Transcript Chapter 2: DC Circuit Analysis

CHAPTER 2: DC Circuit Analysis
and AC Circuit Analysis
Methods of Circuit
Analysis and Circuit
Theorems
Introduction




Motivation(1) & Motivation(2)
Nodal Analysis (Node-Voltage
Method)
Mesh Analysis (Mesh-Current
Method)
Thevenin’s Theorem
Motivation (1)
If you are given the following circuit, how can we
determine (1) the voltage across each resistor,
(2) current through each resistor. (3) power
generated by each current source, etc.
What are the things which we need to
know in order to determine the answers?
Motivation (2)
Things we need to know in solving any
resistive circuit with current and
voltage sources only:
• Kirchhoff’s Current Laws (KCL)
• Kirchhoff’s Voltage Laws (KVL)
• Ohm’s Law
How should we apply these laws to
determine the answers?
Introduction (Continued…)

1.
2.
3.
4.
There are four ways of solving
simultaneous equations:
Cramer’s rule
Calculator (real numbers only)
Normal substitution and elimination
(not more than two equations)
Computer program packages: mathcad,
maple, mathematica etc.
Circuit Analysis Method
NODAL ANALYSIS
(Node-Voltage
Method)
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.
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.
Step to determine Node
Voltages:
1.
2.
3.
Mark all essential nodes and assign proper
voltage designations except for the
appointed reference node.
Apply KCL to each non-reference nodes. Use
Ohm’s law to formulate the equation in
terms of node voltages. Assume all
unknown currents are directing out of the
nodes.
Solve the resulting simultaneous equations
to obtain the unknown node voltages.
Applying Nodal Analysis on
Simple Circuit
Example 1 (3 unknowns)
Is2
R1
Is1
R2
R3
R4
Solution:
Step 1:
 Mark all essential
nodes
V1
 Assign unknown
I
node voltages
 Indicate the
reference node.
s1
Is2
R1
V2
R3
R2
V3
R4
Solution (continued…)

Step 2: Perform KCL at each marked
non-reference nodes using Ohm’s law to
formulate the equations in terms of the
node voltages.
Solution (continued…)

KCL V1:
or
I s1  I s 2
V1  V2

R1
(1)
I s 1  I s 2  G1 (V1  V2 )

KCL V2:
V2  0 V2  V3
V2  V1


 0 (2)
R1
R3
R2

KCL V3:
I s2 
V3  V2
V 0
 3
0
R2
R4
(3)
Solution (continued…)

Step 3: Solve the
resulting
simultaneous
equations from step
2 above.
3 mA
10 k
V1
2 mA
V2
5 k
4 k
V3
2 k
Solution (continued…)

KCL V1:
2m  3m 
Simplify to

KCL V2:
Simplify to

KCL V3:
Simplify to
V1  V2
10k
V1 - V2 = 50
(1)
V2  V1 V2  0 V2  V3


0
10k
4k
5k
-2V1 + 11V2 - 4V3 = 0
V3  V2 V3  0
3m 

0
5k
2k
-2V2 + 7V3 = -30
(2)
(3)
Solution (continued…)

Cramer’s rule: Put the equations in matrix forms.
Col-1
a1 b1 c1  V1  d1 
a b c  V   d 
 2 2 2  2  2
a3 b3 c3  V3  d3 
=
Col-2
Col-3
1  1 0  V1   50 
 2 11  4 V    0 

  2 

0  2 7  V3   30
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
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Middle Matrix:
Right Matrix:
Unknown variables i.e. V1, V2 and V3
Constants i.e. d1, d2 and d3
Solution (continued…)
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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:
1 1 0 1
c2   2 11  4 -2
0 2 7 0
c3
a1 b1 c1
  a 2 b2
a3 b3
-1
11
-2
  (1)(11)(7)  (1)( 4)(0)  (0)( 2)( 2) 
(0)(11)(0)  (2)(4)(1)  (7)(2)(1)}
 77  (22)  55
Solution (continued…)

Determinant 1 when coefficients
for V1 is replaced by the constants.
50  1 0 50
-1
c2  0 11  4 0 11
-30 -2
c3  30  2 7
d1 b1 c1
1  d 2 b 2
d 3 b3
1  (50)(11)(7)  (1)( 4)( 30)  (0)(0)( 2) 
(30)(11)(0)  (2)(4)(50)  (7)(0)(1)}
 3730  (400)  3330
Solution (continued…)

Determinant 2 when coefficients
for V2 is replaced by the constants.
a1 d1 c1
 2  a2 d 2
a3 d 3
1 50 0
c2   2 0  4
0  30 7
c3
1
50
-2
0
0 -30
 2  (1)(0)(7)  (50)( 4)(0)  (0)( 2)( 30) 
(0)(0)(0)  (30)(4)(1)  (7)(2)(50)}
  (580)  580
Solution (continued…)

