Physics Investigation: Example IA

Physics Investigation: Example IA

RESEARCH QUESTION

What are the factors that affect resistance in conductor?
Would a thick copper wire give less resistance than a thin copper wire?

BACKGROUND INFORMATION

Electrical current is the flow of electrical charge in an electrical conductor. Current flows only when there is voltage present across a conductor. Voltage (measured in volts), also known as electrical potential difference is the difference in electrical potential energy between two points; in other words, voltage is the ability to move current between two points. Therefore, current can only travel in a conductor if voltage is present.[1] Electrical current is carried by moving electrons in a conductor. The SI[2] unit for measuring electric current, which is the flow of electric charges through a conductor, is amperes.

An electron’s path in through a conductor is a zigzag path. This is a result of countless collisions of charged electrons with the atoms of the conductor. While the voltage across two units encourages the flow of electrons, these collisions discourage it. These collisions of electrons and a conductor’s atoms[3] result in a loss of electric current. These collisions are called the resistance in a conductor. Each loss of current due to resistance is referred to as voltage drop.

The amount of current in a conductor is dependent on the amount of voltage available to encourage the charged electrons and also the amount of resistance in the conductor to oppose to the current flow. In a conductor, if voltage is increased, then current increases too. If resistance in increased then current decreases. This theory is calculated using Ohm’s law. The law states that voltage (V) is directly proportionate to current (I): V∝ I. This law is valid for most coductors. Through these proportional values, Ohm’s law concludes that

R = V
I in which I is the current through the conductor, measured in amperes, V is the voltage across the conductor measured in volts and R is the resistance of the conductor measured in ohms.

Electric resistance quantifies how strongly a conductor opposes to current flow.There are three to four factors that affect the resistance in a conductor. One of which is the Cross Sectional Area of the conductor, which is the conductor’s circumference, also called its thickness. For instance if we take a wire as our conductor then it can be assumed that the greater the wire’s circumference, the higher the rate at which electrons can flow through it. This can be attributed to the lesser amount of resistance in the conductor (wire), a result of its greater circumference. A conductor with lesser circumference (a smaller cross section area will give more resistance to the flow of electrons, due its greater resistance.

PURPOSE

The purpose of this investigation is to investigate whether variations in cross section areas of a wire (conductor) will affect its resistance. I will be performing this investigation in the physic’s lab using wires of different diameter.

HYPOTHESIS

General Prediction: I predict that that wire with a larger circumference will give less resistance to current than a wire with lesser circumference.

Specific Hypothesis:

“If the cross section area of a wire is increased then its resistance will decrease“

As electrons travel in a wire, collisions with atoms impede their flow. When the diameter of the wire increases, the electrons have larger space to travel, and there is a decrease in these collisions. Thus if two pairs of wires of the same material and length and of different cross section areas are tested at room temperature, using a circuit consisting of a battery eliminator and ampmeter, the wire with the lesser cross section area will give greater resistance when compared to the wire with larger cross section area.VARIABLES

Table 1.1: Table of Variables

TABLE OF CONSTANTS
Table 1.2: Contstants of the emperiment

In order to determine if the cross section area of a wire affects the resistance in a wire, other factors that might affect resiatnce have to be kept controlled. Therefore, the length and material of the wire, as well as the temperature of the environment the experiment is conducted are controlled. The output ftom the battery eliminator is controlled manually, however it will give a fixed amount of voltage, enabling me to caluculate resistancem therefore battery eliminator output is controlled. In order to have fair test, it is also important that the instruments used remain the same for the entrie course of the emperiment because other instruments may have an error which must then be checked everytime. Hence, the instruments used will not change and will remain controlled for the entrie experiment.

MATERIALS
Table 1.3: Table of materials with quantity and their function.

PLAN OF ACTION

In order to test my hypothesis, I will construct two circuits:

1.  Using wires of cross sectional area 0.001 m2

2.  Using wires of cross sectional area 0.004 m2

The circuit with the higher reistance in both will show the cross sectional area with resistance quotient. However, in order for fair test, I will have to ensure that the following factors are in order:

·  The wires being tested are of the same length and material.

·  Both circuits constructed entail the same battery eliminator, voltmeter, ammeter.

·  The experiment is performed in the same room (in order to ensure uniform atmosphere for the entire experiment)

PROCEDURE

Step 1: Gather all material required for experiment.

Step 2: In order to verify the cross section area of both wires, measure diameter of each wire using a vernier calliper

Step 3: Using formula for area of a circle (πr2) calculate the cross sectional area for both wires. Check with printed cross sectional areas on wires. Confirm accuracy in order to proceed.

Step 4: Using scissors cut wire of cross sectional area 0.001 m2 into two equal pices (0.35 m each). Do the same with the wire of cross sectional area 0.004 m2.

