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Engineering
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Lab Report
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# Engineering Lab Report About Internal Resistance (Lab Report Sample)

Instructions:

The task was to write a lb report on internal resistance of electrical components.

source..
Content:

Student’s Name
Professor’s Name
Course
Date
Internal Resistance
* Abstract
This experiment was performed so as to investigate the concept of internal resistance which is an inherent resistance present in all electrical devices and is mainly ignored in most of the circuit analysis. The objective of the analysis was achieved by first connecting two 100 k Ohms resistors to a 10 V power supply them then obtaining the current through them. Then a Simpson meter was connected parallel to the second resistor and using the provided DMM, the voltage drop across the second resistor was obtained and the internal resistance of the Simpson meter calculated. It was observed that the value as the value of internal resistance increased, the current through the meter decreased.
* Introduction
Electrical resistance is a quantity that is employed to measure how an electrical component reduces the amount of current flowing through it and is measured in Ohms (Ω). The magnitude of resistance of an electrical device is calculated from the Ohms’ law relationship which suggests that current through a conductor is directly proportional to the voltage across it.
V = IR (1)
Where;
V = voltage
I = current
R = resistance
Making (R) to be the formula, then the above equation can be rewritten as;
R = VI (2)
Most electrical instruments have an inherent resistance, and in most circuit analysis, this resistance is ignored which can be valid for short wire connection. However, for power lines, this value has to be determined to avoid errors. The concept of internal resistance can also, be developed from resistivity which relates the length and the cross – sectional area of the material and the resistance of that particular material. Thus, resistance of a material can be obtained using another expression as shown below;
R = ρ * LA (3)
Where;
ρ = resistivity
R = resistance
L = length of the conductor
A = cross – sectional area of the conductor
For this particular experiment, the concept of internal resistance was investigated sing Simpson meter which is used to measure voltages.
* Lab Description and Procedure
First, the Simpson meter was connected to a voltage source at 5 V, and then the students practiced on how to read different scales of the meter. Then a similar circuit to the one shown below was constructed without a Simpson meter, and the value of IAB and IBC were determined using the provided DMM. Afterward, the Simpson meter was connected parallel to the second resistor, and the voltage drop across the resistor noted as displayed by the Simpson meter. Once the current through the meter was determined from Kirchhoff’s law and recorded as I ERROR then, the internal resistance of the Simpson’s meter was calculated. The above procedure was repeated with the Simpson meter at high scales i.e. 50 V, 250 V and 500 V. The entire experiment described above was repeated using resistors of 2.2 KΩ instead of 100 KΩ.
Figure1. Experimental set – up
* Results Analysis
When 100 kΩ resistors were used;
IAB = 50.13 μA
IAB = 50.03 μA
These values are almost similar; the variation might be due to the internal resistance of wires.
With Simpson at 10 V;
IAB = 60.25 μA
IAB = 40.21 μA
VR2 = 4.3 V
IERROR obtained using Kirchhoff’s current law which states that current entering a junction should be equal to current leaving a junction.
IERROR = 60.25 – 40.21 = 20.04 μA
R internal resistance = VR2I ERROR = 4.3 V0.00002004 = 214.570 KΩ
With Simpson at 50 V;
IAB = 52.59 μA
IAB = 47.82 μA
VR2 = 6 V
IERROR obtained using Kirchhoff’s current law which states that current entering a junction should be equal to current leaving a junction.
IERROR = 52.29 – 47.82 = 4.77 μA
R internal resistance = VR2I ERROR = 6V 0.00000477 = 1257.861 KΩ
With Simpson at 250 V;
IAB = 56.70 μA
IAB = 49.71 μA
VR2 = 0.5 V
IERROR obtained using Kirchhoff’s current law which states that current entering a junction should be equal to current leaving a junction.
IERROR = 56.70 – 49.71 = 6.99 μA
R internal resistance = VR2I ERROR = 0.5 V0.00000699 = 715.308 KΩ
With Simpson at 500 V;
IAB = 50.45 μA
IAB = 49.95 μA
VR2 = 0.25 V
IERROR obtained using Kirchhoff’s current law which states that current entering a junction should be equal to current leaving a junction.
IERROR = 50.45 – 49.95 = 0.5 μA
R internal resistance = VR2I ERROR = 0.25 V0.0000005 = 500 KΩ
When 2.2 kΩ resistors were used;
IAB = 2.1687 mA
IAB = 2.1690 mA
These values are almost similar, the variation might be due to the internal resistance of wires.
With Simpson at 10 V;
IAB = 2.1817 mA
IAB = 2.1564 mA
VR2 = 5.25 V
IERROR obtained using Kirchhoff’s current...
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