Determinant 3 when coefficients
for V3 is replaced by the constants.
1 1
50
d 2   2 11
0
0  2  30
d3
a1 b1 d1
 3  a2 b2
a3 b3
1
-2
0
-1
11
-2
 3  (1)(11)(30)  (1)(0)(0)  (50)(2)(2) 
(0)(11)(50)  (2)(0)(1)  (30)(2)(1)}
  130  (60)   70
Solution (continued…)
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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)/R1
PR1 = IR12R1 = VR12/R1
VR2= V2 – V3
IR1 = (V2– V3)/R2
PR2 = IR22R2 = VR22/R2
**VR3= V2
IR1 = V2/R3
PR3 = IR32R3 = VR32/R3
**VR4= V3
IR4 = V32/R4
PR4 = IR42R3 = 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 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 non-reference essential
nodes.
R
E
P
U
S
Nonreference
essential node
V1
Supernode
Equation:
E
D
NO
Vs
V2
Nonreference
essential node
Vs  V1  V2
Case 2: Voltage source between
a reference essential node and a
non-reference essential node.
E
D
NO
N GE
W
O LTA
N
K VO
Nonreference
essential node
V1
Known node voltage:
Vs
0V
Reference
essential node
V1  Vs
Case 3: Voltage source between
an essential node and a nonessential node.
Vs
Nonreference
essential node
V1
FIN
NO D N
N- OD
ES E
TE SE VO
RM NT LT
S O IAL AG
VO
N E
LT F NO OD AT
Va AG D E IN
E E
Non-essential
node
Node voltage at
non-essential
node:
Va  V1  Vs
Solution Case 2 (Known
node voltage)

Step 1: Mark essential nodes and assign
unknown node voltages and indicate the
reference node.
10 V
i
5
8
50 
30 
3A
V2
KNOWN NODE
VOLTAGE V1
40 
Checklist:
3 essential nodes –
1 ref node
– 1 known node
voltage
= 1 KCL Eqn.
Solution (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:
V1  10 V
V2  10 V2
KCL V2:

 3
80
8
11V2   230
(1)
Solution (Continued…)

Step 3: Solve the resulting simultaneous
equations which have been simplified in step 2
above.
Solving Eqn. (1) yields,
 230
V2 
  20.91 V
11
Finding current through the voltage source,
KCL at V1:
V1 V1  V2

i0
40
80
 10 10  20.91
i

  0.636 A
40
80
Solution (Continued…)
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
Hence,
P10-V = Vi= (10)(-0.636) = -6.36 W.
(Delivering energy)
Circuit Analysis Method
MESH ANALYSIS
(Mesh-Current
Method)
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
through any element or branch in
the circuit.
Assumptions

KVL is performed in clockwise direction.
Voltage rise – negative (subtract)
Voltage drop – positive (add)
Mesh Analysis Procedures:
1.
2.
3.
Label all independent meshes and
assign proper unknown mesh currents
in clockwise direction. Do the checklist.
Formulate KVL/Supermesh/Constraint
Equation.
Solve the resulting simultaneous
equations to obtain the unknown mesh
current.
Applying Mesh Analysis on Simple
Circuit

Example (2 unknowns)
Question: Find power dissipated in 12resistor and 3-resistor using mesh
analysis.
2
9
12 
12 V
4
3
8V
Solution:
1.
Label all independent meshes and assign proper
unknown mesh currents in clockwise direction.
2
12 V
I1
12 
4
9
I2
3
Checklist:
2 meshes = 2 KVL
8 V 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/
12  12
1 
I2 = 2/
8
24
18  12

 12 24
 (18)( 24)  (12)( 12)
 288
 (12)( 24)  (12)( 8)
 192
18 12
2 
 12  8
 (18)( 8)  (12)(12)
0
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.
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 Value
R1
R2
Is
I2
I1
Immediately known
mesh current,
R3
12 V
I3
I3 = -Is
R4
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 current source.
Formulate supermesh equation – express the
relationship between mesh currents that form
the supermesh and current source that it
encloses.
TIPS: Nodal versus Mesh Analysis
To select the method that results in the smaller
number of equations. For example:
1. Choose nodal analysis for circuit with fewer nodes
than meshes.
*Choose mesh analysis for circuit with fewer
meshes than nodes.
*Networks that contain many series connected
elements, voltage sources, or supermeshes are
more suitable for mesh analysis.
*Networks with parallel-connected elements,
current sources, or supernodes are more
suitable for nodal analysis.
2. If node voltages are required, it may be expedient
to apply nodal analysis. If branch or mesh
currents are required, it may be better to use
mesh analysis.
Circuit Theorem
SOURCE
TRANSFORMATION
What benefits from source
transformation?


Another tool to simplify circuit – the
simpler the circuit, the easier will be the
solution.
How to simplify? – rearrange the
resistors/sources by Source
Transformation so that they end up with
series/parallel connections.
Definition


An equivalent circuit is one whose v-i
characteristics are identical with the
original circuit.
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
Is 
Rs
(a) Independent source transform
Vs  I s Rs
(b) Dependent source transform
• The arrow of the
current source is
directed toward
the positive
terminal of the
voltage source.
• The source
transformation is
not possible when
R = 0 for voltage
source and R = ∞
for current source.
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


It states that a linear two-terminal circuit (Fig.
a) can be replaced by an equivalent circuit
(Fig. b) consisting of a voltage source VTH in
series with a resistor RTH,
where
• VTH is the open-circuit voltage at the
terminals.
• RTH is the input or equivalent
resistance at the terminals when the
independent sources are turned off.
Procedures to obtain VTh
and RTh



Step 1: Preliminary – 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 – short circuit (s.c.)
Current sorce – open circuit (o.c.)
Step 3: Find VTh – calculate VTh by returning
all sources back to their original positions.
Find the open circuit voltage between the
marked terminals using the method which
takes least effort.
Exercise 1:

Find the Thevenin equivalent between terminal
a-b. (Ans: VTh=32V, RTh=8)
5
4
a
25 V
20 
3A
b
Exercise 2:

Find the Thevenin equivalent circuit at the
terminal a-b of the circuit below. (Ans: VTh=-
4.8V, RTh=2.4)
4
a
6
3
b
8V
2