CONCLUSION The experiment is all about “Superposition Theorem and Linearity”. This experiment is to illustrate the principle of linearity. In performing this experiment, we were able to investigate and learn the effects of multiple active linear source in a network. We were also able to know how to verify 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. The experiment no. 7 explains the principle of superposition. This states that the response in a linear circuit having more than one independent source can be obtained by adding the responses caused the separate independent sources acting alone. By reviewing the principle of superposition, I learned that superposition can be applied only in linear circuits. Linear circuits are composed entirely of independent sources, linear dependent sources and linear elements. I can say that after conducting the experiment, we were able to finish it correctly and successfully. We easily understood the procedure but we experienced a little confused in connecting the wires, whether the current that will flow in a resistor will be positive or negative. Great understanding and showing precision in every step we do are the things we considered while doing the experiment.
ANSWERS TO QUESTIONS AND PROBLEMS 1. A negative response in superposition impU lies that the assumed direction of the current is wrong. 2. The number of responses are based from the number of independent source. 3. It is possible to eliminate dependent sources on superposition. 4. It is not possible to apply superposition theorem directly to determine power associated with an element. 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 of sources. 5. Solution:
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10= 4I1-2I2 0= -2I1+10I2 I2= 0.625 A I1= 2.8125 A
0= 4I1-2I2 + -15= -2I1+10I2 I2= -1.667 A I1= -0.833 A 6. Solution: Vb= 10A Va= (1/2+1/2+1/3) = V b/3 4/3Va= 10/3 Vx= 2.5
