Exprt 6 3d6o5b

08 - 2

Mapua Institute of Technology School of EECE Electrical Engineering Department

Experiment # 6 The Superposition Theorem and Linearity

INDUCTIVO, Nathalie C. EE20L / B17

Grade Engr. Ronaldo Cabuang PROFESSOR

Final Data Sheet

Table 7.1 (Measured Values) Measured Resistances

R1 8Ω

R2 2Ω

10 V 3 V ON ON Calculated Values ON OFF OFF ON Superposition values

ON

ON

R3 2Ω

R4 4Ω

V1

V2

V3

V4

6.5V

1.625V

1.875V

-0.5V

V5

R5 5Ω

I1

I2

I3

-0.625V 0.8125A -0.125A 0.9375A

V1’

V2’

V3’

V4’

V5’

I1’

6.872V

1.718V

1.406V

0.624V

0.78V

0.859A

V1’’

V2’’

V3’’

V4’’

V5’’

I1’’

I2’

I3’

0.156A 0.703A

I2’’

I3’’

-0.375V -0.0938V 0.468V -1.124V -1.405V -0.0469A -0.281A 0.234A

V1

V2

V3

V4

V5

6.497V

1.624V

1.874V

-0.5V

-0.625V

I1

I2

I3

0.812A -0.125A 0.937A

Table 7.2 (Simulated Values) Optional 10 V 3 V ON ON Calculated Values ON OFF OFF ON Superposition values

ON

ON

V1

V2

V3

V4

6.5V

1.63V

1.88V

-0.5V

V5

I1

V1’

V2’

V3’

V4’

V5’

I1’

6.87V

1.72V

1.41V

0.625V

0.781V

0.859A

V1’’

V2’’

V3’’

V5’’

I1’’

-0.375V -0.0938V 0.469V

V1

V2

6.495V

1.626V

V3

V4’’ -1.12V

V4

I2

I3

-0.625V 0.8125A -0.125A 0.9375A

I2’

I3’

0.156A 0.703A

I2’’

I3’’

-1.41V -0.0469A -0.281A 0.234A

V5

1.879V -0.495V -0.629V

I1

I2

I3

0.812A -0.125A 0.937A

Graphs/Curves Comparison of the Calculated values, Superposition values, and Simulated values of Voltage and Current Responses 7

6.5 6.497 6.5

Calculated Values Superposition Values

6

Simulated values

5

4

3

1.625 1.624 1.626

1.875 1.874 1.88

2

0.9375 0.937 0.9375 -0.125 -0.125 -0.125

0.8125 0.812 0.8125

1

-1

V2

V3

V4

V5 -0.625 -0.625 -0.625

V1

-0.5 -0.5 -0.5

0

I1

I2

I3

Data Analysis/Interpretation of Results In this experiment, electric circuit presented on Fig 7.1 with more than one independent source was analyzed and solved using superposition theorem. First, Table 7.1 was filled in by calculations using mesh analysis and superposition theorem. The calculated values, where both independent sources (10 V and 3 V) are active, were computed using mesh analysis. This would be the basis if the obtained superposition values were correct. The next calculated values were attained using the superposition theorem. From the circuit given, it was observed that there are two independent sources present; thus, we could say that there are two responses that can be obtained. Consequently, the responses caused by a certain independent source was indicated by putting an apostrophe (‘) on a desired current or voltage and the number of apostrophe also indicate the sequence of the responses caused by the independent sources. At first, the 3V independent source was set to zero by short circuiting it. The circuit was important to be redrawn for a clearer illustration of the new circuit diagram and to avoid of confusion. For an easy calculation, mesh analysis was used; hence, getting the current responses and multiplying the resistors to its corresponding current responses to get the voltage responses. Next was the 10V independent source was set to zero by short circuiting it also, then same procedures was done. I obtained the calculated values V1’’= 6.872V, V2’’= 1.718V, V3’’= 1.406V, V4’’= 0.624V, V5’’= 0.78V, I1’’= 0.859A, I2’’= 0.156A, and I3’’= 0.703A when 10V was active and V1’’’= -0.375V, V2’’’= -0.0938V, V3’’’= 0.468V, V4’’’= -1.124V, V5’’’= -1.405V, I1’’’= -0.0469A, I2’’’= -0.281A, and I3’’’= 0.234A when 3V was active. After getting these values, the superposition values were computed by the definition of superposition theorem, the response is equal to the sum of responses caused by each independent source acting alone. Thus, attaining these values: V1 = 6.497V, V2 = 1.624V, V3 = 1.874V, V4 = -0.5V, V5 = -0.625V, I1 = 0.812A, I2 = -0.125A, and I3 = 0.937A. By comparing these data to the calculated values using mesh analysis, the values were the equivalent. For further comparison, the circuit was simulated using Tina Pro. At first, the two sources were still present and the responses were taken down notes. Next, the 3V source was short circuited while the 10V was activated; its current and voltage responses were noted. Also with the 10V source that was short circuited and 3V that was activated, its current and voltage responses were obtained. Finally, the responses in the circuit were computed by adding the responses caused by the independent sources acted alone. For comparison with the calculated and superposition values, the simulated values were the same; there were just minor discrepancy because of the rounding off. Conclusion After successfully conducted this experiment, I was able to attain the objectives and understood the concepts behind superposition theorem and linearity. I was able to investigate the effects of multiple active linear sources in a network. Also, I was able to whether the linear response at any point in a linear circuit having several independent linear sources is

