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Homework Help: Electric current, resistance and circuits

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    See attachment pls.
    for a) I really need to know if this is right(well I know its right but it may not be what they want??).
    for b) i need some pointers for (ii) and (iii)
    very grateful if you would check a) and i) too... thanks to all future helpers

    2. Relevant equations

    3. The attempt at a solution
    Ohm = Volt / Ampere = Volt second / Coulomb
    Farad = Coulomb / Volt

    Ohm Farad = Volt second / Coulomb * Coulomb/Volt = second
    i) U=Q^2/2C
    U=7.1J (Please check this if you can)
    no current though - not sure where to go

    Attached Files:

    • Q8.png
      File size:
      52.2 KB
  2. jcsd
  3. Jul 21, 2010 #2


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    Yup. That's one way to do it (there are other ways too, btw.) :approve:
    'Looks good to me. :approve:
    Don't get discouraged. :smile: So you have the equation
    Ploss = I2R.
    But don't forget about Ohm's law,
    V = IR. Or, rewriting it, I = V/R.

    Can you think of a way to combine those to get rid of the I?
  4. Jul 21, 2010 #3
    ah so P_loss=V^2R
    then I still dont have V though????
    and iii) I am at a complete loss for

    thanks for your help so far.
  5. Jul 21, 2010 #4


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    Not quite! :eek:

    If Ploss = I2R, and I = V/R then,

    Ploss = V2/R

    where V is the voltage across the terminals of the resistor R.
    You know that C = Q/V. Rearrange and solve for V. :smile:
    Part iii) is really just a rehash of what what you've already done. Start by halving the answer you found in i), and then work your way through it again with the new numbers. :wink:
  6. Jul 21, 2010 #5
    ok thanks for that, ill post my answers back for you to check if you dont mind.
  7. Jul 21, 2010 #6
    hey collinsmark, for ii

    P_loss=Q^2/(C^2 R)
    P_loss=(8.10*〖10〗^(-3) )^2/((4.62*〖10〗^(-6) )^2*850)
    seems big????

    for iii)
    C=5.73*〖10〗^(-3) C
    P_loss=Q^2/(C^2 R)
    P_loss=((5.73*〖〖10〗^(-3))〗^2)/((4.62*〖10〗^(-6) )^2*850)
    I dont think that that is right since it is exactly half of the power dissipated at full charge and i didn't think that the charging v P_loss curve would be linear??
  8. Jul 22, 2010 #7


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    Yeah, it may seem pretty big. :biggrin: But I also think it's pretty correct. :approve:

    The voltage is really high on this puppy.
    V = Q/C = (8.10 x 10-3 [C])/(4.62 x 10-6 [F]) = 1753.25 Volts.

    P = V2/R = (1753.25 [V])2/(850 [Ω]) = 3616.34 Watts

    Everything checks out. :approve:

    What's not given in this problem (perhaps you have to solve for it in your next step) is the specific RC time constant for this circuit (although part (a) alludes to this). Yes, the initial power is huge! But you'll find that it doesn't last very long at all. :smile:
    Also looks good to me. :approve:
    Careful, some things are linear and some things are not. Ploss vs. V is definitely not linear, as you have said. (P is proportional to the square of V). (And nothing here is linear with respect to time -- that might come in the next problem so I won't talk more about that now.)

    But using
    U = (1/2)CV2, and
    Ploss = V2/R, or rewriting as [itex] V = \sqrt{P_{loss}R}[/itex],
    find the relationship between U and Ploss (i.e. check whether they are proportional to one another). You'll be pleased with the result. :cool:
  9. Jul 22, 2010 #8
    thanks for all your help.
    I have another question that is concerned with RC time constant.
    If you have time could you please give me some pointers with this question
    I dont have any actual values for this so I dont know what to do.

    Attached Files:

    • RC.png
      File size:
      23.5 KB
  10. Jul 22, 2010 #9


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    I suspected a question like that was coming. :rofl:

    You should start by looking in your textbook for a relevant equation (or if you know basic differential equations, you can derive the relationship yourself -- but I'm guessing your textbook has the relationship already in there for you). Then give it a shot at solving the problem. Per the forum rules, we can't help unless you demonstrate that you've attempted to solve the problem (including showing your work).

