ELE 111
Electrical Circuits
Dr. Ahmed Nader
Professor
Faculty of Engineering, Galala University
Spring 2024
1
Techniques of Circuit Analysis
 Nodal Analysis
 Mesh Analysis
2
Methods of Solving DC Circuits
• Step by Step Simplification
• Applying KCL and KVL
• Nodal analysis and Mesh analysis
3
Simplification of Electrical Circuits
• Resistors in series can be replaced by one resistor
N
Req   Ri
i 1
• Resistors in parallel can be replaced by one resistor
N
Geq   Gi
i 1
• Voltage sources in series can be replaced by one source
N
Veq   Vi
i 1
• Current sources in parallel can be replaced by one source
N
I eq   I i
i 1
4
Overview
• With Ohm’s and Kirchhoff’s law established, they may
now be applied to circuit analysis.
• Two techniques will be presented in this chapter:
 Nodal analysis, which is based on Kirchhoff current law (KCL).
 Mesh analysis, which is based on Kirchhoff voltage law (KVL).
• Any linear circuit can be analyzed using these two
techniques.
• The analysis will result in a set of simultaneous equations
which may be solved by Cramer’s rule or computationally
(using MATLAB for example).
• Circuit simulator such as MultiSim or Pspice can be also
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used.
Nodal Analysis
• Analyze circuits using node voltages as circuit variables.
• Given a circuit with n nodes, without voltage sources, the
nodal analysis is accomplished via three steps:
1. Select a node as the reference node. Assign voltages v1 , v2 ,
…., vn-1 to the remaining (n-1) nodes. Voltage of other nodes
are referred to the reference node.
2. Apply KCL to each of the (n-1) non-reference nodes. Use
Ohm’s law to express the branch currents in terms of node
voltages.
3. Solve resulting (n-1) simultaneous equations to obtain the
unknown node voltages (elimination, matrix, software, etc.)
Note: Symbol of reference commonly referred to as the ground
since its voltage is by default zero.
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Reference Node
• In shown circuit v1 is actually v10 (voltage difference
between node 1 and node 0)
• Note that voltage difference between node 1 & node 2
v12 = v10-v20 = v1-v2
3
1
2
v1
v2
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Nodal Analysis
• This circuit has a node that is
designed as ground. We will use
that as reference node (node 0).
• The remaining two nodes are
designed 1 and 2 and assigned
voltages v1 and v2.
• Applying KCL at each nonreference node
• 𝐼1 = 𝐼2 + 𝑖1 + 𝑖2
• 𝐼2 + 𝑖2 = 𝑖3
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Nodal Analysis
• Applying Ohm’s law and keep in mind that current flows
from high voltage to low voltage
 𝑖1 =
𝑣1 −0
𝑅1
 𝑖2 =
𝑣1 −𝑣2
𝑅2
 𝑖3 =
𝑣2 −0
𝑅3
= 𝐺1 𝑣1
= 𝐺2 𝑣1 − 𝑣2
= 𝐺3 𝑣2
• 𝐼1 = 𝐼2 + 𝑖1 + 𝑖2 ⇒ 𝐼1 = 𝐼2 + 𝐺1 𝑣1 + 𝐺2 (𝑣1 − 𝑣2 )
• 𝐼2 + 𝑖2 = 𝑖3 ⇒ 𝐼2 + 𝐺2 𝑣1 − 𝑣2 = 𝐺3 𝑣2
• Rearranging the previous equations:
 𝐼1 − 𝐼2 = 𝐺1 + 𝐺2 𝑣1 − 𝐺2 𝑣2
 𝐼2 = −𝐺2 𝑣1 + (𝐺2 + 𝐺3 )𝑣2
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Nodal Analysis By Inspection
𝐺1 + 𝐺2
−𝐺2
𝑣1
−𝐺2
𝐼1 − 𝐼2
=
𝑣
𝐼2
𝐺2 + 𝐺3 2
In general
𝐼𝑡1
𝐺11 −𝐺12 𝑣1
=
𝑣
𝐼𝑡2
−𝐺21 𝐺22
2
– Gkk is the summation of conductances
connected to node k
– Gkj = Gjk is the summation of conductances
connecting nodes j and k
– vk is the unknown voltage at node k
– Itk summation of all independent current
sources directly connected to node k with
current entering the node treated as
positive
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Example
V1
V2
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Solving Linear Equations
Obtain the currents
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Nodal Analysis By Inspection
3/4 −1/4 𝑣1
5
•
=
−1/4 5/12 𝑣2
5
• Using Cramer's rule
• 𝑣1 = Δ1 /Δ, 𝑣2 = Δ2 /Δ
3/4
• Δ=
−1/4
−1/4
= 0.25
5/12
5 −1/4
• Δ1 =
= 3.33
5 5/12
3/4
• Δ2 =
−1/4
• 𝑣1 =
Δ1
Δ
5
=5
5
= 13.33 V, 𝑣2 =
Δ2
Δ
= 20𝑉
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Nodal Analysis By Inspection I
• There is a faster way to construct a matrix for solving a
circuit by nodal analysis.
