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RC Circuits - R vs. tau

  1. Nov 17, 2007 #1
    I'm working on an RC circuit lab and I can't figure out if my graph of R vs. tau is supposed to go through the origin. According to the equation tau = RC, it seems like the graph definitely should pass through the origin. However, given the definition of tau as the time constant equal to the amount of time it takes for the charge on the capacitator to go from 0 to 63.2%, it seems like tau should never be zero. And don't the wires themselves have some degree of resistance, meaning that R can also never really be zero?

    Thanks for your help!
    J
     
  2. jcsd
  3. Nov 17, 2007 #2
    i am not sure what you mean by the amount of time it take for a charge to on the capacitator to go from 0 to 63.2%.....


    but yes practically it is not possible to get zero resistance unless you get super-resistors.....
     
  4. Nov 17, 2007 #3

    ranger

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    Well in your simple lab setup, you will never get a zero time constant. Your wires have resistance. Even though it is small, it keeps your time constant greater than zero even when you don't have a resistor in your circuit. Keep in mind that nothing happens instantaneously. So even with RC = 0, it doesn't make sense to quantify charge accumulation on the plates as happening in zero time.
     
  5. Nov 17, 2007 #4
    Also, can anyone talk to me about the relationship between V(C), V(R) and total V in an RC circuit? Why is delta V(R) negative when the circuit is discharging and positive when it is charging? What is the relationship between V(C) + V(R) while the circuit is charging, and V(C) and V(R) while the circuit is discharging?
    Thanks
     
  6. Nov 17, 2007 #5

    ranger

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    These questions sound like homework. Why don't you tell us what you think the answers are and then someone will correct if needed.
     
  7. Nov 17, 2007 #6
    ok, sounds good. From what I know, V(C) + V(R) = V(battery), For the results I got for the charging portion of the experiment, this makes sense. The output of the power supply was 5V, and total V(C) + V(R), at mulitple different times = approximately 5V, with the V(C) value, going up as the capacitor stores energy and less current is flowing through the resistor.

    I get stuck on the discharging section. As the switch is opened and the circuit discharges, the capacitor gradually loses it's stored energy and V(C) begins to drop. With V(R) however, the values drop to negative values, near opposite those of V(C). I'm not sure why they go negative. Does this indicate that the current is now moving in the opposite direction?
     
  8. Nov 17, 2007 #7

    ranger

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    Sounds good.
    Yes, you are correct again. Charge is now moving in the opposite direction.
     
  9. Nov 18, 2007 #8
    The reason the current reverses is that when you disconnect the battery, and now (presumably*) connect that end of the resistor to ground, the capacitor is now the source of the voltage, and drives current back through the resistor.

    *If you just open circuit the switch, the circuit will (in theory) remain in a stable state, with no change in voltages, and no current flowing.
     
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