Superposition Theorem, Thesis for Power Electronics
Nasiruzzaman
Nasiruzzaman

Superposition Theorem, Thesis for Power Electronics

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To study Superposition theorem practically
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1

University of Technology

Laser and Optoelectronics Engineering Department

Laser Engineering Branch

Power electronics 2010-2011

Experiment No.12

Superposition Theorem

Aim of experiment: To study Superposition theorem practically.

Apparatus 1. DC circuit training system 2. Set of wires.

3. DC Power supply

4. Digital A.V.O. meter

Theory

The superposition theorem is very useful for finding the voltages and currents

in a circuit with two or more sources of supply, and is usually easier to use than

Kirchoff ’s law equations. One supply is selected and the circuit is redrawn to

show the other supply (or supplies) short-circuited (leaving only the internal

resistance of each supply). The voltage and current caused by the first supply can

then be calculated, using V = RI methods together with the rules for combining

series and parallel resistors. Each supply is treated in turn in the same way, and

finally the voltages and currents caused by each supply are added.

Hence, this theorem may be state as follows:

Example: In the network shown, find the voltage across the 2.2 k resistor.

In a network of linear resistances containing more than one generator (or

source of e.m.f.), the current which flows at any point is the sum of all the

currents which would flow at that point if each generator were considered

separately and all the other generators replaced for the time being by

resistances equal to their internal resistances.

2

University of Technology

Laser and Optoelectronics Engineering Department

Laser Engineering Branch

Power electronics 2010-2011

Procedure

1. Connect the circuit shown below. 2. Measure values of ( I1 , I2 , I3 ) and record it in the table

I1 (mA) I2 (mA) I3 (mA)

In this network, there are

two generators and three

resistors. The generators

might be batteries,

oscillators, or other signal

sources.

To find the voltage caused by

the 6V generator, replace the

4V generator by its internal

resistance of 0.5k. Using

Ohm’s law, and the potential

divider equation: V = 1.736V.

To find the voltage caused by

the 4 V generator, the 6 V

generator is replaced by its

1k internal resistance. In

this case: V = 2.315 V.

Now the total voltage in the

original circuit across the

2.2k resistor is simply the

sum of these: 4.051 V

3

University of Technology

Laser and Optoelectronics Engineering Department

Laser Engineering Branch

Power electronics 2010-2011

3. Connect the circuit below, when V1 = on and V2 = short

4. Measure values of ( I1 , I2 , I3 ) and record it in the table

I1 (mA) I2 (mA) I3 (mA)

100 82

5 V

5 0

9 V

A

I2

A

I1

A I3

100 82

5 V

5 0

A

I2

A

I1

A I3

4

University of Technology

Laser and Optoelectronics Engineering Department

Laser Engineering Branch

Power electronics 2010-2011

5. Connect the circuit below, when V1 =short and V2 = on

5. Measure values of ( I1 , I2 , I3 ) and record it in the table

I1 (mA) I2 (mA) I3 (mA)

6. From results, calculate the current pass through each resistor and

voltage across each resistor.

Discussion 1. Compare between the theoretical and practical results.

2. Comment on your results. 3. Find ( Ia ) by using superposition theorem for the circuit below.

100 82

5 0

9 V

A

I2

A

I1

A I3

5

University of Technology

Laser and Optoelectronics Engineering Department

Laser Engineering Branch

Power electronics 2010-2011

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