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EEE1012
Introduction to Electrical &
Electronics Engineering
Chapter 2: Circuit Analysis Techniques
by Muhazam Mustapha, July 2010
Learning Outcome
By the end of this chapter students are
expected to:
• Understand and perform calculation on
circuits with mesh and nodal analysis
techniques and superposition
• Be able to transform circuits based on
Thevenin’s or Norton’s Theorem as
necessary
Chapter Content
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Mesh Analysis
Nodal Analysis
Linearity and Superposition
Source Conversion
Thevenin’s Theorem
Norton’s Theorem
Mesh
Mash Analysis
Mesh Analysis
Steps:
• Assign a distinct current in clockwise direction
to each independent closed loop of network.
• Indicate the polarities of the resistors depending
on individual loop.
• [*] If there is any current source in the loop path,
replace it with open circuit – apply KVL in the
next step to the resulting bigger loop. Use back
the current source when solving for current.
Mesh Analysis
Steps: (cont)
• Apply KVL on each loop:
– Current will be the total of all direction
– Polarity of the sources will maintained
• Solve the simultaneous equations.
Mesh Analysis
Example: [Boylestad 10th Ed. E.g. 8.11 - modified]
R2
R1
R3 4Ω
2Ω
I1
2V
a
Ia
I3
1Ω
b
I2
Ib
6V
Mesh Analysis
Example: (cont)
Loop a: 2 = 2Ia+4(Ia−Ib) = 6Ia−4Ib
Loop b: −6 = 4(Ib−Ia)+Ib = −4Ia+5Ib
After solving: Ia = −1A, Ib = −2A
Hence: I1 = 1A, I2 = −2A, I3 = 1A
Nodal
Noodle Analysis
Nodal Analysis
• Determine the number of nodes.
• Pick a reference node then label the rest with
subscripts.
• [*] If there is any voltage source in the branch,
replace it with short circuit – apply KCL in the
next step to the resulting bigger node.
• Apply KCL on each node except the reference.
• Solve the simultaneous equations.
Nodal Analysis
Example: [Boylestad 10th Ed. E.g. 8.21 - modified]
a
I2
R2
b
12Ω
I3
I1
4A
R1
2Ω
6Ω
R3
2A
Nodal Analysis
Example: (cont)
Node a:
Va Va  Vb
4 
 7Va  Vb  48
2
12
Node b:  2  Vb  Va  Vb  V  3V  24
a
b
6
12
After solving: Va = 6V, Vb = − 6A
Hence: I1 = 3A, I2 = 1A, I3 = −1A
Mesh vs Nodal Analysis
• Mesh: Start with KVL, get a system of
simultaneous equations in term of current.
• Nodal: Start with KCL, get a system of
simultaneous equations on term of voltage.
• Mesh: KVL is applied based on a fixed loop
current.
• Nodal: KCL is applied based on a fixed node
voltage.
Mesh vs Nodal Analysis
• Mesh: Current source is an open circuit and it
merges loops.
• Nodal: Voltage source is a short circuit and it
merges nodes.
• Mesh: More popular as voltage sources do exist
physically.
• Nodal: Less popular as current sources do not
exist physically except in models of electronics
circuits.
Linearity and Superposition
Linearity Concept of Circuit
Elements
• Due to Ohm’s Law, the effect of voltage across
a circuit element is linear.
– Can be added linearly depending on how much
potential is applied to each of them.
• This is true for the effect of current too.
Superposition Theorem
Statement:
The current through, or voltage across,
an element is equal to the algebraic sum
of the currents or the voltages produced
independently by each source
Superposition Theorem
• Choose one power source to consider, then
switch off other sources:
– Voltage source: remove it and replace with short
circuit
– Current source: remove it and replace with open
circuit
• Calculate the voltages and currents in the
elements of concern based on the resulting
circuit.
• Do the above for all sources, then sum the
respective voltages or currents by considering
the polarities.
Superposition Theorem
Example: [Boylestad 10th Ed. E.g. 9.5 - modified]
I2
I1
4Ω
R1
3A
2Ω
12V
6V
R2
Superposition Theorem
Example: [Boylestad 10th Ed. E.g. 9.5 - modified]
Consider only the 12V source:
I2a
I1a
4Ω
R1
2Ω
12V
R2
12
I1a  
24
 2A
I 2 a  2A
Superposition Theorem
Example: [Boylestad 10th Ed. E.g. 9.5 - modified]
Consider only the 6V source:
I2b
I1b
4Ω
R1
2Ω
6V
R2
6
I1b 
24
 1A
I 2b  1A
Superposition Theorem
Example: [Boylestad 10th Ed. E.g. 9.5 - modified]
Consider only the current source:
I2c
I1c
4Ω
R1
2Ω
3A
R2
4
I1c   3
6
 2A
I 2c  1A
Superposition Theorem
Example: [Boylestad 10th Ed. E.g. 9.5 - modified]
Hence:
I1 = I1a + I1b + I1c = 1A
I2 = I2a + I2b + I2c = 2A
Thevenin’s Theorem
Thevenin’s Theorem
Statement:
Network behind any two terminals of
linear DC circuit can be replaced by an
equivalent voltage source and an
equivalent series resistor
• Can be used to reduce a complicated network
to a combination of voltage source and a series
resistor
Thevenin’s Theorem
• Calculate the Thevenin’s resistance, RTh, by
switching off all power sources and finding the
resulting resistance through the two terminals:
– Voltage source: remove it and replace with short
circuit
– Current source: remove it and replace with open
circuit
• Calculate the Thevenin’s voltage, VTh, by
switching back on all powers and calculate the
open circuit voltage between the terminals.
Thevenin’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
Convert the following network into its
Thevenin’s equivalent:
3Ω
6Ω
9V
Thevenin’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
RTh calculation:
3Ω
6Ω
RTh  3 6  2
Thevenin’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
VTh calculation:
3Ω
6Ω
9V
6
VTh 
 9  6V
3 6
Thevenin’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
Thevenin’s equivalence:
2Ω
6V
Norton’s Theorem
Norton’s Theorem
Statement:
Network behind any two terminals of
linear DC circuit can be replaced by an
equivalent current source and an
equivalent parallel resistor
• Can be used to reduce a complicated network
to a combination of current source and a parallel
resistor
Norton’s Theorem
• Calculate the Norton’s resistance, RN, by
switching off all power sources and finding the
resulting resistance through the two terminals:
– Voltage source: remove it and replace with short
circuit
– Current source: remove it and replace with open
circuit
• Calculate the Norton’s voltage, IN, by switching
back on all powers and calculate the short
circuit current between the terminals.
Norton’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
Convert the following network into its
Norton’s equivalent:
3Ω
6Ω
9V
Norton’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
RN calculation:
3Ω
6Ω
RN  3 6  2  RTh
Norton’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
IN calculation:
3Ω
6Ω
9V
9
I N   3A
3
Norton’s Theorem
Example: [Boylestad 10th Ed. E.g. 9.6 - modified]
Norton’s equivalence:
OR,
3A
2Ω
We can just take the
Thevenin’s equivalent and
calculate the short circuit
current.
Maximum Power Consumption
An element is consuming the maximum power out
of a network if its resistance is equal to the
Thevenin’s or Norton’s resistance.
Source Conversion
Use the relationship between Thevenin’s and
Norton’s source to convert between voltage and
current sources.
2Ω
3A
6V
V = IR
2Ω