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.

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.

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

• 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..

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

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

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

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

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.

• 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

12

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

'

13

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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