Question

Two conducting wires have identical lengths \(L_{1}=L_{2}=\) \(L=10.0 \mathrm{~km}\) and identical circular cross sections of radius \(r_{1}=r_{2}=r=1.00 \mathrm{~mm} .\) One wire is made of steel (with resistivity \(\rho_{\text {steel }}=40.0 \cdot 10^{-8} \Omega \mathrm{m}\) ); the other is made of copper (with resistivity \(\left.\rho_{\text {copper }}=1.68 \cdot 10^{-8} \Omega \mathrm{m}\right)\) a) Calculate the ratio of the power dissipated by the two wires, \(P_{\text {copper }} / P_{\text {steel }},\) when they are connected in parallel; a potential difference of \(V=100 . \mathrm{V}\) is applied to them. b) Based on this result, how do you explain the fact that conductors for power transmission are made of copper and not steel?

Short answer

Answer: Copper is preferred over steel for power transmission because it has a lower resistivity and resistance, which allows more current to flow and dissipates less power as heat compared to steel. This makes copper more efficient for power transmission, as less energy is wasted as heat.

Step-by-step solution

Calculate the resistance of each wire

To calculate the resistance of a wire, we use the formula: \(R = \frac{\rho \cdot L}{A}\) where \(R\) is the resistance, \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area. Since both wires have identical lengths and cross-sectional areas, we can calculate their respective resistances using the values given for their resistivities. For the steel wire: \(R_{steel} = \frac{\rho_{steel} \cdot L}{A_{steel}} = \frac{40.0 \cdot 10^{-8} \Omega m \cdot 10.0 \times 10^{3} m}{\pi \cdot (1.00 \times 10^{-3} m)^2} \) For the copper wire: \(R_{copper} = \frac{\rho_{copper} \cdot L}{A_{copper}} = \frac{1.68 \cdot 10^{-8} \Omega m \cdot 10.0 \times 10^{3} m}{\pi \cdot (1.00 \times 10^{-3} m)^2} \)

Calculate the equivalent resistance of the parallel connection

Now that we have the resistance of both wires, we can find the equivalent resistance of their parallel connection using the formula: \(\frac{1}{R_{eq}} = \frac{1}{R_{steel}} + \frac{1}{R_{copper}}\)

Find the total current through the parallel connection

Now, with the equivalent resistance, we can find the total current through the parallel connection using Ohm's law. The potential difference \(V = 100V\) is applied to the parallel connection: \(I_{total} = \frac{V}{R_{eq}}\)

Calculate the current through each wire

Since the wires are connected in parallel, the voltage across each wire is the same. We can use Ohm's law again to find the current through each wire: \(I_{steel} = \frac{V}{R_{steel}}\) \(I_{copper} = \frac{V}{R_{copper}}\)

Calculate the power dissipated by each wire

To calculate the power dissipated by each wire, we can use the formula: \(P = IV\) For the steel wire: \(P_{steel} = I_{steel} \cdot V\) For the copper wire: \(P_{copper} = I_{copper} \cdot V\)

Calculate the ratio of the power dissipated by the two wires

Now, we can find the ratio of the power dissipated by the copper wire to that of the steel wire: \(P_{copper}/P_{steel} = \frac{P_{copper}}{P_{steel}}\)

Explain the preference of copper over steel for power transmission

Based on the calculated power dissipation ratio, we can discuss why copper is preferred over steel for power transmission. Copper has a lower resistivity than steel, which means it has a lower resistance for the same length and cross-sectional area. Consequently, a copper conductor will allow more current to flow and dissipate less power as heat compared to a steel conductor. This makes copper more efficient for power transmission, as less energy is wasted as heat.

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering that relates voltage, current, and resistance in an electrical circuit. According to Ohm's Law, the voltage (V) across a conductor between two points is directly proportional to the current (I) flowing through it, with the proportionality factor being the resistance (R). The mathematical expression for Ohm's Law is: \[ V = I \times R \] This means if you know any two of these quantities, you can calculate the third. In our exercise, the potential difference applied is 100 V, and knowing the resistance allows us to calculate current using this formula. This is crucial for determining how much current passes through each wire when they are connected in parallel, ensuring efficient power distribution.

Power Dissipation

Power dissipation in an electrical circuit refers to the loss of electrical energy as heat in circuit elements. The power dissipated (P) by a component can be calculated using the formula: \[ P = I \times V \] where \( I \) is the current through the component and \( V \) is the voltage across it. Additionally, using Ohm's Law, power dissipation can also be expressed as: \[ P = I^2 \times R = \frac{V^2}{R} \] These equations highlight that increased resistance leads to higher power loss. In our exercise, when comparing copper and steel wires, it's observed that copper, having lower resistance, dissipates less power, making it more efficient for power transmission. This is another reason why copper is favored over steel: it minimizes energy loss.

Material Properties

The properties of materials, particularly their resistivity, play a crucial role in determining electrical performance. Resistivity is an intrinsic property of a material that indicates how strongly it opposes the flow of electric current. Materials with low resistivity are good conductors, while high resistivity materials are poor conductors.

• Copper: Known for its excellent electrical conductivity with a resistivity of \(1.68 \times 10^{-8} \Omega \text{m}\).
• Steel: Has a higher resistivity of \(40.0 \times 10^{-8} \Omega \text{m}\), making it a less efficient conductor than copper.

These differences in resistivity result in significant variations in the resistance of wires made from these materials, affecting how much power is lost as heat. This exercise demonstrates why copper is preferred in circumstances where efficiency and minimal energy loss are critical, such as in power lines.

Resistance Calculation

Resistance is a measure of how much a material opposes the flow of electric current. It can be calculated using the formula: \[ R = \frac{\rho \cdot L}{A} \] where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area. For our given wires:

• The length \( L = 10.0 \text{ km} \)
• The radius \( r = 1.00 \text{ mm} \)
• The cross-sectional area \( A = \pi \cdot r^2 \)

Using these values, we calculate the resistance for both copper and steel wires, acknowledging their distinct resistivities. Copper, with its lower resistivity, will have lower resistance, leading to lesser power dissipation when connected in a circuit. This calculation not only reinforces the importance of using materials with low resistivity for efficient power transmission but also illustrates practical applications of physics in real-world engineering decisions.