superposition circuits
TRANSCRIPT
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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
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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
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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
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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
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Set 8A active 20V = 0
+
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
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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``
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BACK
Set 8A active 25V = 0
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
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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