V and I of a capacitor in RC circuit

Discussion in 'Homework Help' started by Hitman6267, May 1, 2010.

  1. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Capture1.PNG
    I'm having a little trouble with this so I'm going to explain my thought process.

    Finding V(0) (the question in the attached picture)
    ok so first thing, I know at t=0 the capacitor acts like an open circuit.
    In other exercises usually we do one of two things. Say that the V of capacitor is equal to the voltage source because it has been connected for a long time or say that the V is equal to the voltage of something parallel to it. I can't find how any of them apply here.

    So what's the first step ?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Remove the cap. With the switch open, what will be the potentials at A & B? Place the cap back in circuit, switch still open. Wait several seconds. What will the potentials be at A & B?
     
  3. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    This is my attempt at calculating the potential at A B with the cap removed.and the switch open

    Calculating I debited from the source: I = V/Req = 12/8.5 = 1.41
    Now I need to know the current in the branches so I can calculate the voltages of the resistors.

    Current divider:
    i= 2/3 * 1.41 = 0.94 that's for R1 and R2 = 1.41-0.94= 0.47

    This is where something is off. If I calculate V (using V= RI) of all each resistors they all would have V=2.82. What is my mistake ?
     
    Last edited: May 1, 2010
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    Your figure for It is okay. Remember that the current divides into the parallel branches.
     
  5. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Sorry I didn't get what you said. You're saying that in principle my calculations are correct?
    But they can't be. And how would I be able to calculate VAB from them
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    It's a voltage source with a series resistance with two parallel voltage dividers.

    The voltage dividers have total resistance each of 6+3 = 9 kilohms.

    In parallel you get 4.5 kilohms.
    The voltage across the voltage dividers therefore, is:

    12\frac{4.5}{4+4.5} = 6.35 V

    Or, using your approach:
    I = \frac{12}{8.5} = 1.41 mA\\<br />
x = 12 - 1.41(4) = 6.35 V


    Node A then, is:
    6.35\frac{6}{3+6} = 4.23 V

    Node B:
    6.35\frac{3}{3+6} = 2.12 V

    Voltage across the capacitor is node A - node B

    Alternatively, the voltage dividers take half the supply current:
    1.41 / 2 = 0.7 mA

    Node A will be 0.7 * 6 = 4.2 V
    Node B will be 0.7 * 3 = 2.1 V
     
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  7. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    wow, it's correct thank you :)
    But why is node A 6.35 * 6/(3+6) and not 6.35 * 3(3+6)
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    For Node A we have R2 = 6 kilohms as the output resistor while for node B we have R1 = 3 kilohms as the output resistor.

    Also look at the current version... node A is 0.7 mA * 6 kilohms

    FYI 6.35 / (3+6) = 0.7 mA
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    What is meant by Vab(0+)?

    Does it mean just after the switch closes or after several minutes have passed?
     
  10. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    oh I was looking at the wrong resistors (the top ones) I should have looked at the bottom two.

    Last question for today (I promise :p ) - I can't think any more any way
    They want V(0.001) - same circuit -
    So I'm using this equation:
    V = Is R+ ( Vo - IsR) e^ -(t/RC)
    V= (1)(4)+ (2.12 -4) e^- (0.001/(4*5))

    What is wrong ?

    @mik3 I'm thinking just after the switch closes.
     
  11. mik3

    Senior Member

    Feb 4, 2008
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    If Vab(0+) is just after the switch closes, then Vab=0V because the uncharged capacitor acts a short circuit at the instant the switch closes.
     
  12. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    The 2.12 answer is correct so I don't know where that leaves us. But I don't see why the cap would be uncharged, it is connected to a voltage source (there are 2 voltage sources in the circuit)
     
  13. mik3

    Senior Member

    Feb 4, 2008
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    Ohh yes, I didn't realized that the right voltage source is after the switch.

    Therefore, after a few calculations the answer is Vab=2.12V.
     
  14. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Indeed it is, any idea on my last question ?
     
  15. mik3

    Senior Member

    Feb 4, 2008
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    You have to explain what Is and R represent.

    I don't know if your formula is correct.

    The formula I know is:

    Vc=V[1-exp(t/RC)]+Vo*exp(t/RC)

    Vc=capacitor voltage
    V=voltage source
    RC=time constant
    Vo=initial capacitor voltage
     
  16. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Is is the current debited from source and R is the thevenin equivalent resistor.
    I copied the formula from my book so it's correct.
     
  17. mik3

    Senior Member

    Feb 4, 2008
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    I get an answer of 4V.

    I think you mistake is the calculation of the time constant RC.

    You have RC=4*5 but the correct is RC=4*5*10^-6, see the capacitor value.
     
  18. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    The answer 4 is not correct.
    Here's something that might help.
    1st question they asked. What is the value of iC(0+) in mA ? answer 0.47
    2nd question they asked. What is the value of vab(0+) in V? answer 2.12

    Now we need to calculate Vab(0.001).
    I think my mistake other than the one you mentioned (because that was a typo, I used the correct value in my calculations) is the possibility of Is changing when the switch closes. Does it ?
     
  19. mik3

    Senior Member

    Feb 4, 2008
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    Yes, Is is not the same with the switch open and closed.

    With the switch closed Is increases.
     
  20. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    In that case how can I calculate it ?
    I thought about source transformations but I don't see how I can do that.
     
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