Question

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.6 V. You cut off a 20.0-m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a 40.0-m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

Short answer

The resistance per meter is approximately 0.0825 ohms.

Step-by-step solution

Understanding Ohm's Law

First, recall Ohm's Law, which states that the voltage \( V \) across a resistor is equal to the current \( I \) through it times its resistance \( R \): \[ V = I imes R \] Here, since we measure the voltage supplied by the battery, we can use this formula to find the resistance of different lengths of wire.

Calculating Resistance for the 20.0-meter Wire

For the 20.0-meter wire, we know the voltage \( V = 12.6 \) volts and the current \( I = 7.00 \) amperes. Using Ohm's Law:\[ R_{20} = \frac{V}{I} = \frac{12.6}{7.00} \approx 1.80 \, \text{ohms} \]Hence, the total resistance for the 20.0-meter wire is approximately 1.80 ohms.

Calculating Resistance for the 40.0-meter Wire

For the 40.0-meter wire, the voltage is still \( V = 12.6 \) volts but the current is \( I = 4.20 \) amperes. Again, we use Ohm's Law:\[ R_{40} = \frac{V}{I} = \frac{12.6}{4.20} \approx 3.00 \, \text{ohms} \]Therefore, the total resistance for the 40.0-meter wire is approximately 3.00 ohms.

Finding Resistance per Meter

We have the resistance for the 20.0-meter wire (1.80 ohms) and the 40.0-meter wire (3.00 ohms). To find the resistance per meter, we can calculate it for each case:For the 20.0-meter wire:\[ R_{\text{per meter (20)}} = \frac{1.80 \, \text{ohms}}{20.0 \, \text{meters}} = 0.090 \, \text{ohms per meter} \]For the 40.0-meter wire:\[ R_{\text{per meter (40)}} = \frac{3.00 \, \text{ohms}}{40.0 \, \text{meters}} = 0.075 \, \text{ohms per meter} \]Since we expect a constant resistance per meter, check the calculations or average them if there's a discrepancy.

Verifying Consistency of Resistance

The two calculated values for resistance per meter are not the same. This could be due to measurement error or assumptions in the equipment's ideal behavior. To avoid these errors, find an average of the two values:\[ R_{\text{per meter}} = \frac{0.090 \, \text{ohms per meter} + 0.075 \, \text{ohms per meter}}{2} = 0.0825 \, \text{ohms per meter} \]This result averages the discrepancies between the wire length measurements, assuming uniform resistance throughout.

Key concepts

Resistance measurement

Resistance measurement is an important aspect when working with electrical devices, as it helps to determine how much a material opposes the flow of electric current. In this exercise, we used Ohm's Law to measure resistance, since we did not have a direct tool like an ohmmeter. With Ohm's Law, resistance is calculated using the formula: \( R = \frac{V}{I} \)where \( V \) is the voltage across the wire, and \( I \) is the current flowing through the wire.
To find the resistance of a specific length of wire, you connect it to a known voltage source and measure the resulting current. By substituting these values into the formula, the wire's resistance can be calculated. This method of determining resistance relies on accurate measurements of voltage and current.
While ideal equipment would eliminate all external factors, different conditions and equipment precision can introduce discrepancies into the measurements, hence the difference in the resistance per meter values in the example.

Electrical circuits

An electrical circuit is a path through which electric current flows. When measuring resistance, understanding the circuit configuration is crucial.
In our exercise, we set up a simple circuit that included a battery, an ammeter, and a wire section. The battery acts as the voltage source, pushing the current through the circuit, while the ammeter measures the current's flow. The simple circuit's basic structure allowed us to apply Ohm's Law effectively.
Key elements in an electrical circuit include:

• Voltage Source: Provides the necessary energy to move the electrons through the circuit.
• Conductor (the Wire): Connects various components and carries the electrical current.
• Ammeter: Measures the amount of current flowing through the circuit, a crucial step in finding resistance.

By understanding how these components interact, one can better assess how changes, like those in wire length, affect overall resistance.

Wire resistance calculation

Calculating wire resistance involves determining how much opposition a wire presents to the flow of electric current, which is essential for various applications.
To calculate the resistance in wire, you first need to establish the current and voltage conditions using a given wire length, as shown in the calculations from the exercise. For multiple wire lengths, consistent measurement approaches are vital for valid results.For this specific exercise:

• The resistance for a 20.0-meter wire was found by dividing 12.6 volts by 7.00 amperes, resulting in about 1.80 ohms.
• For a 40.0-meter wire, dividing the same voltage by 4.20 amperes yielded approximately 3.00 ohms.

To determine resistance per meter, divide the total resistance by the wire length: \( R_{\text{per meter}} = \frac{R_{\text{total}}}{L_{\text{wire}}} \).However, if the resistance values don't perfectly align due to measurement variables, averaging these values can provide a reasonable estimate, as was used in this solution to address discrepancies.