dc circuits: chapter 3 det 101/3 basic electrical circuit 1
TRANSCRIPT
DC CIRCUITS:
CHAPTER 3
DET 101/3 Basic Electrical Circuit 1
Methods of Circuit Analysis and Circuit Theorems:
Nodal Analysis (Node-Voltage Method) Mesh Analysis (Mesh-Current Method) Superposition Theorem Source Transformation Thevenin’s Theorem Norton’s Theorem Maximum Power Transfer
Introduction
Direct methods are not suitable to solve complex and large circuits or as we demand many unknowns.
To aid the analysis of complex circuit structures-need to develop more powerful techniques from the basic laws by systematic approaches: Nodal Analysis and Mesh Analysis.
These two techniques can be used to solve almost any kind of circuit analysis problems by means of a set of simultaneous equations.
Introduction (Continued…)
Four ways of solving simultaneous equations:
1. Cramer’s rule2. Calculator (real numbers only)3. Normal substitution and elimination
(not more than two equations)4. Computer program packages:
matcad, maple, mathematica etc.
Introduction (Continued…)
Circuit analysis problems in this course will be limited to three linear simultaneous equations for conventional hand solutions.
Circuit theorems merely developed to simplify circuit analysis applicable to linear circuits such as Thevenin’s and Norton’s theorems, superposition theorem, source transformation and maximum power transfer.
Introduction (Continued…)
Circuit theorems are not analysis techniques, rather they add up to the list of simplifying/reduction techniques such as the series-parallel reductions and -Y transformations.
Although many computer aids facilitate us as effective mathematics tools to solve engineering problems they cannot replace the compulsory needs to study the circuit theories govern circuit behavior in performing both circuit analysis and design.
Circuit Analysis Method
NODAL ANALYSIS
Concept
Developed based on the systematic approach of Kirchhoff’s current law (KCL) to find all circuit variables without having to sacrifice any of the elements.
General procedure which is making use of node voltages in circuit analysis as key solutions.
Importance terms
Node Voltage: Potential difference between a marked node and the selected reference node.
Element Voltage: Potential difference across any element or branch in the circuit.
When Node Voltage = Element Voltage?
Why use Node Voltage?
Further reduce the number of equations to be solved simultaneously.
No of independent equations = No of the marked nodes exclusive of the reference node.
Element voltages and currents can be obtained in few steps using the solved node voltages.
Assumptions
KCL is performed with current going out from a node as positive (+ve) while current entering a node as negative (-ve).
in – negative (subtract) out – positive (add)
All unknown currents assumed to be leaving a particular node.
Nodal Analysis Procedures:
1. Mark all essential nodes and assign proper voltage designations except for the appointed reference node.
2. Apply KCL to each nonreference nodes. Use Ohm’s law to formulate the equation in terms of node voltages. Assume all unknown currents are directing out of the nodes.
3. Solve the resulting simultaneous equations to obtain the unknown node voltages.
Applying Nodal Analysis on Simple Circuit
Example 1 (3 unknowns)
R1
R4R3Is1
R2
Is2
Solution
Step 1: Mark all essential
nodes Assign unknown
node voltages Indicate the
reference node.
R1
R4R3Is1
R2
Is2
V1
V2 V3
Solution (continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of the node voltages.
Solution (continued…)
KCL V1:
KCL V2:
KCL V3:
1
2121 R
VVII ss
)( 21121 VVGII ss or
(1)
00
2
32
3
2
1
12
R
VV
R
V
R
VV (2)
00
4
3
2
232
R
V
R
VVI s (3)
Solution (continued…)
Step 3: Solve the resulting simultaneous equations from step 2 above.
10 k
2 k4 k
3 mA
5 kV1
V2V3
2 mA
Solution (continued…)
KCL V1:
KCL V2:
KCL V3:
k
VVmm
1032 21
Simplify to (1)
054
0
1032212
k
VV
k
V
k
VV
-2V1 + 11V2 - 4V3 = 0 Simplify to (2)
V1 - V2 = 50
02
0
53 323
k
V
k
VVm
Simplify to -2V2 + 7V3 = -30 (3)
Solution (continued…)
Cramer’s rule: Put the equations in matrix forms.
