STEADY STATE ANALYSIS 9–-Phasor– Sinusoidal steady state response concepts of impedance and admittance– Analysis of simple circuits– Power and power factors–– Solution of three phase balanced circuits and three phase unbalanced circuits–-Power measurement in three phase circuits.
Methods of Analysis95
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Sinusoidal Steady State Sinusoidal Steady State ResponseResponse
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Sinusoidal Steady State Sinusoidal Steady State ResponseResponse
1. Identify the frequency, angular frequency, peak value, RMS value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and complex impedances.3. Compute power for steady-state ac circuits.
4. Find Thévenin and Norton equivalent circuits.
5. Determine load impedances for maximum power transfer.
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The sinusoidal function v(t) = VM sin t is plotted (a) versus t and (b) versus t.
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The sine wave VM sin ( t + ) leads VM sin t by radian
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T 2
f 2
Frequency T
f 1 Angular frequency
tt
tt
zz
o
o
o
cos)90sin(
)90cos(sin
)90cos(sin
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Representation of the two vectors v1and v2.
In this diagram, v1 leads v2 by 100o + 30o = 130o, although it could also be argued that v2 leads v1 by 230o.
Generally we express the phase difference by an angle less than or equal to 180o in magnitude.
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4-6
Euler’s identity
tfAtAf
t
2coscos2
In Euler expression,
A cos t = Real (A e j t )A sin t = Im( A e j t )
Any sinusoidal function canbe expressed as in Euler form.
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The complex forcing function Vm e j ( t + ) produces the complex response Im e j (t + ).
Applying Euler’s Identity
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The sinusoidal forcing function Vm cos ( t + θ) produces the steady-state response Im cos ( t + φ).
The imaginary sinusoidal input j Vm sin ( t + θ) produces the imaginary sinusoidal output response j Im sin ( t + φ).
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Re(Vm e j ( t + ) ) Re(Im e j (t + ))
Im(Vm e j ( t + ) ) Im(Im e j (t + ))
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Phasor Definition
111 cos :function Time θtωVtv
tdroppingbye
e
V
j
tj
)Re(
)Re(
:Phasor
)(1
)(1
111
1
1
V
V
V
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A phasor diagram showing the sum of
V1 = 6 + j8 V and V2 = 3 – j4 V,
V1 + V2 = 9 + j4 V = Vs
Vs = Ae j θ
A = [9 2 + 4 2]1/2
θ = tan -1 (4/9)
Vs = 9.8524.0o V.
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Phasors Addition
Step 1: Determine the phase for each term.Step 2: Add the phase's using complex arithmetic.Step 3: Convert the Rectangular form to polar form.
Step 4: Write the result as a time function.
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Conversion of rectangular to polar form
)30cos(10)(
)45cos(20
2
1
ttv
ttv
3010
4520
2
1
V
V
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7.39cos97.29 ttvs
7.3906.2314.19tan,96.29)14.19(06.23
V
122
A
Aes j
7.3997.2914.1906.23
5660.814.1414.1430104520
21s
jjj
VVV
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Phase relation ship
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COMPLEX IMPEDANCE
LL Lj IV
90 LLjZL
LLL Z IV
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(b)(a)
(c)
In the phasor domain,
(a) a resistor R is represented by an impedance of the same value;
(b) a capacitor C is represented by an impedance 1/jC;
(c) an inductor L is represented by an impedance jL.
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ZcCj
Cj
CVejdt
VedCdt
dC
Ve
tjtj
tj
1IVVI
VI
V
Zc is defined as the impedance of a capacitor
)90(cos,cos,VI tCVithentVvifCjAs
The impedance of a capacitor is 1/jC.
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MM
MM
CVItCVithentVvifCjAs
)90(cos,cos,VI
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L
tjtj
tj
ZLjLj
LIejdtIedL
dtdL
Ie
IVIV
IV
I
ZL is the impedance of an inductor.
