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Current as a function of time in an RC circuit

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG


    How to determine the current as a function of time that flows in 10 ohm resistor, 100 ohm resistor, and 15 microFarad capacitor?
    Please help

    2. Relevant equations
    I = V/R
    I_capacitor = I e^(-t/RC)

    3. The attempt at a solution

    I only know the equation V=IR and the capacitor current function. Please help
     
  2. jcsd
  3. Feb 26, 2017 #2

    gneill

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    Staff: Mentor

    If you are studying RC circuits you must have already covered other circuit analysis methods, laws, and formulas.

    You should be able to make some attempt either by writing the differential equation for the circuit and solving, or by determining initial and final conditions and applying the known response functions that govern RC circuits in general.

    I note that your problem doesn't specify a value for the supply voltage. Is this intentional?
     
  4. Feb 26, 2017 #3
    The voltage source is 5V.
    Actually I just want to know how to get the function, not about the value.
    Please guide me :\
     
  5. Feb 26, 2017 #4

    gneill

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    No, show an attempt first. That's the rule here.
     
  6. Feb 26, 2017 #5
    Alright,

    I'll use KVL on loop 1, let the 10 ohm resistor be R1, and 100 ohm resistor be R2
    the current goes through r1 is i1
    the current through r2 is i2
    the current through capacitor is i3

    For loop 1 (assume current is clockwise)
    V - i1 R1 - i2 R2 = 0

    For loop 2 (assume current is clockwise)
    -i2 R2 + (integral i3 dt)/C = 0 (is it correct? im not sure about the positive sign, but I think it is since there will be voltage rise across the capacitor)

    Is it correct?
     
  7. Feb 26, 2017 #6

    gneill

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    I see that you've chosen to write the differential equations. Okay, what you've done so far is fine. You''ve got three unknowns though (i1, i2, and i3) so you'll need three equations. You've got two so far.
     
  8. Feb 26, 2017 #7
  9. Feb 26, 2017 #8
    Please look at my reply above. I forgot to quote you before
     
  10. Feb 26, 2017 #9

    gneill

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    It is true that the potential across the capacitor will increase over time, but this is not what "rise" or "drop" terms refer to in this context. They refer to the change in potential seen when you sum the potential changes around the circuit according to KVL.

    If you do a "KVL walk" around the loop in a clockwise direction (the direction of the current flow) there's a potential rise across the battery, then potential drops across the resistor and capacitor.
     
  11. Feb 26, 2017 #10
    So, the voltage drop refers to the voltage of something - voltage source?
    Is it correct?
    I mean, even if potential across capacitor is increasing, but since the value of it is still less than the voltage source, so we can say that it is voltage drop, right?
     
  12. Feb 26, 2017 #11

    gneill

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    The value of the voltage source doesn't matter. It's the polarity of the change in potential when you "walk over" the component that matters. If you did your KVL walk in the counterclockwise direction (against the current flow in this case) then there would be potential rises on the the capacitor and resistor and a potential drop across the voltage source.
     
  13. Feb 26, 2017 #12
    Oh I see.. Thanks a lot

    P_20170227_084348.jpg
    P_20170227_084251.jpg

    Is this differential equation correct?
    Please help
     
  14. Feb 26, 2017 #13

    gneill

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    Somehow the ##V_1## (source voltage) disappeared from the last two lines of your work. Better check those steps.

    Edit: Never mind! I see that it disappeared with the differentiation. It'll come back again when you supply the initial condition for ##i_2## when you solve the differential equation.
     
    Last edited: Feb 26, 2017
  15. Feb 26, 2017 #14

    gneill

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    @terryds, See my edit to my last post.

    Your differential equation looks fine. You can simplify that first term a bit more; You should see a familiar expression involving resistors when you do.
     
  16. Feb 27, 2017 #15
    It becomes

    (R1*R2/(R1+R2)) di2/dt - i2/C1 = 0

    Let (R1*R2)/(R1+R2) be R_parallel

    And then, I get

    ##i_2 = k e^{\frac{t}{C_1 R}}##

    Is it correct?
    But, how to know the value of k?
     
    Last edited: Feb 27, 2017
  17. Feb 27, 2017 #16

    gneill

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    It's the initial current. Go back to the circuit and find the initial current value for ##i_2##.
     
  18. Feb 27, 2017 #17

    NascentOxygen

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    When t=0, this simplifies to ##i_2 = k##
    So k will be the initial value of ##i_2## before current has had a chance to add charge to the capacitor plates.
     
  19. Feb 27, 2017 #18
    Using current divider, I get

    ##I_{2i} = \frac{VR_1}{(R_1+R_2)^2}##

    So,
    ##i_2 = \frac{VR_1}{(R_1+R_2)^2} e^{\frac{t}{C_1 R}}##
    Is it right?
     
  20. Feb 27, 2017 #19

    gneill

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    No. At time = 0+, what does an uncharged capacitor "look like" in a circuit? What is it equivalent to?
     
  21. Feb 27, 2017 #20
    It is short circuit.

    So i2 initial is zero, since all the current will rather go through the capacitor than r2 right?
    But, it's very funny to say that k is 0
     
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