Left Matrix:Col-1: coefficients of V1 i.e. a1, a2 and a3
Col-2: coefficients of V2 i.e. b1, b2 and b3
Col-3: coefficients of V3 i.e. c1, c2 and c3
Middle Matrix: Unknown variables i.e. V1, V2 and V3 Right Matrix: Constants i.e. d1, d2 and d3
3
2
1
3
2
1
333
222
111
d
d
d
V
V
V
cba
cba
cba =
Col-1 Col-2 Col-3
30
0
50
720
4112
011
3
2
1
V
V
V
Solution (continued…)
For third-order determinants, we use shorthand methods of expansion solution.
Shorthand method consists of repeating the first two columns of the determinant to the right of the determinant and then summing the products along the specific diagonals as shown below.
Solution (continued…)
Use determinants to solve for each variable:
720
4112
011
333
222
111
cba
cba
cba 1 -1-2 11 0 -2
55)22(77
)}1)(2)(7()1)(4)(2()0)(11)(0(
)2)(2)(0()0)(4)(1()7)(11)(1(
Solution (continued…)
Determinant 1 when coefficients for V1 is replaced by the constants.
7230
4110
0150
333
222
111
1
cbd
cbd
cbd
3330)400(3730
)}1)(0)(7()50)(4)(2()0)(11)(30(
)2)(0)(0()30)(4)(1()7)(11)(50(1
50 -1 0 11-30 -2
Solution (continued…)
Determinant 2 when coefficients for V2 is replaced by the constants.
7300
402
0501
333
222
111
2
cda
cda
cda 1 50-2 00 -30
580)580(
)}50)(2)(7()1)(4)(30()0)(0)(0(
)30)(2)(0()0)(4)(50()7)(0)(1(2
Solution (continued…)
Determinant 3 when coefficients for V3 is replaced by the constants.
3020
0112
5011
333
222
111
3
dba
dba
dba 1 -1-2 110 -2
70)60(130
)}1)(2)(30()1)(0)(2()50)(11)(0(
)2)(2)(50()0)(0)(1()30)(11)(1(3
Solution (continued…)
V1 = 1/ = 3330/55= 60.55 V
V2 = 2/ = 580/55 =10.55 V
V3 = 3/ = -70/55 =-1.27 V
You should verify your answers using calculator for three unknowns.
Solution (continued…)
Knowing all the node voltages, we can obtain the element voltages, element currents and even power dissipated by the passive elements such as:
VR1= V1 – V2 IR1 = (V1 – V2)/R1PR1 = IR12R1 = VR1
2/R1
VR2= V2 – V3 IR1 = (V2– V3)/R2 PR2 = IR22R2 = VR2
2/R2
**VR3= V2 IR1 = V2/R3 PR3 = IR32R3 = VR3
2/R3
**VR4= V3 IR4 = V32/R4 PR4 = IR4
2R3 = VR42/R4
**In these two cases, the element voltages identical to node voltages because one of its terminals is at reference node.
Can you find the power dissipated by the 10 k resistor?
Need to find the element voltage of 10-k resistor because power calculation formula uses element voltage.
P10k = (V1 – V2)2/10k = (60.55 –10.55)2/10k = 502/10k = 0.25 W
Applying Nodal Analysis on Simple Circuit
Example 2 (2 unknowns)Q: Find the power dissipated in the
20- resistor?
50
50 20
50
10 mA
9 mA
Solution
Step 1: Mark all
essential nodes Assign unknown
node voltages Indicate the
reference node.
50
50 20 50
10 mA
9 mA
V1
V2
Two nonreference nodes
reference node
Solution (Continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages.
KCL V1: mVVV
1050100
211
Hence 123 21 VV
mmVVV
1095020
122
Hence 1.072 21 VV
KCL V2:
(2)
(1)
Solution (Continued…)
Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above using Cramer’s rule.
