Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.

html

ET/ Module-5 / DC CIRCUITS / Mesh and Nodal Analysis

Problem Set

P1 Q). Obtain the current in each branch of the network shown in Fig
using Mesh Current analysis and find power loss in 6Ω

resistor.

P2 Q). Determine the current in the 8Ω resistor in the circuit given below
using Super mesh analysis

P3 Q). Obtain the current in each branch of the network shown in Fig using

1 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Node voltage analysis.

P4 Q). Determine the voltage across 8 Ω resistor for the circuit shown in
Figure given below

Examples

SAQ 1). Obtain the current in each branch of the network shown in Fig
using Mesh current analysis

2 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Solution:

The currents I1and I2 are chosen as shown on the circuit diagram. Applying
KVL around the left loop, starting at point ,

-20 + 5I1 + 10(I1 - I2) = 0

And around the right loop, starting at point ,

8 + 10(I2 - I1) + 2I2 = 0

Rearranging terms,

15I1 -10I2 = 20 --> (1)

-10I1 +12I2 = -8 --> (2)

Solving (1) and (2) simultaneously results in I1 = 2A and I2 =1 A. The
current in the center branch, shown dotted, is I1 - I2 = 1 A.

SAQ 2). Determine the current in the 5Ω resistor in the circuit given below
using Super mesh analysis

3 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Solution: From the first mesh, i.e. abcda, we have

50 = 10(I1 - I2) + 5(I1 – I3)

or 15I1 – 10I2 – 5I3 = 50 --> (1)

From the second and third meshes, we can form a super mesh

10(I2 – I1) + 2I2 + I3 + 5(I3 – I1) = 0

or - 15I1 + 12I2 + 6I3 = 0 -->(2)

The current source is equal to the difference between II and III mesh
currents, i.e.

I2 – I3 = 2A --> (3)

Solving equations (1), (2) and (3), we have

I1 = 19.99 A, I2 = 17.33 A, and I3 = 15.33 A

The current in the 5Ω resistor = I1 – I3

= 19.99 – 15.33 = 4.66 A

The current in the 5Ω resistor is 4.66 A

SAQ 3). Solve the circuit shown in figure using Node voltage analysis.

4 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Solution:

The circuit is redrawn as below figure. With two principle nodes, only one
equation is required.

Assuming the currents, all are directed out of the upper node and bottom
node is the reference,

((V1 – 20)/5) + (V1/10) + ((v1 – 8)/2) = 0

From which V1 = 10V. Then,

I1 = (10 – 20)/5 = -2A (the negative sign indicates that current I1 flows into
node 1);

I2 = (10 – 8)/2 = 1A;

I3 = 10/10 = 1A

SAQ 4). Determine the current in the 5 Ω resistor for the circuit shown in
Fig.

5 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Solution: --> At node 1

10 = V1/3 + (V1 – V2)/2 or

0.83V1 – 0.5V2 – 10 = 0 --> (1)
-->At node 2 and 3, the super node equation is
(V2 – V1)/2 + V2/1 + (V3 – 10)/5 + V3/2 = 0 or

-0.5V1 + 1.5 V2 + 0.7V3 – 2 = 0 --> (2)

-->The Voltage between the nodes 2 and 3, is given by

V2 – V3 = 20 --> (3)
--> The current in the 5Ω resistor I5 = (V3-10)/5
Solving Equations (1), (2), and (3), we obtain
V3 = -8.42 V

Current I5 = (-8.42 – 10)/5 = -3.68 A (Current towards node 3)

The current flows towards node 3.

NOTES:
Mesh Current and Node Voltage Analysis
Introduction: The basic circuit analysis techniques discussed in the previous lesson
are difficult to use if the circuit becomes complicated. This lesson deals with two most
powerful techniques for the analysis of the electric circuit i.e. node voltage and mesh
current analysis which are based on the basic laws of KCL and KVL respectively.

Mesh Current Analysis:

In this method first we have to identify all the meshes present in the circuit. Assign
mesh currents corresponding to each mesh. Now by applying KVL for each mesh one
can obtain algebraic equations in terms of mesh currents and can be solved to obtain
the unknowns.

6 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

Consider a simple DC circuit given below

Here we have to find the unknown mesh currents.

Applying KVL around mesh (loop-1)

(1)

Applying KVL around mesh (loop-2)

(2)

Applying KVL around mesh (loop-3)

(3)

In general KVL for ith mesh can be written as

Vii = -Ri1I1 - Ri2I2 - ............+ RiiIi - ...............-RinIn

Where

Vii is the algebraic sum of all voltage sources around the ith mesh.

Rij is the sum of all the resistances common to the ith mesh and j th mesh.

