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 – division – 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 & 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 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 : 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  i2

I 2  i2  i3

v  v i  higher lower 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 G  G  G v I  I   1 2 2  1    1 2    G2 G2  G3 v2   I2 

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 v  v v  0  5  1 2  1 4 2

43 Methods of Analysis Example 3.1

 At node 2

i2  i4  i1  i5 v  v v  0  5  2 1  2 4 6

44 Methods of Analysis Example 3.1

 In matrix form:

1 1 1    2 4 4 v1  5    1 1 1 v  5     2     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 v  v v  v  3  1 3  1 2 4 2

48 Methods of Analysis Example 3.2

 At nodeix  i2 2i3 v  v v  v v  0  1 2  2 3  2 2 8 4

49 Methods of Analysis Example 3.2

 At node 3

i1  i2  2ix v  v v  v 2(v  v )  1 3  2 3  1 2 4 8 2

50 Methods of Analysis Example 3.2

 In matrix form:  3 1 1    4 2 4 v 3   1    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  v  v v  v v  0 v  0 1 2  1 3  2  3 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i2  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, v  v v  v v 3 2 10  1 4  1 6 3 2

v1  v2  20

57 Methods of Analysis Example 3.4

 At supernode 3-4, v  v v  v v v 1 4  3 2  4  3 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

R i V  R (i  i )  0  For mesh 22,2 2 3 2 1

 R3i1  (R2  R3 )i2  V2

64 Methods of Analysis  Solve for the mesh currents.

R  R  R i V  1 3 3  1    1    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, I2  i2 , I3  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

6i  4i 10(i  i ) 10  0  For mesh2 2, 2 2 1 i1  2i2 1

I1  i1, I2  i2 , I3  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, 4 I 0  12 ( i 3  i1 )  4 ( i 3  i 2 )  0

At node A, I 0  I 1  i 2 ,

4 ( i1  i 2 )  12 ( i 3  i1 )  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 6i1  14 i2  20 equations are needed; one is obtained using KVL and Ohm’s law to the supermesh and the i1  i2  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

i 2  i 3   i 4

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  I2  G1v1  G2 (v1  v2 ) (3.7)

I2  G2 (v1  v2 )  G3v2 (3.8)  MATRIX G  G  G v I  I  1 2 2  1    1 2    G2 G2  G3 v2   I2 

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   v 1   i1   G G  G   v   i  Gv = i  21 22 2 N   2    2                     G N 1 G N 2  G NN   v N   i 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

R  R  R i v  1 3 3  1    1    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   v 1   R R  R   i   v  or simply  21 22 2 N   2    2               R R  R   i   v  Rv = i  N 1 N 2 NN   N   N  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    0 . 3 , G     1 . 325 11 5 10 22 5 8 1 1 1 1 1 1 1 G     0 . 5 , G     1 . 625 33 8 8 4 44 8 2 1

 The off-diagonal terms are 1 G    0.2, G  G  0 12 5 13 14 1 1 G  0.2, G    0.125 , G    1 21 23 8 24 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   v 1   3    0.2 1 .325  0.125  1   v    3     2      0  0.125 0 .5  0.125   v 3   0         0  1  0.125 1 .625   v 4   6 

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

v 1  4 , v 2  10  4  6 ,

v 3   12  6   6 , v 4  0 , v 5   6

 The mesh-current equations are

 9  2  2 0 0   i1   4    2 1 0  4  1  1   i     2   6    2  4 9 0 0   i     6  3    0  1 0 8  3   i  0    4      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

sin z  cos(z  90o ) sint  cos(t  90o ) sin(t  90o )  cost Representation of the two vectors v1 and v2.

In this diagram, v1 leads v2 by 100o + 30o = 130o, although it o could also be argued that v2 leads v1 by 230 . Generally we express the phase difference by an angle less than or equal to 180o in magnitude. Euler’s identity

  t   2 f Acos t  Acos 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

j ( t + ) The complex forcing function Vm e produces the complex j (t + ) response Im e . 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). j ( t + ) j (t + ) Re(Vm e )  Re(Im e ) j ( t + ) j (t + ) Im(Vm e )  Im(Im e ) 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  20cos(t  45 )  v2(t) 10cos(t  30 )

 V1  20  45  V2 10  30 Vs  V1  V2  2045 1030 14.14 j14.148.660 j5  23.06 j19.14  29.9739.7 Vs  Ae j 19.14 A  23.062  (19.14)2  29.96,   tan 1  39.7 23.06

 vs 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 LI L (a) (b) In the phasor domain, (a) a resistor R is represented by an impedance of the same value; (b) a C is represented by an impedance 1/jC; (c) an L is represented by an impedance jL.

