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**1. Basic Concepts**

**Kirchhoffâ€™s Voltage Law (KVL):**The sum of voltages around a closed loop circuit is equal to zero.**Kirchhoffâ€™s Current Law (KCL):**The algebraic sum of electrical current that merge in a common node of a circuit is zero.

âˆ‘ I_{IN}= âˆ‘I_{OUT}**NODAL ANALYSIS**

The procedure for analysing a circuit with the node method is based on the following steps:

Clearly, label all circuit parameters and distinguish the unknown parameters from the known.

Identify all nodes of the circuit after that Select a node as the reference node also called the ground and assign to ix a potential of 0 Volts. All other voltages in the circuit are measured with respect to the reference node.

Label the voltages at all other nodes & Assign and label polarities.

Apply KCL to each node and express the branch currents in terms of the node voltages.

Solve the resulting simultaneous equations for the node voltages.

Now that the node voltages are known, the branch currents may be obtained from Ohmâ€™s law.

**Example:** Find out the nodal voltage at each node & current in each loop by using the Nodal method?

**Solution:** First of all we have labeled all elements and identified all relevant nodes in the circuit.

There are a few general guidelines that we need to remember as we make the selection of the reference node.

- A useful reference node is one which has the largest number of elements connected to it.
- A useful reference node is one which is connected to the maximum number of voltage sources.

For the next step we assign current flow and polarities

For node n1 voltage of the voltage source is known so v_{1} = V_{s} â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

& KCL at node n2 associated with voltage v2 gives: i_{1} = i_{2} + i_{3} â€¦.â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2)

The currents i_{1}, i_{2}, i_{3} are expressed in terms of the voltages v_{1}, v_{2}, v_{3} as follows

â€¦â€¦â€¦.. (3)

From the relation (1) (2) & (3) we getâ€¦.

â€¦â€¦â€¦â€¦â€¦ (4)

Rewrite the above expression as a linear function of the unknown voltages v_{2 }and v_{3} gives.

â€¦â€¦â€¦. (5)

KCL at node n3 associated with voltage v_{3} gives:

or ....â€¦â€¦. (6)

Now we can write equation (5) & (6) in matrix form for the node voltage v_{2} & v_{3}.

Or

â€¦â€¦â€¦ (7)

In defining the set of simultaneous equations, we want to end up with a simple and consistent form. The simple rules to follow and check are

- Place all sources (current and voltage) on the right hand side of the equation, as inhomogeneous drive terms.
- The terms comprising each element on the diagonal of the matrix must have the same sign.
- If you arrange so that all diagonal elements are positive, then the off-diagonal elements are negative and the matrix is symmetric: A
_{ij}= A_{ji}. If the matrix does not have this property there is a mistake somewhere.

Once we put the equations in matrix form and perform the checks detailed above the solutions then there is a solution if the the unknown voltage VK is given by:

â€¦â€¦â€¦â€¦ (8)

Whereis the matrixwith the k-th column replaced by the vector V.

So for our example the voltages v_{2} and v_{3} are given by:

â€¦â€¦â€¦ (9)

â€¦â€¦. (10)

**Nodal Analysis with Floating Voltage Sources. (THE SUPERNODE)**

If a voltage source V_{2} is not connected to the reference node it is called a floating voltage source and special care must be taken when performing the analysis of the circuit. In the circuit of given figure below the voltage source V_{2} is not connected to the reference node and thus it is a floating voltage source. Here v_{2} is the node voltage while V_{2} is the source voltage between node n_{2} & n_{3}.**Circuit with a Supernode**

The part of the circuit enclosed by the dotted ellipse is called a supernode. Kirchhoffâ€™s current law may be applied to a supernode in the same way that it is applied to any other regular node. This is not surprising considering that KCL describes charge conservation which holds in the case of the supernode as it does in the case of a regular node.

In our example application of KCL at the supernode gives I_{1} = i_{2} + I_{3} â€¦â€¦â€¦. (11)

In term of the node voltages Equation (11) becomes:

â€¦â€¦â€¦â€¦ (12)

The relationship between node voltages v_{1} and v_{2} is the constraint that is needed in order to completely define the problem. The constraint is provided by the voltage source V_{2}.

V_{2} = v_{3} - v_{2} â€¦â€¦â€¦(13)

From equation (12) & (13)â€¦..

and ..â€¦â€¦..(14)

Example- Nodal Analysis with supernode.

Determine the node voltages v_{1}, v_{2}, and v_{3} of the circuit in Figure below?**Solution:** We have applied the first five steps of the nodal method and now we are ready to apply KCL to the designated nodes. In this example, the current source Is constraints the current i3 such that i_{3}.

KCL at node n_{2} givesâ€¦

I_{1} = I_{2}+I_{s} â€¦â€¦â€¦â€¦â€¦.. (1)

And with the application of Ohmâ€™s law

â€¦â€¦â€¦â€¦ (2)

Where we have used v_{1} = V_{s} at node n1.

