Experiment 6Superposition Principle 6.1 ObjectiveThe objective of this experiment is to verify experimentally the superposition principle as applied on dc circuits. The following circuit will be examined during this experiment.6.2 EquipmentBreadboardResistors (R1=560Ω, R2=1kΩ, R3=470Ω, R4=2,2kΩ, R5=1kΩ)DC Power SupplyMultimeter6.3 Theory A linear system obeys the principle of superposition, which states that whenever a linear system is driven by more than one independent source, the total response is the sum of the individual responses. An individual response is the result of an independent source acting alone. To determine the contribution of each individual source, all of the other sources must be set to zero by ; Replacing all other independent voltage sources with a short circuitReplacing all other independent current sources with an open circuitFinally, for each source (voltage or current) the resultant responses are added. 6.4 Preliminary WorkCalculate every current in the given circuit by using superposition principle.( IR1C=?, IR2C=?, IR3C=?, IR4C=?, and IR5C=?)6.5 Procedure Construct the circuit in the Figure 6.1 ( V1 =12V and V2 = 5V)Measure the values of the resistors provided to you and fill Table 1Measure the values of the voltage drops across the resistors and fill Table 1 Table6.1 Figure 6.1Remove V2.Short-circuit the terminals and redraw theMeasure the currents (IR1’, IR2’, IR3’, IR4’, and IR5’) and the voltage drops ( VR1’, VR2’, VR3’, VR4’, VR5’) across the resistors and fill Table 2.Table 6.2 Reconnect V2 and remove V1. Repeat steps 5 and 6 and complete Table 6.3Table 6.3 Combine currents I’ and I’’ to find the true values and complete Table 4. Be careful with the direction of currents.I = I’ + I’’ Table 6.4 Combine voltages V’ and V’’ on the resistor to find the true values and complete Table 5 Table 6.5 Calculate voltage drops (VR1C =?, VR2C =?, VR3C =?, VR4C =?, VR5C =? ) across the resistors by using superpositionCalculate the %Difference between the theoretical and experimental results, to prove the validity of the Comment on your results 6.6 Question Discuss the advantages and disadvantages of using Mesh current analysis, Nodal Analysis and Superposition Principle, for analyzing electrical circuits. R1R3R2R4RsV2Figure 6.14. Remove V2.5. Short-circuit the terminals and redraw the circuit.6. Measure the currents (Ir1', l82', Ira', Ira', and IRs') and the voltage drops ( VR1', Vr2', VR3', VrA , Vrs')across the resistors and fill Table 6.2.Table 6.2Ig2' =Iga' =Ira' =VR1' =Vra' =IRs' =Vr2' =VRzVra =Vrs =%3D7. Reconnect V2 and remove V,. Repeat steps 5 and 6 and complete Table 6.3Table 6.3Ir1" =Ir2" =Ia" =IRA" =%3D%3DVRi"Vr2" =Vr3" =Vra" =VRs" =8. Combine currents I' and I" to find the true values and complete Table 6.4. Be careful with thedirection of currents.| = l + l"Table 6.4IR1 =I82 =Ir3 =%3DIra =IRAIRs =9. Combine voltages V' and V" on the resistor to find the true values and complete Table 6.5Table 6.5VR1 =VR2 =VR3 =%3D%3D%3DVRAVr4 =VRS =%3D210. Calculate voltage drops (VRIC =?, VR2c =?, VR3C =?, VRAc =?, VRsc =? ) across the resistors by usingsuperposition principle.11. Calculate the %Difference between the theoretical and experimental results, to prove the validityof the principle. Comment on your results% DifferenceIR1IR2IR3IRs6.6 Question1. Discuss the advantages and disadvantages of using Mesh current analysis, Nodal Analysis andSuperposition Principle, for analyzing electrical circuits. Experiment 6Superposition Principle6.1 ObjectiveThe objective of this experiment is to verify experimentally the superposition principle as applied ondc circuits. The following circuit will be examined during this experiment.6.2 Equipment• BreadboardResistors (R1=560N, R3=1kN, R3=4700, R4=2,2kN, Rs=1k0)DC Power SupplyMultimeter6.3 TheoryA linear system obeys the principle of superposition, which states that whenever a linear system isdriven by more than one independent source, the total response is the sum of the individualresponses. An individual response is the result of an independent source acting alone. To determinethe contribution of each individual source, all of the other sources must be set to zero by ;Replacing all other independent voltage sources with a short circuitReplacing all other independent current sources with an open circuitFinally, for each source (voltage or current) the resultant responses are added.6.4 Preliminary Work1. Calculate every current in the given circuit by using superposition principle.( Iric=?, Ir2c=?, IR3c=?,IRac=?, and Irsc=?)%3D6.5 Procedure1. Construct the circuit in the Figure 6.1 ( V1 =12V and V2 = 5V)%3D2. Measure the values of the resistors provided to you and fill Table 6.13. Measure the values of the voltage drops across the resistors and fill Table 6.1Table 6.1R =VR1 =R2 =Vr2 =R3 =Rg =R4 =%3DVR2Vr3 =Vr4 =VRS =%3D%3D

In superposition principle we have to calculate responses of circuit with individual sources acting…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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6.1 Objective

The objective of this experiment is to verify experimentally the superposition principle as applied on dc circuits. The following circuit will be examined during this experiment.

