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.

1 Methods of Analysis
 Resistance

2 Methods of Analysis
 Ohm’s Law

3 Methods of Analysis
 Resistors & Passive Sign Convention

4 Methods of Analysis
 Example: Ohm’s Law

5 Methods of Analysis
 Short Circuit as Zero Resistance

6 Methods of Analysis
 Short Circuit as Voltage Source (0V)

7 Methods of Analysis
 Open Circuit

8 Methods of Analysis
 Open Circuit as Current Source (0 A)

9 Methods of Analysis
 Conductance

10 Methods of Analysis
 Circuit Building Blocks

11 Methods of Analysis
 Branches

12 Methods of Analysis
 Nodes

13 Methods of Analysis
 Loops

14 Methods of Analysis
 Overview of Kirchhoff’s Laws

15 Methods of Analysis
 Kirchhoff’s Current Law

16 Methods of Analysis
 Kirchhoff’s Current Law for
Boundaries

17 Methods of Analysis
 KCL - Example

18 Methods of Analysis
 Ideal Current Sources: Series

19 Methods of Analysis
 Kirchhoff’s Voltage Law - KVL

20 Methods of Analysis
 KVL - Example

21 Methods of Analysis
 Example – Applying the Basic Laws

22 Methods of Analysis
 Example – Applying the Basic Laws

23 Methods of Analysis
 Example – Applying the Basic Laws

24 Methods of Analysis
 Resistors in Series

25 Methods of Analysis
 Resistors in Parallel

26 Methods of Analysis
 Resistors in Parallel

27 Methods of Analysis
 Voltage Divider

28 Methods of Analysis
 Current Divider

29 Methods of Analysis
 Resistor Network

30 Methods of Analysis
 Resistor Network - Comments

31 Methods of Analysis
 Delta  Wye Transformations

32 Methods of Analysis
 Delta  Wye Transformations

33 Methods of Analysis
 Example –
Delta  Wye Transformations

34 Methods of Analysis
35 Methods of Analysis
 Methods of Analysis

• 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

36 Methods of Analysis
 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.
37 Methods of Analysis
 Figure 3.1

Common symbols for indicating a reference node,

(a) common ground, (b) ground, (c) chassis.

38 Methods of Analysis
 Figure 3.2

Typical circuit for nodal analysis

39 Methods of Analysis
 I 1  I 2  i1  i 2
I 2  i 2  i3

vhigher  vlower
i
R
v1  0
i1  or i1  G1v1
R1
v1  v2
i2  or i2  G2 (v1  v2 )
R2
v2  0
i3  or i3  G3v2
R3
40 Methods of Analysis
 v1 v1  v2
 I1  I2  
R1 R2
v1  v2 v2
I2  
R2 R3
 I1  I 2  G1v1  G2 (v1  v2 )
I 2  G2 (v1  v2 )  G3v2

 G1  G2  G2   v1   I1  I 2 
 
  G2 G2  G3  v2   I 2 

41 Methods of Analysis
 Example 3.1

 Calculus the node voltage in the circuit

shown in Fig. 3.3(a)

42 Methods of Analysis
 Example 3.1

 At node 1

i1  i2  i3
v1  v2 v1  0
5 
4 2

43 Methods of Analysis
 Example 3.1

 At node 2
i2  i4  i1  i5
v2  v1 v2  0
5 
4 6

44 Methods of Analysis
 Example 3.1

 In matrix form:

1 1 1 

2 4    v  5
4 1
 
 1   
1 1 v2  5
   
 4 6 4
45 Methods of Analysis
 Practice Problem 3.1

Fig 3.4

46 Methods of Analysis
 Example 3.2

 Determine the voltage at the nodes in Fig.

3.5(a)

47 Methods of Analysis
 Example 3.2

 At node 1,

3  i1  ix
v1  v3 v1  v2
3 
4 2

48 Methods of Analysis
 Example 3.2

 At node
ix  i2 2i3
v1  v2 v2  v3 v2  0
  
2 8 4

49 Methods of Analysis
 Example 3.2

 At node 3
i1  i2  2ix
v1  v3 v2  v3 2(v1  v2 )
  
4 8 2

50 Methods of Analysis
 Example 3.2

 In matrix form:

 3 1 1
 4   
2 4  v1  3
 1 7 
1    
    v2    0 
 2 8 8
 3 9 3  v3  0

 4 8 8 
51 Methods of Analysis
 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.
52 Methods of Analysis
 3.3 Nodal Analysis with Voltage
Sources

i1  i4  i2  i3 
v1  v2 v1  v3 v2  0 v3  0
  
2 4 8 6
 v2  v3  5

Fig. 3.7 A circuit with a supernode.

