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Another DC Circuit (But more complicated with a SWITCH and Capacitor!)

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    For the circuit shown below, the capacitors are initially uncharged. At t=0, the switch S is closed.

    a) Determine the current in each resistor immediately after the switch is closed.
    b) Determine the current in each resistor a very long time after the switch is closed.
    c) Determine the voltage across the capacitor a very long time after the switch is closed.

    After the switch has been closed for a very long time, it is reopened.

    d) Determine the initial current I0 through the 100 ohm resistor, as a function of time after the switch is reopened.

    e) Determine I(t), the current through the 100 ohm resistor, as a function of time after the switch is reopened.

    f) How long after the switch is reopened does the charge on the capacitor fall to 10% of its fully charged state?

    [​IMG]



    2. Relevant equations
    V=IR
    I through a capacitor is I = Cdv/dt = Q/C
    Kirchoff's Loops Rules

    3. The attempt at a solution

    a) Immediately after the switch is closed, the capacitor behaves ls a short circuit and hence no current flows through the 100 ohm resistor.

    I in the 200 ohm resistor = V/R=15V/200ohms=.075A

    b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors.

    Left Loop Clockwise:
    15V-200ΩI1-100ΩI1=0
    Right Loop Counter clockwise: Q/C-100ΩI1-0

    I1 is the only current passing through both the resistors

    I= 20A

    as for the rest of the parts, I want to make sure I get A and B right before I move on.
     
  2. jcsd
  3. Apr 10, 2010 #2

    ehild

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    I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A?

    ehild
     
  4. Apr 11, 2010 #3
    I = C dv/dt is correct.

    But I = Q/C is not. Don't use it. V = Q/C is the correct equation.
     
  5. Apr 11, 2010 #4
    I solved the top equation and that's what I got. Did I label my currents right?
     
  6. Apr 11, 2010 #5

    ehild

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    Do you mean 15V-200ΩI1-100ΩI1=0? That is 15 V = 300Ω * I1. Do you mean I1= 20 A?

    ehild
     
  7. Apr 11, 2010 #6
    Yeah, that's what I meant.
     
  8. Apr 11, 2010 #7

    ehild

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    Amazing. Tell me please, how much is 300*20???

    ehild
     
  9. Apr 11, 2010 #8
    A lot, 6000. Which, is wrong.

    I'm asking for your help because I don't understand what I'm doing wrong, ehild.
     
  10. Apr 11, 2010 #9

    ehild

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    You have an equation. 300 I = 15. This equation is correct.
    Solve it for I. Is not it 15/300 A?

    ehild
     
  11. Apr 11, 2010 #10
    Oh. Now I see what I was doing wrong. Simple Math error. I = 0.05A
     
  12. Apr 11, 2010 #11
    Moving on to part C now.

    When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor.

    V= IR=(0.05A)(100Ω)=5V
     
  13. Apr 11, 2010 #12
    That's right. After a very long time, the capacitor becomes essentially an open circuit.
     
  14. Apr 11, 2010 #13
    Thanks Melawrghk. :)
     
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