The current source provides a constraint for the voltage v_{3} at node n_{3}.

V_{3} = I_{S}R_{3} â€¦â€¦â€¦. (3)

Now combining the equation (2)& (3).

â€¦â€¦.. (4)**MESH ANALYSIS **

A mesh is defined as a loop which does not contain any other loops The procedure for obtaining the solution is similar to that followed in the Node method and the various steps are given below.

- Clearly, label all circuit parameters and distinguish the unknown parameters from the known.
- Identify all meshes of the circuit & assign mesh currents and label polarities.
- Apply KVL to each mesh and express the voltages in terms of the mesh currents.
- Solve the resulting simultaneous equations for the mesh currents.
- Now that the mesh currents are known, the voltages may be obtained from Ohmâ€™s law.

EXAMPLE- Find out the mesh current i_{1} & i_{2} for mesh 1 & mesh 2?**Solution: **Our circuit example has three loops but only two meshes as show,the meshes of interest are mesh1 and mesh2.

In the next step we will assign mesh currents, define current direction and voltage polarities. The direction of the mesh currents I_{1} and I_{2} is defined in the clockwise direction as shown in the next figure.

The branch of the circuit containing resistor R_{2} is shared by the two meshes and thus the branch current (the current flowing through R_{2}) is the difference of the two mesh currents.**Considering mesh1.** For clarity we have separated mesh1 from the circuit in doing this, care must be taken to carry all the information of the shared branches. Here we indicate the direction of mesh current I_{2} on the shared branch.

Apply KVL to mesh1. Starting at the upper left corner and proceeding in a clock-wise direction the sum of voltages across all elements encountered is

I_{1}R_{1}+ (I_{1}-I_{2}) R_{2}-VS = 0 â€¦â€¦â€¦â€¦. (1)**Similarly, consideration of mesh2 :** we have indicated the direction of the mesh current I_{1} on the shared circuit branch.

Apply KVL to mesh2:

I_{2} (R_{3} + R_{4}) + (I_{2}-I_{1}) R_{2} = 0 â€¦â€¦â€¦â€¦â€¦â€¦. (2)

From equation (1) & (2)â€¦..

I_{1}(R_{1}+ R_{2}) - I_{2} R_{2} = VS â€¦â€¦â€¦â€¦. (3)

-I_{1}R_{2} + I_{2} (R_{2}+R_{3}+R_{4}) = 0 â€¦â€¦â€¦â€¦ (4)

In matrix form equations (3) & (4) becomesâ€¦..

â€¦â€¦â€¦â€¦.. (5)

Equation (5) may now be solved for the mesh currents I_{1} and I_{2}.

Note: It is evident from Figure next figure below that the branch currents are i_{1} , i_{2} & i_{3} are obtained from the mesh currents I_{1} & I_{2} such as.

I_{1} = i_{1} i_{2} = I_{1} â€“ I_{2} i_{3} = I_{2}**MESH ANALYSIS WITH CURRENT SOURCES**

1. If a current source exists only in one mesh.

(i) The mesh current is defined by the current source.

(ii) Number of variables is reduced.

2. If a current source exists between two meshes.

(i) The two nodes form a Supermesh.

(ii) Use one current variable for both meshes. The current difference between these two meshes is known.

(iii) Apply KVL to the Supermesh.

Example: Find out the unknown mesh current I_{1}?**Solution:** Consider the circuit in the figure which contains a current source. The application of the mesh analysis for this circuit does not present any difficulty once we realize that the mesh current of the mesh containing the current source is equal to the current of the current source:

i.e. I_{2} = I_{S} â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

In defining the direction of the mesh current, we have used the direction of the current I_{S}. We also note that the branch current I_{3} = I_{S}.

Applying KVL around mesh1 we obtain

I_{1}R_{1} + (I_{1} + I_{S}) R_{2} = V_{S} â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

The above equation simply indicates that the presence of the current source in one of the meshes reduces the number of equations in the problem.

The unknown mesh current is**PRACTICE PROBLEMS WITH ANSWERS.Q.1.** Determine the currents in the given circuits with reference to the indicated direction?

**Q.2.** Determine the currents in the given circuits with reference to the indicated direction?**Answer: **i1 = 3.31A, i2 = 1.68A, i3 =1.63A, i4 = 0.627A, v2 = 8.39V, v3 = 6.51V

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30 videos|20 docs|27 tests

### Test: Network Equation(mesh And Node) Part - 2

- Test | 7 ques | 30 min

- KCL, KVL, Series & Parallel Networks, Source Transformation
- Video | 36:01 min
- Test: Mesh & Nodal Analysis
- Test | 20 ques | 20 min
- Test: Network Equation(mesh And Node) Part - 1
- Test | 10 ques | 30 min
- Test: Kirchhoffâ€™s Current And Voltage Law
- Test | 20 ques | 20 min