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Experiment to show the application of Kirchhoffs Voltage Law & Kirchhoffs Current Law in series, parallel and combination circuits.

Extracts from this document...

Introduction

INTRODUCTION

2

THEORY

3

OBJECTIVES

5

EQUIPMENT AND COMPONENTS

5

PROCEDURE

5

EXPERIMENT PART A

5

EXPERIMENT PART B

8

EXPERIMENT PART C

10

DISCUSSION

12

CONCLUSION

12

REFERENCES

12

CONTENTS


Introduction

In 1845, a German physicist, Gustav Kirchoff developed a pair or set of rules or laws which deal with the conservation of current and energy within electrical circuits. These two rules are commonly known as: Kirchoffs Circuit Laws with one of Kirchof’s laws dealing with the current flowing around a closed circuit, Kirchoffs Current Law, (KCL) while the other law deals with the voltage sources present in a closed circuit, Kirchoffs Voltage Law, (KVL).

Kirchoff’s first law that is KCL  states that the total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node. In other words the algebraic sum of all the currents entering and leaving a node must be equal to zero. This idea by Kirchoff is commonly known as the Conservation of Charge.

The term node in an electrical circuit generally refers to a connection or junction of two or more current carrying paths or elements such as cables and components.

...read more.

Middle

R1 = 1kΩ

R2 = 4.7kΩ

Voltage

1.75

8.25

Set 2:

Resistant

R1 = 100 Ω

R2 = 2kΩ

Voltage

0.47

9.52

Comment: For set 1 ,the total voltage drop of a circuit which is VR1 , 1.75 v and VR2 , 8.25 v is    

                  10.0 v .So, its not accomplished to KVL.This is because,the the total voltage drop is

                  Not equal to voltage source,10.02 v.Meanwhile,for set 2,same like set 1 the total        

                   voltage drop of of a circuit which is VR1 , 0.47 v and VR2 , 9.52v is 9.99 v .

                   So, its not  accomplished to KVL.This is because,the the total voltage drop not equal  

                    to voltage source,10.02 v.

Results from simulation:

Set 1:

Resistant

R1 = 1kΩ

R2 = 4.7kΩ

Voltage

1.75

8.25

Set 2:

Resistant

R1 = 100 Ω

R2 = 2kΩ

Voltage

0.47

9.52

Comment: For set 1 ,the total voltage drop of a circuit which is VR1 , 1.75 v and VR2 , 8.25 v is    

                  10.0 v .So, its not accomplished to KVL.This is because,the the total voltage drop is

                  Not equal to voltage source,10.02 v.Meanwhile,for set 2,same like set 1 the total        

                   voltage drop of of a circuit which is VR1 , 0.47 v and VR2 , 9.52v is 9.99 v .

                   So, its not  accomplished to KVL.This is because,the the total voltage drop not equal  

                    to voltage source,10.02 v.

Part B

2.        Parallel circuit was patched same as the Diagram 2.

3.        Positive voltage control was set to 10 V.

4.        The current drop across R1 and R 2 was measured using voltmeter by  setting it series to the resistor and labeled IR1 and IR 2.

5.        The above step was repeated using R1 = 100 Ωand R 2 = 2kΩ and the reading was recorded.

6. Results from experiments was compared with the simulation result.

Results from experiment:

Set 1:

Resistant

R1 = 1kΩ

R2 = 4.7kΩ

Current

10 mA

2.2 mA

Set 2:

Resistant

R1 = 100 Ω

R2 = 2kΩ

Current

0.1 A

0.48 mA

Comment: For set 1 ,the result is IR1 is 10mA and IR2 2.2mA. The total current is 12.2m A.So,its

                  accomplished to the KCL,which is total current in is equal to current out.Meanwhile,

                 for set 2,the results for IR1 is 0.1A and IR2 0.48mA.The total current is 100.48mA.

                So,its not accomplished to the KCL.This is because current in is not equal to current  

                  Out.

Results from simulation:

Set 1:

Resistant

R1 = 1kΩ

R2 = 4.7kΩ

Current

10 mA

2.126 mA

...read more.

Conclusion

              e= Yn- Xn                                                                     where   e=absolute error

                                                                                       Yn= expected value

                                                                                       Xn=measured  value

         e=10.02-10.0

               =0.02v

  Error in percentage:

    % error=  e/Yn (100)                      = 0.02/10.02 (100) = 1.99%

2.Beside that, when we run our experiment, we always have different thinking to get better result.Thus, the gross error was occurred. Example, we have incorrect reading when we measured the multimeter. Also, parallax error should be taken into consideration. Parallax error is the error that human make when taking a reading from any device where our eyes are not perpendicular to the scale reading. To overcome this problem, we carefully read the scale reading to minimize any fault.

3.Next, the problem that our group have come to face is incorrect reading of the multimeter, voltmeter and ammeter. We suspected that the fault was caused by the instrument itself. To overcome this problem, we take at least three separate reading to obtain better reading.

Conclusion

From the experiment that we had conducted, we had practically use Kirchoff Law to determine the voltage drop, voltage supplied, current for each resistor in series and parallel circuit.

References

  1. http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
...read more.

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