53 Methods of Analysis
  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
54 voltage source. Methods of Analysis
 Example 3.3
 For the circuit shown in Fig. 3.9, find the node
voltages.
2  7  i1  i 2  0
v v
27 1  2  0
2 4
v1  v2  2

i1 i2

55 Methods of Analysis
 Example 3.4
Find the node voltages in the circuit of Fig. 3.12.

56 Methods of Analysis
 Example 3.4

 At suopernode 1-2,
v3  v2 v1  v4 v1
 10  
6 3 2
v1  v2  20

57 Methods of Analysis
 Example 3.4

 At supernode 3-4,
v1  v4 v3  v2 v4 v3
  
3 6 1 4
v3  v4  3(v1  v4 )

58 Methods of Analysis
 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

59 Methods of Analysis
 Fig. 3.15

(a) A Planar circuit with crossing branches,

(b) The same circuit redrawn with no crossing branches.
60 Methods of Analysis
 Fig. 3.16

A nonplanar circuit.

61 Methods of Analysis
  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.

62 Methods of Analysis
 Fig. 3.17

A circuit with two meshes.

63 Methods of Analysis
  V1  R1i1  R3 (i1  i2 )  0
 Apply KVL to each mesh. For mesh 1,
( R1  R3 )i1  R3i2  V1

 For mesh i2  V2  R3 (i2  i1 )  0

R22,
 R3i1  ( R2  R3 )i2  V2

64 Methods of Analysis
  Solve for the mesh currents.

 R1  R3  R3   i1   V1 

  R3 R2  R3  i2   V2 
 Use i for a mesh current and I for a branch
current. It’s evident from Fig. 3.17 that
I1  i1 , I 2  i2 , I 3  i1  i2
65 Methods of Analysis
 Example 3.5

 Find the branch current I1, I2, and I3 using

mesh analysis.

66 Methods of Analysis
 Example 3.5
 15  5i1  10(i1  i2 )  10  0
 For mesh 1,
3i1  2i2  1

6i2  4i2  10(i2  i1 )  10  0

 For mesh 2,
i1  2i2  1
I1  i1 , I 2  i2 , I 3  i1  i2
 We can find i1 and i2 by substitution method
or Cramer’s rule. Then,
67 Methods of Analysis
 Example 3.6

 Use mesh analysis to find the current I0 in the

circuit of Fig. 3.20.

68 Methods of Analysis
 Example 3.6

 Apply KVL to each mesh. For mesh 1,

 24  10(i1  i2 )  12(i1  i3 )  0
11i1  5i2  6i3  12

 For mesh 2,
24i2  4(i2  i3 )  10(i2  i1 )  0
 5i1  19i2  2i3  0

69 Methods of Analysis
 Example 3.6
 For mesh 3, 4I 0  12 ( i 3  i 1 )  4 ( i 3  i 2 )  0
At node A, I 0  I 1  i2 ,
4 ( i 1  i 2 )  12 ( i 3  i 1 )  4 ( i 3  i 2 )  0
 i1  i 2  2 i 3  0

 In matrix from Eqs. (3.6.1) to (3.6.3) become

 11 5  6   i1  12 
 5 19  2   i2    0 
1 1 2   i3   0 

we can calculus i1, i2 and i3 by Cramer’s rule,
and find I0.
70 Methods of Analysis
 3.5 Mesh Analysis with Current
Sources

Fig. 3.22 A circuit with a current source.

71 Methods of Analysis
  Case 1 i1   2 A
– 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.

72 Methods of Analysis
 Fig. 3.23

 a supermesh results when two meshes have

a (dependent , independent) current source
in common.

73 Methods of Analysis
 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.

74 Methods of Analysis
 Example 3.7

 For the circuit in Fig. 3.24, find i1 to i4 using

mesh analysis.

75 Methods of Analysis
  If a supermesh consists of two meshes, two 6 i1  14 i 2  20
equations are needed; one is obtained using
KVL and Ohm’s law to the supermesh and the i1  i 2   6
other is obtained by relation regulated due to
the current source.

76 Methods of Analysis
  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.

77 Methods of Analysis
 2 i1  4 i 3  8 ( i 3  i 4 )  6 i 2  0
i1  i 2   5
i2  i3   i4
8 ( i 3  i 4 )  2 i 4  10  0

78 Methods of Analysis
 3.6 Nodal and Mesh Analysis by
Inspection

The analysis equations can be

obtained by direct inspection
(a) For circuits with only resistors and independent
current sources
(b) For planar circuits with only resistors and
independent voltage sources

79 Methods of Analysis
  In the Fig. 3.26 (a), the circuit has two
nonreference nodes and the node equations
I1  I 2  G1v1  G2 (v1  v2 ) (3.7)
I 2  G2 (v1  v2 )  G3v2 (3.8)
 MATRIX
G1  G2  G2   v1   I1  I 2 

  G2 G2  G3  v2   I 2 
80 Methods of Analysis
  In general, the node voltage equations in
terms of the conductance is
or simply  G 11 G 12  G 1 N   v1   i1 
G G 22  G 2 N  v  i 
Gv = i  21   2    2 
           
G  v  i 
 N 1 G N 2  G NN   N   N 
where G : the conductance matrix,
v : the output vector, i : the input vector
81 Methods of Analysis
  The circuit has two nonreference nodes and
the node equations were derived as

 R1  R3  R3   i1   v1 

  R3 R2  R3  i2   v2 

82 Methods of Analysis
  In general, if the circuit has N meshes, the
mesh-current equations as the resistances
term is  R 11 R 12  R 1 N   i1   v1 
R R 22  R 2 N   i 2  v 
or simply  21
    2 
           
R  R NN  i  v 
 N1 RN  N   N 
Rv = i 2

where R : the resistance matrix,

i : the output vector, v : the input vector
83 Methods of Analysis
 Example 3.8

 Write the node voltage matrix equations in

Fig.3.27.

