EE231
Electrical Circuits 1
Lecture 2 – Methods of Analysis
Outline:
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Nodal Analysis
Nodal Analysis with Voltage Sources
Mesh Analysis
Mesh Analysis with Current Sources
Nodal vs. Mesh Analysis
Notes:
With both Ohm’s law and Kirchhoff’s law, we can now develop more powerful
techniques in analysing circuits: nodal analysis which is based on KCL and mesh
analysis which is based on KVL.
Using the two techniques, we can analyse any linear circuits by obtaining a set
of linear equations that are then solved by to obtain required voltage and current
values.
a. Nodal Analysis
Nodal analysis provides a general procedure for analysing circuits using
node voltages as the circuit variables. Here, we are interested in finding the
node voltages. Given a circuit with n nodes, the nodal analysis of the circuit
consists of three steps:
1. Select a node as the reference node (𝑣
= 0 V). Assign voltages, 𝑣 , 𝑣 ,…,
𝑣
to the remaining 𝑛 − 1 nodes. The voltages are referenced with respect
to the reference node.
2. Apply KCL to each of the 𝑛 − 1 nonreference nodes. Use Ohm’s law to
express the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknow node
voltages.
Using the steps above, let’s try to analyse the circuit given below using nodal
analysis:
We select node 0 as the reference node (𝑣 = 0 V). The other nodes, 1 and 2,
we assign node voltages 𝑣 and 𝑣 respectively. We then assume current
directions for the passive elements and redraw the circuit to the one below:
We then apply KCL analysis to the nonreference nodes. Applying KCL to node
1, we get
𝐼 =𝐼 +𝑖 +𝑖
At node 2,
𝐼 +𝑖 =𝑖
We now then apply Ohm’s law to express the unknown currents 𝑖 , 𝑖 , and 𝑖
in terms of node voltages. Key idea to bear in mind, is that resistance is a
passive element, by the passive sign convention:
Current flows from a higher potential to a lower potential in a resistor.
Therefore, we can express this principle as
𝑖=
𝑣
−𝑣
𝑅
Therefore, the current values will become,
𝑖 =
𝑣 −0
𝑅
𝑖 =
𝑣 −𝑣
𝑅
𝑖 =
𝑣 −0
𝑅
Substituting the above equations to the KCL ones in the previous page, we
get
𝐼 =𝐼 +
𝐼 +
𝑣
𝑣 −𝑣
+
𝑅
𝑅
𝑣 −𝑣
𝑣
=
𝑅
𝑅
Where the above equations can be reduced to two equations two unknowns
system.
Example 1
Given the circuit below, solve for the voltages at the three nonreference
nodes.
b. Nodal Analysis with Voltage Sources
Case 1: If a voltage source is connected between the reference node and a
nonreference node, we simply set the voltage at the nonreference node equal
to the voltage of the voltage source.
For example, the portion of the circuit below, 𝑣 = 10 V
Thus, this simplifies our analysis, since the node voltage is already known.
Case 2: If the voltage source (dependent or independent) in connected between
two nonreference nodes. The two nonreference nodes, forms a supernode.
For example, the circuit below shows one supernode:
A supernode is formed by enclosing a (dependent or independent) voltage
source connected between two nonreference nodes and any elements
connected in parallel with it.
Like any other node, KCL must be satisfied at the supernode. For the above
supernode, we get the KCL equation
𝑖 +𝑖 =𝑖 +𝑖
Or
𝑣 −𝑣
𝑣 −𝑣
𝑣 −0 𝑣 −0
+
=
+
2
4
8
6
We can redraw the circuit to the one below
Applying KVL to the loop
−𝑣 + 5 + 𝑣 = 0
→
𝑣 −𝑣 =5
From the equations above, we can then obtain the node voltages.
Note the following properties of a supernode:
1. The voltage source inside the supernode provides a constraint equation
needed to solve for the node voltages.
