Millman's Theorem - Free MCQ Practice Test with solutions, GATE EE Network

Applying KCL at node A.

32 V – 240 + 20 V – 200 + 5 V = 0
57 V = 440
V = 440/57
= 0.96 A

Millman’s Theorem:
(1) voltage source: It is states that – when a number of voltage sources (V₁, V₂, V₃……… Vn) are in parallel having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R. 


The equivalent circuit parameter

(2) Current source: It is state that – when a number of current sources (I₁, I₂, I₃......In) are in series having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent current source I in parallel with an equivalent series resistance R.


The equivalent circuit parameter

R = R₁ + R₂ + R₃ ...... R_n

Millman’s theorem:

• It is a very useful theorem to find out the voltage across the load and the current through the load.
• It is also called as parallel generator theorem.
• It is helpful to reduce a mixture of voltage and current sources connected in parallel to a single equivalent voltage or current source.

1. voltage source

It is stated that – when a number of voltage sources (V₁, V₂, V₃……… V_n) are in parallel having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R. 

The equivalent circuit parameter

2. Current source:

It is state that – when a number of current sources (I₁, I₂, I₃......In) are in series having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent current source I in parallel with an equivalent series resistance R.


The equivalent circuit parameter
I = I₁ + I₂ + I₃ ....... I_n
R = R₁ + R₂ + R₃ ...... R_n

Milliman's Theorem:

Millman's Theorem states that it is possible to replace numbers of voltage sources with finite internal resistance operating in parallel with a single voltage source with a series of equivalent resistance.

According to Ohm's Law:

V = IR

Concept:
Millman’s Theorem:
It is stated that – when a number of voltage sources (V₁, V₂, V₃……… Vn) are in parallel having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R.


The equivalent circuit parameter

V = equivalent voltage
R = equivalent resistance
Calculation:
From Millman's theorem, the above circuit can be redrawn as

Where

Millman’s Theorem:
(1) voltage source: It is states that – when a number of voltage sources (V₁, V₂, V₃……… Vn) are in parallel having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R. 


The equivalent circuit parameter

(2) Current source: It is state that – when a number of current sources (I₁, I₂, I₃......In) are in series having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent current source I in parallel with an equivalent series resistance R.


The equivalent circuit parameter

R = R₁ + R₂ + R₃ ...... R_n

Concept:
Let us have a circuit as shown in the below figure a

Here V₁, V₂, and V₃ are voltages of respectively 1st, 2nd, and 3rd branch, and R₁, R₂, and R₃ are their respective resistances.

I_L, R_L, and V_T are load current, load resistance, and terminal voltage respectively.

Now, this complex circuit can be reduced easily to a single equivalent voltage source with a series resistance with the help of Millman’s Theorem as shown in figure b.

The value of equivalent voltage VE is specified as per Millman’s theorem will be

This VE is nothing but Thevenin’s voltage and Thevenin’s resistance RTH can be determined by shorting the voltage source.
So RTH will be obtained as

Now load current and terminal voltage can be easily found by

Calculation:
Given circuit diagram

 

R_TH = 1/0.135 = 7.40 Ω 

RL = 20 Ω 
Load current I will be as follows

Therefore the current I flowing in the circuit shown below in amperes is 0 A

It is states that – when a number of voltage sources (V₁, V₂, V₃……… Vn) are in parallel having internal resistance (R₁, R₂, R₃………….R_n) respectively, the arrangement can replace by a single equivalent voltage source V in series with an equivalent series resistance R. 


The equivalent circuit parameter

V = equivalent voltage
R = equivalent resistance
Calculation:
From Millman's theorem, the above circuit can be redrawn as

where

∴ Load current 

Applying Millman's theorem we get

= 420 V


IL = 420/70
= 6 A