84 Methods of Analysis
 Example 3.8

 The circuit has 4 nonreference nodes, so

1 1 1 1 1
G 11    0 .3 , G 22     1 . 325
5 10 5 8 1
1 1 1 1 1 1
G 33     0 .5 , G 44     1 . 625
8 8 4 8 2 1

 The off-diagonal terms are

1
G 12     0 . 2 , G 13  G 14  0
5
1 1
G 21   0 . 2 , G 23     0 . 125 , G 24     1
8 1
G 31  0 , G 32   0 . 125 , G 34   0 . 125
G 41  0 , G 42   1, G 43   0 . 125
85 Methods of Analysis
 Example 3.8
i1  3 , i 2   1  2   3 , i 3  0 , i 4  2  4  6
 The input current vector i in amperes

 The node-voltage equations are

 0 .3  0.2 0 0   v1   3
 0.2 1 .325  0.125  1  v   3
   2    
 0  0.125 0 .5  0.125  v3   0
 0  1  0.125 1 .625  v   6 
   4  

86 Methods of Analysis
 Example 3.9

 Write the mesh current equations in Fig.3.27.

87 Methods of Analysis
 Example 3.9

 The input voltage vector v in volts

v1  4 , v 2  10  4  6 ,
v 3   12  6   6 , v 4  0, v5  6

 The mesh-current equations are

 9  2  2 0 0   i1 
 4 
 2 10  4 1  1   i2 
 6 
  
 2  4 9 0 0   i3    6
 
 0 1 0 8  3   i4 
 0 
    6
 0 1 0  3 4   i 5   

88 Methods of Analysis
 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.
89 Methods of Analysis
 BJT Circuit Models

(a) dc equivalent model.

(b) An npn transistor,
90 Methods of Analysis
 Example 3.13

 For the BJT circuit in Fig.3.43, =150 and

VBE = 0.7 V. Find v0.

91 Methods of Analysis
 Example 3.13

 Use mesh analysis or nodal analysis

92 Methods of Analysis
 Example 3.13

93 Methods of Analysis
 3.10 Summary

1. Nodal analysis: the application of KCL at

the nonreference nodes
– A circuit has fewer node equations
2. A supernode: two nonreference nodes
3. Mesh analysis: the application of KVL
– A circuit has fewer mesh equations
4. A supermesh: two meshes

94 Methods of Analysis
 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.

95 Methods of Analysis
Sinusoidal Steady State
Response
Sinusoidal Steady State
Response

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.

The sinusoidal function v(t) =
VM sin  t is plotted (a) versus
 t and (b) versus t.
The sine wave VM sin ( t + ) leads VM sin
 t by  radian
 1 2
Frequency f  Angular frequency 
T T

  2f
o
sin z  cos( z  90 )
o
sin t  cos(t  90 )
o
sin(t  90 )  cos t
 Representation of the two vectors v1
and 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.
 Euler’s identity

  t
  2 f
A cos  t  A cos 2 f t
In Euler expression,

A cos t = Real (A e j t )
A sin t = Im( A e j t )

Any sinusoidal function can

be expressed as in Euler
form.

4-6
 Applying Euler’s Identity

The complex forcing function Vm e j ( t + ) produces the complex

response Im e j (t + ).
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).
Re(Vm e j ( t + ) )  Re(Im e j (t + ))
Im(Vm e j ( t + ) )  Im(Im e j (t + ))
Phasor Definition

Time function : v1 t   V1 cosωt  θ1 

Phasor: V1  V11
j (t 1 )
V1  Re(e )
j (1 )
V1  Re(e ) by dropping t
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 = 9.8524.0o V.
 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.

Conversion of rectangular to
polar form


v1 t   20 cos(t  45 )

v2 (t )  10 cos(t  30 )


V1  20  45

V2  10  30
 Vs  V1  V2  
 20  45 10  30
 14.14 j14.14 8.660 j 5
 23.06 j19.14
 29.97  39.7
Vs  Ae j
 19.14
2 2
A  23.06  ( 19.14)  29.96,   tan1
 39.7
23.06

v s t   29.97 cost  39.7 


Phase relation ship
COMPLEX IMPEDANCE

VL  jL  I L

Z L  jL  L90

VL  Z L I L
(a) (b)
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/jC;
(c) an inductor L is represented by an
impedance jL.
(c)
 j t
V  Ve
dV d Ve j t j t
IC C  j  CVe
dt dt
V 1
I  j C V    Zc
I j C
Zc is defined as the impedance of a capacitor

The impedance of a capacitor is 1/jC.

