LECTURE 2
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Circuit Analysis Techniques
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Circuit Analysis Techniques
•
•
•
•
•
•
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Voltage Division Principle
Current Division Principle
Nodal Analysis with KCL
Mesh Analysis with KVL
Superposition
Thévenin Equivalence
Norton’s Equivalence
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Circuit Analysis using Series/Parallel Equivalents
1. Begin by locating a combination of resistances that are
in series or parallel. Often the place to start is farthest
from the source.
2. Redraw the circuit with the equivalent resistance for
the combination found in step 1.
3. Repeat steps 1 and 2 until the circuit is reduced as far
as possible. Often (but not always) we end up with a
single source and a single resistance.
4. Solve for the currents and voltages in the final
equivalent circuit.
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Working Backward
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Voltage Division
R1
v1  R1i 
v total
R1  R2  R3
R2
v 2  R2 i 
v total
R1  R2  R3
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Application of the VoltageDivision Principle
R1
v1 
vtotal
R1  R2  R3  R4
1000

15
1000  1000  2000  6000
 1.5V
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Current Division
R2
v
i1 

itotal
R1 R1  R2
R1
v
i2 

itotal
R2 R1  R2
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Application of the Current-Division Principle
R2 R3
30  60
Req 

 20
R2  R3 30  60
Req
20
i1 
is 
15  10A
R1  Req
10  20
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•Voltage division
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•Voltage division and
•current division
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•Current division
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Although they are very
important concepts,
series/parallel equivalents
and
the current/voltage division
principles are not sufficient
to solve all circuits.
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Node Voltage (Nodal) Analysis
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Definitions of node and Supernode
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Obtain values for the unknown voltages
across the elements in the circuit below.
At node 1
At node 2
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v1
v v
 1 2  3.1
2
5
v2
v2  v1

 - (-1.4)
1
5
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Writing KCL Equations in Terms of the Node
Voltages for Figure 2.16
v1  v s
node 1
node 2
v 2  v1 v2 v 2  v3


0
R2
R4
R3
node 3
v3  v1 v3 v3  v2


0
R1
R5
R3
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node 1
v1 v1  v 2

 is  0
R1
R2
node 2
v 2  v1 v 2 v 2  v3


0
R2
R3
R4
node 3
v3 v3  v2

 is
R5
R4
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No. of unknown: v1, v2, v3
No. of linear equation : 3
Setting up nodal equation with
KCL at Node 1, Node 2, Node 3
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No. of unknown: v1, v2, v3
No. of linear equation : 3
Setting up nodal equation with
KCL at Node 1, Node 2, Node 3
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Problem with node 3, it is rather
hard to set the nodal equation at
node 3, but still solvable. Why?
As there is no way to determine
the current through the voltage
source, but v3=Vs
v3
Problem with node 3, it is rather
hard to set the nodal equation at
node 3 but still solvable.
Same as before.
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May Not Be That Simple
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Circuits with Voltage Sources
We obtain dependent equations if we use all
of the nodes in a network to write KCL
equations.
Any branch with a voltage source:
• define SUPERNODE, sum all current either
in or out at the supernode with KCL
• use KVL to set up dependent equation
involving the voltage source.
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(a) The circuit of Example 4.2 with a
22-V source in place of the 7-
resistor. (b) Expanded view of the
region defined as a supernode; KCL
requires that all currents flowing
into the region must sum to zero, or
we would pile up or run out of
electrons.
At node 1:
83 
v1  v2 v1  v3

3
4
At the “supernode:”
v2  v1 v3  v1 v3 v2

 
3
4
5 1
v3  v2  22
3  25 
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A
B
Summing all the current out
from the supernode A
For supernode A, EXCLUDE THE SOURCE
v1 v1  (15)


R2
R1
v2 v2  (15)

0
R4
R3
Why?
As the current via the 10V source is equal to the current via
R4 plus the current via R3
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B
For supernode B, EXCLUDE THE SOURCE
v1 v1   15 v 2 v2   15



0
R2
R1
R4
R3
Summing all the current into the supernode B
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v1 v1  (15)


R2
R1
v2 v2  (15)

0
R4
R3
-v1 -10 + v2 = 0
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Any branch with a voltage source:
• define supernode, sum all current either in or out at the
supernode with KCL
• use KVL to set up dependent equation involving the
voltage source.
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v1 v1  v3 v2 v 3


1
R1
R2
R3
 v1  10  v2  0
v3  v1 v3  v 2
v3


0
R2
R3
R4
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v1 v3

1
R1 R4
 v1  10  v2  0
v1 v3

1
R1 R4
v1 v1  v3 v2 v 3


1
R1
R2
R3
v3  v1 v3  v 2
v3


0
R2
R3
R4
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Node-Voltage Analysis with a
Dependent Source
•First, we write KCL equations at each node, including
the current of the controlled source just as if it were
an ordinary current source.
•Next, we find an expression for the controlling
variable ix in terms of the node voltages.
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v1  v 2
 is  2i x
R1
v3  v 2 v3