Solution (Continued…)
17
)2)(2()7)(3(
72
23
8.6
)1.0)(2()7)(1(
71.0
211
7.1
)1)(2()1.0)(3(
1.02
132
Hence,V1 = 1/ = 6.8/17 = 0.4 VV2 = 2/ = 1.7 /17 = 0.1 VP20 = V2
2/20 = 0.12(20) = 0.2 W #
Applying Nodal Analysis on Circuit with Voltage Sources
Three different effects depending on placement of voltage source in the circuit.
Does the presence of a voltage source complicate or simplify the analysis?
Case 1: Voltage source between two nonreference essential nodes.
Vs
Nonreferenceessential node
Nonreferenceessential nodeV1 V2
Supernode Equation: 21 VVVs
Case 2: Voltage source between a reference essential node and a nonreference essential node.
Nonreferenceessential node
Referenceessential node0 V
Vs
V1
Known node voltage: sVV 1
Case 3: Voltage source between an essential node and a non-essential node.
Nonreferenceessential node
Non-essentialnode
Va
Vs
V1
Node voltage at non-essential node: sa VVV 1
Example 3(Supernode or Known node voltage)
Q: Find the power of the 10-V voltage source? Is it supplying energy to the circuit or absorbing energy from the circuit? Show your work according to the nodal analysis procedure.
40
50 8 25
3 A
10 V
5
Solution 1 (Supernode)
Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node.
40
50 8 30
3 A
10 V
5 V1
V2
iChecklist:3 essential nodes – 1 ref node – 1 supernode = 1 KCL Eqn. + 1 Supernode Eqn.
Solution 1 (Continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages.
KCL supernode V1/V2: 388021 VV
Hence 24010 21 VV
1021 VVSupernode Equation:
(1)
(2)
Solution 1 (Continued…)
Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above.
Solving Eqn. (1) and (2) simultaneously yields,
V1 = 30.91 V and V2 = 20.91 V (You can check this answer by
calculator or Cramer’s rule).
Solution 1 (Continued…)
Finding current through the voltage source,
KCL at V1:
Hence, P10-V = Vi= (10)(-0.636) = -6.36 W.
(Delivering energy)
04080
211
iVVV
Ai 636.040
91.2091.30
80
91.30
Solution 2 (Known node voltage)
Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node.
40
50 8 30 3 A
10 V
5
V1
V2
KN
OW
N N
OD
EV
OLT
AG
E
iChecklist:3 essential nodes – 1 ref node – 1 known node voltage = 1 KCL Eqn.
Solution 2 (Continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages.
Immediately known node voltage at V1: VV 101
KCL V2: 3880
10 22 VV
23011 2 V (1)
Solution 2 (Continued…)
Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above.
Solving Eqn. (1) yields, VV 91.2011
2302
Finding current through the voltage source,KCL at V1:
08040
211
iVVV
Ai 636.080
91.2010
40
10
Solution 2 (Continued…)
Hence, P10-V = Vi= (10)(-0.636) = -6.36 W.
(Delivering energy)
Example 4 (One of the terminals not an essential node)
Q: Find the current through the 10-k resistor. Show your work according to the nodal analysis procedure.
15 V
25 V
1 k10 k
5 k
4 k
8 mA
4 k
Solution Step 1: Mark essential nodes and assign
unknown node voltages and indicate the reference node. For voltage sources, indicate the node voltages at both ends with respect to the assigned unknown node voltages at the essential nodes
15 V
25 V
1 k 10 k5 k
4 k
8 mA
4 k
Va = V2 + 15
Vb = -25 V
V1
V2
V3
i
Checklist:4 essential nodes – 1 ref node = 3 KCL Eqns.
Solution (Continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages.
KCL V1: mk
VV
k
V8
4
)15(
1211
Hence 175 21 VV
044
)15(
10
25 32122
k
VV
k
VV
k
V
250101810 321 VVV
KCL V2:
Hence
(1)
(2)
Solution (Continued…)
KCL V3: mk
VV
k
V8
45233
Hence 16095 32 VV
(3)
Solution (Continued…)
Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above.
Solving Eqn. (1) till (3) simultaneously yields,
V1 = -5.43 V, V2 = -10.17 V and
V3 = 12.13 V (You can check this answer by
calculator and Cramer’s rule).
