Millman's Theorem is a fundamental concept in electrical engineering. It provides a viable tool for analyzing electrical networks. This theorem is named after Jacob Millman. It enables engineers to simplify complex networks into simpler equivalent circuits. This makes it easier to analyze and understand the behavior of networks. By applying Millman's Theorem, engineers can calculate voltage and current in a complex network. As such, it is an essential tool for circuit analysis and design. Millman's Theorem plays a vital role in the field of electrical engineering.

This blog shall enumerate Millman’s Theorem and the steps to use it for simplification of complicated networks. This topic is crucial for upcoming examinations pertaining to Electrical engineering, including SSC JE, GATE Electrical Exam, etc.

Millman’s Theorem Statement

Millman's Theorem in network theory states that in a network with multiple parallel branches, the voltage across each branch is equal to the ratio of the sum of individual branch currents multiplied by their respective resistances, divided by the sum of all branch resistances. This theorem simplifies complex electrical networks. It represents such networks as equivalent circuits with simpler voltage sources and resistors.

The Millman’s Equation applied to any circuit drawn as a set of parallel-connected branches is given as follows:

\(frac{\frac{E_{B 1}}{R_{1}}+\frac{E_{B 2}}{R_{2}}+\frac{E_{B 3}}{R_{3}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}}=\text { Voltage across all branches }\)

Millman’s Theorem involves the redrawing of a circuit as a parallel network composed of branches, where each branch consists of a resistor or a series combination of a battery and a resistor. It is important to note that Millman's Theorem is applicable only to circuits that can be reconfigured in this manner. Once the circuit is redrawn accordingly, Millman's Theorem can be applied for analysis. To illustrate this, let us take the following circuit diagram as a precedent.

The modified form of the circuit assumes the following form:

By taking into account the voltage source within each branch and the resistance within each branch, Millman's Theorem allows us to determine the voltage across all branches. It is worth mentioning that in the circuit, the battery in the rightmost branch has been labeled as "B3" to clearly indicate its position in the third branch, even though there is no "B2" in the circuit.

By using the Millman’s Equation for the above circuit, we get:

\(frac{\frac{28 \mathrm{~V}}{4 \Omega}+\frac{0 \mathrm{~V}}{2 \Omega}+\frac{7 \mathrm{~V}}{1 \Omega}}{\frac{1}{4 \Omega}+\frac{1}{2 \Omega}+\frac{1}{1 \Omega}}=8 \mathrm{~V}\)

This voltage is across all the parallel branches.

Solving for Resistor Voltage Drops

In order to determine the voltage drops across resistors, compare the Millman voltage across the parallel network with the voltage source in each branch. By applying the principle of voltages adding in series, you can determine both the magnitude and polarity of the voltage across each resistor.

\(E_{\mathrm{R} 1}=8 \mathrm{~V}-28 \mathrm{~V}=-20 \mathrm{~V} \quad \text{ (negative on top)}\)

\(E_{\mathrm{R} 2}=8 \mathrm{~V}-0 \mathrm{~V}=8 \mathrm{~V} \quad \text{ (positive on top)}\\\)

\(E_{R3}=8 V -7V=1 V (positive on top)\)

Solving for Branch Currents

Millman's theorem is not directly used to solve for branch currents in a circuit. It is primarily used to determine the voltage across each branch in a parallel network. However, once the voltage drops across the resistors are known, you can use Ohm's law (I = E/R) to calculate the branch currents by dividing the voltage drop across each resistor by its respective resistance.

\({}\mathrm{I}_{\mathrm{R} 1}=\frac{20 \mathrm{~V}}{4 \Omega}=5 \mathrm{~A} \)

\({I}_{\mathrm{R} 2}=\frac{8 \mathrm{~V}}{2 \Omega}=4 \mathrm{~A} \)

\(\mathrm{I}_{\mathrm{R} 3}=\frac{1 \mathrm{~V}}{1 \Omega}=1 \mathrm{~A}\)

Steps for Applying Millman’s Theorem

By following a set of simple steps, we can apply Millman's Theorem to analyze an electrical network effectively. Here are the summarized steps:

Step 1: The circuit converts all voltage sources to current by dividing their potential by their internal resistance, Where G represents the conductance of the circuit.

