Electric Currents and Resistance I NEED HELP

In summary, the conversation discusses an example problem involving a copper cable designed to carry a current of 400A with a power loss of 4 W/m. The formula P=I^2R is used to compute the required radius of the cable, and the variables R, A, L, and ρ are defined. It is suggested to use 1m for L and to ensure that the value for ρ is in ohms/m. The final answer will depend on whether the diameter or radius is solved for, and the units will depend on the units of the value used for ρ. The conversation also emphasizes the importance of solving the problem in terms of variables before plugging in numbers.
  • #1
phystudent
11
0
Our physics teacher gaves us online homework, and he was gone on friday and it is due by midnight tonight. Here is an example problem that we are having trouble with:

A copper cable is designed to carry a current of 400A with a power loss of 4 W/m. What is the required radius of this cable?
 
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  • #2
Start with P=I2R

You have the information you need to compute the resistance of 1m of wire.

Use

[tex] R= \frac {\rho L } A [/tex]

where [tex] \rho [/tex] is the resistivity of copper.

Can you finish it?
 
Last edited:
  • #3
All i know what to do is to plug in the amperage, but i don't know where the other numbers are substituted for what letter, and i have no idea what Latex is for this equation.
 
  • #4
P = power = 4 w
R = resistance
A = area
L = length (use 1m)
[tex] \rho [/tex] is the resistivity be sure your value is in ohms/m

Edit:
Oh yeah... I = current
 
  • #5
Is the answer from those numbers going to be the diameter or the radius? And if I used 1.56e-8 for the resisitivity, would the answer be in m or cm?
 
  • #6
Power, as Integral said, is given by:
[tex]P = I^2R[/tex]
Subtitute R and you get:
[tex]P = \frac{I^2\rho L}{A} = \frac{I^2\rho L}{\pi r^2}[/tex]
Solve for r:
[tex]r = \sqrt{\frac{I^2\rho L}{\pi P}}[/tex]
Since the question says the power per length unit is 4W/m, L/P is 0.25m/W. The rest you should have.
 
  • #7
The answer that I came up with was .01409, now is that in "m" or "cm", and is it the radius?
 
  • #8
Chen said:
Power, as Integral said, is given by:
[tex]P = I^2R[/tex]
Subtitute R and you get:
[tex]P = \frac{I^2\rho L}{A} = \frac{I^2\rho L}{\pi r^2}[/tex]
Solve for r:
[tex]r = \sqrt{\frac{I^2\rho L}{\pi P}}[/tex]
Since the question says the power per length unit is 4W/m, L/P is 0.25m/W. The rest you should have.
Chen, we really in courage people to the complete solution on their own. this is simply doing his homework for him. While we appreicate your efforts please let them do some of the work :)

BTW:
Note how he completely solved the problem in terms of the the variables before plugging in a single number. This is the right way to do physics problems.
 
  • #9
phystudent said:
The answer that I came up with was .01409, now is that in "m" or "cm", and is it the radius?

If you solve for the diameter then your answer will be the diameter, if you solve for the radius then your answer will be the radius.

What are the units of the value you used for resistivity? That will determine the units of your answer.
 

1. What is electric current?

Electric current is the flow of electric charge through a conductor. It is measured in amperes (A) and is represented by the symbol I. Electric current is essential for the operation of many electrical devices.

2. What factors affect the resistance of a conductor?

The resistance of a conductor is affected by its length, cross-sectional area, and material. Longer conductors have higher resistance, while wider conductors have lower resistance. Different materials have different resistivities, which also affect resistance.

3. How is resistance calculated?

Resistance is calculated using Ohm's Law, which states that resistance (R) is equal to the ratio of voltage (V) to current (I). This can be written as R = V/I. Resistance is measured in ohms (Ω).

4. What is the difference between AC and DC currents?

AC (alternating current) and DC (direct current) are two types of electric currents. AC current changes direction periodically, while DC current flows in one direction. AC is used for long-distance power transmission, while DC is used for electronic devices.

5. How can the resistance of a circuit be reduced?

The resistance of a circuit can be reduced by using conductors with larger cross-sectional areas, shorter lengths, and lower resistivities. Adding more conductors in parallel can also decrease resistance. Additionally, reducing the temperature of the circuit can decrease resistance.

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