Soumitra Mondal

Lecturer
Raghunathpur Govt. Polytechnic, Purulia
Electrical Circuit & Network
Unit No: 3
Mesh and Node Analysis

Mesh Analysis
Mesh is a loop that doesn’t consists of any other loop inside it. Mesh analysis technique, uses
mesh currents as variables, instead of currents in the elements to analyse the circuit.
Therefore, this method absolutely reduces the number of equations to be solved. Mesh
analysis applies the Kirchhoff’s Voltage Law (KVL) to determine the unknown currents in a
given circuit. Mesh analysis is also called as mesh-current method or loop analysis. After
finding the mesh currents using KVL, voltages anywhere in a given circuit can be determined
by using Ohms law.

Steps to analyse the mesh analysis technique

1) Check whether there is a possibility to transform all current sources in the given circuit to
voltage sources.
2) Assign the current directions to each mesh in a given circuit and follow the same direction
for each mesh.
3) Apply KVL to each mesh and simplify the KVL equations.
4) Solve the simultaneous equations of various meshes to get the mesh currents and these
equations are exactly equal to the number of meshes present in the network.
Explain:
Consider the below DC circuit to apply the mesh current analysis, such that currents in
different meshes can be found. In the below figure there are three meshes present as ACDA,
CBDC and ABCA but the path ABDA is not a mesh. As a first step, the current through each
mesh is assigned with the same direction as shown in figure.

Secondly, for each mesh we have to apply KVL. By applying KVL around the first loop or
mesh we get
V1 − V3 − R2 ( I1 − I 3 ) − R4 ( I1 − I 2 ) = 0

V1 − V3 = I1 ( R2 + R4 ) − I2R4 − I3R2 ………………(1)

Similarly, by applying KVL around second mesh we get,

 −V2 − R3 ( I 2 − I 3 ) − R4 ( I 2 − I1 ) = 0

− V2 = − I1R4 + I 2 ( R3 + R4 ) − I 3 R3 ………………………(2)

And by applying KVL around third mesh or loop we get,

V3 − R1I 3 − R3( I 3 − I 2 ) − R2( I 3 − I1 ) = 0

V3 = − I1R2 − I2R3 + I3(R1 + R2 + R3) ………………………(3)

Therefore, by solving the above three equations we can obtain the mesh currents for each
mesh in the given circuit.

Example problems on mesh analysis:

Example 1:
Consider the below example in which we find the voltage across the 12A current source
using mesh analysis. In the given circuit all the sources are current sources.

Step 1: In the circuit there is a possibility to change the current source to a voltage source on
right hand side source with parallel resistance. The current source is converted into a voltage
source by placing the same value of resistor in series with a voltage source and the voltage in
that source is determined as
Vs = Is Rs

= 4× 4 = 16 Volts

Step 2: Assign the branch currents as I1 and I2 to the respective branches or loops and
represent the direction of currents as shown below.
Step 3: Apply the KVL to each mesh in the given circuit
Mesh -1:
Vx − 6 × (I1 − I 2) − 18 = 0

Substituting I1 = 12 A

Vx + 6I2 = 90…………………… (1)

Mesh – 2:

18 − 6 × (I 2 − I1) − 4 × I 2 − 16 = 0

2 – 10 × I2 + 6(12) = 0

I2 = 74/ 10

= 7.4 Amps

Substituting in equation 1 we get

Vx = 90 – 44.4

= 45.6 Volts

Example 2:
Consider the below circuit where we determine the voltage across the current source and a
branch current Iac. Assign the directions as shown below and note that current is assigned
opposite to the source current in second loop.

By applying KVL to the first mesh we get

V1 − R2 ( I1 − I 3 ) − R4 ( I1 − I 2 ) = 0
 4 – 2 I1 − 2I3 − 4I1 − 4I2 = 0

-6I1 − 2I3 = 4 ……………(1)

By applying KVL to the second mesh we get

−Vc − R4( I 2 − I1 ) − R3 ( I 2 − I 3 ) = 0

– Vc = 4I2 − 4I1 + 2I2 − 2I3 = 0

– Vc = – 4I1 + 6I2 – 2I3

But I2 = -2 A, then

– Vc = – 4I1 – 12 – 2 I3 ………………….(2)

By applying KVL to the third mesh we get

− R1 I 3 − R3 ( I 3 − I 2 ) − R2 ( I 3 − I1 ) = 0

−4 I3 − 2I3 + 2I2 − 2I3 + 2I1 = 0

− 8I3 − 4 + 2I1 = 0 (by substituting I2 = -2 A)

2I1 − 8I3 = 4 …………………(3)

By solving 1 and 3 equations we get I3 = -0.615 and I1 = 4.46

Therefore, the voltage Vc = 4 (4.46) + 12 + 2(-0.615)

Vc = 28.61 V

And the branch current Iac = I1- I3

Iac = 5.075 amps

Super Mesh Analysis

As we seen in the example 2, it contains current source in one of its branches. And before the
application of the mesh analysis to that circuit, we assumed the unknown voltage across the
current source and then mesh analysis is applied. This is a quite difficult approach and this
can be overcome by applying super mesh technique.

A super mesh is formed when two adjacent meshes share a common current source and none
of these (adjacent) meshes contains a current source in the outer loop. Consider the below
circuit in which super mesh is formed by the loop around the current source.

The current source is common to the meshes 1 and 2 and hence it must be analysed
independently. To achieve this, assume the branch that contains current source is open
circuited and create a new mesh called super mesh.

