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Time Dependent Current

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    The current in amps through a resistor with a resistance of 90 Ohm varies according to I=2.8e^(−7.6t) when t is in seconds. What is the total energy dissipated in the resistor from 0.5 to 1.4 seconds, written in Joules?


    2. Relevant equations
    I(t)=dQ/dt

    dU=dQ*dV

    P=dU/dt=I*I*R


    3. The attempt at a solution

    So I've tried writing it as dQ=2.8e^(−7.6t)dt and integrating from 0.5 to 1.4 seconds to find dQ. Then squaring that and dividing by (1.4-0.5)^2 to find I^2 and then P=R*I^2 and then U=P*dt. I'm not sure if there is a way I can use dQ directly to find dU. Actually I did just try rewriting the second equation as dU=dQ*(dQ/dt)*R and I got the same number, 0.006778 J which does seem pretty small
     
  2. jcsd
  3. Oct 8, 2013 #2

    gneill

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    Staff: Mentor

    The correct result will be not too much larger than what you're finding. About an order of magnitude or so.

    Rather than trying to decompose the current into individual charge movements, make use of the the power relationship that you quoted:

    P = I2R

    R is constant but you're given I as a function of time. So construct P(t). What do you get when you integrate power over time?
     
  4. Oct 8, 2013 #3
    Ah, yes thank you I got it. I was over complicating it I guess
     
  5. Oct 4, 2014 #4
    I have the same problem (a whole year later! Woah they really change it up!). I can't figure it out. I've tried integrating the I function, straight plugging in, everything, with no luck. Any help (and you can use the equation and numbers above. I'll throw in the proper numbers after)?
     
  6. Oct 4, 2014 #5

    gneill

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    Staff: Mentor

    Show us an example of what you've tried.
     
  7. Oct 4, 2014 #6
    P=I^2*R
    I=5[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3A.png8e−^6[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/70/char3A.png6t [Broken]
    R=84
    U=P*dt

    I've integrated the I function from 0.5 to 1.7 (my range of times), squared it, multiplied by 84, and multiplied that by (1.7-0.5)^2/2 (since in reality, dt=x right, and you need to integrate?).

    That was one. I thought maybe not integrating would work too? Still nothing . . .
     
    Last edited by a moderator: May 7, 2017
  8. Oct 4, 2014 #7

    gneill

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    Staff: Mentor

    From your description you integrated something and then squared it afterwords. Is that correct? Here t is the variable that is being integrated over. So your integral will have a 'dt' not a 'dx'.

    How about typing out your integrand so I can take a look.
     
  9. Oct 5, 2014 #8
    5.8e^-6.6tdt from 0.5 to 1.7

    'Cause we just integrate the I function, correct (and sorry for the format . . . did this forum change??)
     
  10. Oct 5, 2014 #9

    gneill

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    Okay, you want to integrate the power, not the current. The power is I2R, so that is your integrand (replacing the "I" with your given current function, of course).

    Yes, the forum system software was upgraded recently. It's a slightly new look but is functionally much the same as before. There are new features which you can find out about in the Lounge area.
     
  11. Oct 5, 2014 #10
    This look was confusing me for a sec . . . I was trying to find the forum that's saved me at least once a unit since starting college.

    So it should look something like this:

    P = (I(1.7)-I(0.5))^2*R
    U = P*dt

    So I get a power number and basically multiply it by time (so dt will just become t, so P*(1.7-0.5))?
     
  12. Oct 5, 2014 #11

    gneill

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    Staff: Mentor

    You'll need to do the integration before plugging in any numbers. You're integrating
    $$\int_{t_o}^{t_f} 84 \left( 5.8 e^{-6.6 t}\right)^2 dt $$
     
  13. Oct 5, 2014 #12
    Ah, so multiply through the constants (so in this case, R and the coefficient of "e"), than integrate. And the time becomes that dt!
     
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