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Homework Help: Electric Current problem.

  1. Mar 4, 2006 #1
    I did all the word but my answers are not matching my books?

    1) Can a 2.5 mm-diameter copper wire have the same resistence as a tungsten wire of the same length? Give numerical detials. Answer, yes, 4.6 mm

    Here is what I did:
    ( p = resistivity )

    R = p(L/A) so I set R(Cu) = R(tug) and since L is the same it cancels out.

    Area = 1/2pie2^2 ---> Cu area = (.5)(3.14)(1.25mm) ---> A = .393 mm

    So p(A) = p(A) ---> (1.68 x 10^-8)(.393 mm) = (5.6 x 10^-8)(A)
    A = .1179 mm

    therefore diameter would be... .1179 mm = 2(pie)r^2
    r = .2729 mm which mean diameter = .546 mm... so what did I do incorrect?

    2) A small immersion heater can be used in a car to heat a cup of water. If the heater can heat 120 mL of water from 25C to 95 C in 8.0 minutes, (a) How much current does it draw from the car's 12 V battery, (b) What is the resistence? Assume the manufacturer's clain of 60% effciency. Answers: 10 A, 1.2 Olhms

    What I did...

    Used specific heat equation ---> Q = mcT where I changed 120 mL of water into .120 kg ---> Q = (.120 kg)(4185 J/kg C)(95C-25C)
    Q = 35154 J

    Then I used the equation PE = 1/2Q(where Q is the charger)V
    35151 J = (.5)(Q)(12 V)
    Q = 5859 C

    Finally I used I(current) = Q/T(time) ---> 5859C/480s
    I = 12 A

    And for Resistence I used V/I = R ---> 12 V/12 A = 1 Olhm

    I am off my two units for some odd reason in my current and off by .2 in my resistence. Did I make a mistake?
     
  2. jcsd
  3. Mar 4, 2006 #2

    Integral

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    Last edited by a moderator: Apr 22, 2017
  4. Mar 4, 2006 #3

    Integral

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    After sorting through your work, I see several things wrong. What is the area of a circle? Where did you get the number for the area of the copper? Re examine your work you have make some very basic errors.

    It is essential that you make a habit of working with symbols until you have isolated the parameter you need a number for. In this case you have.

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

    I will use a T subscript for Tungsten and a C subscript for copper.

    [tex] R_T = R_C [/tex]

    so

    [tex] \frac {\rho_T L_T} {A_T} = {\frac {\rho_C L_C} {A_C} [/tex]

    further you are given that
    [tex] L_T = L_C [/tex]

    so now we have:
    [tex] \frac {\rho_T } {A_T} = \frac {\rho_C } {A_C} [/tex]

    [tex] A_T = \frac {\rho_C } { \rho_T A_C } [/tex]

    Now can you complete the problem in this same manner until you have isolated the radius of the tungsten?

    I have found that it is often necessary to repeat calculations multiple times. Do it until you can get the same answer several times in a row.
     
    Last edited: Mar 4, 2006
  5. Mar 5, 2006 #4
    That does stil not work. Area of copper will be 4.91 mm^2. If you then use.

    [tex] A_T = \frac {\rho_C } { \rho_T A_C } [/tex]

    You get Area of T = .0611

    and then if we use the equation of a circle for this area we get d = .27 mm.
     
  6. Mar 5, 2006 #5
    Now I see. You actually manipulated the equation wrong. The Area of T is equal to AcPt/Pc. That gives the correct answer. Thanks for a good try though.

    Does number two look okay?
     
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