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Homework Help: One more capacitor question

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img231.imageshack.us/img231/7518/physicsdw3.jpg [Broken]

    2. Relevant equations

    Q=Qfinal(1-e^(-t/rc) I=Io(e^-t/rc) V=q/c T=RC

    3. The attempt at a solution

    A.) I think that'd it would be brighter since there is more current through the circuit but have no clue...

    B.) Qfinal is .1 and Io = .2 I thought that T=RC though and I get .5, but the answer in the book RC = to -2....

    D.) I thought I'd work backward for this problem. I found Ceq equal to 1200 uF. Then I did Q = V/C so Q = .0328. I know from the answer ke that Q(t) = .1(1-e^-2t) so then I get .0328 = .1(1=e^-2t) so e^-2t = .672 and then =2t = ln.672 but then i get t = .199 and the answer key says .416...
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 26, 2008 #2
    Let's start with part a. There are a few things you should know about the behavior of capacitors. In a short time, that is, for a time much less than the time constant [tex] \tau = RC [/tex], the capacitor behaves like a closed switch. For a long time, or a time much great than the time constant, it will behave like an open switch. Knowing this, what can you say about the brightness of the bulb immediately after the switch is closed? A long time after it is closed?
  4. Mar 26, 2008 #3
    Immediately after not much would happen and a long time after the bulb would get brighter?
  5. Mar 26, 2008 #4
    If the capacitor behaves like a closed switch in the short time (immediately after the switch is closed)..think about it..
  6. Mar 26, 2008 #5
    I really have no idea, if it acted ilke a closed switch that would mean it would get brighter immediately???? But then wouldn't that mean that after a long time it would start to dim out?
  7. Mar 26, 2008 #6
    You've got it. After a long time the charge begins to build up on the capacitor, which slows the current until the current is zero. Think about Ohm's Law. No current, no voltage drop across the light bulb, right?

    For part b, I dont have a calculator in front of me, but you have the correct equations for the current and charge as functions of time. The charge is gathering at an exponential rate on the capacitor, and the current is decreasing as mentioned above. Does that part make sense?
  8. Mar 26, 2008 #7
    Well i sovled for Qfinal and Io but I can't figure out how to solve for RC. I use [tex] \tau = RC [/tex] so T = 500x1000uF but that's .5 and the answer in the book has it at 2.
  9. Mar 26, 2008 #8
    It has 2 seconds for the time constant? If it does, I would disagree, I get 0.5 seconds for tau as well.
  10. Mar 26, 2008 #9
    yah the answer in the book is I(t)=.2e^-2t Q(t)=.1(1-e^-2t)
  11. Mar 26, 2008 #10
    OHH haha. That's not the time constant, they just simplified. They have:

    [tex] e^{\frac{1*t}{.5}} = e^{2t} [/tex]

    Keep in mind that the time constant tau is in the denominator.
  12. Mar 26, 2008 #11
    oh duh lol cool, thanks for the help, so what am I doing wrong for part d?
  13. Mar 26, 2008 #12
    Hmm, I'm getting .346 seconds. I'll work through it one more time..just a sec.
  14. Mar 26, 2008 #13
    well the answer is .416 so we both suck lol
    Last edited: Mar 26, 2008
  15. Mar 26, 2008 #14
    Ok I found the answer. You won't need the charge equation, you want the current equation. You're going to have a new time constant, do you see where I'm going with this?
  16. Mar 26, 2008 #15
    Ok so I did V=IR 50V=.2R so R = 250 ohm. Then t=RC so the new RC = .25. So I(t)=.2e^-4t. then i'm lost...
  17. Mar 26, 2008 #16
    Well lets be careful about what we're doing here, we need to establish a few constant using Ohm's Law, and an additional constant, the time constant. We want I when the voltage drop across the resistor is 50V, and the initial current, when the voltage drop across the resistor is 100V, and tau.

    [tex] \tau = (.0012F)(500 \Omega) = blank [/tex]

    I at the time Vdrop is 50V: I = V/R = fill in blank

    Find I when Vdrop is 100V using Ohm's Law above.

    Now you should be able to set up your equation, and solve for t.
  18. Mar 26, 2008 #17
    I'm not sure why you are using I(t) to calculate the time it takes for the voltage to drop to 50V.

    You can use:


    You know the new capacitance, you know the voltage drop across the capacitors (which is the same as that of the resistor), so just calculate t.
  19. Mar 26, 2008 #18
    It's the same thing except you're not using Ohm's Law to find constants in the equation.
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