2. A supernode has no voltage of its own
3. A supernode requires the application of both KCL and KVL.
Example 2
Find 𝑖 and 𝑣 in the circuit given below:
c. Mesh Analysis
Mesh analysis provides another general procedure for analysing circuits, using
mesh currents as the circuit variables. Using mesh currents instead of element
currents as circuit variables is convenient and reduces the number of equations
that must be solved simultaneously. Recall that 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.
Mesh analysis can only be applied to a circuit that is planar – is one that can be
drawn in a plane with no branches crossing one another. The circuit to the right
below is a planar one, while the one to the left is nonplanar.
A mesh is a loop that does not contain any other loops within it. Given the
circuit below:
The meshes can be that paths abefa and bcdeb. But the path abcdefa is not a
mesh. The current through a mesh is known as mesh current. In mesh analysis,
we are interested in applying KVL to find the mesh current in a given circuit.
Steps to determine Mesh Currents:
1. Assign mesh currents 𝑖 , 𝑖 , …, 𝑖 to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express the
voltages in terms of mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents
For the network above, let’s try to solve for the values of the mesh currents 𝑖
and 𝑖 . Applying KVL to mesh 1, we get
−𝑉 + 𝑅 𝑖 + 𝑅 (𝑖 − 𝑖 ) = 0
Or
(𝑹𝟏 + 𝑹𝟑 )𝒊𝟏 − 𝑹𝟑 𝒊𝟐 = 𝑽𝟏
For mesh 2, applying KVL gives us
𝑅 𝑖 + 𝑉 + 𝑅 (𝑖 − 𝑖 ) = 0
Or
−𝑹𝟑 𝒊𝟏 + (𝑹𝟐 + 𝑹𝟑 )𝒊𝟐 = −𝑽𝟐
The two resulting equations can serve as a shortcut for the mesh equation. So
long as the mesh currents have the same directions, the shortcut method holds.
The circuit can now be solved for two equations two unknowns. Notice that the
branch currents are different from the mesh currents unless the mesh is
isolated. It is evident from the network that
𝐼 =𝑖
𝐼 =𝑖
𝐼 =𝑖 −𝑖
Example 3
Using mesh analysis, find 𝐼 in the circuit given below:
d. Mesh analysis with current sources
Case 1: When a current source exists only in one mesh: Consider the circuit
given below:
We set 𝑖 = −5 A and write a mesh equation for the other mesh in the usual
way; that is
−10 + 10𝑖 − 6𝑖 = 0
→
𝑖 = −5 A
Case 2: When a current source exists between two meshes, we create a
supermesh by excluding the current source and any elements connected in
series with it. A supermesh results when two meshes have a (dependent or
independent) current source in common.
Applying KVL to the supermesh, we get
−20 + 6𝑖 + 10𝑖 + 4𝑖 = 0
Or
6𝑖 + 14𝑖 = 20
We then apply KCL to a node where the two meshes intersect. Applying KCL
to node 0
𝑖 =𝑖 +6
Solving the two equations above, we get
𝑖 = −3.2 A
𝑖 = 2.8 A
Note the following properties of a supermesh:
1. The current source in the supermesh provides the constraint equation
necessary to solve for the mesh current.
2. A supermesh has no current on its own.
3. A supermesh requires the application of both KVL and KCL.
Example 4:
Use mesh analysis to determine 𝑖 , 𝑖 , and 𝑖 in the circuit shown below
e. Nodal vs. Mesh Analysis
When to use what?
Mesh – networks that contain many series-connected elements, voltage
sources, or supermeshes are more suitable for mesh analysis. Circuits with
fewer meshes than nodes are better analysed with mesh analysis.
Nodal – networks that contain many parallel-connected elements, current
source, or supernodes are more suitable for nodal analysis. Also, circuits with
fewer nodes than meshes is better analysed for is better analysed with nodal
analysis.
The key is to select the method that results in the smaller number of
equations.