1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Few RC Circuits/Electricity Problems

  1. Feb 9, 2008 #1
    1. The problem statement, all variables and given/known data

    1. What electric field strength is needed to create a 5.0 A current in a 2.0-mm-diameter iron wire?

    2. You need to design a 1.0 A fuse that "blows" if the current exceeds 1.0 A. The fuse material in your stockroom melts at a current density of 500 A/cm^2. What diameter wire of this material will do the job?

    3.A car battery is rated at 90 Ahr, meaning that it can supply a 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery? Express your answer in Coulombs to three significant figures

    4. The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q = 4t - t^2 , where t is in seconds and t>=0.

    Which is an expression for the current in the wire at time t.


    2. Relevant equations

    E = [tex]\frac{I}{(\sigma)(\pi)r^2}[/tex]

    J = [tex]\frac{I}{(\pi)(r^2)}[/tex]

    Q = [tex]\frac{I}{t}[/tex]

    3. The attempt at a solution

    1. I know [tex]\sigma[/tex] is 10^7 so

    E = [tex]\frac{5}{(10^7)(\pi)(0.001^2)}[/tex] = 0.159 N/c

    2. 500x10^-2=5 A/m^2

    r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)5}}[/tex] = 0.252 m and to get diameter I multiply it by two to get 0.504. The question asks me to express it in mm so I got 504mm.

    3. 1 hour = 3600 seconds

    Q = [tex]\frac{I}{t}[/tex] = Q = [tex]\frac{90}{3600}[/tex] = 0.025 C

    4. Current is the derivative of charge so it would be the third equation.

    Did I make a math (or any other kind) error?
    Last edited by a moderator: Apr 23, 2017 at 10:41 AM
  2. jcsd
  3. Feb 11, 2008 #2
    #1 and #4 are correct, but I cannot get the correct answers for #2 and #3. Does anyone know why?
  4. Feb 11, 2008 #3
    If you have 500 amps for every square CENTImeter, should you have a lot less per square meter, or, say, a LOT more?

    #3, assuming you did it right, has only 2 significant figures

    it's always the easy ones ^_^
  5. Feb 11, 2008 #4
    so 1 Square Centimeter = 0.0001 Square Meters therefore 500 A per Square Centimeters = 0.05 A per Square Meters

    r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)0.05}}[/tex] = 5.046 m = 5046 mm

    and to change 0.025 from 2 significant digits to 3, I'll need to use scientific notation so 0.025 = 2.50 x 10^-2
  6. Feb 11, 2008 #5
    ehh, you kinda missed my point for #2, here's a hint


    Make sure you can get that to work right. If I can fix x things into a square centimeter, I'd better be able to fit a helluva lot more into a square meter, which is notably bigger than a cm^2
  7. Feb 11, 2008 #6
    Ah, now I get it thanks. Let's give this another try:

    r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)(5000000)}}[/tex] = 0.000252 m and to get the diameter I multiply it by two to get 5.05 x 10^-4 m = 0.505 mm.

    And I also figured out why #3 is wrong. It's not Q = I/t, it's Q = I*t so:

    Q = I*t = (90)(3600) = 324000 C

    Just tried them, and it said I was correct. Thank you blochwave.
    Last edited: Feb 11, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: A Few RC Circuits/Electricity Problems
  1. RC Circuit Problem (Replies: 16)

  2. RC circuit problem (Replies: 6)