As I  j C V, if v  V cos  t , then i  CV cos ( t  90 )

As I  j C V, if v  VM cos  t , then i   CVM cos ( t  90 )
I M   CVM
 I  Ie j t
dI d Ie j t j t
VL L  j  LIe
dt dt
V
V  j L I   j L  Z L
I
ZL is the impedance of an inductor. The impedance of a inductor is jL

As V  j C I, if i  I cos  t , then v   LI cos ( t  90 )

or i  I cos ( t  90 ), and v   LI cos  t.
As V  jCI, if i  IM cost, thenv   LIM cos(t  90 )
ori  IM cos(t  90 ), and v   LIM cost, VM   LIM
Complex Impedance in Phasor
Notation

VC  ZC I C
1 1 1 
ZC  j    90
jC C C
VL  Z L I L
Z L  jL  L90

VR  RI R


 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.
Ztotal  100  j(150  50)
 100  j100  141.445
V 10030  
I   0.70730  45
Ztotal 141.445
 0.707  15
VR  100  I  70.7  15 , vR (t )  70.7 cos(500t  15 )
VL  j150  I  15090  0.707  15  106.0590  15
 106.0575 , vL (t )  106.05cos(500t  75 )
VC   j50  I  50  90  0.707  15  35.35  90  15
 35.35  105 , vL (t )  35.35cos(500t  105 )
 1 1
Z RC  
1 / R  1 / Zc 1 / 100  1 /(  j100)
1 1 j j 0.01
As    2
 j0.01
 j100  j100 j  j
1 10 
Z RC    70 . 71  45
0.01  j 0.01 0.0141445
 50  j50
 Z RC
Vc  Vs ( voltage division)
Z L  Z RC
 
70. 71  45 70.71  45
 10  90  10  90
j100  50  j50 50  j50

70.71  45
 10  45  10   135

70.7145
vc (t )  10 cos (1000t  135 )   10 cos 1000t V
 Vs
I
Z L  Z RC
10  90 10  90
 
j100  50  j50 50  j50
10  90 
  0 .414  135
70.7145
i (t )  0.414 cos (1000t  135 )
 VC
IR 
R
10  135
  0.1  135
100
iR (t )  0.1 cos (1000t  135 )
VC 10  135 10  135 
IC     0 .1  45
Zc  j100 100  90
iR (t )  0.1 cos (1000t  45 ) A
Solve by nodal analysis

V1 V1  V2 
  2  90   j 2 eq(1)
10  j5
V2 V2  V1
  1.50  1.5 eq(2)
j10  j5
V1 V1  V2 V2 V2  V1
  2  90   j 2 eq(1)   1.50  1.5 eq(2)
10  j5 j10  j5
From eq (1)
1 1 j j 1
0.1V1  j0.2V1  j0.2V2   j 2 As      j, j
j j j 1 j
(0.1  j0.2)V1  j0.2V2   j 2
From eq (2)
 j0.2V1  j0.1V2  1.5
SolvingV1 by eq(1)  2  eq(2)
(0.1  j0.2)V1  3  j 2
3  j2 3.6  33.69  
V1    16.1  33.69  63.43
0.1  j0.2 0.2236  63.43
 16.129.74
v1  16.1cos(100t  29.74 ) V
Vs= - j10, ZL=jL=j(0.5×500)=j250

Use mesh analysis,

 Vs  V R  V Z  0
 (  j10 )  I  250  I  ( j 250 )  0
 j10 10   90 
I   
 0 . 028   90   45 
250  j 250 353 .33  45
I  0 . 028   135 
i  0 . 028 cos( 500 t  135  ) A
V L  I  Z L  ( 0 . 028   135  )  250  90 
 7   45 
v L ( t )  7 cos( 500 t  45  ) V
V R  I  R  ( 0 . 028   135  )  250  7   135 
v R ( t )  7 cos( 500 t  135  )
AC Power Calculations

P  Vrms I rms cos 

PF  cos 
  v  i

Q  Vrms I rms sin 

 apparent power  Vrms I rms
2
P  Q  Vrms I rms 
2 2

2
PI 2
R V Rrms
rms P
R
2
QI 2
X V Xrms
rms Q
X
THÉVENIN EQUIVALENT
CIRCUITS
The Thévenin voltage is equal to the open-circuit
phasor voltage of the original circuit.

Vt  Voc

We can find the Thévenin impedance by

zeroing the independent sources and
determining the impedance looking into the
circuit terminals.
The Thévenin impedance equals the open-circuit
voltage divided by the short-circuit current.

Voc Vt
Z t 
I sc I sc
I n  I sc
 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.
 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.