.cos),90(cos)90(cos,cos,IV
tLIvandtIiortLIvthentIiifCjAs
The impedance of a inductor is jL
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MMMM
MM
LIVtLIvandtIiortLIvthentIiifCjAs
,cos),90(cos)90(cos,cos,IV
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90
90111
LLjZZ
CCj
CjZ
Z
L
LLL
C
CCC
IV
IV
RR RIV
Complex Impedance in Phasor Notation
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Kirchhoff’s Laws in Phasor Form
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.
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)105500cos(35.35)(,10535.35
159035.3515707.09050I50
)75500cos(05.106)(,7505.106
159005.10615707.090150I150)15500cos(7.70)(,157.70 I100
15707.0
4530707.0454.141
30100I
454.141100100)50150(100
ttvjV
ttvjV
ttvV
ZV
jjZ
L
C
L
L
RR
total
total
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5050
4571.704501414.0
0101.001.0
1
01.001.01001
1001
)100/(1100/11
/1/11
2
jj
Z
jj
jjj
jjAs
jZcRZ
RC
RC
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V1000cos10)1351000(cos10)(
135104571.704571.704510
50504571.709010
50501004571.709010
)(
tttv
jjj
divisionvoltageZZ
ZVsVc
c
RCL
RC
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)1351000(cos414.0)(
135414.04571.709010
50509010
50501009010
tti
jjj
ZZVsI
RCL
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A)451000(cos1.0)(
451.090100
13510100
13510
)1351000(cos1.0)(
1351.0100
13510
ttijZc
VI
tti
RVI
R
CC
R
CR
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Solve by nodal analysis
)2(5.105.1510
)1(2902510
122
211
eqjVV
jV
eqjjVVV
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V)74.29100cos(1.1674.291.16
43.6369.331.1643.632236.0
69.336.32.01.0
2323)2.01.0(
)2(2)1(5.11.02.0
)2(22.0)2.01.0(
1,1
1122.02.01.0
)1(
)2(5.105.1510
)1(2902510
1
1
1
1
21
21
211
122211
tv
jjV
jVjeqeqbyVSolving
VjVjeqFrom
jVjVj
jj
jjjj
jjAsjVjVjV
eqFrom
eqjVV
jVeqj
jVVV
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Vs= - j10, ZL=jL=j(0.5×500)=j250
Use mesh analysis,
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)135500cos(7)(1357250)135028.0(
V)45500cos(7)(457
90250)135028.0(A)135500cos(028.0
135028.0
4590028.04533.353
9010250250
100)250(250)10(
0
ttvRIV
ttv
ZIVti
Ij
jI
jIIjVVVs
R
R
L
LL
ZR
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AC Power Calculations
cosrmsrmsIVP
cosPF iv
sinrmsrmsIVQ
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rmsrmspower apparent IV
2rmsrms22 IVQP
RIP 2rms
XIQ 2rms
RVP R
2rms
XVQ X
2rms
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THÉVENIN EQUIVALENT CIRCUITS
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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.
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The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
scsc
oc
IV
IV t
tZ
scII n
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Maximum Power Transfer
•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.
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UNIT III–NETWORK THEOREMS (BOTH AC AND DC CIRCUITS) 9– Superposition theorem– The venin’s theorem- Norton’s theorem-Reciprocity theorem- Maximum power transfer theorem.
161
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9.1 – Introduction
This chapter introduces important fundamental theorems of network analysis. They are theSuperposition theoremThévenin’s theoremNorton’s theoremMaximum power transfer theoremSubstitution TheoremMillman’s theoremReciprocity theorem
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9.2 – Superposition Theorem
Used to find the solution to networks with two or more sources that are not in series or parallel.
The current through, or voltage across, an element in a network is equal to the algebraic sum of the currents or voltages produced independently by each source.
Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources.
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Superposition Theorem
The total power delivered to a resistive element must be determined using the total current through or the total voltage across the element and cannot be determined by a simple sum of the power levels established by each source.
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9.3 – Thévenin’s Theorem
Any two-terminal dc network can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.
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Thévenin’s Theorem
Thévenin’s theorem can be used to:Analyze networks with sources that are not in series or
parallel.Reduce the number of components required to
establish the same characteristics at the output terminals.
Investigate the effect of changing a particular component on the behavior of a network without having to analyze the entire network after each change.
