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Voltage drop for a charged capacitor

  1. Jul 16, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the voltage drop across the 1000 Ohm resistor when capacitor is fully charged?

    2. Relevant equations
    [itex] V(t) = {V_i}*e^{-t/RC} = {V_i}*e^{-t/Tau} [/itex]

    3. The attempt at a solution
    In RC circuit experiment, initial is 9V and final is 2V. The voltage drop should be difference between final and initial voltages, so 7V but I think this is wrong. If initial and final are given, why do we need decay equation?
     
  2. jcsd
  3. Jul 16, 2015 #2

    TSny

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    Hello and welcome to PF!

    The phrase "voltage drop across a resistor" means the potential difference between a point on one side of the resistor and a point on the other side of the resistor, measured at the same instant of time. In your case, the instant of time is when the capacitor is fully charged.

    So, you need to consider the law that specifies the potential difference across a resistor. Hint: It involves the current in the resistor.

    You haven't described the circuit, so it leaves us a little bit in the dark as to what you are working with.
     
  4. Jul 16, 2015 #3

    mfb

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    If you know the initial voltage drop, finding the initial voltage drop seems to be pointless.
    Please make sure to include the full problem statement, including images if present. The first line in your post is certainly not everything you were given.
     
  5. Jul 16, 2015 #4
    This is full problem statement sorry, no pictures given, nothing. @TSny so I use V=IR to find potential difference?
     
  6. Jul 16, 2015 #5

    TSny

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    That would be one way, assuming that you can determine the current at the time of interest.
    But, there could be other ways such as using what you know about the sum of all potential drops around a closed loop.

    There must be more to your problem than what you stated. Can you describe the problem in more detail? Are you dealing with a single circuit loop with a resistor and a capacitor in series? Is there a battery in the circuit? Is the capacitor being charged or is the capacitor being discharged? When you refer to 9V, what voltage is this? Is it the voltage across the capacitor when the capacitor is fully charged? What does the 2V refer to?
     
  7. Jul 16, 2015 #6
    Setup is single circuit loop with resistor and capacitor in series. 9V is the voltage across capacitor when it is fully charged. 2V is final voltage, and it is almost done discharging.
     
  8. Jul 16, 2015 #7

    TSny

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    If the only circuit elements in the loop are the resistor and the capacitor, then there is a simple relation between the voltage drop across the resistor and the voltage drop across the capacitor. That should make it easy to answer the question.
     
  9. Jul 16, 2015 #8

    mfb

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    How do you know that, if the single sentence in the first post was the full problem statement?
     
  10. Jul 16, 2015 #9
    It's RC circuit experiment.
     
  11. Jul 16, 2015 #10

    mfb

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    ... and the 2 and 9 Volts?

    Sorry, I'm out of this thread, getting the relevant information we need to help you (at least the full problem statement) is way too annoying.
     
  12. Jul 16, 2015 #11
    All info in first post is what I'm given.
     
  13. Jul 17, 2015 #12

    CWatters

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    I agree with the others. The problem statement is wrong and/or missing some info.

    The only way I can make even partial sense of it all is to assume that the capacitor starts off at 9V and discharges through the resistor until it reaches 2V when the experiment stops.

    Capacitors in isolation don't have a concept of "fully charged". You can charge them until their breakdown voltage is exceeded. Capacitors in a circuit are considered "fully charged" when the circuit is unable to put any more charge into the capacitor. For example it's hard to charge a capacitor over 9V if the maximum voltage available in the circuit (the power supply) is 9V. It can be done but not with a simple RC circuit.

    So without more info you can only deduce that the capacitor starts off fully charged at 9V and that's the same voltage across the resistor. But that makes the question trivial.

    If your teacher will be giving you the answer I'd be interested to see it.
     
    Last edited: Jul 17, 2015
  14. Jul 17, 2015 #13

    CWatters

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    Is the question one of a series of questions that all refer to the same circuit?
     
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