16
1
9.1 – Introduction

 This chapter introduces important

fundamental theorems of network analysis.
They are the
Superposition theorem
Thévenin’s theorem
Norton’s theorem
Maximum power transfer theorem
Substitution Theorem
Millman’s theorem
Reciprocity theorem
 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.
 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.
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.
 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.
Thévenin’s Theorem

 Procedure to determine the proper values of RTh and

ETh
 Preliminary
1. 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.
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.)
 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.)
 Thévenin’s Theorem

 Conclusion: Insert Figure 9.26(b)

5. Draw the Thévenin equivalent
circuit with the portion of the
circuit previously removed
replaced between the
terminals of the equivalent
circuit. This step is indicated
by the placement of the
resistor RL between the
terminals of the Thévenin
equivalent circuit.
 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.
 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

 Vth

17
3
 Rth

17
4
 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.
 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.
 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.
17
8
 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
 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
 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
UNIT IV - TRANSIENT RESPONSE FOR DC CIRCUITS
–– Transient response of RL, RC and RLC
–– Laplace transform for DC input with sinusoidal input.
Solution to First Order Differential Equation
Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt
Let the initial condition be x(t = 0) = x( 0 ), then we solve the differential
equation:

dx(t )
  x(t )  K s f (t )
dt

The complete solution consists of two parts:

• the homogeneous solution (natural solution)
• the particular solution (forced solution)
The Natural Response

Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt
Setting the excitation f (t) equal to zero,

dx N (t ) dx N (t ) x N (t ) dx N (t ) dt
  x N (t )  0 or  , 
dt dt  x N (t ) 
dx N (t ) dt
   , x N (t )   e t / 
x N (t ) 

It is called the natural response.

 The Forced Response

Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt

Setting the excitation f (t) equal to F, a constant for t 0

dxF (t )
  xF (t )  K S F
dt
xF (t )  K S F for t  0

It is called the forced response.

 The Complete Response
Solve for ,

Consider the general Equation

dx(t )
  x(t )  K s f (t ) for t  0
dt
The complete response is: x(t  0)  x(0)    x()
• the natural response +   x(0)  x()
• the forced response
The Complete solution:
x  x N (t )  x F (t ) x (t )  [ x(0)  x()] e t /   x ()
  e t /   K S F
[ x(0)  x()] e t / called transient response
  e  t /   x ( )
x () called steady state response
TRANSIENT RESPONSE

Figure 5.1
 A general model of the transient

Circuit with switched DC excitation analysis problem

Figu
re
5.2,
5.3
For a circuit containing energy storage element

Figure 5.5, 5.6

 (a) Circuit at t = 0

(b) Same circuit a long time after the switch is closed

Figure

5.9, 5.10

The capacitor acts as open circuit for the steady state condition
(a long time after the switch is closed).
(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).
Reason for transient response

 The voltage across a capacitor cannot be

changed instantaneously.

 
VC (0 )  VC (0 )
• The current across an inductor cannot be
changed instantaneously.

 
I L (0 )  I L (0 )
 Example

Figure

5.12, 5.13

5-6
 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.
 Transients

The solution of the differential equation

represents are response of the circuit. It
is called natural response.

The response must eventually die out,

and therefore referred to as transient
response.
(source free response)
Discharge of a Capacitance through a
Resistance

 i  0, iC  iR  0

ic iR dvC t  vC t 
C  0
dt R
Solving the above equation
with the initial condition
Vc(0) = Vi
 Discharge of a Capacitance through a
Resistance
1
s
dvC t  vC t  RC
C  0
dt R vC t   Ke t RC

dvC t  
RC  vC t   0 vC (0 )  Vi
dt 0 / RC
 Ke
vC t   Ke st
K
st st
RCKse  Ke  0 vC t   Vi e  t RC
vC t   Vi e  t RC

Exponential decay waveform

RC is called the time constant.
At time constant, the voltage is
36.8%
of the initial voltage.

vC t   Vi (1  e t RC )
Exponential rising waveform
RC is called the time constant.
At time constant, the voltage is 63.2%
of the initial voltage.
RC CIRCUIT

for t = 0-, i(t) = 0

u(t) is voltage-step function Vu(t)
RC CIRCUIT

iR  iC
vu (t )  vC dvC
iR  , iC  C
R dt
Vu(t) dvC
RC  vC  V , v u (t )  V for t  0
dt
Solving the differential equation
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.
 Figure
5.17,

5.18
a). Complete, transient and steady state
response
b). Complete, natural, and forced responses
of the circuit

5-8
Circuit Analysis for RC Circuit

Apply KCL

iR  iC
vs  v R dvC
iR  , iC  C
R dt
dvC 1 1
 vR  vs
dt RC RC
vs is the source applied.
Solution to First Order Differential Equation

Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt
Let the initial condition be x(t = 0) = x( 0 ), then we solve the differential
equation:

dx(t )
  x(t )  K s f (t )
dt
The complete solution consist of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)
 The Natural Response

Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt
Setting the excitation f (t) equal to zero,

dx N (t ) dx N (t ) x (t )
  x N (t )  0 or  N
dt dt 
x N (t )   e  t / 

It is called the natural response.

