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Current resistor and time

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    a current of 5 ampere exists in a 10 ohm resistor for 4 minutes

    a) how many coulombs pass through the resistor in this time? should be 1200 coulombs
    b) how many electrons pass through the resistor in this time? should be 7.5*10^21
    c) what potential drop exists across the resistor? should be 50 volts
    d) how much energy is dissipated in the resistor during this time? should be 60000joules


    2. Relevant equations

    resistance R = V/I where V is electric potential, I is current

    capacitance C = Q/V where Q is charge ---> not sure about this one

    current I = nAqv_d where n is number of electrons, A is area, q is charge, v_d is drift velocity

    power P = IV where I is current, V is electric potential.

    work W = 1/2(CV^2) where C is capacitance

    3. The attempt at a solution

    somewhat started part a:

    use first eq to solve for electric potential V --> V = IR = 5(10) = 50 volts

    tried using second eq, C = Q/V but did not have capacitance, i don't even think you can use capacitance in this problem

    i need help with an equation containing time, because t = 4mins = 240seconds. other wise i could use the fourth eq, and let P = W/t where W is work, t is time and solve for t. the work work formula requires capacitance, which i don't have

    as for part b and finding the number of elections i think i use the third eq, hold A as constant but how do i figure out the drift velocity v_d, or do i hold that constant too?

    i haven't tried the other parts yet, looking for some insight...
     
  2. jcsd
  3. Jun 11, 2008 #2

    alphysicist

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    Hi Scholio,

    You're right that you cannot use C=Q/V here. Instead, since you know the current, what is the definition of current?
     
  4. Jun 11, 2008 #3
    current I is the change in Q over the change in time --> I = dQ/dt

    oh i figured it out, i didn't even think of that relationship:

    since V = IR = 50 volts

    V = (dQ/dt)R
    50 = (dQ/240)10
    dQ = 12000/10
    Q = 1200 coulombs

    that was straightforward, thanks!
     
    Last edited: Jun 11, 2008
  5. Jun 11, 2008 #4
    For the number of electrons per coulumb... h t t p : / / en.wikipedia.org/wiki/Electric_current

    That should help.
     
  6. Jun 11, 2008 #5

    alphysicist

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    scholio,
    That's right. Another way to think about it, is that a current of 5A means 5 coulombs per second. Since you know how many seconds the current is running and it's constant, you can just multiply current times time to get coulombs.
     
  7. Jun 11, 2008 #6
    The way I did part A which works without needing the voltage is...

    I = Q/t...
    Q = I*t
    = 5A * 4min*(60s/1min)
    = 1200 Coul
     
  8. Jun 11, 2008 #7
    thanks, i got part a now.

    how would i got about handling part b using I = nAqv_d, do i hold A area, and drift velocity v_d constant? the q in this equation is not the 1200 i found in part a is it?

    anyways i tried it and it didn't work out: i took A and v_d constant

    5 = n(1200)
    n = 4.16*10^-3 ---> not right, need 7.5*10^21 electrons
     
  9. Jun 11, 2008 #8

    alphysicist

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    I don't think you need to use that equation. What's the charge on an electron? Be sure to include the units.

    (You can argue that A and v_d are constant, but they still have to be in the equation!)

    Edit: Also, the n in that equation is not the number, it is the number density (particles / volume).
     
    Last edited: Jun 11, 2008
  10. Jun 11, 2008 #9
    oh okay, the charge of an electron is 1.6*10^-19 coulombs, is that what i substitute in for q in the equation?

    there is another equation involving n, q and v_d. its for current density J.

    J = nqv_d where J = sigma(E) where E is electric field, and sigma is conductivity. the material is not specified though.

    electric field E = F/q where F is force and q is charge

    is this along the right lines now?
     
  11. Jun 11, 2008 #10

    alphysicist

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    No, it's more like changing units. There is one electron in [itex]1.6\times 10^{-19}[/itex] coulombs. How many electrons are in 1200 coulombs?
     
  12. Jun 11, 2008 #11
    oh, again i overanalyzed the problem. 1200coulombs/(1.6*10^-19 coulombs) = 7.5*10^21 electrons

    thanks again

    as for part c, finding the potential drop that exists along the resistor, did i already solve for that in part a? am i correct to say that the V in V = IR is potential difference?

    then isn't V = 50 volts then?
     
  13. Jun 11, 2008 #12

    alphysicist

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    That's right. Did you get part d?
     
  14. Jun 11, 2008 #13
    no not yet, still need to finish up part c, any tips before i start part d?
     
    Last edited: Jun 11, 2008
  15. Jun 11, 2008 #14

    alphysicist

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    I'll let you work on it first. You have everything you need (just don't use the equations with capacitance).
     
  16. Jun 11, 2008 #15
    i got part c and d.

    part c:
    what potential drop exists across the resistor?

    QV/t = I^2R where Q is 1200 coulombs, I is 5 amp, R is 10 ohms, t = 240 sec
    V = I^2RT/Q = 50 votls

    part d:
    how much energy is dissipated in the resistor during this time?

    energy lost/t = I^2R where t = 240 sec, I = 5amp, R = 10 ohms

    energy lost = I^2Rt = 60000 joules


    did i do it correctly? i got the answers i was looking for.
     
  17. Jun 11, 2008 #16

    alphysicist

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    It looks right, but I don't think you need to do quite all that. Your original approach V=IR in the first post is exactly what's needed.

    (In the equation you have above, notice that Q/t =I, when the current is constant, so:

    QV/t = I^2R ----> I V = I^2 R ------> V = I R
     
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