equivalent to the algebraic sum of individual responses produced by each independent source acting alone. Furthermore, I was able to illustrate the principle of linearity. There are certain electric circuits that contain multiple independent sources. One of the methods to solve this kind of circuits is by the superposition theorem. It is a consequence of linearity. A linear circuit is a circuit that consists entirely of independent sources and linear elements. Thus, the response is proportional to the source so there is N number of response that can be obtained from an N number of independent sources present in a given circuit. A linear element is a ive element that has a linear voltage-current relationship or doesn’t have any power of some current or voltage variable in the circuit. The principle of superposition states that the response, whether a desired current or voltage, in a linear circuit having more than one independent source is equal to the sum of the responses caused by each independent sources acting alone. In setting a source to zero, a voltage source is equivalent to a short circuit and a current source is equivalent to an open circuit. There are instances that a negative response is obtained, it only means that the assumed direction of the current is opposite the actual current flow.

Answers to Questions and Problems 1) The negative response in superposition implies that the assumed direction of the current is opposite the actual current flow. 2) There is N number of responses obtained from an N number of independent sources present in a given circuit. 3) No, it isn’t possible to eliminate dependent sources on superposition because their values are not constant and always depend on other elements.

4) The possible limitation of the superposition theorem is for linear circuits only. This theorem can be applied to circuits with multiple independent sources and linear dependent sources and elements. Also, since it is a consequence of linearity, superposition cannot be applied in circuits where the output current or voltage of the dependent source is proportional to the square of some current or voltage variable in the circuit. In addition, application of superposition theorem does not normally lead to simplification of analysis. It is not the best technique to determine all currents and voltages in a circuit, driven by multiple sources. 5)

By source transformation: V = IR = 10A (4+1) = 50V By superposition theorem and mesh analysis: Let 10V=active; 15V = 50V = 0 m1: 10 = 4I1’ - 2I2’ m2: 0 = -2I1’ + 10I2’ I1’ = 2.778 A I2’ = 0.556 A Ix’’ = I1’ - I2’ = 2.778A - 0.556A = 2.222A Vx’’ = Ix’’R = (2.222A)(2) = 4.444V Let 15V = active; 10V = 50V = 0 m1: 0 = 4I1’’ - 2I2’’ m2: -15 = 2I1’’ - 10I2’’ I1’’ = 0.833 A I2’’ = 1.667 A Ix’’ = I2’’ - I1’’ = 1.667A - 0.833A = 0.834A Vx’’” = Ix’’”R = (0.834A)(2) = 1.668V

Let 50V= active; 10V = 15V =0 m1: 0 = 4I1’’’ - 2I2’’’ m2: -50V = 2I1’’’ - 10I2’’’ I1’’’ = 2.778A I2’’’ = 5.556A Ix’’’=I2’’’ - I1’’’=5.556 A - 2.778 A = 2.778A Vx’’’ = Ix R = (2.778A)(2) = 5.556V

Vx = Vx’ + Vx” + Vx’’’ Vx = 4.444V + 1.668V + 5.556V

6)

Let 12V = active; 7A = 0 7 (4 R = 2Ω 7 4 R = 4.545Ω 2 I = = 2.64 4.545Ω (2.64 )(4Ω) I = 7Ω 4Ω I = 0.96 = 0.96 (5Ω = 4.8 Let 7A = active; 12V=0 ( 7 (2Ω) I = 5Ω .333Ω 2Ω I = .68 = ( .68 )(5Ω) = 8.405 = = 4.8

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