In electrial engineering, Millman's Theorem (or the parallel generator theorem) is a useful method to simplify the solution of a circuit. It is named after Jacob Millman, who proved the theorem. A similar method, known as Tank's method, had already been commonly used before Millman's proof.

In Millman's Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination. Millman's Theorem is applicable only to those circuits which can be re-drawn accordingly.Millman's theorem gives a very quick method to compute the voltage at the ends of a circuit made up of only branches in parallel.

Let ek be the voltage generators and am the current generators.

Let Ri be the resistances on the branches with no generator.

Let Rk be the resistances on the branches with voltage generators.

Let Rm be the resistances on the branches with current generators.

Then Millman states that the voltage at the ends of the circuit is given by:

It can be proved by considering the circuit as a single supernode. Then, according to Ohm and Kirchhof, the voltage between the ends of the circuit is equal to the total current entering the supernode divided by the total equivalent conductance of the supernode.

The total current is the sum of the currents flowing in each branch.

The total equivalent conductance of the supernode is the sum of the conductance of each branch, since all the branches are in parallel. When computing the equivalent conductance all the generators have to be switched off, so all voltage generators become short circuits and all current generators become open circuits. That's why the resistances on the branches with current generators do not appear in the expression of the total equivalent conductance.

Example

When re-drawn for the sake of applying Millman's Theorem the circuit becomes

By considering the supply voltage within each branch and the resistance within each branch, Millman's Theorem will tell us the voltage across all branches. Please note that I've labeled the battery in the rightmost branch as "B3" to clearly denote it as being in the third branch, even though there is no "B2" in the circuit!

Millman's Theorem is nothing more than a long equation, applied to any circuit drawn as a set of parallel-connected branches, each branch with its own voltage source and series resistance:

Substituting actual voltage and resistance figures from our example circuit for the variable terms of this equation, we get the following expression:

The final answer of 8 volts is the voltage seen across all parallel branches like this.

The polarity of all voltages in Millman's Theorem are referenced to the same point. In the example circuit above, bottom wire of the parallel circuit is taken reference point, and so the voltages within each branch (28 for the R1 branch, 0 for the R2 branch, and 7 for the R3 branch) were inserted into the equation as positive numbers. Likewise, when the answer came out to 8 volts (positive), this meant that the top wire of the circuit was positive with respect to the bottom wire (the original point of reference). If both batteries had been connected backwards (negative ends up and positive ends down), the voltage for branch 1 would have been entered into the equation as a -28 volts, the voltage for branch 3 as -7 volts, and the resulting answer of -8 volts would have told us that the top wire was negative with respect to the bottom wire (our initial point of reference).

With the result we get we can calculate the resistor voltage dops and branch currents like this:

ER1 = 8V - 28V = -20V (negative on top)

ER2 = 8V - 0V = 8V (positive on top)

ER3 = 8V - 7V = 1V (posiive on top)

To solve for branch currents, each resistor voltage drop can be divided by its respective resistance (I=E/R):

IR1 = 20V/4 = 5A

IR2 = 8V/2 = 4A

IR3= 1V/1 = 1A

The direction of current through each resistor is determined by the polarity across each resistor, not by the polarity across each battery, as current can be forced backwards through a battery, as is the case with B3 in the example circuit. This is important to keep in mind, since Millman's Theorem doesn't provide as direct an indication of "wrong" current direction as does the Branch Current or Mesh Current methods. You must pay close attention to the polarities of resistor voltage drops as given by Kirchhoff's Voltage Law, determining direction of currents from that.

Millman's Theorem is very convenient for determining the voltage across a set of parallel branches, where there are enough voltage sources present to preclude solution via regular series-parallel reduction method. It also is easy in the sense that it doesn't require the use of simultaneous equations. However, it is limited in that it only applied to circuits which can be re-drawn to fit this form. It cannot be used, for example, to solve an unbalanced bridge circuit. And, even in cases where Millman's Theorem can be applied, the solution of individual resistor voltage drops can be a bit daunting to some, the Millman's Theorem equation only providing a single figure for branch voltage.

