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Network Analysis X Millman’s Theorem Analysis

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Page 1: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Network Analysis X

Millman’s Theorem Analysis

Page 2: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Concept of Millman’s

• All about the parallel configuration

– Cannot have more than one source in a branch

– Cannot have more than one resistance in a branch

Page 3: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Converted to proper configuration for Millman’s

• Combine resistances in a branch to one value

• Combine voltages in a branch to one value

Page 4: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Formula for calculation

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

+⋯+𝑉𝑁𝑅𝑁

1

𝑅1+

1

𝑅2+

1

𝑅3+⋯+

1

𝑅𝑁

=𝐼

𝐺

(𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =1

𝐺)

Page 5: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

First circuit

• In the figure shown, apply Millman’s theorem to solve for the voltage VXY.

Page 6: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solution

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

1

𝑅1+

1

𝑅2+

1

𝑅3

=𝐼

𝐺 (𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =

1

𝐺)

• 𝑉𝑋𝑌 =−50V

1kΩ+

20V

1kΩ+

0

2kΩ1

1kΩ+

1

1kΩ+

1

2kΩ

=

•−50mA+20mA+0mA

0.001mhos+0.5mmhos+0.001mhos=

•−30mA

0.0025mhos= −12V

Page 7: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅1 = −12V − −50V= −12V + 50V= 38V

-12V

Page 8: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅3 = −12V

-12V

Page 9: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅2 = −12V − 20V= −32V

-12V

Page 10: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

First circuit revised

• If the polarity of V2 is reversed, recalculate the voltage VXY in the figure shown.

Page 11: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solution

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

1

𝑅1+

1

𝑅2+

1

𝑅3

=𝐼

𝐺 (𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =

1

𝐺)

• 𝑉𝑋𝑌 =−50V

1kΩ−

20V

1kΩ+

0

2kΩ1

1kΩ+

1

1kΩ+

1

2kΩ

=

•−50mA−20mA+0mA

0.001mhos+0.5mmhos+0.001mhos=

•−70mA

0.0025mhos= −28V

Page 12: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅1 = −28V − −50V= −28V + 50V= 22V

-28V

Page 13: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅3 = −28V

-28V

Page 14: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅2 = −28V − −20V= −28V + 20V= −8V

-28V

Page 15: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Second circuit

• In the figure shown, apply Millman’s theorem to solve for the voltage VXY.

Page 16: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solution

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

1

𝑅1+

1

𝑅2+

1

𝑅3

=𝐼

𝐺 (𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =

1

𝐺)

• 𝑉𝑋𝑌 =6V

3Ω+

18V

9Ω+

0

18Ω1

3Ω+

1

9Ω+

1

18Ω

=

•2A+2A+0A

333.333mmhos+55.556mmhos+111.111mmhos=

•4A

500mmhos= 8V

Page 17: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅1 = 8V − 6V = 2V

8V

Page 18: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅3 = 8V

8V

Page 19: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅2 = 8V − 18V= −10V

8V

Page 20: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Second circuit revised

• If the polarity of V2 is reversed, recalculate the voltage VXY in the figure shown.

Page 21: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solution

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

1

𝑅1+

1

𝑅2+

1

𝑅3

=𝐼

𝐺 (𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =

1

𝐺)

• 𝑉𝑋𝑌 =6V

3Ω−

18V

9Ω+

0

18Ω1

3Ω+

1

9Ω+

1

18Ω

=

•2A−2A+0A

333.333mmhos+55.556mmhos+111.111mmhos=

•0A

500mmhos= 0V

Page 22: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅1 = 0V − 6V = −6V

0V

Page 23: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅3 = 0V

0V

Page 24: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅2 = 0V − −18V= 0 + 18V = 18V

0V

Page 25: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Third circuit

• In the figure shown, apply Millman’s theorem to solve for the voltage VXY.

Page 26: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solution

• 𝑉𝑋𝑌 =

𝑉1𝑅1

+𝑉2𝑅2

+𝑉3𝑅3

1

𝑅1+

1

𝑅2+

1

𝑅3

=𝐼

𝐺 (𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐼 × 𝑅 𝑠𝑖𝑛𝑐𝑒 𝑅 =

1

𝐺)

• 𝑉𝑋𝑌 =12V

10Ω−

18V

15Ω+

24

40Ω1

10Ω+

1

15Ω+

1

40Ω

=

•1.667A−1.2A+575mA

100mmhos+66.667mmhos+25mmhos=

•191.667mA

191.667mmhos= 1V

Page 27: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅1 = 1V − 12V= −11V

1V

Page 28: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅2 = 1V − −18V= 1V + 18V = 19V

1V

Page 29: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Solve for individual values

∴ 𝑉𝑅3 = 1V − 24V= −23V

1V

Page 30: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

Any questions?

• Contact us at:

– 1-800-243-6446

– 1-216-781-9400

• Email:

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Page 31: Network Analysis VII - Electronics · Concept of Millman’s •All about the parallel configuration –Cannot have more than one source in a branch –Cannot have more than one resistance

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