    (By the way, e is the mathematical constant 2.71828... The natural log of e is 1. The solution to this problem involves "e to the power of something" and also performing natural logarithms.)
  11. Jul 22, 2010 #10
    Hey guys can you explain this formula please? P_loss=Q^2/(C^2 R) I don't know where it comes from/what it means :'(

    The way I went about this question was the the same to find energy but then diverged so:

    -found energy and halved it, rearranged to find charge Q=5.3*10^-3C
    -then used the discharging in an RC circuit formula to find the time taken (t) to reach the above charge.
    -then used deltaq/deltat (8*1*10^-3 - 5.727*10^-3)/0.00168s to find the current I (1.41)
    -then used power dissipation by resistor I^2*R to find power loss

    I came up with a different answer like 1700W. Can you guys tell me what assumptions I have made incorrectly? I know I diverged by trying to find current, but I didn't know any other way to get the dissipation.
    Last edited: Jul 22, 2010
  12. Jul 22, 2010 #11
    q(t)=qmax (1-e(-t/RC) )
    is q_max something???? im pretty sure that this should go down further but i cant think of what else to sub in???
  13. Jul 22, 2010 #12
    thats where the formula comes from
  14. Jul 22, 2010 #13
    I took q_max to be the initial charge stored in the capacitor 8.1*10^-3, and q(t) to be the amount of charge stored in the capacitor at the half-energy point. RC is given in the question. I was solving for t so I could get current to use in my P_loss formula (I^2R.

    I just dunno if I can do it this way (i'm guessing not?), but wanna know why I can't/where am I making mistakes.

    Finding electricity stuff heaps harder than mechanics!
  15. Jul 22, 2010 #14
    how did you know q at half energy
  16. Jul 22, 2010 #15
    dont worry i see how you knew (exactly how I knew) - not sure why it doesnt work, im sure someone will enlighten us.
  17. Jul 22, 2010 #16
    On collecting some rounding errors by using my calculator to solve t i'm getting 1700W as the final answer using an I = 1.41 found from deltaq/deltat. Maybe I should write it all out and scan it in or something so you guys can find what I'm doing wrong?

    edit:yeah hopefully
  18. Jul 22, 2010 #17
    I dont know if I can tell where your going wrong (I could be wrong and you right) I would be interested in seeing what your doing though.
  19. Jul 22, 2010 #18
    Okay, attached is my part for part iii) but as mentioned i did it a different way. Hope someone can find where I have gone wrong (I think I have assumed something that isn't correct).

    -Does the rate of change of current vary with respect to time when discharging from a capacitor? and is that why my deltaq/deltat isn't working? Because they're supposed to be taken in the limit as deltat approaches zero (the instantaneous rate of change) and i've used an average over the 0.00168s period? I'm looking at a graph of current vs time and it appears it doesn't discharge at a linear rate.
    Have I answered my own question why my method hasn't worked?

    Attached Files:

    Last edited: Jul 22, 2010
  20. Jul 22, 2010 #19
    Yeah im thinking what you said could be right. I is C/s how many electrons per 1 sec and when charging or discharging a cap it is certainly not a linear curve. sorry for the late reply - was playing squash - no one on PF reply at this time prbly all americans
  21. Jul 22, 2010 #20


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    Hello louza8,

    Part of your calculations show:
    Check your math. Something is not right with your final value of t.

    Then later you assume that
    But your following calculations assume that charge Q is linear with time t. But I assure you it is not! Charge Q is an exponentially decaying function with time!

    But if you must take this approach (which I don't necessarily recommend doing by the way -- pat666's solution is much easier), you can do it by finding the instantaneous current.

    You already know that

    [tex] q(t) = q_0 e^{\frac{-t}{RC}} [/tex]

    So take the derivative with respect to t,

    [tex] \frac{d}{dt}q(t) = -I(t) = q_0 \frac{d}{dt} \left( e^{\frac{-t}{RC}} \right) [/tex]

    then use that to find the instantaneous current at time t1/2 (but again, you'll have to fix your calculation of this t1/2 value, because there is a mistake in your previous calculation of it, above).

    Finally, plug that into Ploss = I2R, and you'll get the correct answer.

    But as I said before, although this works, there is a much easier way to handle this problem that doesn't involve calculating any derivatives or anything. :wink:
  22. Jul 22, 2010 #21
    Thanks heaps collinsmark, explains it thoroughly.
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