For more details on this topic visit the folloing sites : http://www.analyzethat.net/87_millman_theorem.php www.allaboutcircuits.com/vol_1/chpt_10/5.html

http://en.wikipedia.org/wiki/Millman

www.ibiblio.org/kuphaldt/socratic/output/millman.pdf --- Exercise Problems

In Millman's Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination. Millman's Theorem is applicable only to those circuits which can be re-drawn accordingly.Millman's theorem gives a very quick method to compute the voltage at the ends of a circuit made up of only branches in parallel.

Let ek be the voltage generators and am the current generators.

Let Ri be the resistances on the branches with no generator.

Let Rk be the resistances on the branches with voltage generators.

Let Rm be the resistances on the branches with current generators.

Then Millman states that the voltage at the ends of the circuit is given by:

It can be proved by considering the circuit as a single supernode. Then, according to Ohm and Kirchhof, the voltage between the ends of the circuit is equal to the total current entering the supernode divided by the total equivalent conductance of the supernode.

The total current is the sum of the currents flowing in each branch.

The total equivalent conductance of the supernode is the sum of the conductance of each branch, since all the branches are in parallel. When computing the equivalent conductance all the generators have to be switched off, so all voltage generators become short circuits and all current generators become open circuits. That's why the resistances on the branches with current generators do not appear in the expression of the total equivalent conductance.

Example

When re-drawn for the sake of applying Millman's Theorem the circuit becomes

By considering the supply voltage within each branch and the resistance within each branch, Millman's Theorem will tell us the voltage across all branches. Please note that I've labeled the battery in the rightmost branch as "B3" to clearly denote it as being in the third branch, even though there is no "B2" in the circuit!

Millman's Theorem is nothing more than a long equation, applied to any circuit drawn as a set of parallel-connected branches, each branch with its own voltage source and series resistance:

Substituting actual voltage and resistance figures from our example circuit for the variable terms of this equation, we get the following expression:

The final answer of 8 volts is the voltage seen across all parallel branches like this.

The polarity of all voltages in Millman's Theorem are referenced to the same point. In the example circuit above, bottom wire of the parallel circuit is taken reference point, and so the voltages within each branch (28 for the R1 branch, 0 for the R2 branch, and 7 for the R3 branch) were inserted into the equation as positive numbers. Likewise, when the answer came out to 8 volts (positive), this meant that the top wire of the circuit was positive with respect to the bottom wire (the original point of reference). If both batteries had been connected backwards (negative ends up and positive ends down), the voltage for branch 1 would have been entered into the equation as a -28 volts, the voltage for branch 3 as -7 volts, and the resulting answer of -8 volts would have told us that the top wire was negative with respect to the bottom wire (our initial point of reference).

With the result we get we can calculate the resistor voltage dops and branch currents like this:

ER1 = 8V - 28V = -20V (negative on top)

ER2 = 8V - 0V = 8V (positive on top)

ER3 = 8V - 7V = 1V (posiive on top)

To solve for branch currents, each resistor voltage drop can be divided by its respective resistance (I=E/R):

IR1 = 20V/4 = 5A

IR2 = 8V/2 = 4A

IR3= 1V/1 = 1A

The direction of current through each resistor is determined by the polarity across each resistor, not by the polarity across each battery, as current can be forced backwards through a battery, as is the case with B3 in the example circuit. This is important to keep in mind, since Millman's Theorem doesn't provide as direct an indication of "wrong" current direction as does the Branch Current or Mesh Current methods. You must pay close attention to the polarities of resistor voltage drops as given by Kirchhoff's Voltage Law, determining direction of currents from that.

Millman's Theorem is very convenient for determining the voltage across a set of parallel branches, where there are enough voltage sources present to preclude solution via regular series-parallel reduction method. It also is easy in the sense that it doesn't require the use of simultaneous equations. However, it is limited in that it only applied to circuits which can be re-drawn to fit this form. It cannot be used, for example, to solve an unbalanced bridge circuit. And, even in cases where Millman's Theorem can be applied, the solution of individual resistor voltage drops can be a bit daunting to some, the Millman's Theorem equation only providing a single figure for branch voltage.

For more details on this topic visit the folloing sites : http://www.analyzethat.net/87_millman_theorem.php www.allaboutcircuits.com/vol_1/chpt_10/5.html

http://en.wikipedia.org/wiki/Millman

www.ibiblio.org/kuphaldt/socratic/output/millman.pdf --- Exercise Problems

## 1 comment:

Hi madam it is nice even a poor student can understand the theorem well and even the display ie example and graphics are nice.

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