6.2 Equipment

Breadboard
Resistors (R₁=560Ω, R₂=1kΩ, R₃=470Ω, R₄=2,2kΩ, R₅=1kΩ)
DC Power Supply
Multimeter

6.3 Theory

A linear system obeys the principle of superposition, which states that whenever a linear system is driven by more than one independent source, the total response is the sum of the individual responses. An individual response is the result of an independent source acting alone. To determine the contribution of each individual source, all of the other sources must be set to zero by ;

Replacing all other independent voltage sources with a short circuit
Replacing all other independent current sources with an open circuit
Finally, for each source (voltage or current) the resultant responses are added.

6.4 Preliminary Work

Calculate every current in the given circuit by using superposition principle.( I_R1C=?, I_R2C=?, I_R3C=?, I_R4C=?, and I_R5C=?)

6.5 Procedure

Construct the circuit in the Figure 6.1 ( V₁ =12V and V₂ = 5V)
Measure the values of the resistors provided to you and fill Table 1
Measure the values of the voltage drops across the resistors and fill Table 1

Table6.1

Figure 6.1

Remove V₂.
Short-circuit the terminals and redraw the
Measure the currents (I_R1’, I_R2’, I_R3’, I_R4’, and I_R5’) and the voltage drops ( V_R1’, V_R2’, V_R3’, V_R4’, V_R5’) across the resistors and fill Table 2.

Table 6.2

Reconnect V₂ and remove V₁. Repeat steps 5 and 6 and complete Table 6.3

Table 6.3

Combine currents I’ and I’’ to find the true values and complete Table 4. Be careful with the direction of currents.

I = I’ + I’’

Table 6.4

Combine voltages V’ and V’’ on the resistor to find the true values and complete Table 5

Table 6.5

Calculate voltage drops (V_R1C =?, V_R2C =?, V_R3C =?, V_R4C =?, V_R5C =? ) across the resistors by using superposition
Calculate the %Difference between the theoretical and experimental results, to prove the validity of the Comment on your results

6.6 Question

Discuss the advantages and disadvantages of using Mesh current analysis, Nodal Analysis and Superposition Principle, for analyzing electrical circuits.

R1
R3
R2
R4
Rs
V2
Figure 6.1
4. Remove V2.
5. Short-circuit the terminals and redraw the circuit.
6. Measure the currents (Ir1', l82', Ira', Ira', and IRs') and the voltage drops ( VR1', Vr2', VR3', VrA , Vrs')
across the resistors and fill Table 6.2.
Table 6.2
Ig2' =
Iga' =
Ira' =
VR1' =
Vra' =
IRs' =
Vr2' =
VRz
Vra =
Vrs =
%3D
7. Reconnect V2 and remove V,. Repeat steps 5 and 6 and complete Table 6.3
Table 6.3
Ir1" =
Ir2" =
Ia" =
IRA" =
%3D
%3D
VRi"
Vr2" =
Vr3" =
Vra" =
VRs" =
8. Combine currents I' and I" to find the true values and complete Table 6.4. Be careful with the
direction of currents.
| = l + l"
Table 6.4
IR1 =
I82 =
Ir3 =
%3D
Ira =
IRA
IRs =
9. Combine voltages V' and V" on the resistor to find the true values and complete Table 6.5
Table 6.5
VR1 =
VR2 =
VR3 =
%3D
%3D
%3D
VRA
Vr4 =
VRS =
%3D
2
10. Calculate voltage drops (VRIC =?, VR2c =?, VR3C =?, VRAc =?, VRsc =? ) across the resistors by using
superposition principle.
11. Calculate the %Difference between the theoretical and experimental results, to prove the validity
of the principle. Comment on your results
% Difference
IR1
IR2
IR3
IRs
6.6 Question
1. Discuss the advantages and disadvantages of using Mesh current analysis, Nodal Analysis and
Superposition Principle, for analyzing electrical circuits.

Experiment 6
Superposition Principle
6.1 Objective
The objective of this experiment is to verify experimentally the superposition principle as applied on
dc circuits. The following circuit will be examined during this experiment.
6.2 Equipment
• Breadboard
Resistors (R1=560N, R3=1kN, R3=4700, R4=2,2kN, Rs=1k0)
DC Power Supply
Multimeter
6.3 Theory
A linear system obeys the principle of superposition, which states that whenever a linear system is
driven by more than one independent source, the total response is the sum of the individual
responses. An individual response is the result of an independent source acting alone. To determine
the contribution of each individual source, all of the other sources must be set to zero by ;
Replacing all other independent voltage sources with a short circuit
Replacing all other independent current sources with an open circuit
Finally, for each source (voltage or current) the resultant responses are added.
6.4 Preliminary Work
1. Calculate every current in the given circuit by using superposition principle.( Iric=?, Ir2c=?, IR3c=?,
IRac=?, and Irsc=?)
%3D
6.5 Procedure
1. Construct the circuit in the Figure 6.1 ( V1 =12V and V2 = 5V)
%3D
2. Measure the values of the resistors provided to you and fill Table 6.1
3. Measure the values of the voltage drops across the resistors and fill Table 6.1
Table 6.1
R =
VR1 =
R2 =
Vr2 =
R3 =
Rg =
R4 =
%3D
VR2
Vr3 =
Vr4 =
VRS =
%3D
%3D

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R1 R3 Ra Rs CR2 V13 V2 Figure 6.1 +