Superposition theorem
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Transcript of Superposition theorem
![Page 1: Superposition theorem](https://reader033.fdocuments.in/reader033/viewer/2022061408/587185271a28ab2c198b4bb3/html5/thumbnails/1.jpg)
Superposition Theorem
• The superposition theorem extends the use of Ohm’s Law to circuits with multiple sources.
• Statement:- “The current through, or voltage across, an element in a linear bilateral network equal to the algebraic sum of the currents or voltages produced independently by each source.”
• Superposition theorem is very helpful in determining the voltage across an element or current through a branch when the circuit contains multiple number of voltage or current sources.
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In order to apply the superposition theorem to a network, certain conditions must be met :
• All the components must be linear, for e.g.- the current is proportional to the applied voltage (for resistors), flux linkage is proportional to current (in inductors), etc.
• All the components must be bilateral, meaning that the current is the same amount for opposite polarities of the source voltage.
• Passive components may be used. These are components such as resistors, capacitors, and inductors, that do not amplify or rectify.
• Active components may not be used. Active components include transistors, semiconductor diodes, and electron tubes. Such components are never bilateral and seldom linear.
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Procedure for applying Superposition Theorem
Circuits Containing Only Independent Sources:-• Consider only one source to be active at a time.• Remove all other ideal voltage sources by short circuit & all
other ideal current sources by open circuit.
Voltage source is replaced by a Short Circuit
Current source is replaced by a Open Circuit
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• If there are practical sources, replace them by the combination of ideal source and an internal resistances.
• After that, short circuit the ideal voltage source & open circuit the ideal current source..
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Example :1
R1R2
R3V1 V2
100 W 20 W10 W
15 V 13 V
R1 R2
R3V1
100 W 20 W10 W
15 VV2 shorted
REQ = 106.7 W, IT = 0.141 A and IR3 = 0.094 A
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R1 R2
R3V1 V2
100 W 20 W
10 W
15 V 13 V
REQ = 29.09 W, IT = 0.447 A and IR3 = 0.406 A
R1 R2
R3V2
100 W 20 W
10 W
13 VV1 shorted
Example :1
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R1 R2
V1 V2
100 W 20 W15 V 13 V
Adding the currents gives IR3 = 0.5 A
REQ = 106.7 W, IT = 0.141 A and IR3 = 0.094 A
REQ = 29.09 W, IT = 0.447 A and IR3 = 0.406 A
With V2 shorted
With V1 shorted
0.094 A 0.406 A
Example :1
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R1 R2
R3V1 V2
100 W 20 W
10 W
15 V 13 V
With 0.5 A flowing in R3, the voltage across R3 mustbe 5 V (Ohm’s Law). The voltage across R1 musttherefore be 10 volts (KVL) and the voltage across R2
must be 8 volts (KVL). Solving for the currents in R1
and R2 will verify that the solution agrees with KCL.
0.5 A
IR1 = 0.1 A and IR2
= 0.4 A
IR3 = 0.1 A + 0.4 A = 0.5 A
Example :1
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Procedure for applying Superposition Theorem
Circuits Containing Independent & Dependent Sources :-
• Consider only one source to be active at a time.• Remove all other ideal independent voltage sources by
short circuit & all other ideal independent current sources by open circuit - as per the original procedure of superposition theorem.
• NEITHER SHORT CIRCUIT NOR OPEN CIRCUIT THE DEPENDENT SOURCE. LEAVE THEM INTACT AND AS THEY ARE.
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• Dependent Current Source:- A current source whose parameters are controlled by voltage/current else where in the system
v = αVx VDCS(Voltage Dependent Current source)
v = βix CDCS(Current Dependent Current source)
• Dependent Voltage Source:-
v = µVx VDVS(Voltage Dependent Voltage source)
v = ρix CDVS(Current Dependent Voltage source)
A voltage source whose parameters are controlled by voltage/current else where in the system
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Example :3Find i0 in the circuit shown below. The circuit involves a dependent source. The current may be obtained as by using superposition as :
''0
'00 iii
i’0 is current due to 4A current source
i’’0 is current due to 20V voltage source
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To obtain i’0we short circuit the 20V sources
i1
i2
i3 A. 4i1
0)(152)(3 320'
212 iiiiii
For loop 2
For loop 1
For loop 3
045)(1)(5 3'02313 iiiiii
310 iii'
For solving i1, i2, i3 A1752i 0
'
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To obtain i’’0 , we open circuit the 4A sources
i4
i5
05ii6i ''054
For loop 4
052010i- ''054 ii
For loop 5
A1760i ''
0
For solving i4 and i5
A178
1760
1752
iii Therefore, ''0
'00
5''
0 ii
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