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Thévenin’s Theorem
Procedure to determine the proper values of RTh and ETh
Preliminary1. Remove that portion of the network across which the
Thévenin equation circuit is to be found. In the figure below, this requires that the load resistor RL be temporarily removed from the network.
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Thévenin’s Theorem
2. Mark the terminals of the remaining two-terminal network. (The importance of this step will become obvious as we progress through some complex networks.)
RTh:3. Calculate RTh by first setting all sources to zero (voltage
sources are replaced by short circuits, and current sources by open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.)
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Thévenin’s Theorem
ETh:4. Calculate ETh by first returning all sources to their original
position and finding the open-circuit voltage between the marked terminals. (This step is invariably the one that will lead to the most confusion and errors. In all cases, keep in mind that it is the open-circuit potential between the two terminals marked in step 2.)
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Thévenin’s Theorem
Conclusion:5. Draw the Thévenin equivalent
circuit with the portion of thecircuit previously removedreplaced between theterminals of the equivalentcircuit. This step is indicatedby the placement of theresistor RL between theterminals of the Théveninequivalent circuit.
Insert Figure 9.26(b)Insert Figure 9.26(b)
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Experimental Procedures Two
popular experimental procedures for determining the parameters of the Thévenin equivalent network: Direct Measurement of ETh and RTh
For any physical network, the value of ETh can be determined experimentally by measuring the open-circuit voltage across the load terminals.
The value of RTh can then be determined by completing the network with a variable resistance RL.
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Thévenin’s Theorem Measuring VOC and ISC
The Thévenin voltage is again determined by measuring the open-circuit voltage across the terminals of interest; that is, ETh = VOC. To determine RTh, a short-circuit condition is established across the terminals of interest and the current through the short circuit (Isc) is measured with an ammeter.
Using Ohm’s law:
RTh = Voc / Isc
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Vth
173
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Rth
174
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Norton’s Theorem Norton’s theorem states the following:
Any two-terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a current and a parallel resistor.
The steps leading to the proper values of IN and RN.
Preliminary steps:1. Remove that portion of the network across which the
Norton equivalent circuit is found.2. Mark the terminals of the remaining two-terminal
network.
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Norton’s Theorem Finding RN:
3. Calculate RN by first setting all sources to zero (voltage sources are replaced with short circuits, and current sources with open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.) Since RN = RTh the procedure and value obtained using the approach described for Thévenin’s theorem will determine the proper value of RN.
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Norton’s Theorem Finding IN :
4. Calculate IN by first returning all the sources to their original position and then finding the short-circuit current between the marked terminals. It is the same current that would be measured by an ammeter placed between the marked terminals.
Conclusion:5. Draw the Norton equivalent circuit with the portion of
the circuit previously removed replaced between the terminals of the equivalent circuit.
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178
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9.5 – M Maximum Power Transfer Theorem
The maximum power transfer theorem states the following:
A load will receive maximum power from a network when its total resistive value is exactly equal to the Thévenin resistance of the network applied to the load. That is,
RL = RTh
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Maximum Power Transfer Theorem
For loads connected directly to a dc voltage supply, maximum power will be delivered to the load when the load resistance is equal to the internal resistance of the source; that is, when:RL = Rint
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Reciprocity Theorem
The reciprocity theorem is applicable only to single-source networks and states the following: The current I in any branch of a network, due to a
single voltage source E anywhere in the network, will equal the current through the branch in which the source was originally located if the source is placed in the branch in which the current I was originally measured. The location of the voltage source and the resulting current
may be interchanged without a change in current
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UNIT IV - TRANSIENT RESPONSE FOR DC CIRCUITS–– Transient response of RL, RC and RLC –– Laplace transform for DC input with sinusoidal input.
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Solution to First Order Differential Equation
)()()( tfKtxdt
tdxs
Consider the general Equation
Let the initial condition be x(t = 0) = x( 0 ), then we solve the differential equation:
)()()( tfKtxdt
tdxs
The complete solution consists of two parts: • the homogeneous solution (natural solution)• the particular solution (forced solution)
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The Natural Response
/)(,)()(
)()(,)()(0)()(
tN
N
N
N
NNNN
N
etxdttxtdx
dttxtdxtx
dttdxortx
dttdx
Consider the general Equation
Setting the excitation f (t) equal to zero,
)()()( tfKtxdt
tdxs
It is called the natural response.