(c) V  Ve jt d V d Ve jt I  C  C  jCVe jt dt dt V 1 I  j CV    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 d I d Ie jt V  L  L  jLIe jt dt dt V V  j LI   j L  Z I L

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.  AsV jCI, if i  IM cost,thenv LIM cos(t 90 )  ori  IM cos(t 90),andv LIM cost,VM LIM

Complex Impedance in Phasor Notation

VC  ZCIC 1 1 1 Z    j    90 C jC C C

VL  ZLIL  ZL  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)  70.7cos(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 j0.01 As     j0.01  j100  j100 j  j2 1 10 Z    70.71  45 RC 0.01 j0.01 0.0141445  50  j50 Z Vc  Vs RC (voltage division) ZL  ZRC 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 )  10cos1000t V Vs I  ZL  ZRC 10  90 10  90   j100  50  j50 50  j50 10  90   0.414 135 70.7145 i(t)  0.414 cos (1000t  135 ) V I  C R R 10 135   0.1 135 100  iR (t)  0.1 cos (1000t  135 ) V 10 135 10 135 I  C    0.1  45 C Zc  j100 100  90  iR (t)  0.1 cos (1000t  45 ) A

Solve by nodal analysis

V V V 1  1 2  2  90   j2 eq(1) 10  j5 V V V 2  2 1  1.50  1.5 eq(2) j10  j5 V V V V V V 1  1 2  290   j2 eq(1) 2  2 1 1.50 1.5 eq(2) 10  j5 j10  j5 Fromeq (1) 1 1 j j 1 0.1V  j0.2V  j0.2V   j2 As      j,  j 1 1 2 j j j 1  j

(0.1 j0.2)V1  j0.2V2   j2 Fromeq (2)

 j0.2V1  j0.1V2 1.5

SolvingV1 by eq(1)  2eq(2)

(0.1 j0.2)V1  3 j2 3 j2 3.6 33.69 V   16.133.69  63.43 1 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  VZ  0  ( j10 )  I  250  I  ( j250 )  0  j10 10  90  I    0.028   90   45  250  j250 353 .33 45  I  0.028   135  i  0.028 cos( 500 t  135  )A   VL  I  Z L  (0.028   135 )  250 90  7  45   v L (t)  7 cos( 500 t  45 ) V   VR  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 2 2 P  Q  Vrms I rms 

2 2 VRrms P  I rms R P  R

2 2 VXrms Q  I rms X 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  Isc Isc

In  Isc

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

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 f (t) dt s Setting the excitation f (t) equal to zero,

dxN (t) dxN (t) xN (t) dxN (t) dt   xN (t)  0 or   ,   dt dt  xN (t) 

dxN (t) dt t /     , xN (t)   e xN (t) 

It is called the natural response. The Forced Response

Consider the general Equation dx(t)   x(t)  K f (t) dt s

Setting the excitation f (t) equal to F, a constant for t 0 dx (t)  F  x (t)  K F dt F S xF (t)  KS F for t 0

It is called the forced response. The Complete ResponseSolve for ,

Consider the general Equation dx(t)   x(t)  K f (t) dt s for t  0 The complete response is: x(t  0)  x(0)   x() • the natural response +   x(0)  x() • the forced response The Complete solution: x  xN (t)  xF (t) x(t)  [x(0)  x()]et /  x() t /   e  K F t / S [x(0)  x()]e 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.

  IL (0 )  IL (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 dv t v t C C  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 t RC dt R vC t  Ke

dvC t  RC  v t  0 vC (0 )  Vi dt C  Ke0/ RC v t  Kest C    K st st RCKse  Ke  0 t RC vC t  Vie t RC vC t Vie

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

t RC vC t Vi (1 e ) 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)  v dv i  C , i  C C R R C dt dv Vu(t) RC C  v  V , vu(t)  V for t 0 dt C

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 v  v dv i  s R ,i  C C R R C dt dv 1 1 C  v  v dt RC R RC s

vs is the source applied. Solution to First Order Differential Equation

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

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 f (t) dt s

Setting the excitation f (t) equal to zero,

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

It is called the natural response. The Forced Response

Consider the general Equation dx(t)   x(t)  K f (t) dt s

Setting the excitation f (t) equal to F, a constant for t 0 dx (t)  F  x (t)  K F dt F S xF (t)  KS 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 f (t) dt s x(t  0)  x(0)   x() The complete response is:   x(0)  x() • the natural response + • the forced response The Complete solution: x(t)  [x(0)  x()]et /  x() x  xN (t)  xF (t) t / t /   e  KS F [x(0)  x()]e called transient response   et /  x() x() called steady state response Example

