Superposition and Millman’s Theorems
Exercise 2: Millman’s Theorem
EXERCISE OBJECTIVE
When you have completed this exercise, you will be able to solve a circuit by applying Millman’s theorem.
You will verify your results by comparing calculated and measured data.
DISCUSSION
The superposition method is used to individually determine the effects of each source on a common
circuit element and then combine them algebraically.
Millman’s theorem uses the sum of the branch currents and the sum of the conductances to help you
determine the voltage across the branches.
To use Millman’s theorem, the circuit used in Exercise 1 is redrawn to show each of its branches. No
electrical changes are made.
R1 and its series voltage source form a branch. R2 and its series voltage source form a branch.
R3 is placed across (in parallel with) the other branches to form the third branch.
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Superposition and Millman’s Theorems
The objective is to determine the common branch voltage, which is the voltage drop of R3. Once known,
this voltage leads to a solution of all circuit currents and voltage drops.
Millman’s theorem uses branch currents and conductance (G) to solve the circuit.
a.
b.
c.
d.
VS2 + VR2
VS1 + VR1
VR3
All of the above
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Superposition and Millman’s Theorems
Are the branch voltages equal?
a. yes
b. no
According to Millman’s theorem, divide the sum of the parallel currents (branch currents) by the sum of
the parallel conductances (1/R) to determine the voltage drop of R3.
VR3
VR1 VR2 VR3
+
+
= R1 R2 R3
1
1
1
+
+
R1 R2 R3
Sum of the branch currents
Sum of the branch conductants
BRANCH 1 current is the voltage of VS1 divided by ohmic value of R1 (VS1/R1); voltage over resistance
yields current.
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Superposition and Millman’s Theorems
VS1 is negative with respect to circuit common.
BRANCH 2 current equals VS2/R2.
BRANCH 3 current equals VS3/R3. Because BRANCH 3 does not have a voltage source, its current is
zero (0/R3 = 0).
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Superposition and Millman’s Theorems
PROCEDURE
฀
Locate the SUPERPOSITION circuit block, and connect the circuit shown.
Adjust each variable voltage source to 10.0 Vdc.
฀
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The Millman equivalent circuit schematic is illustrated. VR3 equals the sum of the branch
currents divided by the sum of the circuit conductances.
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Superposition and Millman’s Theorems
฀
What is the total conductance of the circuit?
GT =
฀
What is the value of R1 branch current? Use VS1/R1.
IR1 =
฀
millisiemens (Recall Value 1)
mA (Recall Value 2)
Based on the Millman equivalent circuit, what is the R3 branch current?
IR3 =
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mA (Recall Value 3)
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Superposition and Millman’s Theorems
฀
What is the value of R2 branch current? Use VS2/R2.
IR2 =
฀
mA (Recall Value 4)
Based on your calculated values and the given formula, what is the value of VR3?
GT =
mS (Step 3, Recall Value 1)
IR1 =
mA (Step 4, Recall Value 2)
IR2 =
mA (Step 6, Recall Value 4)
VR3 = (IR1 + IR2)/GT
VR3 =
฀
Vdc (Recall Value 5)
Based on the Millman solution for VR3, can you determine the actual circuit currents and
voltage drops?
a. yes
b. no
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Superposition and Millman’s Theorems
฀
Use your voltmeter to measure VR3. Are your results consistent with Millman’s theorem?
a. yes
b. no
฀
Are your results consistent with the Millman solution?
a. yes
b. no
CONCLUSION
•
Use Millman’s theorem to calculate the voltage across a common circuit element.
•
Millman’s theorem takes the form of current divided by conductance.
•
Current divided by conductance (I/G), or multiplied by resistance (I x R), yields voltage.
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Superposition and Millman’s Theorems
REVIEW QUESTIONS
1. Based on Millman’s theorem, what is the sum of the branch currents?
a.
b.
c.
d.
4.08 mA
–4.08 mA
±4.08 mA
None of the above
2. Based on Millman’s theorem, what is the sum of the branch conductances?
a.
b.
c.
d.
5.74 mS
–5.74 mS
±5.74 mS
All of the above
3. With respect to circuit common, what is the
voltage drop across R3?
a.
b.
c.
d.
0.7 V
–0.7 V
Either of the above
None of the above
4. The sum of the conductances
a.
b.
c.
d.
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increases as the source voltage is increased.
does not change with changes in source voltage.
decreases as the source voltage is decreased.
equals zero when the source voltage is zero.
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Superposition and Millman’s Theorems
5. For the circuit shown,
a.
b.
c.
d.
Millman’s theorem cannot be applied because common is not in the proper place.
VS1 is negative when applied to Millman’s theorem.
VS2 is positive when applied to Millman’s theorem.
Millman’s theorem does not affect the polarity of VS1 or VS2 when they are measured with respect
to circuit common.
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