 The Forced Response

Consider the general Equation

dx(t )
  x(t )  K s f (t )
dt
Setting the excitation f (t) equal to F, a constant for t 0
dxF (t )
  xF (t )  K S F
dt
xF (t )  K S F for t  0
It is called the forced response.
 The Complete Response

Consider the general Equation Solve for ,

dx(t ) for t  0
  x(t )  K s f (t ) x(t  0)  x(0)    x()
dt
The complete response is:   x(0)  x()
• the natural response +
• the forced response The Complete solution:

x  x N (t )  x F (t ) x (t )  [ x(0)  x()] e t /   x ()

  e t /   K S F [ x(0)  x()] e t / 
called transient response
t / 
 e  x ( )
x () called steady state response
Example

Initial condition Vc(0) = 0V

iR  iC
vs  vC dvC
iR  , iC  C
R dt
dvC
RC  vC  vs
5
dt  6 dvC
10  0.01 10  vC  100
dt
3 dvC
10  vC  100
dt
Example

Initial condition Vc(0) = 0V

 3 dvC
10  vC  100
dt
dx(t ) t
  x(t )  K s f (t ) 
dt vc  100  Ae 10 3
and As vc (0)  0, 0  100  A
x  x N (t )  x F (t ) A  100
  e t /   K S F t

 e t / 
 x ( ) vc  100  100e 10 3
Energy stored in capacitor

dv
p  vi  Cv
dt
t dv
t o pdt  tt Cv dt  C tt vdv
o dt o

1

 C v ( t ) 2  v ( t o ) 2
2

If the zero-energy reference is selected at to, implying that the
capacitor voltage is also zero at that instant, then

1 2
wc (t )  Cv
2
RC CIRCUIT

Power dissipation in the resistor is:

pR = V2/R = (Vo2 /R) e -2 t /RC

Total energy turned into heat in the resistor

 Vo2 0 e 2t / RC dt

WR  0 p R dt 
R
1
 Vo2 R( )e  2t / RC |
0
2 RC
1
 CVo2
2
RL CIRCUITS

Initial condition
i(t = 0) = Io

di
vR  vL  0  Ri  L
dt
L di
i  0
R dt
Solving the differential equation
 RL CIRCUITS

i(t)
di R
 i  0
dt L
- + di R i(t) di t R
VR R L VL
i
 
L
dt ,  Io i
 
o

L
dt
+ -
R
ln i |iI o   t |ot
L
Initial condition R
i(t = 0) = Io
ln i  ln I o   t
L
i ( t )  I o e  Rt /L
 RL CIRCUIT

Power dissipation in the resistor is:

i(t)
pR = i2R = Io2e-2Rt/LR
Total energy turned into heat in the resistor
- +
 
VR R L VL 2  2 Rt / L

+ -
W R   0
p R dt  I o R  0
e dt
L  2 Rt / L
 I o2 R (  )e | 0
2R
1
LI o2
2
It is expected as the energy stored in the inductor is 1 2
LI o
2
 RL CIRCUIT Vu(t)

i(t)

+
Vu(t)
R
L
+
VL  L
_
-
i ( 0 )  0 , thus k   ln V
R
di L
Ri  L  V  [ln( V  Ri )  ln V ]  t
dt R
Ldi V  Ri
 dt  e  Rt / L or
V
V  Ri
V V  Rt / L
Integrating both sides, i  e , for t  0
R R
L
 ln(V  Ri )  t  k where L/R is the time constant
R
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.
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.
 Any passive electric circuit will resonate if it has an inductor
and capacitor.

Resonance is characterized by the input voltage and current

being 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.
21
8
 Consider the series RLC circuit shown below.

R L
+
V _ C
V = VM 0 I

The input impedance is given by:

1
Z  R  j ( wL  )
wC
The magnitude of the circuit current is;
Vm
I  | I |
21 1 2
R 2  ( wL  )
wC
9
 Resonance occurs when,

1
wL 
wC
At resonance we designate w as w o and write;

1
wo 
LC

This is an important equation to remember. It applies to both series

And parallel resonant circuits.

22
0
 Series Resonance

The magnitude of the current response for the series resonance circuit
is as shown below.

Vm
R
|I|
Vm
2R

Half power point

w1 wo w2 w

Bandwidth:

22 BW = wBW = w2 – w1
1
 Series Resonance

The peak power delivered to the circuit is;

.

2
V
P m
R Vm
I
The so-called half-power is given when
2R

We find the frequencies, w1 and w2, at which this half-power

occurs by using;

22 1 2
2 R  R 2  ( wL  )
2 wC
 After some insightful algebra one will find two frequencies at which
the previous equation is satisfied, they are:

2
R  R  1
w1      
2L  2 L  LC
an
d
2
R  R  1
w2     
2L  2 L  LC

The two half-power frequencies are related to the resonant frequency by

wo  w1 w2
22
3
 The bandwidth of the series resonant circuit is given by;

R
BW  wb  w2  w1 
L
We define the Q (quality factor) of the circuit as;

wo L 1 1 L
Q    
R wo RC R  C 

Using Q, we can write the bandwidth as;

wo
BW 
Q
These are all important relationships.
22
4
 An Observation:

If Q > 10, one can safely use the approximation;

BW BW
w1  wo  and w2  wo 
2 2

These are useful approximations.

22
5
 By using Q = woL/R in the equations for w1and w2 we have;

 1  1 
2 
w1  wo      1
 2Q  2Q  
 
and

  
2 
1 1
w2  wo      1
 2Q  2Q  
 

22
6
 An Example Illustrating Resonance:

The 3 transfer functions considered are:

Case 1: ks
s 2  2 s  400
Case 2:
ks
s 2  5s  400
Case 3:
ks
s 2  10s  400

22
7
 An Example Illustrating Resonance:

The poles for the three cases are given below.

