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Homework Help: RC Circuit Discharge

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

    A circuit consists of 3 components connected between 2 terminals: A 6 V source, a 60 ohms resistor and a 0.02 F capacitor. The capacitor is initially uncharged. The voltage source is turned on. What is the current flowing through the above capacitor after a very long time?

    2. Relevant equations

    I(t) = E/R *e^(-t/time constant)
    time constant = (R)(C)

    3. The attempt at a solution
    time constant = 1.2 s

    Because the time is a long time, I thought that I wouldn't have to enter anything in for time. I then used the I= E/R*e^(-1/time constant) to solve for the current. I got I = (6 V/60 ohms)*e^(-1/1.2s) = 4.35E-1. I'm pretty sure I'm wrong here so any help to tell me what I am doing wrong would be greatly appreciated.
  2. jcsd
  3. Jul 1, 2008 #2


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    Hi purduegirl,

    But you did enter something here for time--you entered 1, so you found the current at 1 second.

    When they say the time is a very long time, what does that mean? Or alternatively, what is [itex]e^{-t/(\mbox{time constant})}[/itex] for very large t (if you plot it, for example)?
  4. Jul 1, 2008 #3
    They didn't give a specified time for " a long time". That's why I'm not sure what they want here.
  5. Jul 1, 2008 #4


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    No, there's not a specified number; that's not what they want here.

    But try answering my second question; the problem is understanding the behavior of the exponential function, which is important for understanding RC circuits. Make a plot of [itex]\exp(-t/1.2)[/itex]. What happens to the value of that as t gets larger and larger?

    Once you know what happens to the exponential function, you'll know what happens to the current, since it's just a constant multiplied by the exponential function.
  6. Jul 1, 2008 #5
    If this is a series circuit, remember that capacitors are technically breaks in the circuit. After a long time, the circuit is in steady state, and the current across branches involving capacitors can set to 0A. When referring to a long time, take the limit as t approaches infinite.
  7. Jul 2, 2008 #6
    So from what I just read in my textbook, long after the switch is closed, the potential difference across the capacitor is nearly equal to the emf and the current is small. Looking at a graph of current as a function of time, I can see that the longer the circuit is flowing, the smaller the current gets.
  8. Jul 2, 2008 #7
    Unforunatly, the only exponential examples my book gives are for finding the voltage across the capacitor (0.632*Voltage of Battery) and for finding the current when the time equals the time constant (0.368*(Voltage of Battery/Resisitivity). I tried the last equation to no avail.
  9. Jul 2, 2008 #8


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    That's right; and as t gets larger and larger (goes to infinity), what would you say the current is? That is the answer they are looking for here.
  10. Jul 2, 2008 #9


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    What could they mean by 'after a very long time'? There's no number to put in corresponding to a 'very long time' is there? I think you've just stated the answer, metaphorically.
  11. Jul 2, 2008 #10
    Thanks to all! My prof gave us homework on RC circuits, but never lectured over it. Thanks so much for the help!
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