Solution (Continued…)
Finding current through the 10-k resistor,
KCL at V2: 044
)15( 3212
k
VV
k
VVi
mAkk
i 01.34
13.1217.10
4
43.51517.10
Applying Nodal Analysis on Circuit with Dependent Sources
Circuits contain dependent sources either VCVS, CCVS, VCCS or CCCS.
The presence of the dependent sources require ‘Constraint Equation’ (CE).
CE describes the dependent term of the dependent sources in relation to the assigned unknown node voltages or known values at the essential nodes.
Example 5 (Circuit with dependent sources)
Q: Use the node-voltage method to find both dependent terms iO and Vx of the dependent sources of the circuit in Figure below.
10 2
8
12 V
3 Vx
+ Vx -
10 io
io
Solution
Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node.
10 2
8
12 V
3 Vx
+ V
x -
10 io
io
V1
V2 V3
KN
OW
NChecklist:4 essential nodes – 1 ref node – 1 s/node – 1 known= 1 KCL Eqn. + 1 s/node Eqn. + 2 contraint Eqns.
Solution (Continued…)
Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages.
Known node voltage: VV 123 KCL s/node V2
:oi
VV10
8
12
222
Hence 12805 2 oiV
S/node equation: 123 1 VVx
(1)
(2)
Solution (Continued…)
Constraint equations: 2VVx
(3)
and oo iVV
i 10108
12 12
(4)Hence 120108720 21 VVio
Substituting Eqn. (3) into (2) yields
123 21 VV
(2’)
Solution (Continued…)
Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above.
Solving Eqn. (1), (2’) and (4) simultaneously yields,
V1 = -6.51 V, V2 = 1.83 V and io = -0.264 A
(You can check this answer by calculator and Cramer’s rule).
Chapter 3, Problem 16.
Determine voltages v1 through v3 in the circuit of Fig. 3.64 using nodal analysis. (Ans:V1=18.86V, v2=6.29V, V3=13V)
Figure 3.64
Chapter 3, Problem 30.
Using nodal analysis, find vo and io in the circuit of Fig. 3.78. (Ans: Vo=-1.344kV, io=-5.6A)
Figure 3.78
Chapter 3, Problem 31.
Find the node voltages for the circuit in Fig. 3.79. (Ans: V1=4.97V, V2=4.85V, V3=-0.1212V)
Figure 3.79
Chapter 3, Problem 32.
Obtain the node voltages v1, v2, and v3 in the circuit of Fig. 3.80. (Ans:V1=2V, V2=12V, V3=-8V)
Figure 3.80
5 k
10 k4 mA
10 V 20 V
12 V
V1
V2
V3
Circuit Analysis Method
MESH ANALYSIS
Concept
Similar to nodal analysis. Developed based on the systematic
approach of Kirchhoff’s voltage law (KVL) to find all circuit variables without having to sacrifice any of the elements.
General procedure which is making use of mesh current in circuit analysis as key solutions.
Importance terms
Mesh Current: Assigned unknown current flows around the perimeter of the particular mesh/loop.
Element Current: Actual current thru any element or branch in the circuit.
When Mesh Current = Element Current?
Assumptions
KVL is performed in clockwise direction.Voltage rise – negative (subtract)Voltage drop– positive (add)
Mesh Analysis Procedures:
1. Label all independent meshes and assign proper unknown mesh currents in clockwise direction. Do the checklist.
2. Formulate KVL/Supermesh/Constraint Equation.
3. Solve the resulting simultaneous equations to obtain the unknown mesh current.
Applying Mesh Analysis on Simple Circuit
PP 3.5 (2 unknowns)Q: Find power dissipated in 12-resistor
and 3-resistor using mesh analysis.
12 V 8 V
2 9
12
4 3
Solution
1. Label all independent meshes and assign proper unknown mesh currents in clockwise direction.
12 V 8 V
2 9
12
4 3
I1 I2
Checklist:2 meshes = 2 KVL Eqns.