Writing KVL to the super mesh we get

V = I1R1 + (I2 – I3) R3

= I1R1 + I2R3 – I3R3

Applying KVL to the Mesh 3 we get

(I3 – I2) R3 + I3R4 = 0

And the difference between the two mesh currents gives the current from the current source.
Here the current source direction is in the loop current direction I1. Hence I1 is more than I2,
then

I = I1 – I2

Thus, by using these three mesh equations we can easily find the three unknown currents in
the network.

Example on super mesh analysis

1. Consider the below example in which we have to find the current through the 10
ohms resistor.

By applying the KVL to the mesh 1 we get

1I1 + 10 (I1 – I2) = 2

11I1 – 10 I2 = 2 …………………………. (1)

The meshes 2 and 3 consist of 4A current source and hence form a super mesh. The current
from 4A current source is in the direction of I3 and thus the super mesh current is given as

I = I3 – I2

I3 – I2 = 4…………………………. (2)

By applying KVL to the outer loop of the super mesh we get,

 – 10 (I2 – I1) – 5I2 – 15I3 = 0

10I1 – 15I2 – 15I3 = 0…………………….. (3)

By solving 1, 2 and 3 equations, we get

I1 = –2.35 A

I2 = –2.78 A

I3 = 1.22 A

Hence the current through the 10 ohms resistor is I1 – I2

= –2.35 + 2.78 A

= 0.43 A

Node:
In electrical engineering, a node is any point on a circuit where the terminals of two or
more circuit elements meet. In circuit diagrams, connections are ideal wires with zero
resistance, so a node may consist of the entire section of wire between elements, not just a
single point.
Nodal Voltage Analysis complements the previous mesh analysis in that it is equally
powerful and based on the same concepts of matrix analysis. As its name implies, Nodal
Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage
potentials around the circuit.
So by adding together all these nodal voltages the net result will be equal to zero. Then, if
there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these
alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchhoff’s first law equation, that is: “the currents entering
a node are exactly equal in value to the currents leaving the node” then express each current
in terms of the voltage across the branch. For “n” nodes, one node will be used as the
reference node and all the other voltages will be referenced or measured with respect to this
common node.
Basic Steps Used in Nodal Analysis

1. Select a node as the reference node. Assign voltages V1, V2… Vn-1 to the remaining
nodes. The voltages are referenced with respect to the reference node.
2. Apply KCL to each of the non-reference nodes.
3. Use Ohm’s law to express the branch currents in terms of node voltages.

Problem: 1
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three nodes are
assumed to have voltages, Va, Vb and Vc with respect to node D. For example;

As Va = 10v and Vc = 20v , Vb can be easily found by:

Again is the same value of 0.286 amps, we found using Kirchhoff’s Current Law in the
previous tutorial.
Problem: 2
Example: Calculate Node Voltages in following circuit

In the following circuit we have 3 nodes from which one is reference node and other two are
non-reference nodes – Node 1 and Node 2.
Step I. Assign the nodes voltages as v1 and v2 and also mark the directions of branch currents
with respect to the reference nodes

Step II. Apply KCL to Nodes 1 and 2

KCL at Node 1

KCL at Node 2

Step III. Apply Ohm’s Law to KCL equations

• Ohm’s law to KCL equation at Node 1

Simplifying the above equation we get,

• Now, Ohm’s Law to KCL equation at Node 2

Simplifying the above equation we get

Step IV. Now solve the equations 3 and 4 to get the values of v1 and v2 as,

Using elimination method

And substituting value v2 = 20 Volts in equation (3) we get-

Hence node voltages are as v1 = 13.33 Volts and v2 = 20 Volts.

Nodal Analysis with Voltage Sources
Case I. If a voltage source is connected between the reference node and a non-reference node,
we simply set the voltage at the non-reference node equal to the voltage of the voltage source
and its analysis can be done as we done with current sources. v1 = 10 Volts.
Case II. If the voltage source is between the two non-reference nodes then it forms a super
node whose analysis is done as following

Super node Analysis

Definition of Super Node
Whenever a voltage source (Independent or Dependent) is connected between the two non-
reference nodes then these two nodes form a generalized node called the Super node. So,
Super node can be regarded as a surface enclosing the voltage source and its two nodes.

In the above Figure 5V source is connected between two non-reference nodes Node – 2 and
Node – 3. So here Node – 2 and Node – 3 form the Super node.
Properties of Super node

 Always the difference between the voltage of two non-reference nodes is known at
Super node.
 A super node has no voltage of its own
 A super node requires application of both KCL and KVL to solve it.
 Any element can be connected in parallel with the voltage source forming the super
node.
 A Super node satisfies the KCL as like a simple node.
How Solve Any Circuit Containing super node

Let’s take a example to understand how to solve circuit containing super node

Here 2V voltage source is connected between Node-1 and Node-2 and it forms a Supernode
with a 10Ω resistor in parallel.

Note – Any element connected in parallel with the voltage source forming Super node
doesn’t make any difference because v2– v1 = 2V always whatever may be the value of
resistor. Thus 10 Ω can be removed and circuit is redrawn and applying KCL to the
supernode as shown in figure gives,

Expressing and in terms of the node voltages.

From Equation 5 and 6 we can write as

Hence, v1 = – 7.333V and v2 = – 5.333V which is required answer.

Reference:
1. Circuits and systems; By Dr. K. M. Soni; Katson Publication.
2. Internet Google.