Step 5: In order to check for error in battery eliminator output, assemble a circuit using copper wires (cross sectional area : 0.001 m2), battery eliminator and voltmeter.

Step 6: Make sure that the wires connect from positive terminal to positive and negative terminal to negative.

Step 7: Changeand increase battery output, simulteniously recording voltmeter readings at every change.

Step 8: Replace voltmeter with ammeter.

Step 9: Ensure wires connected and ciruit are in normal conditions (not heated up, not disconnected)

Step 10: Ensure that wires are connected correctly; positive terminal to positive and negative terminal to negative.

Step 11: Change and gradually increase the battery eliminator output while simulteniously recording the ammeter readings resultant of the battery output.

Step 12: Using readings from voltmeter and ammeter, calculate the resistance for each output value using equation derived from Ohm’s Law:

R=VI (Ω)

Step 11: Assemble a similar circuit,this time replacing wire of cross sectional area 0.001 m2 with wires of cross sectional area 0.004 m2

Step 12: Repeat steps 7 through 10 for second set of wires (W2)

Step 13: Calculate average resistance for both wires.

SAFETY PRECAUTIONS

The following are safety precautions one must always keep in mind when working with electricity and wires etc.

v  Never touched charged wires.

v  Always turn off the battery eliminator when reassembling or altering circuit.

v  Always check the capacity of the voltmeter and ammeter before assembling the circuit to prevent the components from fusing due to high voltage input

v  Never assemble circuit near wet surfaces

v  Do not handle any component of circuit with wet hands

CIRCUIT DIAGRAMS

Diagram 1.1: Basic Circuit (with ammeter and voltmeter)

Diagram 1.2: Circuit when checking charge output from battery

Diagram 1.3: Circuit when measuring current in wires of cross sectional area 0.1 cm2 and 0.4 cm2 respectively

MEANS OF REPRESENTATION

The best way tor epresent the data that will be collected in this investigation is a voltage-current (V-I) characteristics graph, wherein the voltage is plotted on the x axis and its corresponding current values are plotted on the y axis.

Diagram 1.4: A V-I Characteristics Graph

Here the slope or gradient of the red line will give us the resistance of the conductor the V-I characteristics graphwas charted for.

VERIFYING OHM’S LAW

Table 1.3: Verfying Ohm’s Law

This graph shows voltage-current (V-I) characteristics of a selected conductor. We can observe how the potential difference (voltage) is proportional to the current passing through it. The values have minute but negligible differences between each other. We can therefore conclude that Ohm’s Law can be applied to this graph, thus proing that the wire and its material (copper) used, is ‘ohmic’ (obeys Ohm’s Law).

DATA COLLECTION

Table 1.4: Diameters of both wires

Table 1.5: Resultant volt meter readings of battery elimnator output values
(Checking for error)

Table 1.5: Ammeter readings for corresponing voltage for wires of cross sectional area 0.001 m2 (W1)

Table 1.6: Ammeter readings for corresponding voltage for wires of cross sectional area 0.004 m2 (W2)

DATA PROCESSING

Table 1.7: Diameter and calculated cross sectional areas of both wires

Table 1.8: Calculated resistance for wire of cross sectional area area 0.001 m2 (W1)

Table 1.9: Calculated resistance for wire of cross sectional area area 0.004 m2 (W2)

Table 2.0: Comparison Table of resistance of W1 and W2. Also displays average resistance of each circuit.

DATA REPRESENTATION AND ANALYSIS

Graph 1.2: V-I Characteristics: Wire with cross sectional area 0.001 m2 (W1)

This graph displays the relationship between coltage output from the battery eliminator and the corresponding ammeter readings for the wire with cross sectional area 0.001 m2(W1). The slope or gradient of this graph represents the reciproacal of the resistance in wire. It can be observed that the points do not form a straight line and therefore are not constant. Thus conlcuding that resitance is not constant in W1. This may be a result of errors in the experiment. One can observe an obvious increase in current passing through the wire at voltage 3.8 (V)- 2.6 (I).

Graph 1.3: V-I Characteristics: Wire with cross sectional area 0.004 m2 (W2)

This graph displays the relationship between coltage output from the battery eliminator and the corresponding ammeter readings for the wire with cross sectional area 0.004 m2(W2) . There is an clear increase in current passing through the wire at voltage 3.8 (V)- 3.4 (I). The slope or gradient of this graph represents the reciproacal of the resistance in the wire. The points do not form a straight line, thus conlcuding that resitance is not constant in W2 . This may be a result of errors in the experiment. Nevertheless when the values of Graph 1.3 are compared to those of Graph 1.2 , the (I) values of Graph 1.3 are greater, indicating that the resistance of in this graph, and thereby W2 was lesser.