Case 1:

s 2  2 s  400  ( s  1  j19.97)( s  1  j19.97)

Case 2:

s 2  5s  400  ( s  2.5  j19.84)( s  2.5  j19.84)

Case 3:

s 2  10s  400  ( s  5  j19.36)( s  5  j19.36)

22
8
 Comments:

Observe the denominator of the CE equation.

2R 1
s  s
L LC
Compare to actual characteristic equation for Case 1:

2
s  2s  400
2
wo  400 w  20 rad/sec

R wo
BW   2 rad/sec Q  10
L BW

22
9
 Poles and Zeros In the s-plane:
jw axis
( 3) (2)
(1) x 20
x x

s-plane

 axis
0
-5 -2.5 -1
Note the location of the poles
0
for the three cases. Also note
there is a zero at the origin.

x x x -20
23 ( 3) (2)
0 (1)
 Comments:

The frequency response starts at the origin in the s-plane.

At the origin the transfer function is zero because there is
a
zero 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
starts
to 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 response

has the above characteristics, using a graphical approach.
23
1
 1

0 .9
Q = 10, 4, 2
0 .8

0 .7

0 .6
Amplitude

0 .5

0 .4

0 .3

0 .2

0 .1

0
0 10 20 30 40 50 60
w ( r a d /s e c )
23
2
 Next Case: Normalize all responses to 1 at w o

0 .9
Q = 10, 4, 2

0 .8

0 .7

0 .6
A mplitude

0 .5

0 .4

0 .3

0 .2

0 .1

23 0
0 10 20 30 40 50 60
w ( r a d /s e c )

3
 Three dB Calculations:

Now we use the analytical expressions to calculate w 1 and w2.

We will then compare these values to what we find from the
Matlab simulation.
Using the following equations with Q = 2,

2
1  1  
w ,w w  w 
1 2 o
 1
o
 2Q  2Q  
we find,
w1 = 15.62 rad/sec
23 w2 = 21.62 rad/sec
4
 Parallel Resonance
Consider the circuits shown below:

I
1 1 
R L C I  V   jwC  
 R jwL 

R L
 1 
V C V  I  R  jwL  
I
 jwC 
23
5
 Duality

1 1   1 
I  V   jwC   V  I  R  jwL  
 R jwL   jwC 
We notice the above equations are the same provided:

I V
IfIf we
we make
makethe
theinner-change,
inner-change,
thenone
then oneequation
equationbecomes
becomes
1 the same
the sameas
asthe
theother.
other.
R
R For such
For suchcase,
case,we
wesaysaythe
the one
one
circuitisisthe
circuit thedual
dualofof
thethe other.
other.
L C
23
6
 What
Whatthisthismeans
this means
means isisis
that
thatfor
that all
allthe
forfor all equations
thethe
equations we
wehave
equations have
we have
derived
derivedfor
derived forthe
forthe
theparallel
parallel
parallel resonant
resonant
resonant circuit,
circuit,we
wecan
circuit, we use
can can
use use
for
forthe
for theseries
the series
seriesresonant
resonant
resonant circuit
circuit provided
circuitprovided we
providedwemake
make
we make
the
thesubstitutions:
the substitutions:
substitutions:

1
R replaced be
R
L replaced by C
C replaced by L
23
7
 Parallel Resonance Series Resonance

1
1 w 
O
w  wL LC Q  w RC
O
LC Q O
o

R
R ww
,w 1
BW  ( w  w )  w 
1 2
BW  w  BW
L
2 1 BW
RC
2
 R  R 1   1  1  1 
2

w ,w  
1 2
     w ,w       
 2 L  2 L  LC 
1 2
 2 RC  2 RC  LC 
2
1  1   2
1  1  
w ,w  w     1 w ,w  w   
1 2 o
 1
 2Q  2Q 
1 2 o

 2Q  2Q   
23
8
 Example 1: Determine the resonant frequency for the circuit below.

1
jwL ( R  ) 2
jwC ( w LRC  jwL )
Z  
IN
1 (1  w LC )  jwRC 2

R  jwL 
jwC

23 At resonance, the phase angle of Z must be equal to zero.

9
 Analysis 2
( w LRC  jwL)
2
(1  w LC )  jwRC
For zero phase;

wL wRC
 2 2
( w LCR) (1  w LC
This gives;
2 2 2 2
w LC  w R C 1
or
1
w
o 2 2
( LC  R C )
24
0
 Parallel Resonance
Example 2:

A parallel RLC resonant circuit has a resonant frequency admittance of

2x10 -2 S(mohs). The Q of the circuit is 50, and the resonant frequency is
A series RLC resonant circuit has a resonant frequency admittance of
10,000 rad/sec. Calculate
2x10-2 S(mohs). the circuit
The Q of the valuesisof50,
R, L,
andand
theC.resonant
Find thefrequency
half-power
is
frequencies and the bandwidth.
10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power
frequencies and the bandwidth.

