Calculating Current and Power in an Aluminium Wire: A Homework Guide

  • Thread starter Thread starter KTTxx
  • Start date Start date
  • Tags Tags
    Current
Click For Summary

Discussion Overview

The discussion revolves around calculating the current and power loss in a 38 m aluminium wire with a specified diameter and resistivity, under a potential difference of 9000 Volts. It includes theoretical considerations and practical applications related to electrical resistance and power equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Participants discuss the calculation of resistance using the formula R = pL / A, where p is the resistivity, L is the length, and A is the cross-sectional area of the wire.
  • There is a query about the correct method to calculate current using I = V / R, with some participants expressing difficulty in obtaining the correct answer.
  • Multiple participants mention alternative power equations, including P = I * V, P = I²R, and P = V²/R, as valid methods for calculating power loss.
  • One participant provides a specific current value of 70194473.39 A and a resistance value of 1.23e(-4), suggesting that these values are consistent with the problem as stated.
  • There is a suggestion to use scientific notation for clarity and to consider significant figures in the reported values.

Areas of Agreement / Disagreement

Participants generally agree on the formulas to use for resistance and power calculations, but there is no consensus on the correctness of the calculated current value, as some participants express uncertainty about their results.

Contextual Notes

Participants have not resolved the discrepancies in their calculations, and there are indications of confusion regarding significant figures and scientific notation in reporting results.

Who May Find This Useful

This discussion may be useful for students or individuals interested in electrical engineering concepts, particularly those dealing with resistance, current, and power calculations in conductive materials.

KTTxx
Messages
2
Reaction score
0

Homework Statement


You are working for the power company and wonder why we use a set voltage of power lines instead of a set current. You test your system on 38 m of aluminium wire with a diameter of 10.0 cm. The resistivity of aluminium is 2.65e-8 Ohm*meters. You place a 9000 Volt potential difference across the wire.
a. What is the current through the wire?
b. What is the power lost from the wire?[/B]

Homework Equations

The Attempt at a Solution


a. First, i need to find the Resistance of the wire by using equation R = pL /A
p = 2.65 e-8, L = 38 m, A= pi * (.05 m) ^2
Then i find the Current by I = V / R. I plug in the number, and i keep get the wrong answer.

For the power lost: Can i use the equation P = I * V ?[/B]
 
Physics news on Phys.org
KTTxx said:
1. Homework Statement
You are working for the power company and wonder why we use a set voltage of power lines instead of a set current. You test your system on 38 m of aluminium wire with a diameter of 10.0 cm. The resistivity of aluminium is 2.65e-8 Ohm*meters. You place a 9000 Volt potential difference across the wire.
a. What is the current through the wire?
b. What is the power lost from the wire?

Homework Equations

The Attempt at a Solution


a. First, i need to find the Resistance of the wire by using equation R = pL /A
p = 2.65 e-8, L = 38 m, A= pi * (.05 m) ^2
Then i find the Current by I = V / R. I plug in the number, and i keep get the wrong answer.
What answers do you get for resistance and current?
For the power lost: Can i use the equation P = I * V ?[/B]
You can, yes. There are also two other forms of power equation that you can use: P = I2R and P = V2/R.
 
gneill said:
What answers do you get for resistance and current?

You can, yes. There are also two other forms of power equation that you can use: P = I2R and P = V2/R.

For the current, i get 70194473.39 A... , R = 1.23e(-4)
 
KTTxx said:
For the current, i get 70194473.39 A... , R = 1.23e(-4)
The values are good for the problem as stated, although you might want to use scientific notation for the current value and show only a reasonable number of significant figures.
 

Similar threads

Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K