Pls help me with my Resistivity Question in pHysics due tomorrow morning.

In summary, the conversation revolves around different scenarios involving copper and iron wires and their resistance. The first question asks for the length of a 20 mills diameter copper wire to have a resistance of 5.0Ω. The second question discusses the minimum diameter of a copper wire to ensure a voltage drop not exceeding 2% when an electric motor draws 10 A from a 240-V source 120 ft away. The third question asks for the potential difference between two points in a 1.0 mm diameter copper wire that carries a current of 12 A and is 100m apart. The fourth question deals with finding the resistance of a copper wire at 0oC and 80oC when its resistance at
  • #1
emjez15
5
0
Pls help me..thanks

Homework Statement



17. How long should a copper wire 20 mills in diameter be for it to have a resistance of 5.0Ω?

20. An electric motor draw 10 A from a 240-V source 120 ft away. What is the minimum diameter of the copper wire in mills if the voltage drop is not to exceed 2%?

22. A copper wire 1.0 mm in diameter carries a current of 12 A. Find the potential difference between two points in the wire that are 100m apart.

24. The resistance of a copper wire is 100 Ω @ 20oC. Find its resistance at 0oC and at 80oC

25. An iron wire has a resistance of 2.00 Ω at 0oC and a resistance of 2.46 Ω at 45oC. Find the temperature coefficient of resistivity of the wire
 
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  • #2
What is it you don't understand?

You need to show some effort and identify where you are stuck.
 
  • #3
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Hello, I understand that you need help with your resistivity questions in physics. It is important to start by understanding the formula for resistivity, which is p = RA/L, where p is the resistivity, R is the resistance, A is the cross-sectional area of the wire, and L is the length of the wire.

For question 17, you can use the formula to solve for L, which would be L = RA/p. Plug in the given values for R, A, and p to solve for the length of the wire in meters. Remember to convert the given diameter of 20 mills to meters before solving.

For question 20, you can use the formula V = IR, where V is the voltage drop, I is the current, and R is the resistance. Plug in the given values for V and I, and use the calculated value of R to solve for the minimum diameter of the wire in meters.

For question 22, you can use the formula V = IR, where V is the potential difference, I is the current, and R is the resistance. Again, plug in the given values for I and R, and use the calculated value of V to solve for the potential difference in volts.

For question 24, you can use the formula R2 = R1(1 + αΔT), where R2 is the resistance at the new temperature, R1 is the resistance at the initial temperature, α is the temperature coefficient of resistivity, and ΔT is the change in temperature. Plug in the given values for R1 and ΔT, and solve for R2 at the desired temperatures of 0°C and 80°C.

For question 25, you can use the formula α = (R2 - R1)/(R1ΔT), where α is the temperature coefficient of resistivity, R2 is the resistance at the new temperature, R1 is the resistance at the initial temperature, and ΔT is the change in temperature. Plug in the given values for R1, R2, and ΔT to solve for α.

I hope this helps you with your questions. Remember to show your work and double check your units to ensure accurate answers. Good luck with your assignment!
 

FAQ: Pls help me with my Resistivity Question in pHysics due tomorrow morning.

1. What is resistivity and how does it affect electrical conductivity?

Resistivity is a measure of a material's ability to resist the flow of electrical current. It is a property of the material itself and is dependent on factors such as temperature and impurities. Higher resistivity means a material is less conductive, while lower resistivity means a material is more conductive.

2. How is resistivity calculated?

Resistivity is calculated by dividing the resistance (in ohms) of a material by its cross-sectional area (in square meters) and its length (in meters). The resulting unit is ohm-meters (Ω·m).

3. How does pH affect resistivity?

pH has a significant impact on the resistivity of a solution. As the pH of a solution decreases (becomes more acidic), the resistivity decreases as well, meaning the solution becomes more conductive. Conversely, as the pH increases (becomes more basic), the resistivity increases, making the solution less conductive.

4. What are some common applications of resistivity in physics?

Resistivity is used in a variety of applications, including determining the conductivity of materials, measuring the purity of substances, and studying the electrical properties of different solutions. It is also essential in the design and development of electronic devices and circuits.

5. How can I use the resistivity equation to solve problems?

To solve problems involving resistivity, you can use the resistivity equation (ρ = R x A / L) to calculate the resistivity of a material or solution. You can also use it to determine the resistance of a material given its resistivity, length, and cross-sectional area. It is important to make sure all units are consistent and to use the correct value for the material's temperature when using the equation.

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