• It requires that all current sources within the circuit be
independent.
• In general, for a circuit with N non-reference nodes, the
node-voltage equations may be written as:
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Nodal Analysis By Inspection II
• Each diagonal term on the conductance matrix is the sum of
conductances connected to the node indicated by the matrix
index.
• The off diagonal terms, Gjk are the negative of the sum of all
conductances connected between nodes j and k with j  k.
• The unknown voltages are denoted as vk
• The sum of all independent current sources directly
connected to node k are denoted as ik. Current entering the
node are treated as positive and vice versa.
• This matrix equation can be solved for the unknown values
of the nodal voltages.
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Example
4 −2 0 𝑣1
6
−2 7 −4 𝑣2 = −9
0 −4 6 𝑣3
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Solution:
 v1 = 1.2V
 v2 = -0.6V
 v3 = 0.6V
 v4 = 0V
V1
V2
V4
V3
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What is a Circuit Simulator?
• Circuit simulation is a process in which a model of an electronic
circuit is created and analyzed using various software algorithms,
which predict and verify the behavior and performance of the circuit.
• Since fabrication of electronic circuits, especially integrated circuits
(ICs), is expensive and time-consuming, it is faster and more costeffective to verify the behavior and performance of the circuit using a
circuit simulator before fabrication.
• There are different types of circuit simulators:
• Analog simulators: solve accurate representations of the
electronic circuits. They offer high accuracy and are commonly
used to simulate small circuits.
• Digital simulators: use functional representations of electronic
circuits, typically described using hardware description languages
(HDL). These offer the highest performance and capacity, but at
relatively lower levels of accuracy. Simulate very large circuits.
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Dependent Current Sources
V1
V2
V3
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Solution
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Including Voltage Sources
Depending on what nodes the source is connected to,
approach varies.
Between the reference node and a non-reference node:
• Set the voltage at the non-reference node to the
voltage of the source.
• In the example circuit v1 = 10V.
Between two non-reference nodes:
• The two nodes form a supernode.
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Nodal Analysis Including
Independent Voltage Sources
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Including Dependent Sources
Use the nodal method to find the power dissipated in
the 5Ω resistor in the circuit.
3
1
v1
2
4
v2
Node 1:
Node 2:
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Supernode
•
A super node is formed by enclosing a voltage source
(dependent or independent) connected between two nonreference nodes and any elements connected in parallel with it.
•
Why?
 Nodal analysis requires applying KCL.
 The current through the voltage source cannot be known
in advance (Ohm’s law does not apply).
 By lumping the nodes together, the current balance can
still be described.
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Supernode Analysis
𝑖1 + (−𝑖2 ) = 𝑖
𝑖 = 𝑖3 + (−𝑖4 )
𝑖1 + (−𝑖2 ) = 𝑖3 + (−𝑖4 )
𝑖1 + 𝑖4 = 𝑖2 + 𝑖3
In the example circuit node 2 and 3 form a supernode.
The current balance can be expressed as:
v1  v2 v1  v3 v2  0 v3  0



2
4
8
6
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Example
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Supernode
𝒗𝟏 = 𝟓𝟎𝑽
Node 2:
Node 3:
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Two Supernodes
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Mesh (Loop) Analysis
• Another general procedure for analyzing circuits is to
use the mesh currents as the circuit variables.
• Using mesh currents instead of element currents is
convenient and reduces the number of equations that
must be solved simultaneously.
• Recall:
 A loop is a closed path with no node passed more than
once.
 A mesh is a loop that does not contain any other loop
within it (independent).
