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Current Through A Resistor That Is In Parallel With A Capacitor

  1. Feb 10, 2017 #1
    1. The problem statement, all variables and given/known data
    uqwUIxE.png

    My question is regarding part C of the question.
    2. Relevant equations
    V = IR
    V(t) = V(1-e^(-t/tau))
    3. The attempt at a solution
    My idea is to use Kirchoff's Voltage Law and find the voltage of the capacitor as a function of time, then since the voltage across capacitor is the same as voltage across resistor I can simply divide that by a constant R and obtain current as a function of time.

    The problem I am running into is: I am unsure what to put as R in the time constant.
    To my understanding time constant is the amount of time it takes to charge the capacitor to about 60%, and from my instinct it does not depend on the resistance of light bulb that is in parallel with the capacitor. Therefore Tau(time constant) = 50*Capacitance.

    However, I am unsure of what I said above, and would like to know if there's a more definitive way to find the R value for time constant. I did see one approach which uses Thevenin's Equivalence but it was very confusing.
     
  2. jcsd
  3. Feb 10, 2017 #2

    gneill

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    Staff: Mentor

    Thevenin is the way.

    Show what you've tried and where it gets confusing.
     
  4. Feb 10, 2017 #3
    I am taking first year physics and Thevenin is not in the curriculum so I'm not sure how to calculate it. I will learn how it works then!
     
  5. Feb 10, 2017 #4

    gneill

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    Staff: Mentor

    Okay. After you've done some research come back with any questions.
     
  6. Feb 10, 2017 #5

    cnh1995

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    Homework Helper

    Thevenin would be the easiest and most practical way of doing this problem. But if it is not in your curriculum, I am not sure if you'll get full credit for this question if it is a part of your assignment or exam. After solving this problem using Thevenin, you might want to try the usual mesh analysis KVL method.

    But the Thevenin method will surely be very useful for you to analyse circuits with increased complexity.
     
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