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Kirchoff's Loop Rule Involving 2 Resistors and a Capacitor

  1. Feb 11, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    A 5μF capacitor is connected in parallel with a 2MΩ resistor and the combination is then connected in series with a 1MΩ resistor through a switch to a 12V battery. Find the time taken for the voltage across the capacitor to rise from zero initial value to 2V after the switch is closed.


    From what I understand this requires Kirchoff's loop and junction rules which is okay up until I hit the voltage across the capacitor which is where I get stuck.


    Here's the diagram I made to work from, hopefully it's right:
    yV9S3QY.jpg
     
  2. jcsd
  3. Feb 11, 2016 #2
    Vc(t)=Forced response + Transient response
    forced response"steady state response" is found at time equals infinity at which the capacitor becomes open circuit
    here it is : 12*(2/3)=8 volts .... voltage division
    ____________
    transient response (after solving a differential equation) is : A*exp(-t/taw) where "taw" is the time constant RC
    then you can find A by substituting the initial condition which is the initial voltage of the capacitor at time t<0
    here A is -8 volts
    ___________
    now: Vc(t)=8 - 8*exp(-t/taw)
    and you can easily find the time at which the voltage reaches 2 volts by sitting Vc(t)=2
     
  4. Feb 11, 2016 #3
    Thanks for the help, I was going to go around it that way but my issue was that taw = RC and I'm not sure what taw is since there's a resistor in series and one in parallel
     
  5. Feb 11, 2016 #4
    R is the parallel combination of R1 and R2
     
  6. Feb 11, 2016 #5

    gneill

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    Staff: Mentor

    Hint: Think Thevenin equivalent.
     
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