\(I_{\text{source}} = \frac{V_{\text{source}}}{R_{\text{source}}}\)

\(\begin{array}{l}I=E / R=E G \\G=1 / R\end{array}\)

Step 2: Calculate the circuit’s total current using the following mathematical expression.

\(\begin{array}{l}\text { leq }=E_1 / R_1 + E_2 / R 2 + E_3 / R_3+\ldots+E_N / R_N \\\text { leq }=( \pm E_1 G_1)+( \pm E_2 G_2)+( \pm E_3 G_3)+\ldots+( \pm E_N G_N)\end{array}\)

Step 3: Calculate the conductance of each branch by taking the reciprocal of the resistance (G = 1/R). The conductance of each current source can be calculated using mathematical expressions.

\(\begin{array}{l}1 / \mathrm{R_eq}=1 / \mathrm{R}_1+1 / \mathrm{R}_2+1 / \mathrm{R}_3+\ldots+1/\mathrm{R_N}\\\mathrm{Geq}=\mathrm{G}_1+\mathrm{G}_2+\mathrm{G}_3+\ldots+\mathrm{G_N}\end{array}\)

Step 4: Assign a positive polarity to the voltage source within each branch, indicating its direction.

Step 5: Based on Ohm’s law, we can calculate the voltage at the open end of the parallel circuit.

\(I_{\text {total }}=\frac{E_{B_1}}{R_{1}}+\frac{E_{82}}{R_{2}}+\frac{E_{83}}{R_{3}} \longrightarrow \frac{\frac{E_{B_1}}{R_{1}}+\frac{E_{B_2}}{R_{2}}+\frac{E_{B_3}}{R_{3}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}}\)

Step 6: Calculate the current flowing through the load circuit RL.

\(\mathrm{IL}=\mathrm{Eeq} /(\mathrm{Req}+\mathrm{R_L})\)

Determining The Direction of Current

The current direction across each resistor is determined by the polarity across the resistor itself, rather than the polarity across each battery. It is possible for the current to be forced back through a battery. It is crucial to remember this because Millman's Theorem does not offer as clear an indication of an "incorrect" current direction compared to the Branch Current or Mesh Current methods. It is necessary to closely consider the polarities of resistor voltage drops, as determined by Kirchhoff's Voltage Law, in order to determine the direction of currents accurately.

Applications of Millman's Theorem

Millman's Theorem is widely used in electrical engineering for a range of applications. The applications of Millman’s theorem encompass the following::

• Circuit analysis and design: Millman's Theorem enables engineers to simplify complex electrical circuits with multiple parallel branches, making it easier to analyze and design circuits. It helps determine voltage and current distributions, power dissipation, and overall circuit behavior.
• Power systems: Millman's Theorem plays a crucial role in analyzing and designing power distribution networks. It aids in calculating voltages and currents at different nodes, assessing power flow, and determining the performance of power systems.
• Electronic circuit design: When designing electronic circuits with multiple parallel branches, Millman's Theorem helps simplify the circuit and analyze voltage and current characteristics. It assists in optimizing circuit performance and ensuring proper functionality.
• Control systems: Millman's Theorem is utilized in the analysis and design of control systems, allowing engineers to understand and manipulate voltage and current parameters in parallel branches. This aids in the design of stable and efficient control systems.
• Telecommunications: In telecommunications systems, Millman's Theorem is employed to analyze and optimize network configurations. It helps assess voltage and current distributions in parallel branches, ensuring reliable transmission and reception of signals.

Millman's Theorem Importance

The importance of Millman's Theorem extends to various applications in electrical engineering. It helps engineers to analyze and design circuits. It is used in power systems, telecommunications, control systems, and electronic devices. It helps in checking the performance of circuits. It also helps in calculating the voltage and current distribution. Millman's theorem can also help in determining power dissipation.