Rii is the sum of all the resistances in the ith mesh.

Steps to be followed for mesh current analysis:

1. Draw the circuit on a flat surface with no conductor crossovers.

7 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

2. Label the mesh currents (Ii) carefully in clockwise direction.

3. Write the mesh equations by applying KVL around each mesh.

4. Solve the set of algebraic equations for mesh currents.

Super-Mesh Analysis:

If a current source is in a branch common to two meshes then it introduces the concept
of super-mesh analysis. Since it is difficult to find the voltage across the current source,
it is not possible to apply KVL in the mesh containing a common current source. Here
the meshes containing the common current sources are combined and KVL is applied
to the combined mesh. The difference between the currents in the meshes will be equal
to the source current common to the two meshes.

Node Voltage analysis:

In this method one has to first find out all the nodes in the circuit and then choose one
of the nodes as reference node (having zero potential) and then assign other node
voltages (unknown) with respect to a reference voltage. Here the (n-1) node voltages
are unknowns. In this problem we apply KCL at each node such that we get
simultaneous linear equations in terms of unknown node voltages which can be solved
simultaneously.

Consider a simple dc circuit shown below

Now for solving the above circuit first find the number of nodes and name them. In the
above circuit there are 4 nodes and are named as 1,2,3 and one node is selected as
reference node. Now by applying KCL we can solve for the unknowns which are
required.

KCL equation at Node-1:

8 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

; →

(4)

KCL equation at Node-2:

; →

(5)

KCL equation at Node-3:

; →

(6)

By solving the above equations the unknown node voltages can be found.

In general for a circuit containing (n+1) nodes the KCL equation at ith node can be
written as

Iii = -Gi1V1 - Gi2V2 - ...........+GiiVi - ..........-GinVn

Where,

Iii is the algebraic sum of all the current sources connected at Node-i.

Gij is the sum of conductance connected between the ith Node and j th Node.

Gii is the sum of conductance connected to the ith Node.

Steps to be followed for node voltage analysis:

1. Identify all the nodes in the circuit. Select one node as the reference node and label the
remaining nodes as unknown voltages with respect to the reference node.

2. Assign branch currents in each branch.

3. Express the branch currents in terms of assigned node voltages.

4. Write the node equations at each node.

5. Solve a set of simultaneous algebraic equations for node voltages and then the branch
currents.

Super Node Analysis:

9 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

If a voltage source is connected between two non-reference nodes then it introduces

the concept of super node. If there is a voltage source connected between two
non-reference nodes it is difficult to find the current passing through the voltage source,
because the voltage source is independent of the current passing through it and hence
a super node is formed by enclosing a voltage source connected between two
non-reference nodes and any elements in parallel with it.

Consider the circuit below

It can be seen that a voltage source is connected between the nodes 2 and 3. Now
consider the enclosed area with a dotted oval as the super node and then applying KCL
at the super node (i.e current entering the enclosed area is equal to the current leaving
the enclosed area) we get

;→

and KVL inside the enclosed area

and it can be seen from the circuit that

Since V1 is known the unknown node voltages at nodes 2 and 3 can be computed by
solving the above equations.

Steps to be followed for super node analysis:

1. Identify all the nodes in the circuit. Select one node as the reference node and label the
remaining nodes as unknown voltages with respect to the reference node.

2. Find for any super node in the circuit and label it as super node.

3. Assign branch currents in each branch except the branch in the super node.

4. Express the branch currents in terms of assigned node voltages.

10 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

5. Write the node equations at each node.

6. Write the KVL inside the enclosed super node (i.e. the voltage difference between the
two nodes is equal to that of the voltage source).

7. Solve a set of simultaneous algebraic equations for node voltages and then the branch
currents.

Applying KVL for super-mesh

Applying KVL around mesh (loop-3)

KCL in super mesh gives

The above 3 equations can be solved simultaneously to solve for the unknown loop
currents.

Steps to be followed in super-mesh analysis:

1. Draw the circuit on a flat surface with no conductor crossovers.

2. Identify the meshes and super meshes.

3. Assign the mesh current in the clockwise direction.

4. Write the KVL at each mesh and super-mesh.

5. Apply KCL for super-mesh (i.e. the difference in the mesh currents is equal to the
source current in the branch common to the meshes).

6. Solve for the mesh currents by solving set of linear algebraic equations.

One can make a choice between a mesh current and node voltage analysis based on
the least number of required equations.

11 of 12 07-12-2011 11:24 AM
Subject/Unit Name/Module Name file:///C:/Users/user/Desktop/ET2/ET_1.5/index.html

COPY RIGHT@RGUKT.IN

12 of 12 07-12-2011 11:24 AM