Solution (Continued…)
2. Formulate KVL/Supermesh/Constraint Eq.
KVL I1: 18I1 – 12I2 = 12
KVL I2: -12I1 + 24I2 = -8
(1)
(2)
Solution (Continued…)
3. Solve the resulting simultaneous equations to obtain the unknown mesh current.
I1 = 1/ I2 = 2/
288
)12)(12()24)(18(
2412
1218
192
)8)(12()24)(12(
248
12121
0
)12)(12()8)(18(
812
12182
Solution (Continued…)
Using calculator/Cramer’s rule we obtain:
I1 = 0.667 A and I2 = 0 A P12 = (I1 -I2)2(12) = 5.33 W P3 = I22(3) = 0 W Notice that the branch (3-resistor)
forming the outer most boundary of the circuit will have mesh current = element current.
Circuit with current sources and dependent sources
Two different effects depending on placement of voltage source in the circuit.
Does the presence of a current source complicate or simplify the analysis?
The presence of dependent source in the circuit need to impose constraint equation to describe the r/ship btw. dependent term of the dependent sources in relation to the mesh currents.
Case 1: Current source located at the outer most boundary Connecting mesh current immediately known. No need to apply KVL around that loop/mesh. Mesh Current = Element Current = Current
Source ValueR1
R412 V
R2 R3I1
I3I2Is
Immediately known mesh current,
I3 = -Is
Case 2: Current source located at the boundary between 2 meshes
Enclose the current source and combine the two meshes to form a SUPERMESH.
KVL is performed around the supermesh – do not consider voltage across cur. source.
Formulate s/mesh equation – express the r/ship btw mesh currents that form the s/mesh and cur.source that it encloses.
SUPERMESH
KVL S/Mesh I2/I3:
-12 + I2R2 +I3R3 = I3R4
R1
R412 V
R2 R3
3 mA
I1
I3I2
S/Mesh Eq:I3 – I2 = 3 mA
Practice Problem 3.7 (S/Mesh)
Use the mesh analysis to determine i1, i2 and i3.
Figure 3.25
Solution
Step 1: Checklist.
Checklist:3 meshes – I s/m = 2 KVL Eqns. + 1 s/node Eq.
Solution (Continued…)
Step 2: Formulate KVL/s.mesh equation.
KVL i3: -2i1 – 4i2 + 8i3 = 0
KVL s/mesh i1/i2:
2i1 + 12i2 – 6i3 = 6
S/Mesh Eq: i1 – i2 = 3
(1)
(2)
(3)
Solution (Continued…)
Step 3: Solve the simultaneous equations using Cramer rule or by calculator.
We obtain, i1 = 3.474 A
i2 = 0.4737 A i3 = 1.1053 A
Example 6 ( Known current & dependent source)
Find the voltage of the dependent source (CCCS).
9
8 V
9
12 5 ix
ix
Solution
Step 1: Assign mesh currents in CW direction and perform checklist.
I2
9
8 V
9
12 5 ix
I1
ixChecklist:2 meshes – I known = 1 KVL Eqn. + 1 CE
Solution (Continued…)
Step 2: Formulate KVL/Constraint equations.
Immediately known, I1 = 5ix KVL I2: 21I2 = -8
CE: ix = I2
(1)
(2)
(3)
Solution (Continued…)
Step 3: Solve the simultaneous equations.
Substitute (3) into (1) and solve (1) and (2) simultaneously, we obtain
I1 = 1.9048 A I2 = -0.3810 A
Solution (Continued…)
To find voltage across the CCCS, perform KVL around loop I1.
9
8 V
9
12
ix
+
V
-
I1
KVL I1: -V + 9I1 – 8 = 0 V = 9(1.9048) – 8 = 9.1432 V
Chapter 3, Problem 55.
In the circuit of Fig. 3.97, solve for i1, i2, and i3. (Ans: i1=-1A, i2=0A, i3=2A)
Figure 3.97
Circuit Theorem
SUPERPOSITION
Advantages
Use of superposition theorem: to find solution to circuits with multiple independent sources which are neither series nor parallel.