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The Forced Response
0)(
)()(
tforFKtx
FKtxdt
tdx
SF
SFF
Consider the general Equation
Setting the excitation f (t) equal to F, a constant for t 0
)()()( tfKtxdt
tdxs
It is called the forced response.
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The Complete Response
)(
)()(
/
/
xe
FKe
txtxx
tS
tFN
Consider the general Equation
The complete response is: • the natural response + • the forced response
)()()( tfKtxdt
tdxs
Solve for ,
)()0()()0()0(
0
xxxxtx
tfor
The Complete solution:
)()]()0([)( / xexxtx t
/)]()0([ texx called transient response
)(x called steady state response
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TRANSIENT RESPONSE
Figure 5.1
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Figure
5.2, 5.3
Circuit with switched DC excitation
A general model of the transient
analysis problem
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For a circuit containing energy storage element
Figure 5.5, 5.6
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Figure
5.9, 5.10
(a) Circuit at t = 0
(b) Same circuit a long time after the switch is closed
The capacitor acts as open circuit for the steady state condition(a long time after the switch is closed).
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(a) Circuit for t = 0
(b) Same circuit a long time before the switch is opened
The inductor acts as short circuit for the steady state condition(a long time after the switch is closed).
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Reason for transient response
The voltage across a capacitor cannot be changed instantaneously.
)0()0( CC VV• The current across an inductor cannot be
changed instantaneously.
)0()0( LL II
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Figure
5.12, 5.13
5-6
Example
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Transients Analysis
1. Solve first-order RC or RL circuits.
2. Understand the concepts of transient response and steady-state response.
3. Relate the transient response of first-order circuits to the time constant.
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Transients
The solution of the differential equationrepresents are response of the circuit. Itis called natural response.
The response must eventually die out,and therefore referred to as transientresponse.(source free response)
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Discharge of a Capacitance through a Resistance
ic iR
0,0 RC iii
0R
tvdt
tdvC CC
Solving the above equation with the initial conditionVc(0) = Vi
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Discharge of a Capacitance through a Resistance
0R
tvdt
tdvC CC
0 tvdt
tdvRC CC
stC Ketv
0 stst KeRCKse
RCs 1
RCtC Ketv
KKe
VvRC
iC
/0
)0(
RCtiC eVtv
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RCtiC eVtv
Exponential decay waveformRC is called the time constant.At time constant, the voltage is 36.8%of the initial voltage.
)1( RCtiC eVtv
Exponential rising waveformRC is called the time constant.At time constant, the voltage is 63.2% of the initial voltage.
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RC CIRCUIT
for t = 0-, i(t) = 0u(t) is voltage-step function Vu(t)
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RC CIRCUIT
Vu(t) 0)(,
,)(
tforVtuvVvdt
dvRC
dtdvCi
Rvtvui
ii
CC
CC
CR
CR
Solving the differential equation
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Complete Response
Complete response = natural response + forced response Natural response (source free response)
is due to the initial condition Forced response is the due to the
external excitation.
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Figure 5.17,
5.18
5-8
a). Complete, transient and steady state response
b). Complete, natural, and forced responses of the circuit
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Circuit Analysis for RC Circuit
sRC
CC
RsR
CR
vRC
vRCdt
dvdt
dvCiR
vvi
ii
11
,
Apply KCL
vs is the source applied.
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Solution to First Order Differential Equation
)()()( tfKtxdt
tdxs
Consider the general Equation
Let the initial condition be x(t = 0) = x( 0 ), then we solve the differential equation:
)()()( tfKtxdt
tdxs
The complete solution consist of two parts: • the homogeneous solution (natural solution)• the particular solution (forced solution)
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The Natural Response
/)(
)()(0)()(
tN
NNN
N
etx
txdt
tdxortxdt
tdx
Consider the general Equation
Setting the excitation f (t) equal to zero,
)()()( tfKtxdt
tdxs
It is called the natural response.
