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

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The discussion revolves around calculating the current and power loss in a 38 m long aluminium wire with a diameter of 10.0 cm and a resistivity of 2.65e-8 Ohm*m, subjected to a 9000 Volt potential difference. The resistance is calculated using the formula R = pL/A, leading to a resistance value of approximately 1.23e-4 Ohms. The current is then determined using I = V/R, resulting in a value of about 7.02e+7 A, which should be expressed in scientific notation with appropriate significant figures. Participants confirm that the power lost can be calculated using P = I*V, along with alternative formulas P = I^2R and P = V^2/R. The calculations are deemed correct for the parameters provided in the problem.
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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]
 
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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.
 
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