What is Millman’s Theorem : Circuit & Its Working

For circuit analysis, there are different types of network theorems available. In that, Millman’s theorem is one of the most useful methods for particularly electrical circuits which contain many voltage sources. Based on this theorem, each voltage source as well as their particular resistors generates a current & the amount of all currents is equivalent to the whole current generated through the circuit. The formation of Millman circuits can be done through resistors & there is no involvement of reactance. The circuit’s whole output voltage can be stated through Millman’s theorem that can be stated through the first Ohm’s law. Here, to analyze Millman’s equivalent circuit, the load should be disabled. Similarly, to find the resistance of the Millman equivalent circuit, the voltage sources must be changed with short circuits. This article discusses an overview of Millman’s theorem with examples.

Millman’s Theorem?

The name of Millman’s theorem was stated by the famous professor in the electrical department namely JACOB MILLMAN. The idea of this theorem was proposed by him and this theorem plays a key role in simplifying complex electrical circuits. This theorem is a blend of two theorems like Thevenin’s & Norton’s. An alternate name of this theorem is Parallel Generator Theorem and it is very helpful to discover the flow of current throughout the load and voltage across the load. This kind of theorem is applicable when a circuit includes voltage sources only that are connected in parallel.

Millman’s Theorem Statement

In this Theorem, the circuit diagram can be modified like a parallel network with branches where every branch includes a resistor or a combination of voltage sources with a resistor. Millman’s Theorem is simply applicable to the circuit which can be redrawn accordingly. Here, the example circuit of Millman’s theorem is shown below.

To apply Millman’s Theorem, again this circuit can be redrawn like the following.

When both the voltage supply and resistance are considered in every branch, then this theorem notifies the voltage across all branches In the above circuit, the voltage sources like batteries are denoted with B1, B2 & B3 whereas resistors are denoted with R1, R2 & R3.

Millman’s Theorem Equation

Millman’s Theorem is applied to any electrical circuit which has branches connected in parallel where each branch has its own series resistance & voltage source. The equation of this theorem is shown below.

V = Σ(ek/Rk)/ Σ1/Rk

= (EB1/R1 + EB2/R2 + EB3/R3)/(1/R1+1/R2+1/R3)

In the above equation, ‘ek’ is the voltage generator & ‘Rk’ is the resistance through voltage generators on the branches. This theorem states that the voltage across the circuit can be given like the above equation.

Millman’s Theorem Example Problems

The circuit diagram of Millman’s theorem with values is shown below. This circuit can be solved by using the above formula.

The formula for Millman’s Theorem is

V’ = V1G1+V2G2+—–VnGn/G1+G2…Gn

R1 = 1/G1+G2…Gn

Substitute the values mentioned in the circuit in the above equation, then we will get

(20V/4Ohms + 0V/2Ohms + 6V/1Ohm)/(1/2Ohm+1/2Ohm+1/1Ohm)

(20+0+24/4)/(4/2) = (44/4)/(4/2)=11/2 = 5.5V

Resistor Voltage Drops

The Millman voltage should be evaluated to solve resistor voltage drops against the voltage source in every branch through the principle of the voltage adding in series to conclude the polarity and magnitude of the voltage across every resistor:

ER1 = 5.5V – 20V = 14.5V

ER2 = 5.5V – 0V = 5.5V

ER3 = 5.5V – 5V = 0.5V

Branch Currents

To solve branch current, the voltage drop at each resistor can be separated through its particular resistance like I=E/R

IR1 = 14.5/2 = 7.25 A

IR2 = 5.5/2 = 2.75 A

IR3 = 0.5/1 = 0.5 A

Direction of Current

By using each resistor, the current direction can be determined with the help of the polarity across every resistor, but not through the polarity across every battery, as because the flow of current can be pushed back using a battery.

Millman’s Theorem Example

Using Millman’s theorem, find out the voltage and current across the load resistor RL in the following circuit.

Eeq = (+E1/R1 – E2/R2 + E3/R3)/ (1/R1 + 1/R2 + 1/R3)

The negative sign is mainly used for E2/R2 due to the opposite polarity of the other two. The selected reference way is therefore that of ‘E1’ & ‘E3’. The whole conductance can be not affected through the flow of direction,

Substitute the values in the above equation

Eeq = (+ 10V/5 Ohms) – (16 V/4 Ohms) + (8V/2 Ohms)/ (1/5 + 1/4+ 1/2)

2A – 4A + 4A/0.2S + 0.25S + 0.5S

2 A/0.95 S = 2.11 V

Req = 1/ (1/5 Ω + 1/ 4 Ω + 1 /2 Ω) = 1 / 0.95 S = 1.05 ohms

The final source is shown in the above circuit

IL = 2.11 V / (1.05 Ω + 3 Ω) = 2.11 V / 4.05 Ω = 0.52 A

VL = ILRL = (0.52 A)*(3 Ω) = 1.56 V

Millman’s Theorem Experiment

Aim:

To prove Milliman’s Theorem

Theory/Statement

Millman’s theorem states that in any circuit, if the number of voltage sources is connected in series along with internal resistances in the circuit which are in parallel, afterward these voltage sources may be changed through a single voltage source in series with a resistor.

V’ = V1G1+V2G2+—–VnGn/G1+G2…Gn

R1 = 1/G1+G2…Gn

Required Apparatus

The required components of this experiment include the following but its range mainly depends on the requirement.

• Bread Board
• Connecting Wires
• Voltmeter
• Ammeter
• Regulated Power Supply
• Resistors

Circuit Diagram

The circuit diagram for Milliman’s Theorem includes the following.

Working Procedure

The step-by-step working of this experiment includes the following.

As per the first circuit shown above, give the connections accordingly on the breadboard using connecting wires

Determine the flow of current throughout the ‘R3’ resistor.

Similarly, give the connections as per the second circuit diagram & determine the flow of current throughout the ‘R3’ resistor.

Check the two current readings whether the same or not.

Tabular Form

Note down the theoretical and practical values for the parameters.

Result

So finally, Millman’s theorem can be verified.

Advantages & Applications

The advantages and applications of Millman’s theorem include the following.

• This theorem is used to a circuit that has a set of current and voltage sources as the voltage source is changed into a current source & vice versa
• This Millman’s theorem is applicable for determining the voltage across a set of parallel branches, wherever there are sufficient voltage sources present to prevent a solution using the method like series-parallel reduction.
• It is very easy to utilize because it doesn’t need the utilization of instantaneous equations.
• This theorem is frequently used for circuits that use many op-amps to signify complex circuit topology.

Disadvantages

The disadvantages of Millman’s theorem include the following.

• This theorem is not used for the circuits which include impedances among the independent source.
• This theorem is not used for the circuit that has a dependent source among the independent source.
• This theorem is not applicable when the circuit includes two independent sources.

Thus, this is all about an overview of Millman’s Theorem. This theorem states that the voltage across the parallel connections of the circuit includes above one voltage source so it decreases the circuit complexity. For example, once the number of voltage sources is V1, V2 …+Vn which are connected in parallel with internal resistances like R1, R2 … +Rn) correspondingly. This arrangement can change through a voltage source ‘V’ with resistance ‘R’ in series. Here is a question for you, what are the different types of network theorems used in electrical circuits?