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Calculate the time constant (RC circuit)

  1. Feb 3, 2006 #1
    Hi.
    How am I supposed to calculate the time constant for the enclosed circuit diagram.
    I know time constant t = 1/RC
    In this case do i take only one resistance or the combined value (series value)
    Assume any values. I only want to understand how it is to be done.

    PS: Sorry for the stupid drawing. I used MSPAINT to draw it. I think I should have practised drawing more as a kid. :)
     

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  2. jcsd
  3. Feb 4, 2006 #2

    Hootenanny

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    I can't see the picture but as far as I know you use the combined resistance of the timing circuit using the rules for the sumation of parallel and series resistances.
     
  4. Feb 4, 2006 #3
    Ok i will try to draw it in ASCII.
    __res______
    | | |
    | | |
    bat cap res
    | | |
    |____|____|
    res is resistor
    cap is capacitor
    bat is battery (+ve terminal up)
     
  5. Feb 4, 2006 #4
    Obviously I failed :(
     
  6. Feb 4, 2006 #5

    Hootenanny

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    If I guessed right, the resistor is on the top line between the battery and the capacitor and the resistor is on the leg to the right of the capacitor??
     
  7. Feb 4, 2006 #6
    Yes. You got it right.


    Why is the attachment not being cleared. It's been nearly 24 hrs since I posted it. Anyone listening?
     
  8. Feb 5, 2006 #7

    Hootenanny

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    So now you calculate the total resistance of the circuit. Remember one is in parallel while the other is in series.
     
  9. Feb 5, 2006 #8
    How? Could you show me?
    I know R(series) = R1 + R2
    R(parallel) = 1/R1 + 1/R2
     
  10. Feb 5, 2006 #9

    Hootenanny

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    Correct. You want to know the resistance acting through the capacitor. Think about it logically, if one resistor is in series with the capacitor and one is in parallel with it then could you write an equation of the total resistance acting through the circuit base on what you just said?
     
  11. Feb 6, 2006 #10
    I don't know how? Should I apply Ohm's Law and solve for an expression with time variation of current? But that would be quite lengthy.
     
  12. Feb 6, 2006 #11

    Hootenanny

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    No. Simply use [tex] R_{total} = R_{series} + \frac{1}{R_{parallel}} [/tex]
     
  13. Feb 6, 2006 #12
    So the ans is R1 + R2 + (R1*R2)/(R1+R2)
     
  14. Feb 6, 2006 #13

    Hootenanny

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    Ive given you the answer to calculate the total resistance above.
     
  15. May 5, 2009 #14
    I'm doing the same problem--how did you get that expression for Rtotal?
     
  16. May 6, 2009 #15
    Witchcraft or maybe prayer?:rolleyes:

    Let the battery have emf Vo, and label the resistors R and r.

    Answer these simple questions:

    a) When Vc = 0, what is Ic?

    b) When Ic = 0, what is Vc?

    How long would it take the current in a) (if it stayed constant)
    to charge (hint!) the capacitor C to the voltage in b)?
    This is the time constant.

    David
     
    Last edited: May 6, 2009
  17. May 6, 2009 #16
    Here's a better presented problem.
    (Ignore the random stabs at answers)

    Having said that, the first question should read:

    1) Determine the current in each resistor and charge on the capacitor
    immediately after the switch is closed


    https://www.physicsforums.com/showthread.php?t=312098

    In addition to what you are asked for, answer this:

    What is the time constant during the charging process (switch closed)?
     
    Last edited: May 6, 2009
  18. May 6, 2009 #17

    rl.bhat

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    Let R1 is R series and R2 is the resistance in parallel with capacitor. You can interchange R2 and C with affecting the circuit diagram. At the instant the switch is on, the voltage across R2 is E*R2/(R1 + R2). This voltage charges the capacitor. There is no resistance in series with the capacitor. Since the time constant t = RC ( not 1/RC as you have mentioned.) and R = 0 , the time constant is zero.
     
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