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Simple question about RC circuits

  1. Aug 12, 2009 #1
    can anyone show me the derivation of an expression for current in a series RC circuit with an AC source? I have seen many for DC sources simply using KVL is it the same for AC?

    I would imagine that a two part solution is required: 1) homogeneous solution 2)using undetermined coefficients as the source is equal to 50cos(100t)

    Is the derivation the same as for a DC source?
     
    Last edited: Aug 12, 2009
  2. jcsd
  3. Aug 13, 2009 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. Are you familiar with the relationship between current and voltage for a capacitor? That is where the DiffEq aspect of the circuit comes in. Write the KVL equation for the loop, using the simple V=IR for the resistive drop, and the differential for the voltage drop across the capacitor. Then solve the DiffEq...
     
  4. Aug 13, 2009 #3
    i am familiar. i have seen a lot of examples for DC ccts, but not AC. following your suggestions i would make this:

    E=VR+VC
    E=iR+q/C

    E=dq/dt R + q*1/C

    or

    d/dt E= di/dt R + I/C

    where E=50cos(100t).

    would i solve the first order homogeneous ( could probably just separate) first then add on the solution from undetermined coefficients?

    on a second note the initial condition i have is q(0)=0 but the question wants an expression for loop current. can i solve for q(t) then divide through by C and then divide through by Z=R-jX?
     
  5. Aug 13, 2009 #4

    berkeman

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    That's not quite the equation I meant. For a capacitor, the following equation expresses the relationship between voltage and current:

    [tex]i(t) = C \frac{dV(t)}{dt}[/tex]

    Try re-writing the KVL with that for the capacitor current and voltage...
     
    Last edited: Aug 13, 2009
  6. Aug 13, 2009 #5
    d/dt E= di/dt R + 1/C int[ i(t) dt ] ?
     
  7. Aug 13, 2009 #6

    berkeman

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    Almost, but you mixed a couple steps together...

    It looks like you wrote the correct equation for the voltages in the series circuit, and then went to differentiate to get rid of the integral term on the right. But you only show the differentiation of the left hand side (LHS) and the first term on the RHS, but you haven't differentiated the integral yet to get your final DiffEq...
     
  8. Aug 13, 2009 #7
    i'm confused why i want to replace a current expression with voltage? integrating the expression you gave me will produce an expression for capacitor current i believe.
     
  9. Aug 14, 2009 #8

    berkeman

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    Staff: Mentor

    (keep in mind that I could be wrong, but...) Do you see why I think your equation is not correct yet? Why did you differentiate the LHS term and the first RHS term, but not the 2nd RHS term?

    And as for solving for current in a series RC circuit, I thought that's what you were asking to do?

     
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