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The Forced Response
0)(
)()(
tforFKtx
FKtxdt
tdx
SF
SFF
Consider the general Equation
Setting the excitation f (t) equal to F, a constant for t 0
)()()( tfKtxdt
tdxs
It is called the forced response.
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The Complete Response
)(
)()(
/
/
xe
FKe
txtxx
tS
tFN
Consider the general Equation
The complete response is: • the natural response + • the forced response
)()()( tfKtxdt
tdxs
Solve for ,
)()0()()0()0(
0
xxxxtx
tfor
The Complete solution:
)()]()0([)( / xexxtx t
/)]()0([ texx called transient response
)(x called steady state response
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Example
Initial condition Vc(0) = 0V
sCC
CC
CsR
CR
vvdt
dvRC
dtdvCi
Rvvi
ii
,
10010
1001001.010
3
65
CC
CC
vdt
dv
vdt
dv
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Example
Initial condition Vc(0) = 0V
and
3
3
10
10
100100
1001000,0)0(
100
t
c
c
t
c
ev
AAvAs
Aev)()()( tfKtx
dttdx
s
)(
)()(
/
/
xe
FKe
txtxx
tS
tFN
10010 3 C
C vdt
dv
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Energy stored in capacitor
22 )()(21
o
tt
tt
tt
tvtvC
vdvCdtdtdvCvpdt
dtdvCvvip
ooo
If the zero-energy reference is selected at to, implying that thecapacitor voltage is also zero at that instant, then
221)( Cvtwc
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Power dissipation in the resistor is:
pR = V2/R = (Vo2 /R) e -2 t /RC
2
0/22
0/22
0
21
|)2
1(
o
RCto
RCto
RR
CV
eRC
RV
RdteV
dtpW
RC CIRCUIT
Total energy turned into heat in the resistor
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RL CIRCUITS
Initial condition i(t = 0) = Io
equationaldifferentitheSolving
idtdi
RL
dtdiLRivv LR
0
0
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RL CIRCUITS
LR-
VR
+
+VL
-
i(t)
Initial condition i(t = 0) = Io
LRto
o
to
iI
t
o
ti
I
eIti
tLRIi
tLRi
dtLR
ididt
LR
idi
iLR
dtdi
o
o
/
)(
)(
lnln
||ln
,
0
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RL CIRCUIT
LR-
VR
+
+VL
-
i(t)Power dissipation in the resistor is:
pR = i2R = Io2e-2Rt/LRTotal energy turned into heat in the resistor
2
0/22
0
/22
0
21
|)2
(
o
LRto
LRtoRR
LI
eR
LRI
dteRIdtpW
It is expected as the energy stored in the inductor is 2
21
oLI
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RL CIRCUIT Vu(t)
RL
+VL
-
i(t)
+_ Vu(t)
ktRiVRL
sidesbothgIntegratin
dtRiV
Ldi
VdtdiLRi
)ln(
, 0,
]ln)[ln(
ln,0)0(
/
/
tforeRV
RVi
oreV
RiV
tVRiVRL
VRLkthusi
LRt
LRt
where L/R is the time constant
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DC STEADY STATE
The steps in determining the forced response for RL or RC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
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UNIT V RESONANCE AND COUPLED CIRCUITS 9– Series and parallel resonance– their frequency response– Quality factor and Bandwidth– Self and mutual inductance– Coefficient of coupling– Tuned circuits– Single tuned circuits.
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218
Any passive electric circuit will resonate if it has an inductorand capacitor.
Resonance is characterized by the input voltage and currentbeing in phase. The driving point impedance (or admittance) is completely real when this condition exists.
In this presentation we will consider (a) series resonance, and(b) parallel resonance.
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219
Consider the series RLC circuit shown below.
R L
C+_ IVV = VM 0
The input impedance is given by:
1( )Z R j wLwC
The magnitude of the circuit current is;
2 2
| |1( )
mVI IR wL
wC
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220
Resonance occurs when,
1wLwC
At resonance we designate w as wo and write;
1ow
LC
This is an important equation to remember. It applies to both seriesAnd parallel resonant circuits.
