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EXPERIMENT SIX

Superposition Theorem

I. OBJECTIVES

1. To study and apply Superposition Theorem in the solution of an electric circuit.
   1. To verify experimentally principle of Superposition.

II. THEORY

Some circuits require more than one voltage or current source. For example, certain types of amplifiers require both a positive and a negative voltage source for proper operation. The analysis of those circuits can be simplified by the application of the Superposition theorem. The Superposition method is a way to determine currents and voltages in a circuit that has multiple sources by taking one circuit at a time. The other sources are replaced by their internal resistances. (Recall that the ideal voltage sources has zero internal resistance while an ideal current source has an infinite resistance). In this experiment, all sources will be treated as ideal in order to simplify the coverage. The general statement of the Superposition Theorem is as follows: “ In any linear resistive circuit containing two or more independent sources, any circuit voltage (or current) may be calculated as the algebraic sum of all individual voltages (or currents) caused by each independent source acting alone, i.e. with all other independent sources replaces by their internal resistance .” TECHNOLOGICAL UNIVERSITY OF THE PHILIPPINES College of Engineering - Electrical Engineering Department Ayala Blvd. Ermita, Manila, 1000, Philippines ELECTRICAL CIRCUITS 1 LABORATORY Experiment No. 6 Superposition Theorem GRAJO, JERVIS MARC D. Submitted by: BSEE-2I Instructor : Engr. Jun Date Submitted: February 4, 2022

Figure 7.2Page^ |^60 figure (^) 7.2. d. Turn^ on the power supply E 2.^ Measure^ the^ currents^ I 1 ,^ I 2 ,^ I 3 and^ record^ them in Table 7.1. Turn off the power supply after completing the measurements. e. Reconnect^ the^ power supply E 1.^ Disconnect^ the^ power^ E 2 and^ short circuit its (^) terminals as shown in Figure 7.3. f. Turn^ on the power supply E 1.^ Measure^ the^ currents^ I 1 ,^ I 2 and^ I 3 and^ record them in (^) Table 7.1. Turn off the power supply after completing the measurements. g. Using the circuit in Figure 7.1, calculate or compute for the components and the actual currents in each branch using method of superposition. Record the results in Table 7.1. h. Compute for the percent difference between the measured and the computed values. Use the calculated or computed values as the correct values. Figure 7.

Figure 7.

V. DATA AND RESULTS

I’ I’’ IT

COMPUTED MEASURED COMPUTED MEASURED COMPUTED MEASURED

I 1 47.29 mA^ 47.30mA^ 17.28mA^ 17.3mA^ 64.57mA^ 64.6mA^ 0.046%

I 2 33.46 mA^ 33.50mA^ -31.63mA^ -31.6mA^ 1.83mA^ 1.83mA^ 0%

I 3 13.83mA^ 13.80mA^ 48.91mA^ 48.9mA^ 62.74mA^ 62.7mA^ 0.063%

Table 7.

When the E2 is shorted:

RT = R124 + R3 + R

64.66 22  220  RT = 306.66 I3 = IT IT = VT RT = 15 V 306.66 Ω I3 = 48.91mA -I2 = IT (^ R 1 R 1 + R 2 ) = 48.91 mA ( 183 Ω 183 Ω + 100 Ω ) (-I2 = 31.63mA)- I1 = IT (^ R 1 R 1 + R 2 ) = 48.91 mA ( 100 Ω 100 Ω + 183 Ω ) I1 = 17.28m R35 = R3 + R  R35  R235 R35||R = 242 ||100 R235 = 70.76 RT = R235 + R1 + R = 70.76 33  1500  RT = 253.76 I1 = IT IT = VT RT = 12 V 253.76 Ω I1 = 47.29mA I2 = IT ( R 1 R 1 + R 2 ) = 47.29 mA ( 242 Ω 242 Ω + 100 I2 = 33.46mA I3 = IT ( R 1 R 1 + R 2 ) = 48.91 mA ( 100 Ω 100 Ω + 242 I3 = 13.83mA

VII. CONCLUSION

The superposition theorem is used in circuit analysis to estimate the current and voltage across each circuit's components. When there are multiple sources, this is advantageous. In accordance with the superposition theorem. The reaction of components is a mathematical accumulation of each source's response. To see if we followed the statement above, we calculated the total amount of current that flows through the resistors when one source is compared to the total amount of current observed when two sources are linked in a circuit. We demonstrated that the resultant of the current is indeed equal to the sum of current from each source in this experiment. The percentage difference for each current equals zero in the findings. As a result, the superposition theorem is established.

VIII. GUIDE QUESTIONS

1. What advantages does Superposition Theorem have over Kirchhoff’s Law equations in network solutions? ANSWER: According to the superposition theorem the total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately. All other voltage sources should be replaced with short circuits, and all other current sources should be replaced with open circuits, I1 Computed I1 = I’ + I” I1 = 47.29mA + 17.28mA I1 = 64.57mA I2 Computed I2 = I’ + I” I2 = 33.46mA + (-31.63mA) I2 = 1.83mA I3 Computed I3 = I’ + I” I3 = 47.29mA + 17.28mA I3 = 62.74mA Percentage Difference of I I1 =

V 1 − V 2 V 1 + V 2

2 |^

x100%

I1 =

64.57−64. 64.57+ 64.

x100%

I1 = 0.046% Percentage Difference of I I2 =

V 1 − V 2 V 1 + V 2

2 |^

x100%

I2 =

1.83−1. 1.83+1.

x100%

I2 = 0% Percentage Difference of I I3 =

V 1 − V 2 V 1 + V 2

2 |^

x100%

I3 =

62.74−62. 62.74 +62.

x100%

I3 = 0.063%

IF THE 36V IS SHORTED IF THE 24V IS SHORTED COMBINE THE CURRENTS RT = ( 10 ) (^) ( 5 + ( 15 ) ( 5 ) 20 ) 10 + 5 + ( 15 ) ( 5 ) 20 =4.67 Ω IT’ = 24

=5.14 A I1’ = 5.14(^ 10

)=2.74 A

I2’ = 5.14 – 2.74 = 2.4A

I3’ = 2.74(^ 15 20 )=2.06 A RT = ( 10 ) ( 5 + ( 15 ) ( 5 ) 20 ) 10 + 5 + ( 15 ) ( 5 ) 20 =4.67 Ω IT” = 36

=7.71 A I3” = 7.71(^ 10

)=4.11 A

I2” = 7.71 – 4.11 = 3.6A

I1” = 3.6(^ 15 20 )=2.7 A

I1 = I1’ – I1”

I1 = 2.74 – 2.

I1 = 0.4A

I2 = I2” – I2’

I2 = 3.6 – 2.

I2 = 1.2A

I3 = I3” – I3’

I3 = 4.11 – 2.

I3 = 2.05A