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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 \(\mathrm{V} .\) You cut off a \(20.0-\mathrm{m}\) 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 \(40.0-\mathrm{m}\) 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

Expert verified
The resistance of 1 meter of wire is approximately 0.0825 \( \Omega/\text{m} \).

Step by step solution

01

Understanding Ohm's Law

Ohm's Law relates the voltage (V), current (I), and resistance (R) in a circuit with the equation \( V = I \times R \). When rearranged to find resistance, it becomes \( R = \frac{V}{I} \).
02

Calculate Resistance for 20m Wire

Use Ohm's Law to find the resistance of the 20m wire: \( R_{20} = \frac{12.6\, \text{V}}{7.00\, \text{A}} = 1.80\, \Omega \). This is the total resistance for the 20-meter length of wire.
03

Calculate Resistance per Meter for 20m Wire

Divide the total resistance by the length of the wire to find the resistance per meter: \( R_{per\,meter} = \frac{1.80\, \Omega}{20.0\, \text{m}} = 0.090\, \Omega/\text{m} \).
04

Calculate Resistance for 40m Wire

Again, apply Ohm's Law to find the resistance of the 40m wire: \( R_{40} = \frac{12.6\, \text{V}}{4.20\, \text{A}} = 3.00\, \Omega \). This is the total resistance for the 40-meter length of wire.
05

Calculate Resistance per Meter for 40m Wire

Divide the total resistance by the length of the wire to find the resistance per meter: \( R_{per\,meter} = \frac{3.00\, \Omega}{40.0\, \text{m}} = 0.075\, \Omega/\text{m} \).
06

Analyzing Results

The resistance per meter calculations for the two different lengths of wire (0.090 \( \Omega/\text{m} \) and 0.075 \( \Omega/\text{m} \)) are slightly different, suggesting measurement error or imperfections. To approximate and check for consistency, average the two values to estimate the resistance per meter: \( R_{average} = \frac{0.090\, \Omega/\text{m} + 0.075\, \Omega/\text{m}}{2} = 0.0825\, \Omega/\text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance measurement
Determining the resistance of a wire without a direct resistance measuring tool is an interesting exercise in practical electrical work. In such cases, an understanding of Ohm's Law becomes indispensable. Ohm's Law states that voltage (V), current (I), and resistance (R) in an electrical circuit are related through the equation \( V = I \times R \). From this, we can isolate and calculate resistance with be equation \( R = \frac{V}{I} \). So, even with simple tools like only a voltmeter and an ammeter, you can measure resistance by measuring voltage and current separately.

When working with long pieces of wire, it is common to express their resistance in terms of resistance per unit length. This allows for standardization, making it easier to predict resistance for any length of that wire. Let's calculate: by first measuring resistance over a set length and then dividing by that length to find resistance per meter.

In the exercise, you calculated resistance for 20 meters and 40 meters of wire separately, using the available instruments, while noting that the results weren't perfectly consistent due to potential measurement errors or material imperfections. Averaging these values gave a more reliable estimate of resistance per meter, underscoring the need for careful recording and comparison of results to check consistency.
Circuit analysis
Circuit analysis allows us to determine the values of electrical quantities across components within a circuit. In our exercise, we used a series circuit with a battery, an ammeter, and a length of wire to analyze resistance. A series circuit means that there is only one path for the current to take, and hence, the current flowing through the ammeter is the same as the current through the wire.

To perform the analysis, consider the circuit where the battery provides a known voltage of 12.6 V. The ammeter, which is in series with the wire, measures the current flowing through the circuit. By noting the current for a given length of wire, we can apply Ohm's Law to calculate the total resistance of the wire. In this way, the elements of the circuit—voltmeter to measure voltage, ammeter to measure current—play roles in determining resistance indirectly.

It's crucial to remember that the precise measurements from the voltmeter and ammeter are key in this analysis. For this method to provide reliable results, the tools must have their own characteristics well-understood: a voltmeter should ideally have a high resistance, so it doesn't affect the circuit, and an ammeter a low resistance, to avoid adding substantial resistance itself. Being aware of these ensures the analysis gives accurate results.
Electrical resistance
Electrical resistance is a material's inherent opposition to the flow of electric current. This property is fundamental in determining how much energy is lost as heat in conductors. In the context of our exercise, you sought to determine the resistance of a wire, which acts as a resistor. A key factor contributing to resistance in a wire is its length: as the length increases, so does resistance, since electrons encounter more obstacles along their path.

Other factors include the wire's material—some materials like copper are excellent conductors with low resistance—and its cross-sectional area: thicker wires have less resistance because they provide more paths for current flow.

Once the resistance of a length of wire is known, it can help in designing circuits. For instance, knowing the resistance of the wires helps ensure that they can safely carry the desired current without overheating. This understanding is critical to maintain efficiency and safety in electrical systems. In the exercise, even though imperfections in measurement existed, estimating the average resistance per meter allowed an approximation that is critically useful in real-world applications. Overall, grasping resistance is essential for anyone involved in electrical design or maintenance.

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Most popular questions from this chapter

You apply a potential difference of 4.50 \(\mathrm{V}\) between the ends of a wire that is 2.50 \(\mathrm{m}\) in length and 0.654 \(\mathrm{mm}\) in radius. The resulting current through the wire is 17.6 \(\mathrm{A}\) . What is the resistivity of the wire?

A battery-powered global positioning system (GPS) receiver operating on 9.0 \(V\) draws a current of 0.13 \(A\). How much electrical energy does it consume during \(1.5 \mathrm{~h} ?\)

A 1.50 -m cylindrical rod of diameter 0.500 \(\mathrm{cm}\) is connected to a power supply that maintains a constant potential difference of 15.0 \(\mathrm{V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads \(18.5 \mathrm{A},\) while at \(92.0^{\circ} \mathrm{C}\) it reads 17.2 \(\mathrm{A} .\) You can ignore any thermal expansion of the rod. Find (a) the resistivity at \(20.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

A silver wire 2.6 \(\mathrm{mm}\) in diameter transfers a charge of 420 \(\mathrm{C}\) in 80 \(\mathrm{min}\) . Silver contains \(5.8 \times 10^{28}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

A source with emf \(\mathcal{E}\) and internal resistance \(r\) is connected to an external circuit. (a) Show that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (b) If the external circuit consists of a resistance \(R,\) show that the power output is maximum when \(R=r\) and that the maximum power is \(\mathcal{E}^{2} / 4 r_{1}\)

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