Advantage: no need to solve simultaneous equations (tedious computation for complex cct) in order to find the circuit variables by simplification techniques.
Concept
Concept: each independent source is treated independently and the algebraic sum is found to determine a particular unknown quantity or circuit variable of the circuit under study.
Superposition Theorem
ST states that: “ The current or voltage of any
element in a bilateral circuit is equal to the algebraic sum of the currents or voltages produced independently by each source.”
Principle of Operation
To consider the effect of each source independently requires that source to be removed and replaced without affecting the final results.
To remove voltage source – s.c the terminals.
To remove current source – o.c the terminals.
Any dependent source treated as though they are passive element (must be left intact during the process).
Example 6 (P3.5)
Find the voltage across the 12 resistor using superposition hence the power dissipated by this resistor.
12 V 8 V
2 9
12
4 3
+V-
Solution
i) Consider 12V/removed 8V.
12 V
2 9
12
4 3
+V’-
s.c
V
VxV
6
122412//12
12//12'
Solution (Continued…)
ii) Replace 8V/removed 12V
8 V
2 9
12
4 3
+V’’-
s.c
V
VxV
2
83912//6
12//6''
Solution (Continued…)
Hence, V = V’ + V’ = 6V + 2V = 8V.
P12 = V2/R = 82/12 = 5.33W
Example 7
Find the current in the 23 resistor using the concept of superposistion.(Ans:11.23 A)
200 V
4
27 23
47
20 A
I
Exercise 1
Using superposition, find the voltage V in the circuit? (Ans: 40V)
2R5
20 10
6 A
I2100 V40
+
V
-
Circuit Theorem
SOURCE TRANSFORMATON
What benefits from source transformation?
Another tool to simplify circuit – the simpler the cct, the easier will be the solution.
How to simplify? – rearrange the resistors/sources by S.Trans so that they end up with series/parallel connections.
Principle of Operation
The terminal v-i characteristics must retain before and after transformation as this concept is based on equivalence.
S.Trans also applies to dependent sources.
It does not affect the remaining part of the circuit.
Definition
A Source Transformation is the process of replacing a voltage source Vs in series with resistor Rs by a current source is in parallel with the same resistor Rs or vice versa.
Equivalent Circuits
The connections of each case should be between the same terminals before and after transformation.
Vs
Rs
Is
x
y
Rs
x
y
sss RIV s
ss R
VI
In order for the
circuits in the left and
right sides to be
equivalent:
Example 8
Use series of source transformations to find io in the circuit below.
10 V
2 5
40
4
4 A
10 A
i1
Solution
Transform 4A and 5 into voltage source.
10 V
2
5
40
4
10 A
i1
20 V
Solution (cntd…)
Transform 10 A and 1 into voltage source.
Transform 10 V and 40 into current source.
40
9 i1
20 V
10 V
2 0.25 A
Solution (ctnd…)
Transform 10V and 10 into current source.
40
i
2 10 1 A
0.25 A
Solution (cntd…)
Combine the current sources 2A and 0.25A.
Combine resistors 10 and 40.
Solve for I using CDR.
8 2
i
1.25 A
AAxi 125.110
8
Practice Problem 4.6
Find io in the circuit of Figure 4.19 using source transformation. (Ans:1.78A)
5 V
6 3 7 4
1
5 A
3 A
io
Practice Problem 4.7
Use S.Trans to find ix in the cct shown in Figure 4.22. (Ans:1.176A)
10
5
2ix
ix4 A
Exercise 2
Use STrans to find Vo. (Ans:-135V)
8
80
340 V
5
10
45 5 A
+
Vo
-
20
Circuit Theorem
THEVENIN’S THEOREM
Purpose
Used when we are interested ONLY in the terminal behavior of the circuit particularly where a variable load is connected to.
Provides a technique to replaced the fixed part of the circuit by a simple equivalent circuit.
Avoid the re-do on the analysis of the entire circuit except for the changed load.
Thevenin’s Theorem
States that “ A linear two-terminal circuit can be replaced by an equivalent circuit consisting of an open-circuit voltage source at the terminals, VTh in series with a resistor RTh where RTh is the input or equivalent resistance at the terminals when all independent source are turned off.”