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221
Series Resonance
The magnitude of the current response for the series resonance circuitis as shown below.
mVR
2mVR
w
|I|
wow1 w2
Bandwidth:
BW = wBW = w2 – w1
Half power point
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222
Series Resonance
The peak power delivered to the circuit is;
2mVPR
The so-called half-power is given when 2mVIR
.
We find the frequencies, w1 and w2, at which this half-poweroccurs by using;
2 212 ( )R R wLwC
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223
After some insightful algebra one will find two frequencies at whichthe previous equation is satisfied, they are:
2
11
2 2R RwL L LC
and
2
21
2 2R RwL L LC
The two half-power frequencies are related to the resonant frequency by
1 2ow w w
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224
The bandwidth of the series resonant circuit is given by;
2 1bRBW w w wL
We define the Q (quality factor) of the circuit as;
1 1o
o
w L LQR w RC R C
Using Q, we can write the bandwidth as;
owBWQ
These are all important relationships.
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225
An Observation:
If Q > 10, one can safely use the approximation;
1 22 2o oBW BWw w and w w
These are useful approximations.
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226
By using Q = woL/R in the equations for w1and w2 we have;
2
21 1 1
2 2ow wQ Q
2
11 1 1
2 2ow wQ Q
and
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227
An Example Illustrating Resonance:
The 3 transfer functions considered are:
Case 1:
Case 2:
Case 3:
2 2 400ks
s s
2 5 400ks
s s
2 10 400ks
s s
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228
An Example Illustrating Resonance:
The poles for the three cases are given below.
Case 1:
Case 2:
Case 3:
2 2 400 ( 1 19.97)( 1 19.97)s s s j s j
2 5 400 ( 2.5 19.84)( 2.5 19.84)s s s j s j
2 10 400 ( 5 19.36)( 5 19.36)s s s j s j
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229
Comments:
Observe the denominator of the CE equation.
2 1Rs sL LC
Compare to actual characteristic equation for Case 1:
2 2 400s s 2 400ow 20w
2RBWL
10owQBW
rad/sec
rad/sec
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230
Poles and Zeros In the s-plane:
s-plane
jw axis
axis0
0
20
-20xx
x x x
x
( 3) (2) (1)
( 3) (2) (1)
-5 -2.5 -1Note the location of the polesfor the three cases. Also notethere is a zero at the origin.
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231
Comments:
The frequency response starts at the origin in the s-plane.At the origin the transfer function is zero because there is azero at the origin.As you get closer and closer to the complex pole, which has a j parts in the neighborhood of 20, the response startsto increase.The response continues to increase until we reach w = 20.From there on the response decreases.
We should be able to reason through why the responsehas the above characteristics, using a graphical approach.
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232
0 1 0 2 0 3 0 4 0 5 0 6 00
0 .1
0 .2
0 .3
0 .4
0 .5
0 .6
0 .7
0 .8
0 .9
1
w (ra d /s e c )
Am
plitu
de
Q = 1 0 , 4 , 2
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233
Next Case: Normalize all responses to 1 at wo
0 1 0 2 0 3 0 4 0 5 0 6 00
0 .1
0 .2
0 .3
0 .4
0 .5
0 .6
0 .7
0 .8
0 .9
1
w ( ra d /s e c )
Am
plitu
de
Q = 1 0 , 4 , 2
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234
Three dB Calculations:
Now we use the analytical expressions to calculate w1 and w2.We will then compare these values to what we find from the Matlab simulation.
Using the following equations with Q = 2,
1
21
21,
2
21 Qw
Qwww oo
we find, w1 = 15.62 rad/sec
w2 = 21.62 rad/sec
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235
Parallel ResonanceConsider the circuits shown below:
I R L C
V
I
R L
CV
jwLjwC
RVI 11
jwCjwLRIV 1
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236
jwLjwC
RVI 11
jwCjwLRIV 1
We notice the above equations are the same provided:
VI
RR 1
CL
If we make the inner-change,then one equation becomes the same as the other.
For such case, we say the one circuit is the dual of the other.
Duality
If we make the inner-change,then one equation becomes the same as the other.