Advantages of Millman’s Theorem

Millman's Theorem offers several advantages that make it a valuable tool in electrical engineering:

• Simplifies complex circuits: Millman's Theorem in network theory allows for the simplification of complex electrical networks with multiple parallel branches. By representing them as equivalent circuits with voltage sources and resistors, it simplifies the analysis process and reduces complexity.
• Facilitates circuit analysis: The theorem provides a systematic approach to analyze circuit behavior, including voltage drops, current flows, and power dissipation. It helps engineers gain a better understanding of the circuit's performance and characteristics.
• Saves time and effort: By simplifying complex circuits, Millman's Theorem saves time and effort in analyzing and designing electrical networks. It streamlines the calculations required for circuit analysis, enabling engineers to focus on specific components or branches without losing accuracy.
• Versatile application: Millman's Theorem can be applied to a wide range of electrical circuits, including power systems, control systems, telecommunications, and electronic design. Its versatility makes it a valuable tool in various fields of electrical engineering.
• Accurate results: When applied correctly, Millman's Theorem provides accurate results in analyzing circuit voltages and currents. It helps engineers make informed decisions in circuit design and optimization.
• Enhances understanding: By simplifying complex circuits and providing a clear representation, Millman's Theorem enhances engineers' understanding of circuit behavior. It allows for better troubleshooting, optimization, and overall circuit comprehension.

Limitations of Millman's Theorem

While Millman's Theorem is a useful tool in electrical engineering, it also has some limitations to consider:

• Applicable to specific circuit configurations: Millman's Theorem is applicable to circuits with multiple parallel branches. It may not be directly applicable to circuits with series-parallel combinations or more complex topologies, which require alternative analysis methods.
• Assumes ideal conditions: Millman's Theorem assumes ideal conditions, such as negligible internal resistance in voltage sources and linear components. In real-world circuits, these assumptions may not hold true, leading to deviations from the theoretical calculations.
• Limited to linear circuits: Millman's Theorem is valid for linear circuits, where the relationship between voltage and current is linear. It may not provide accurate results for circuits with non-linear elements such as diodes or transistors.
• Relies on known resistances and voltages: To apply Millman's Theorem, knowledge of the resistances and voltages in the circuit is required. In practical situations where these values are not precisely known, obtaining accurate results can be challenging.
• Ignores mutual coupling effects: Millman's Theorem assumes that parallel branches are isolated from each other and do not interact. It does not consider mutual coupling effects between branches, which can occur in certain circuit configurations and affect the accuracy of results.
• Does not account for transient behavior: Millman's Theorem focuses on steady-state analysis and does not address transient behaviors, such as initial transients or transient response to input changes. For time-varying circuits or circuits with transient phenomena, additional analysis techniques are necessary.

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Millman’s Theorem Solved Examples

Q) Find the voltage across 4 Ohms resistance using Millman’s theorem. Also, find the current through 4 Ohms resistance

Fig 2: Circuit diagram 

Solution: 

Step 1: Remove the load resistance and measure the voltage at the source end.

Fig 2: Simplification 

\({l} V=\frac{\frac{V_{1}}{R_{1}}+\frac{V_{2}}{R_{2}}+\frac{V_{3}}{R_{3}}+\cdots+\frac{V_{k}}{R_{k}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\cdots+\frac{1}{R_{k}}}\\\)

\(V=\frac{\frac{25}{10}+\frac{16}{20}+\frac{20}{20}}{\frac{1}{10}+\frac{1}{20}+\frac{1}{20}}=\frac{50+16+20}{2+1+1}=\frac{86}{4}=21.5 \text { Volts }\)

Fig 4: Simplified Circuit

Current via the 4 Ohm resistor is given by:

\(I=\frac{V}{R_{e}+R_{L}}=\frac{21.5}{5+4}=\frac{21.5}{9}=2.38 \text { Amperes }\)