Replacing linear two-terminal (a-b) circuit by its Thevenin equivalent
Original circuit
Replacing linear two-terminal (a-b) circuit by its Thevenin equivalent (Cntd)
Thevenin equivalent circuit
VTh - Thevenin voltageRTh - Thevenin resistance
Procedures to obtain VTh and RTh
Step 1: Priliminary – Omitting load resistor RL (Not applicable if no load resistor)
Step 2: Find RTh – setting all independent sources to zero. Find the resultant resistance between the marked terminals.
Voltage source – s.cCurrent sorce – o.c
Step 3: Find VTh – calculate VTh by returning all sources back to their original positions. Find the o.c voltage btw the marked terminals using the method which takes least effort.
Example 9
Find the Thevenin equivalent between terminal a-b. (Ans: VTh=32V, RTh=8)
5
20 25 V
4
a
b
3 A
Example 10
Find the Thevenin equivalent circuit at the terminal a-b of the circuit below. (Ans: VTh=-4.8V, RTh=2.4)
a
b
8 V3
2
4
6
Practice Problem 4.8
Use the Thevenin’s theorem to find the equivalent circuit to the left of the terminals a-b in the circuit below. Then find i. (Ans: VTh=6V, RTh=3, i=1.5A)
i
12 V
6 6
1 4 2 A
a
b
Circuit Theorem
NORTON’s THEOREM
Norton’s Theorem
The purpose of its use is similar to the Thevenin’s theorem.
States that “A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a short circuit current source through the terminals, IN in parallel with a resistor RN where RN is the input or equivalent resistance at the terminals when all independent source are turned off.”
Replacing linear two-terminal (a-b) circuit by its Norton equivalent
(a) Original Circuit (b) Norton Equivalent Circuit
Finding Norton current, IN
Procedures to obtain VTh and RTh
Step 1: Priliminary – Omitting load resistor RL (Not applicable if no load resistor)
Step 2: Find RTh – setting all independent sources to zero. Find the resultant resistance between the marked terminals.
Voltage source – s.cCurrent sorce – o.c
Step 3: Find VTh – calculate VTh by returning all sources back to their original positions. Find the o.c voltage btw the marked terminals using the method which takes least effort.
Procedures to obtain IN and RN
Step 1: Priliminary – Omitting load resistor RL (Not applicable if no load resistor)
Step 2: Find RN – setting all independent sources to zero. Find the resultant resistance between the marked terminals.
Voltage source – s.cCurrent sorce – o.c
Step 3: Find IN – calculate IN by returning all sources back to their original positions. Find the o.c s.c current btw the marked terminals using the method which takes least effort.
NORTON EQUIVALENT CIRCUITNORTON EQUIVALENT CIRCUIT
A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance.
Can be derive from a Thevenin equivalent circuit simply by making a source transformation.
Norton current, IN = the short-circuit current at the terminal of interest.
Norton resistance, RN = Thevenin resistance, RTh
Series resistors combined, producing theThevenin equivalent circuit
V32
8a
b
THEVENINEQUIVALENT
CIRCUIT
Producing the Norton equivalent circuit
a
b
A4 8
NORTONEQUIVALENT
CIRCUIT
Example 10
Find the Norton Equivalent circuit with respect to the terminals a-b.
12 9
2
4
a
b
9 A
Practice Problem 4.11
Find the Norton equivalent circuit in Figure below. (Ans: IN=4.5A, RN=3)
15 V
3 3
6 4 A
a
b
RL
Circuit Theorem
THEVENIN & NORTON’s THEOREMS WITH
DEPENDENT SOURCE
Procedures to obtain VTh/IN and RTh/RN
Step 1: Priliminary – Omitting load resistor RL
(Not applicable if no load resistor) Step 2: Find VTh = Vo.c or IN = Is.c using the
method that takes the least effort. Step 3: Find RTh/RN
Method 1:If circuit contains independent source.
Rth = RN = Vo.c/Is.c = VTh/IN Method 2: If circuit contains independent source and without independent source.