 2i x  0
R3
R4
v2  v1 v2 v2 v 3


0
R1
R2
R3
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v1  v 2
 is  2i x
R1
v2  v1 v2 v2 v 3


0
R1
R2
R3
v3  v 2 v3

 2i x  0
R3
R4
v3 v 2
ix 
R3
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Substitution yields
v3 v 2
v1  v 2
 is  2
R1
R3
v2  v1 v2 v2 v 3


0
R1
R2
R3
v3  v 2 v3
v3  v 2

2
0
R3
R4
R3
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Node-Voltage Analysis
1. Select a reference node and assign variables for the unknown node
voltages. If the reference node is chosen at one end of an independent
voltage source, one node voltage is known at the start, and fewer need to be
computed.
2. Write network equations. First, use KCL to write current equations for nodes
and supernodes. Write as many current equations as you can without using all
of the nodes. Then if you do not have enough equations because of voltage
sources connected between nodes, use KVL to write additional equations.
3. If the circuit contains dependent sources, find expressions for the controlling
variables in terms of the node voltages. Substitute into the network equations,
and obtain equations having only the node voltages as unknowns.
4. Put the equations into standard form and solve for the node voltages.
5. Use the values found for the node voltages to calculate any other currents or
voltages of interest.
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Step 1.Reference node
Step 1. v1
Step 2.
Step 1 v2
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v 2 v1  v2

1
10
5
v1  10
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Step 1
v1
v2
Step 1
Step 2
v3
node 1
supernode
Step 1
ref supernode
Step 3
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v1  v3 v1  v2
3

5
10
v1  v2  v2 v1  v3  v3



0
10
2
5
5
v2  v3  2i y
v1  v3
iy 
5
v2  v3 v1  v3

2
5
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Mesh Current Analysis
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Definition of a loop
Definition of a mesh
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Choosing the Mesh Currents
When several mesh currents flow through
one element, we consider the current in
that element to be the algebraic sum of
the mesh currents.
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Writing Equations to Solve
for Mesh Currents
If a network contains only resistors
and independent voltage sources, we
can write the required equations by
following each current around its
mesh and applying KVL.
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For mesh 1, we have
R2 i1  i3   R3 i1  i2   v A  0
For mesh 2, we obtain
R3 i2  i1   R4i2  vB  0
For mesh 3, we have
R2 i3  i1   R1i3  v B  0
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Determine the two mesh currents, i1 and i2, in the circuit below.
For the left-hand mesh,
-42 + 6 i1 + 3 ( i1 - i2 ) = 0
For the right-hand mesh,
3 ( i2 - i1 ) + 4 i2 - 10 = 0
Solving, we find that i1 = 6 A and i2 = 4 A.
(The current flowing downward through
the 3- resistor is therefore i1 - i2 = 2 A. )
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Mesh Currents in Circuits
Containing Current Sources
*A common mistake is to assume the voltages across
current sources are zero. Therefore, loop equation
cannot be set up at mesh one due to the voltage across
the current source is unknown
Anyway, the problem is
still solvable.
i1  2
10(i2  i1 )  5i2  10  0
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As the current source common to two mesh, combine meshes 1
and 2 into a supermesh. In other words, we write a KVL equation
around the periphery of meshes 1 and 2 combined.
It is the supermesh.
i1  2i1i3   4i2  i3   10  0
Mesh 3:
3i3  4i3  i2   2i3  i1   0
i2  i1  5
Three linear equations
and three unknown
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Find the three mesh currents in the circuit below.
Creating a “supermesh” from meshes 1 and 3:
-7 + 1 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0
[1]
Around mesh 2:
1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0
[2]
Finally, we relate the currents in meshes 1 and 3:
i1 - i3 = 7
[3]
Rearranging,
i1 - 4 i2 + 4 i3 = 7
[1]
-i1 + 6 i2 - 3 i3 = 0
[2]
i1
[3]
- i3 = 7
Solving,
i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.
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supermesh of mesh1 and mesh2  20  4i1  6i2  2i2  0
branch current
v x  2i2
current source
vx
 i2  i1
4
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 20  4i1  6i2  2i2  0
v x  2i2
vx
 i2  i1
4
Three equations and three unknown.
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Mesh-Current Analysis
1. If necessary, redraw the network without crossing conductors or elements. Then
define the mesh currents flowing around each of the open areas defined by the network.
For consistency, we usually select a clockwise direction for each of the mesh currents,
but this is not a requirement.
2. Write network equations, stopping after the number of equations is equal to the
number of mesh currents. First, use KVL to write voltage equations for meshes that do
not contain current sources. Next, if any current sources are present, write expressions
for their currents in terms of the mesh currents. Finally, if a current source is common to
two meshes, write a KVL equation for the supermesh.
3. If the circuit contains dependent sources, find expressions for the controlling variables
in terms of the mesh currents. Substitute into the network equations, and obtain
equations having only the mesh currents as unknowns.
4. Put the equations into standard form. Solve for the mesh currents by use of
determinants or other means.
5. Use the values found for the mesh currents to calculate any other currents or voltages
of interest.
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Superposition
• Superposition Theorem – the response of a
circuit to more than one source can be
determined by analyzing the circuit’s
response to each source (alone) and then
combining the results
Insert Figure 7.2
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Superposition
Insert Figure 7.3
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Superposition
• Analyze Separately, then Combine Results
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Use superposition to find the current ix.
Current source is zero – open circuit as I = 0 and solve iXv
Voltage source is zero – short circuit as V= 0 and solve iXv
i X  i Xv  i Xc
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Use superposition to find the current ix.
The controlled voltage source is included in all cases as
it is controlled by the current ix.
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Voltage and Current Sources
Insert Figure 7.7
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Voltage and Current Sources
Insert Figure 7.8
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Voltage and Current Sources
Insert Figure 7.9
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Source Transformation
iL
RS
VS
+
+
_
VL
RL
_
iL
+
IS
RP
VL
RL
_
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Under what condition, the voltage and
current of the load is the same when
operating at the two practical sources?
For voltage source
VS
VL 
RL
R S  RL
For current source
i S RP
,
VL 
RL
RP  RL
We have,
VS
i S RP