Initial condition Vc(0) = 0V

iR  iC v  v dv i  s C ,i  C C R R C dt dvC RC  vC  vs dt dv 105 0.01106 C  v 100 dt C dv 103 C  v 100 dt C Example

Initial condition Vc(0) = 0V dv 103 C  v 100 dt C dx(t) t   x(t)  K f (t)  s 103 dt vc 100  Ae and As vc (0)  0, 0 100  A x  xN (t)  xF (t) A  100   et /  K F t S  t / 103   e  x() vc 100 100e Energy stored in capacitor

dv p  vi  Cv dt t t dv t  pdt   Cv dt  C  vdv to to dt to 1 2 2  C v(t)  v(to )  2

If the zero-energy reference is selected at to, implying that the capacitor voltage is also zero at that instant, then 1 w (t)  Cv2 c 2 RC CIRCUIT

Power dissipation in the resistor is: 2 2 -2 t /RC pR = V /R = (Vo /R) e Total energy turned into heat in the resistor

2  2t / RC  Vo 0 e dt W   p dt  R 0 R R 1  V 2R( )e2t / RC | o 2RC 0 1  CV 2 2 o RL CIRCUITS

Initial condition

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

Power dissipation in the resistor is: i(t) 2 2 -2Rt/L pR = i R = Io e R Total energy turned into heat in the resistor - +   V R L V 2  2 Rt / L R L W R  p R dt  I o R e dt + - 0 0 L  I 2 R (  ) e  2 Rt / L | o 2 R 0 1  LI 2 2 o 1 It is expected as the energy stored in the inductor is LI 2 2 o RL CIRCUIT Vu(t)

i(t)

+ + R Vu(t) L V  L _ 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  Rt / L  dt  e or V  Ri V V V Integrating both sides, i   e  Rt / L , 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 + C V _ I V = VM 0

The input impedance is given by: 1 Z R  j() wL  wC The magnitude of the circuit current is; V II| |  m 1 21 R2() wL  2 9 wC Resonance occurs when,

1 wL  wC

At resonance we designate w as wo and write;

1 w  o 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; .

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

We find the frequencies, w1 and w2, at which this half-power occurs by using;

1 22 2R R2  ( wL  ) 2 2 wC After some insightful algebra one will find two frequencies at which the previous equation is satisfied, they are:

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

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

wo  w1 w 2 22 3 The bandwidth of the series resonant circuit is given by; R BW w  w  w  b 2 1 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; w BW  o Q

These are all important relationships. 22 4 An Observation:

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

BW BW w w  and w  w  1o2 2 o 2

These are useful approximations.

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

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

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

22 6 An Example Illustrating Resonance:

The 3 transfer functions considered are:

Case 1: ks s2 2 s  400

Case 2: ks s2 5 s  400

Case 3: ks s2 10 s  400

22 7 An Example Illustrating Resonance:

The poles for the three cases are given below.

Case 1: s2 2 s  400  ( s  1  j 19.97)( s  1  j 19.97)

Case 2:

s2 5 s  400  ( s  2.5  j 19.84)( s  2.5  j 19.84)

Case 3:

s2 10 s  400  ( s  5  j 19.36)( s  5  j 19.36) 22 8 Comments:

Observe the denominator of the CE equation. R 1 s2  s  L LC

Compare to actual characteristic equation for Case 1:

s2 2 s  400

2 rad/sec wo  400 w  20

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 -1 -5 -2.5 0 Note the location of the poles 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

0 .5

Amplitude 0 .4

0 .3

0 .2

0 .1

0 0 1 0 2 0 3 0 4 0 5 0 6 0 23 w (ra d /s e c ) 2 Next Case: Normalize all responses to 1 at wo

1

0 .9 Q = 10, 4, 2

0 .8

0 .7

0 .6

0 .5

Amplitude 0 .4

0 .3

0 .2

0 .1

0 23 0 1 0 2 0 3 0 4 0 5 0 6 0 3 w ( r a d /s e c ) 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,

2  1  1   w1 ,w2  wo   wo   1 2Q  2Q  

we find, w1 = 15.62 rad/sec 23 w = 21.62 rad/sec 4 2 Parallel Resonance

Consider the circuits shown below:

V

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 wewe makemake the the inner inner-change,-change, thenthen oneone equation equation becomes becomes 1 R thethe samesame as as the the other. other. R ForFor suchsuch case, case, we we say say the the one one circuitcircuit is is the the dual dual of ofthe the other. other. L C 23 6 WhatWhat this thisthis means means means is is isthat that that for for for all all theall the the equations equations equations we we have havewe have derivedderived for for for the the the parallel parallel parallel resonant resonant resonant circuit, circuit, circuit, we we can canwe usecan use use forfor the thethe series series series resonant resonant resonant circuit circuit circuit provided provided provided we we make makewe make thethe substitutions: substitutions:substitutions:

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

1 w  1 w L O w  O LC Q  w RC O Q  o LC R ww,w 1 R 1 2 BW  (w  w )  w  BW  wBW  2 1 BW L RC

2 2  R  R  1   1  1  1  w1 ,w2        w ,w    2L 2L LC 1 2         2RC  2RC  LC 

2 2    1  1   1  1  w ,w  w     1 w1 ,w2  wo     1 1 2 o 2Q 2Q 2Q 2Q 23         8 Example 1: Determine the resonant frequency for the circuit below.

1 jwL(R  ) jwC (w2 LRC  jwL) Z   I N 1 2 R  jwL  (1 w LC) jwRC jwC

23 At resonance, the phase angle of Z must be equal to zero. 9 Analysis (w2 LRC  jwL) (1 w2 LC) jwRC

For zero phase; wL wRC  (w2 LCR) (1 w2 LC This gives; w2 LC w2 R 2C 2 1

or 1 w  o (LC  R 2C 2 ) 24 0 Parallel Resonance

Example 2:

A parallel RLC resonant circuit has a resonant frequency admittance of -2 A2x10 seriesS(mohs). RLC resonant The Q circuitof the circuithas a resonantis 50, and frequency the resonant admittance frequency of is 2x1010,000-2 S(mohs).rad/sec. CalculateThe Q of thethe circuitvalues isof 50, R, L,and and the C. resonant Find the frequency half-power is 10,000frequencies rad/sec. and Calculate the bandwidth. the values of R, L, and C. Find the half-power frequencies and the bandwidth.

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

wO L Second, from Q  , 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

wO R 10,000x50 24 1 Parallel Resonance

Example 2: (continued)

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

V L + R _ + + _ + V R C VC S _ 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 = wo. We know the standard 24 equations for series resonance applies: Q, wBW, 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 w w 1  max o 2Q2

 The above being true, we might ask, what is the frequency at which the voltage across the inductor is a maximum?

24 We answer this question by simulation 4 Series RLC Transfer Functions:

The following transfer functions apply to the series RLC circuit. 1 V() s C  LC V() s 2 R 1 S s s  L LC V() s s2 L  V() s 2 R 1 S s s  L LC

R s V() s 24 R  L V() s 2 R 1 5 S s s  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:

6 VC 4x 10  2 6 VS s1000 s  4 x 10

2 VL s  2 6 VS s1000 s  4 x 10

VR 1000s  2 6 VS s1000 s  4 x 10 24 6 Simulation Results

Q = 2

Rsesponse for RLC series circuit, Q =2 2 . 5

2

VC VL 1 . 5

Amplitude 1

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, w0, wmax, |Vc| at wo, and |Vc| at wmax

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

1 1 Solution: w   2000rad / sec O LC 50x102 x5x106

w L 2x103 x5x102 24 Q O  2 R 50 8 Problem Solution:

1 w  w 1  0.9354w MAX O 2Q 2 o

|VR |at wO  Q |VS |  2x1  2volts( peak)

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

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

Rsesponse for RLC series circuit, Q =10 1 2

1 0

8 VC VL

6 Amplitude

4

2

V R

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 25 w ( r a d /s e c ) 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 QxV .  S For Q>10, the voltages across the 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 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 I V _ L _

 We want to find the frequency, wr, 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 LCs2  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,

wL 1 wRC  1     tan . num  tan   , den 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 2 C or w 2 L 2 C  L  R 2 C

This gives, 1 R2 w   r LC L2

Notice that if the ratio of R/L is small compared to 1/LC, we have 1 wr w o  25 LC 6 Extension of Resonant Circuits

Resonant 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, 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 I V _ L _ 25 7 Rsesponse for Resistance in series with L then P arallel with C 1 8

1 6 2646 rad/sec 1 4 R = 1 o h m 1 2

1 0

8 Amplitude

6

R = 3 o h m s 4

2

0 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 0 25 w ( r a d /s e c ) 8