Lab Report on Resistivity | Physics Homework Help

Objectives

• To use Ohm’s law to determine the proportionality of current and voltage.
• To understand the source of resistance.
• To study resistivity and what affects it.

Abstract

The total length of the wire affects the amount of resistance. The longer the wire is the more resistance there will be. This is because resistance occurs as result of collisions between charge carries. More collision along the wire means more resistance. Secondly the diameter will affect the amount of resistance. A bigger diameter means a wider cross-sectional area, thus less resistance there will be to the flow of electric current. Last but not least temperature affects resistance. An increase in temperature of the wire will lead to an increase in resistance.

Equipment

• Power source with a constant current supply.
• Brass wires of different lengths.
• Copper, Aluminum and steel wires.
• Current meter.
• Capstone software used to observe and collect the current and voltage
• Halogen bulbs.
• Signal generator.

Procedure 1

1. The power source was opened and the current adjusted to specific amperes.
2. The capstone software read the data as the current passed through the wire.
3. Current and voltage were recorded for different materials.
4. The data collected was put in a table as shown below.

Data

Result table

Analysis and Calculations.

Resistance calculations:

1. Brass

V=IR                             V2= 0.009, I2= 0.4

V= 0.004, I= 0.2              R2= 0.009/0.4

R= 0.004/0.2                     R2= 0.02Ω.

R= 0.02 Ω

Resistivity (p) = RA, Where R is the resistance and A is the cross-section area

A= πr², r= 1.019/2= 0.51 mm or 0.00051m

A= 3.14*0.00051^2= 8.15*10^-7m²

P= 8.15*0.02*10^-7= 1.63*10^-8Ωm²

1. Copper

V=0.002,I=0.2

R= 0.002/0.2

R=0.01Ω

Resistivity

A=πr²,r= 1.017/2=0.5085 mm 0r 0.0005085

A= 8.12*10^-7m²

P= AR

P= 8.12*10^-9Ωm²

1. Aluminum

V=0.004, I=0.2

R= 0.004/0.2

R= 0.02Ω

Resistivity

A=πr^2, r= 1.012/2=0.506 mm 0r 0.000506m

A= 8.04*10^-7m²

P= AR

P= 1.61*10^-8 Ωm²

1. Steel

V=0.065, I= 0.2

R=0.065/0.2

R=0.325

Resistivity

A=πr^2, r= 1.014/2= 0.507 mm or 0.000507

A= 8.07*10^-7m²

P= AR

P= 2.62*10^-7Ωm²

From the above calculations and analysis of the data obtained from the experiment, it is clear that steel has the highest resistance and resistivity, thus cannot be used in making electrical wires. On the other hand copper has the lowest resistance and resistivity thus can be used in making electrical wires. This shows that the material of a substance will affect charge flow, Not all materials are created equal in terms of conductive ability. As seen from the experiment some materials are better conductors than others.

Procedure 2, The effects of conductor length on resistance

• The current was fixed at 1A and run througha brass wire of 1.019 mm diameter and the length varied in increments.

As observed from the data above an increase in length leads to a correspondent increase in resistance.

Graph of resistance vs Length.

Procedure 3, The Effects of cross sectional area on resistance

• The data was taken with 4 diameters of brass wire and the length fixed at 24cm and current at 1A.

Data

As observed from the above data analysis an increase in cross-sectional area leads to a decrease in resistance of charge flow through the material.

The y intercept is 0.0843 as seen from the graph, R= 0.878.

Conclusion

The experiment was a success, although the data was not 100% accurate it was near perfect to the theoretical value known. All the experiments followed Ohm’s law.