Using Method 2 to find RTh/RN
Turn off all independent sources but dependent sources left intact because they are controlled by circuit variables.
Because of the presence of the dependent source, we excite the circuit with a voltage source or current source between the terminals.
Set Vo=1V to ease calculation since the circuit is linear.
Goal? To find io so that RTh=RN=1/io Alternatively, we may set io=1A. Goal? To find Vo so that RTh=RN=Vo/1
Illustration of Method 2 to find RTh/RN
Example 11
Find the Thevenin and Norton equivalent circuit for the circuit containing dependent sources below between terminals a-b. (Ans: VTh=-5V, RTh=100, IN=-50mA)
2 k
5 V25
i
3Vo 20io
a
b
+
Vo
-
Example 12
Find the Thevenin and Norton equivalent circuit for the circuit containing dependent source below between terminals a-b. (Ans: VTh=20V, RTh=0.625, IN=32A)
5
40 V 1
2ix
8 A
ix
a
b
Practice Problem 4.9
Find the Thevenin equivalent circuit of the circuit in Figure 4.34 the left of terminals a-b. (Ans: VTh=5.33V, RTh=0.44)
6 V
3 5
4 1.5Ix
a
b
Practice Problem 4.10
Obtain the Thevenin equivalent of the circuit in Figure 4.36. (Ans: VTh=0V, RTh=-7.5)
10
5 15
4Vx
+
Vx
-
a
b
Circuit Theorem
MAXIMUM POWER TRANSFER
Introduction
Power transfer from source to the load can be analyzed and discussed from two basic types of systems:1. Efficiency – eg: power utility systems concerned with generation, transmission and distribution of large quantities of electric power.2. Amount – eg. Comm. & instrumentation sys because in the transmission of info or data via electric signals, the power available at the transmitter or detector is limited or small.
At this moment our concern is on the 2nd type of system that is the amount of maximum power transfer in purely resistive circuit.
Thevenin Equivalent Circuit
The Thevenin equivalent circuit is useful in finding the max. power a linear cct. can deliver to a load.
The entire cct can be replaced by its Thevenin eq. except for the adjustable load.
Thevenin Equivalent Circuit used for maximum power Transfer
VTh and RTh are fixed
LLTh
ThL R
RR
VRip
2
2
Power delivered to the load,
(1)
Power delivered to the load as a function of RL.
Maximum Power Theorem
Maximum Power is transferred to the load when the load resistance, equals to the Thevenin resistance as seen from the load (RL = RTh)
Proving Maximum Power Transfer Theorem
Differentiate p in Eq.(1) with respect to RL and set the result equal to zero,
0)(
2(
)(
)(2)(
3
2
4
22
LTh
LLThTh
LTh
LThLLThTh
L
RR
RRRV
RR
RRRRRV
dR
dp
(2)
Proving Maximum Power Transfer Theorem (Cntd…)
Implies that, 0 = (RTh + RL -2RL) = (RTh – RL) Yields, RL = Rth
Eq (3) gives the maximum power by showing that d2p/dRL
2 < 0.
(3)
Maximum Power Formula
Substituting Eq.(3) into (1) to obtain the maximum power transfer,
Eq.(4) applies only when RL = RTh. When RL ≠ RTh, compute power from Eq.(1)
hT
Th
R
Vp
4
2
max (4)
Practice Problem 4.13
Determine the value of RL that will draw the maximum power from the rest of the circuit. Calculate the maximum power.
(Ans: 4.22, 2.901W)
9 V
2
1 RL
4
3Vx
Vx
Example 13
The variable resistor in the circuit below is adjusted for maximum power. Find the value of RL and the maximum power. (Ans: 5k, 45mW)
18 V
6 k 1k25
10 kRL
a
b
5 mA
Exercise 3 The load resistance in both circuits below are
adjusted until maximum power is delivered. Find the power delivered to the loads and the value of both RL. (Ans: 600, 38.4mW)
RL
500 1 k
I210 V
1 k
2 mA
(a)
Exercise 3
30 20
10 RL50 V
a b
(b) (Ans: 21.7, 0.8W)