superposition circuits

8/7/2019 Superposition Circuits

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SUPERPOSITION

THEOREM


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SUPERPOSITION THEOREM

The response (a desired voltage or a desired current) at

any point in the circuit containing more than one independent

source is equal to the sum of the responses caused by each

independent source acting alone.


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STEPS TO APPLY SUPERPOSITION PRINCIPLE

1. Turn off all independent sources except one source. Find the output

(voltage or current) due to active source using the techniques covered.

2. Repeat step 1 for each of the other independent source.

3. Find the total contribution by adding algebraically all the

contributions due to the independent sources.

* Before adding their individual responses, a current going to right/left is assumed to

have a positive/negative sign convention and a current down/up is assumed to have

a positive/negative sign convention.

EXAMPLESAMPLE PROBLEMS


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V1

V2

-

+

Is

-+

Consider,

BACK

Linear

Network Vo

Vo = Vo` + Vo`` + Vo```


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SAMPLE PROBLEMS:

1. Use the superposition to find v in the circuit.2. Using superposition theorem, find vo in the circuit.

3. Determine I3

4. Solve for Ix

5. Determine I and Ix

6. Solve for I

7. Determine Ix

8. Solve for ix

9. Find io

BACKEND


6/22

BACK

-

+

6 V

8

3 A

4

1.

Ans: v = 10 V

SOLUTION

+

V

-

BACKSOLUTION


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2

3 5

-

+20 V

8 A

2.

Ans: vo = 12 V

+

Vo

-

BACKSOLUTION


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3.

4.2 V

0.02

6.3 V

0.03

5

Ans: I3 = 1 A

I3

BACKSOLUTION


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8 A

7.5

1.25

15

5

-

+

25 V

4.

Ans: Ix = 3.3 A

Ix

BACKSOLUTION


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0.1

0.5 A

0.3

-

+

80 mV

5.

Ix

I

Ans: Ix = -175 mA

I = 175 mA

BACKSOLUTION


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-

+

3 V

-

+

1 V

-

+

2 V

10 40

7.

BACK


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-

+

10 V 3 A

2 1

+

-2ix

8.

Ans: ix = 1.4 A

BACK


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9.

4 A

3

5

1

4

2

-+20 V

Ans: io = -0.4706

+ -

io

BACK


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-

+E1

R1 R2

-

+ E2

R3

EXAMPLE:

+V3

-

+ V1 - - V2 +

I1I2

I3

Let E1

acts alone:Let E2 acts alone:

R1 R2

-

+E2

R3

I3``I1``

I2``

To solve for the currents:

I1 = I1` + (-I1``)

I2 = -I2` + I2``

I3 = I3` + I3``

-

+

E1

R1 R2

R3

I1` I2`I3`

To solve for the voltages:

V1 = V1` + (-V1``)

V2 = -V2 + V2``

V3 = V3` + V3``BACK


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Set 6V active 3A = 0

4

8

-

+

6 V

+

V`-

v` = 4i`

i` = 6V / (8+4)

i` = 0.5 A

v` = 4(0.5)

v` = 2V

Set 3A active 6V = 0

+

V``

-

CDP:

i4 = 3(8 / 8+4)

i4 = 2 A

v`` = 4(2) = 8 V

v = v` + v``

v = 2V + 8V

v = 10V

BACK


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+

vo`-

Ix Iy

CDP:

Ix = 8(5 / 5+5) = 4 A

vo` = 2Ix = 2(4) = 8V

Set 20V active 8A = 0

+

vo``-

VDP:

vo`` = 20(2 / 2+3+5)= 20(2 / 10)

vo`` = 4V

vo = vo` + vo``

= 8V + 4V

vo

= 12V

BACK


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Set 4.2 V active

Req = 0.02 + ((5)(0.03) / 5 + 0.03)

= 0.05

IT = 4.2 / 0.05 = 84 A

I3` = 84(0.03 / 5.03)

I3` = 0.5 A

I3`

Set 6.3 V active

Req = 0.03 + ((5)(0.02) / 5.02)

= 0.05

IT = 6.3 / 0.05 = 126 A

I3`` = 126 (0.02 / 5.02)

I3`` = 0.5 AI3 = I3` + I3``

= 0.5 A +0.5 A

I3 = 1 ABACK

I3``


19/22

BACK


IT = 8 A

Ix` = IT (7.5 / 12.5)

Ix` = 4 .8 A

Ix`

Set 25V active 8A= 0

RT = ((8.75)(15) / 23.75) + 5

RT = 10.52

IT = 25 / 10.52

IT = 2.37 A

Ix`` = 2.37(15 / 23.75)

Ix`` = 1.5 A

Ix``

Ix = Ix` - Ix``

Ix = 4.8 A 1.5 A

Ix = 3.3 A


20/22

BACK

Set 80 mV active 0.5A = 0

I`

RT = 0.1 + 0.3 = 0.4

IT = I` = 0.08 / 0.4

I` = 0.2 A

Set 0.5 A active 80 mV = 0

IT = 0.5 A

I`` = 0.5 (0.3 / 0.4)

I`` = 0.375 A

I = -I` + I``

I = -0.2 + 0.375

I = 0.175 A

I``


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BACK

120 A = active , 10V = 0 , 40 A = 0

IT = 120 A

Ix` = 120 (50 / 200)

Ix` = 30 A

40 A = active , 10V = 0 , 120 A = 0

IT = 40 A

Ix`` = 120 (150 / 200)

Ix`` = 30 A120 A = active , 10V = 0 , 40 A = 0

Ix = -Ix` + Ix`` + Ix```

Ix = -30 + 30 + 0.05

Ix = 0.05 AIT = Ix``` = 10 / 200

Ix``` = 0.05 A

Ix`

Ix``

Ix```


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END

superposition circuits

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