For such case, we say the one circuit is the dual of the other.
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237
What this means is that for all the equations we havederived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions:
RbereplacedR 1
LbyreplacedCCbyreplacedL
What this means is that for all the equations we havederived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions:
What this means is that for all the equations we havederived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions:
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238
Parallel Resonance Series Resonance
RLwQ OLC
wO
1 LC
wO
1
RCwQ o
LRwwwBW BW )( 12 RC
wBW BW
1w
21, ww
LCLR
LRww 1
22,
2
21
LCRCRCww 1
21
21,
2
21
21, ww
1
21
21,
2
21 QQwww o
1
21
21,
2
21 QQwww o
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239
Example 1: Determine the resonant frequency for the circuit below.
jwRCLCwjwLLRCw
jwCjwLR
jwCRjwL
Z NI
)1()(
1
)1(
2
2
At resonance, the phase angle of Z must be equal to zero.
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240
jwRCLCwjwLLRCw
)1()(
2
2Analysis
For zero phase;
LCwwRC
LCRwwL
22 1()(
This gives;
12222 CRwLCw
or
)(1
22CRLCwo
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241
Parallel ResonanceExample 2:
A parallel RLC resonant circuit has a resonant frequency admittance of2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is10,000 rad/sec. Calculate the values of R, L, and C. Find the half-powerfrequencies and the bandwidth.
First, R = 1/G = 1/(0.02) = 50 ohms. Second, from
RLwQ O , we solve for L, knowing Q, R, and wo to
find L = 0.25 H.
Third, we can use FxRw
QCO
10050000,10
50
A parallel RLC resonant circuit has a resonant frequency admittance of2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is10,000 rad/sec. Calculate the values of R, L, and C. Find the half-powerfrequencies and the bandwidth.
A series RLC resonant circuit has a resonant frequency admittance of2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is10,000 rad/sec. Calculate the values of R, L, and C. Find the half-powerfrequencies and the bandwidth.
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242
Parallel ResonanceExample 2: (continued)
Fourth: We can use sec/20050101 4
radxQww o
BW
and
Fifth: Use the approximations;
w1 = wo - 0.5wBW = 10,000 – 100 = 9,900 rad/sec
w2 = wo - 0.5wBW = 10,000 + 100 = 10,100 rad/sec
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243
Extension of Series Resonance
Peak Voltages and Resonance:
VS R
L
C+
_ I
+ + +_ _
_
VR
VL
VC
We know the following:
When w = wo = 1
LC, VS and I are in phase, the driving point impedance
is purely real and equal to R.
A plot of |I| shows that it is maximum at w = wo. We know the standardequations for series resonance applies: Q, wBW, etc.
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244
Reflection:
A question that arises is what is the nature of VR, VL, and VC? A little reflection shows that VR is a peak value at wo. But we are not sureabout the other two voltages. We know that at resonance they are equaland they have a magnitude of QxVS.
max 2
112ow wQ
The above being true, we might ask, what is the frequency at which the voltage across the inductor is a maximum?
We answer this question by simulation
Irwin shows that the frequency at which the voltage across the capacitoris a maximum is given by;Irwin shows that the frequency at which the voltage across the capacitoris a maximum is given by;Irwin shows that the frequency at which the voltage across the capacitoris a maximum is given by;
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245
Series RLC Transfer Functions:
The following transfer functions apply to the series RLC circuit.
