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RC Time Constant

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A circuit with a battery that is in parallel with a resistor R1 followed series by a capacitor, which is in parallel with a resistor R2

    2. Relevant equations

    Time constant = RC

    3. The attempt at a solution


    I do not know if it is correct to calculate RC with 1/Req = 1/R1 + 1/R2 or if I would only use R1 because it is in series with the capacitor. Can anyone explain?
     
  2. jcsd
  3. Mar 6, 2009 #2
    the best way to find out is write the actual equations. Write the Kirchoff law for the entire circuit.....

    u know the definition of time constant??
     
  4. Mar 6, 2009 #3
    I dont see the capacitor charging at all because the charge (current) will continue flowing through the resistor in parallel with it.Imagine that your circuit consisted of just the battery and R1 and a gap in the connecting wires.The wires at each end of the gap along with the gap itself can be considered as a capacitor.What would happen if you now used R2 to bridge the gap?
     
  5. Mar 6, 2009 #4

    George Jones

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    Initially, the uncharged capacitor doesn't even "see" R2, because the uncharged capacitor acts as a short circuit. So, Initially, charge flows into the uncharged capacitor at the rate I = V/R1.

    The rate at which the capacitor charges, if at all, is unequivocally determined by the approach advocated by praharmitra, and by q = VC for capacitors.
     
  6. Mar 6, 2009 #5
    George imagine we had the simplest circuit where the poles of the battery were connected by a single wire.If now the wires are cut in two places and then separated we have,in effect, two capacitors in series.If the gaps are now bridged by R1 and R2 we have the circuit described by phrygian.The difference is that in the original circuit one of the capacitors is large and the other small(for practical purposes negligibly small) whereas in the circuit I describe both capacitors are small.I see that the current is 1/total resistance..By the way,I loved that hilarious thread posted yesterday.
     
    Last edited: Mar 6, 2009
  7. Mar 7, 2009 #6
    I've got agree with George Jones on this one. When the battery is connected, the capacitor acts as a short circuit so the current goes through it. And in steady-state the current goes through R2. But why wouldn't the capacitor charge? It's connected to a battery...

    Phrygian, you can make the circuit simpler by using Thevenin's method.
     
  8. Mar 7, 2009 #7
    I'm not so sure the circuit description is the same as what various parties think it is. Heck, I don't trust it.

    Who's in parallel with who, and what's in series with what? Are all the commas in the right places?

    You would be best advised to provide a drawing or some more concise ,phrygian.
     
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