R S  RL RP  RL
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VS
R P  RS , i S 
RS
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Voltage and Current Sources
• Equivalent Voltage and Current Sources – for every
voltage source, there exists an equivalent current
source, and vice versa
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Thevenin’s Theorem
• Thevenin’s Theorem – any resistive circuit
or network, no matter how complex, can be
represented as a voltage source in series
with a source resistance
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Thevenin’s Theorem
• Thevenin Voltage (VTH) – the voltage
present at the output terminals of the circuit
when the load is removed
Insert Figure 7.18
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Thevenin’s Theorem
• Thevenin Resistance (RTH) – the resistance
measured across the output terminals with
the load removed
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Thévenin Equivalent
Circuits
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Thévenin Equivalent
Circuits
Vt  voc
voc
Rt 
isc
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Thévenin Equivalent Circuits
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Finding the Thévenin
Resistance Directly
When zeroing a voltage source, it becomes
a short circuit. When zeroing a current
source, it becomes an open circuit.
We can find the Thévenin resistance by
zeroing the sources in the original
network and then computing the resistance
between the terminals.
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Computation of Thévenin resistance
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Equivalence of open-circuit and Thévenin voltage
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A circuit and its Thévenin equivalent
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Superposition
As the voltage source does not contribute any output voltage,
Only the current source has the effect.
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Determine the Thévenin and Norton Equivalents of Network A in (a).
Source transformation
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Find the Thévenin equivalent of the circuit shown in (a).
v
As i = -1, therefore, the controlled voltage source is -1.5V.
Use nodal analysis at node v,
v  (1.5) v
  1, v  0.6
3
2
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Thus,
Rth =v/I = 0.6/1 = 0.6 ohms
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Applications of Thevenin’s Theorem
• Load Voltage Ranges – Thevenin’s theorem
is most commonly used to predict the
change in load voltage that will result from
a change in load resistance
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Applications of Thevenin’s Theorem
• Maximum Power Transfer
– Maximum power transfer from a circuit to a
variable load occurs when the load resistance
equals the source resistance
– For a series-parallel circuit, maximum power
occurs when RL = RTH
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Applications of Thevenin’s Theorem
• Multiload Circuits
Insert Figure 7.30
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Norton’s Theorem
• Norton’s Theorem – any resistive circuit or network,
no matter how complex, can be represented as a
current source in parallel with a source resistance
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Norton’s Theorem
• Norton Current (IN) – the current through
the shorted load terminals
Insert Figure 7.35
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Computation of Norton current
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Norton’s Theorem
• Norton Resistance (RN) – the resistance
measured across the open load terminals
(measured and calculated exactly like RTH)
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Norton’s Theorem
• Norton-to-Thevenin and Thevenin-to-Norton
Conversions
Insert Figure 7.39
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Step-by-step Thévenin/NortonEquivalent-Circuit Analysis
1. Perform two of these:
a. Determine the open-circuit voltage Vt = voc.
b. Determine the short-circuit current In = isc.
c. Zero the sources and find the Thévenin
resistance Rt looking back into the
terminals.
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2. Use the equation Vt = Rt In to compute
the remaining value.
3. The Thévenin equivalent consists of a
voltage source Vt in series with Rt .
4. The Norton equivalent consists of a
current source In in parallel with Rt .
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Maximum Power Transfer
The load resistance that absorbs the
maximum power from a two-terminal
circuit is equal to the Thévenin
resistance.
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Power transfer between source
and load
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Graphical representation of
maximum power transfer
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