2
1( )
1( )C
S
V s LCRV s s sL LC
2
2
( )1( )
L
S
V s sRV s s sL LC
2
( )1( )
R
S
R sV s LRV s s sL LC
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246
Parameter Selection:
We select values of R, L. and C for this first case so that Q = 2 and wo = 2000 rad/sec. Appropriate values are; R = 50 ohms, L = .05 H, C = 5F. The transfer functions become as follows:
6
2 6
4 101000 4 10
C
S
V xV s s x
2
2 61000 4 10L
S
V sV s s x
2 61000
1000 4 10R
S
V sV s s x
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247
0 5 0 0 1 0 0 0 1 5 0 0 2 0 0 0 2 5 0 0 3 0 0 0 3 5 0 0 4 0 0 00
0 . 5
1
1 . 5
2
2 . 5
w ( r a d / s e c )
Am
plitu
de
R s e s p o n s e f o r R L C s e r i e s c i r c u i t , Q = 2
V C V L
V R
Simulation Results
Q = 2
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248
Analysis of the problem:
VS R=50
L=5 mH C=5 F
+
_ I
+ ++
_ _
_
VRVL
VC
Given the previous circuit. Find Q, w0, wmax, |Vc| at wo, and |Vc| at wmax
Solution: sec/20001051050
1162
radxxxLC
wO
250
105102 23
xxx
RLwQ O
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249
Problem Solution:
oOMAX wQ
ww 9354.02
112
)(212|||| peakvoltsxVQwatV SOR
)066.2968.02
411
||||
2
peakvolts
Q
VQxwatV SMAXC
Now check the computer printout.
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250
Exnsion of Series Resonance
Problem Solution (Simulation):
1.0e+003 *
1.8600000 0.0020651411.8620000 0.0020652921.8640000 0.0020654111.8660000 0.0020655011.8680000 0.0020655601.8700000 0.0020655881.8720000 0.0020655851.8740000 0.0020655521.8760000 0.0020654871.8780000 0.0020653921.8800000 0.0020652651.8820000 0.0020651071.8840000 0.002064917
Maximum
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Simulation Results:
0 5 0 0 1 0 0 0 1 5 0 0 2 0 0 0 2 5 0 0 3 0 0 0 3 5 0 0 4 0 0 00
2
4
6
8
1 0
1 2
w ( ra d /s e c )
Am
plitu
de
R s e s p o n s e fo r R L C s e r i e s c i r c u i t , Q = 1 0
V C V L
V R
Q=10
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Observations From The Study:
The voltage across the capacitor and inductor for a series RLC circuitis not at peak values at resonance for small Q (Q <3).Even for Q<3, the voltages across the capacitor and inductor areequal at resonance and their values will be QxVS.For Q>10, the voltages across the capacitors are for all practical purposes at their peak values and will be QxVS.
Regardless of the value of Q, the voltage across the resistor reaches its peak value at w = wo.
For high Q, the equations discussed for series RLC resonancecan be applied to any voltage in the RLC circuit. For Q<3, thisis not true.
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Extension of Resonant CircuitsGiven the following circuit:
I
+
_
+
_
VC
R
L
We want to find the frequency, wr, at which the transfer functionfor V/I will resonate.
The transfer function will exhibit resonance when the phase anglebetween V and I are zero.
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The desired transfer functions is;
(1/ )( )1/
V sC R sLI R sL sC
This equation can be simplified to;
2 1V R sLI LCs RCs
With s jw
2(1 )V R jwLI w LC jwR
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Resonant Condition:
For the previous transfer function to be at a resonant point,the phase angle of the numerator must be equal to the phase angleof the denominator.
num dem
1tannumwLR
or,
12tan
(1 )denwRCw LC
, .
Therefore;
2(1 )wL wRCR w LC
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Resonant Condition Analysis:
Canceling the w’s in the numerator and cross multiplying gives,
2 2 2 2 2(1 )L w LC R C or w L C L R C
This gives,2
2
1r
RwLC L
Notice that if the ratio of R/L is small compared to 1/LC, we have
1r ow w
LC
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Extension of Resonant CircuitsResonant Condition Analysis:
What is the significance of wr and wo in the previous two equations?Clearly wr is a lower frequency of the two. To answer this question, considerthe following example.
Given the following circuit with the indicated parameters. Write a Matlab program that will determine the frequency response of the transfer function of the voltage to the current as indicated.
I
+
_
+
_
VC
R
L
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0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0 6 0 0 0 7 0 0 0 8 0 0 00
2
4
6
8
1 0
1 2
1 4
1 6
1 8
w ( ra d /s e c )
Am
plitu
de
R s e s p o n s e fo r R e s i s ta n c e i n s e r i e s w i th L th e n P a r a lle l w i th C
R = 3 o h m s
R = 1 o h m
2646 rad/sec