First, R = 1/G = 1/(0.02) = 50 ohms.

wL
Second, from Q  O , we solve for L, knowing Q, R, and wo to
R
find L = 0.25 H.
Q 50
Third, we can use C  100  F
w R 10,000 x50
O

24
1
 Parallel Resonance
Example 2: (continued)

4
w 1x10
Fourth: We can use w  
BW
 200 rad / sec
o

Q 50
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

24
2
 Extension of Series Resonance
Peak Voltages and Resonance:

VL

VR L
+ _ + _ +
+
VS R C VC
_ I _

We know the following:

1
When w = wo = , VS and I are in phase, the driving point impedance
 LC
is purely real and equal to R.

 A plot of |I| shows that it is maximum at w = w o. We know the standard

24 equations for series resonance applies: Q, w BW, etc.
3
 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 sure
about the other two voltages. We know that at resonance they are equal
and they have a magnitude of QxVS.

 Irwin shows that the frequency at which the voltage across the capacitor
is a maximum is given by;

1
wmax  wo 1 
2Q 2
 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

24
4
 Series RLC Transfer Functions:

The following transfer functions apply to the series RLC circuit.

1
VC ( s ) LC

VS ( s ) s 2  R s  1
L LC
VL ( s ) s2

VS ( s ) s 2  R s  1
L LC

R
s
VR ( s ) L
24 
VS ( s ) s 2  R s  1
5
L LC
 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 = 5F. The transfer functions become as follows:

VC 4 x106
 2
VS s  1000s  4 x106

VL s2
 2
VS s  1000s  4 x106

VR 1000s
 2
VS s  1000s  4 x106
24
6
 Simulation Results

Q=2

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 = 2
2 .5

V C V L
1 .5
Amplitude

V R

0 .5

0
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 0

24 w ( r a d /s e c )

7
 Analysis of the problem:

Given the previous circuit. Find Q, w 0, wmax, |Vc| at wo, and |Vc| at wmax

+ VR _ + VL
_
+ L=5 +
mH C=5 F VC
VS R=50  I
_ _

1 1
Solution: w 
O
 2
 2000 rad / sec
6
LC 50 x10 x5 x10

3 2
w L 2 x10 x5 x10
24 Q O
 2
R 50
8
 Problem Solution:

1
w
MAX
 w 1
O
 0.9354w 2 o
2Q
| V | at w  Q | V |  2 x1  2 volts ( peak )
R O S

Qx | V | 2
| V | at w
C MAX
 S
 2.066 volts  peak ) 
1 0.968
1 2
4Q

24 Now check the computer printout.

9
 Exnsion of Series Resonance
Problem Solution (Simulation):

1.0e+003 *

1.8600000 0.002065141
1.8620000 0.002065292
1.8640000 0.002065411
1.8660000 0.002065501 Maximum
1.8680000 0.002065560
1.8700000 0.002065588
1.8720000 0.002065585
1.8740000 0.002065552
1.8760000 0.002065487
1.8780000 0.002065392
1.8800000 0.002065265
1.8820000 0.002065107
1.8840000 0.002064917

25
0
 Simulation Results:

Q=10
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
12

10

8
VC VL
A mplitude

VR

25 0 500 1000 1500 2000

w ( r a d /s e c )
2500 3000 3500 4000

1
 Observations From The Study:


The voltage across the capacitor and inductor for a series RLC
circuit
is not at peak values at resonance for small Q (Q <3).

Even for Q<3, the voltages across the capacitor and inductor
are
equal 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 = w o.

For high Q, the equations discussed for series RLC resonance


can be applied to any voltage in the RLC circuit. For Q<3, this
25 is not true.
2
 Extension of Resonant Circuits
Given the following circuit:

+
R
+
C V
I
_ L
_

 We want to find the frequency, w r, at which the transfer function

for V/I will resonate.

The transfer function will exhibit resonance when the phase angle

between V and I are zero.

25
3
 The desired transfer functions is;

V (1/ sC )( R  sL)

I R  sL  1/ sC
This equation can be simplified to;

V R  sL

I LCs 2  RCs  1

With s jw

V R  jwL

I (1  w2 LC )  jwR
25
4
 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 angle
of the denominator.  num   dem
or,
1 wL  1 wRC 
 num  tan  ,  den  tan  2  .
 R   (1  w LC ) 
Therefore;
wL wRC

R (1  w2 LC )

25
5
 Resonant Condition Analysis:

Canceling the w’s in the numerator and cross multiplying gives,

L(1  w2 LC )  R 2C or w2 L2C  L  R 2C

This gives,
1 R2
wr   2
LC L
Notice that if the ratio of R/L is small compared to 1/LC, we have

1
wr  wo 
25 LC
6
 Extension of Resonant Circuits
Resonant Condition Analysis:

What is the significance of w r and wo in the previous two equations?

Clearly wr is a lower frequency of the two. To answer this question, consider
the 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.

+
R
+
C V
I
_ L
_

25
7
 R s e s p o n s e f o r R e s i s ta n c e i n s e r i e s w i t h L th e n P a r a lle l w i th C
18

16

2646 rad/sec
14
R =1 ohm
12
A mplitude

10

R = 3 o hm s
4

25 0 1000 2000 3000 4000

w ( r a d /s e c )
5000 6000 7000 8000