1. Universal college of Engineering & Technology
SUBJECT - CIRCUITS & NETWORKS

2. TOPIC - EXPLAIN MILLMAN’S THEOREM. AND COMPENSATION THEOREM
TOPIC - EXPLAIN MILLMAN’S THEOREM AND COMPENSATION THEOREM WITH EXAMPLES
BRANCH & DIV - ELECTRICAL (D)
PREPARED BY:-
Utpal kant ( )
Dhruv sinojiya( )
Shah Pujit ( )
Valera Darshan( )
Guided By: Prof. Naveen Sharma

3. Index
CONCEPT OF MILLIMAN’S THEOREM
COCEPT OF COMPENSATION THEOREM

4. All about the parallel configuration
MILL MAN’S THEOREM
Concept of Mill MAN's
All about the parallel configuration
Cannot have more than one source in a branch.
Cannot have more than one resistance in a branch.
Mill man's theorem (or the parallel generator theorem) is a method to simplify the solution of a circuit. Specifically, Mill man's theorem is used to compute the voltage at the ends of a circuit made up of only branches in parallel.

5. Millman’s Theorem is a theorem which helps in simplifying electrical networks with a bunch of parallel branches. It was invented by the Russian born, American Engineer Jacob Millman. Millman’s Theorem can be used to find the potential difference between two points of a network which contains only parallel branches.
Millman’s Theorem states that:
The total voltage or potential difference between any two terminals in a circuit is equal to:

6. In above statement of Millman’s Theorem the theorem takes into account only the current flowing through or current source in each branch. Millman’s theorem can also be stated taking the Voltage source in each branch into account.
The Millman’s Theorem can also be stated in other words as:
= Voltage source in each branch
= internal admittance of each voltage source

7. Using Millman’s Theorem we can easily find the Norton and Thevenin equivalent circuit of a network  so, Millman’s Theorem  is also some times called the combination of Norton’s and Thevenin’s theorem.
It should be noted that Millman’s Theorem holds true only to the circuits which contains only parallel branches with only one resistance and source in a branch, or which can be reduced to the equivalent form with only parallel branches with only one resistance and source in a branch as shown on the figure below. Millman’s theorem cannot be applied in a complex mesh of parallel and series network.

8. As stated earlier the circuit with only parallel branches with the following requirements can be applied with Millman’s theorem
Sometimes even though a circuit does not full-fill both the requirements the circuit can be converted into an equivalent  circuit which full-fills both of the above requirements and the process of conversion is done as following.

9. It is easier to apply Millman’s theorem to a circuit if all the branches contains same type of source either voltage or current. The theorem can also be applied to a circuit containing both types of sources but often the process requires the use of ohm’s law in each branch and  is confusing and complex.
CONVERTING INTO PARALLEL CIRCUIT WITH ONLY VOLTAGE SOURCES :

10. Let us convert following circuit into it’s equivalent circuit which full-fills both the requirements discussed above for applying Millman’s Theorem.
The parallel equivalent of the above circuit will be:

11. Where We have used the conversion of current source into voltage source to calculate the voltage source in each branch.
Thus:

12. Converting into parallel circuit with only Current sources.
Now Let us again convert the above example to a parallel network which contains only current sources.
The circuit which need to be concerted into parallel equivalent circuit so that Millman’s theorem can be applied to it is:

13. And it’s equivalent circuit with only parallel branches with current source in each branch is:
Here, we have used  the  conversion of voltage source into current source to calculate the current source in each branch.

14. Where,

15. Circuits with Voltage sources
Applying Millman’s Theorem To circuits!
We can apply Millman’s Theorem to circuits as following:
Circuits with Voltage sources
According to Millman’s theorem , in circuits with voltage sources:
The total voltage or potential difference build up between any two points in a circuit is equal to:

16. For example in the following circuit:
Here, The potential difference between X and Y is:

17. Circuits with Current Sources:
According to Millman’s theorem , in circuits with voltage sources:
The total voltage or potential difference between any two terminals in a circuit is equal to:
Where,
i = the current flowing through each branch.

18. For example in the following circuit:
The potential difference between X and Y is:

22. Compensation Theorem
It is one of the important theorems in Network Analysis , which finds it’s application mostly in calculating the sensitivity of electrical networks & bridges and solving electrical networks.
The Compensation Theorem states that :-
For the sake of branch responses calculations ; Any resistance in a branch of an linear bilateral electrical network can be replaced by a voltage source which provides the same voltage as the voltage dropped in the resistance.

23. In any linear bilateral Electrical Network If  in any Branch have it’s initial resistance (or impedance in case of AC) “R” conducting a current of “I” through it, And if the resistance of the branch is changed by a factor of R , with it’s final resistance R+ R  , the final effect in various branches due to the change in the resistance of the branch can be calculated by injecting an extra voltage source  along with the resistance in modified branch.
The above statement can be clarified with the following illustration.

25. In the figure above,
In fig: 2.a , The current “I” flows through R3 when V1 acts upon it. In fig: 2.b , the R3 is changed to R4 where R4=R3+dR , or R3 is increased by dR. This can also be thought of as an extra dR added in series with R3. Now , we don’t know how much current flows through the branch when R3 is increased by dR , so to calculate the current flowing through the branch due to the effect of dR , as per Compensation theorem in fig: 2.c we add an extra V=-I.dR along with R4 and calculate the current flowing through the branch due to the V or dR to be -dI. Now in fig: 2.d we add the currents in fig: 2.a and 2.c using superposition theorem to find the new current to be I-dI.

27. Reference by :
BOOK BY U.A.PATEL

28. THANK YOU