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Mesh Analysis Steps
• Identify independent loops
• Assign current in each independent loop (Clockwise
direction) to the n meshes (i1, i2,…, in)
• Apply KVL to each of the n mesh currents & Use
Ohm’s law to express the branch voltage in terms of
loop current.
• Solve the resulting n simultaneous equations to get
the mesh currents.
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Mesh Method
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Example
METHOD 2:
Cramer’s Rule
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Planar versus Non-Planar
• Mesh analysis is limited in one aspect: It can only
apply to circuits that can be rendered planar.
• A planar circuit can be drawn in a plane such that
there are no crossing branches.
Non-Planar Circuit: The branch with the 13Ω resistor prevents the
circuit from being drawn without crossing branches.
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Planar Circuit
Circuit can be redrawn to avoid crossing branches
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Mesh Analysis By Inspection
𝑣1
(𝑅1 +𝑅3 )
−𝑅3
𝑖𝑎
= −𝑣
−𝑅3
(𝑅2 +𝑅3 ) 𝑖𝑏
2
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Mesh Analysis by Inspection I
• There is a similarly fast way to construct a matrix for solving a
circuit by mesh analysis.
• It requires that all voltage sources within the circuit be
independent.
• In general, for a circuit with N meshes, the mesh-current
equations may be written as:
• Each diagonal term on the resistance matrix is the sum of
resistances in the mesh indicated by the matrix index.
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Mesh Analysis by Inspection II
• Rkk is the sum of all resistances in mesh k
• The off diagonal terms, Rjk are the negative of the sum of all
resistances in common with meshes j and k with j  k.
• The unknown mesh currents in the clockwise direction are
denoted as ik.
• The sum taken clockwise of all voltage sources in mesh k are
denoted as vk. Voltage rises are treated as positive.
• This matrix equation can be solved for the values of the
unknown mesh currents.
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Mesh Analysis with
Dependent Sources
Mesh 1:
Mesh 2:
Mesh 3:
Constraint:
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Example
𝑰𝟎 = 𝑰𝟏 − 𝑰𝟐
38
Mesh Analysis with
Current Sources
• The presence of a current source makes the mesh analysis
simpler in that it reduces the number of equations.
• If the current source is located on only one mesh, the
current for that mesh is defined by the source.
• For example:
• Here, the current i2 is equal to −5A.
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Supermesh
• Similar to the case of nodal analysis where a voltage source is
shared between two non-reference nodes, current sources
(dependent or independent) that are shared by more than
one mesh need special treatment.
• The two meshes must be joined together, resulting in a
supermesh.
• The supermesh is constructed by merging the two meshes and
excluding the shared source and any elements in series with it
• A supermesh is required because mesh analysis uses KVL.
• But the voltage across a current source cannot be known in
advance.
• Intersecting supermeshes in a circuit must be combined to for
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a larger supermesh.
Creating a Supermesh
𝟐𝟎 = 𝟔𝒊𝟏 + 𝟏𝟒𝒊𝟐
• In this example, a 6A current source is shared between
mesh 1 and 2.
• The supermesh is formed by merging the two meshes 
The current source and the 2Ω resistor in series with it
are removed.
• Apply KCL to the branch where the two meshes intersect
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 𝟔 = 𝒊𝟐 − 𝒊𝟏
Supermesh Example
Mesh a:
Mesh c:
Supermesh
Mesh b:
Constraint:
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Example (Continued)

43
Two Supermeshes
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Nodal Vs. Mesh Analysis
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Selecting an Appropriate Approach
• In principle both the nodal analysis and mesh analysis
are useful for any given circuit.
• What then determines if one is going to be more
efficient for solving a circuit problem?
• There are two factors that dictate the best choice:
 The nature of the particular network.
 The second factor is the information required.
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Nodal analysis if…
• If the network contains:
 Current sources
 Many parallel connected elements
 Supernodes  reduced unknowns
 A circuit with fewer nodes than meshes  reduced
equations
• If node voltages are what is being solved for.
• Non-planar circuits can only be solved using nodal
analysis.
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Mesh analysis when…
• If the network contains:
 Voltage sources.
 Many series connected elements.
 Supermeshes  reduced unknowns
 A circuit with fewer meshes than nodes  reduced
equations
• If branch or mesh currents are what is being solved for.
• Mesh analysis is the only suitable analysis for transistor
circuits.
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