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Electrical circuit modelling

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data
    I have 3 pairs of resistors and capacitors, with each pair in parallel. These pairs are in series with another resistor and a voltage source. I have been asked to find the voltage across the resistor in the middle pair as a function of time, assuming that at time t=0 the voltage is switched on.

    2. Relevant equations

    3. The attempt at a solution
    I am not sure how to approach this. In all the books i have the only rc circuits analysed are of 1st degree. I think that i need to use thevenin's theorem. Thanks for any help.
  2. jcsd
  3. Jan 2, 2009 #2


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    Staff: Mentor

    Welcome to the PF, Henry. I'll move this to Intro Physics, and try to give you some guidance.

    As with many problems of this type, you should write the KCL equations for the nodes in the network, and solve the problem from there. In this case, since you are asked to solve for the voltage across the resistor in the middle leg, you should only have to solve for two nodes -- the node at the top of all 3 legs (after the voltage source and its associated Rsource), and the node in the middle leg between the bottom of the resistor and its associated cap. The bottom node for all of this just call "ground", and it's the return from all 3 legs back to the voltage source.

    Show us the KCL for these two nodes, and show us your try at solving them for the associated node voltages, and applying the initial condition(s).
  4. Jan 8, 2009 #3
    Hi, thanks for replying.

    If I am right the circuit has 4 nodes. I changed the voltage source into a current source. Here is my attempt at using KCL.
    Node 1: -I1 + I2 + I3 + I4 = 0
    which is -Vsupply/Rc + (V1 - V2)/Rp + C x d(V1 - V2)/dt + V1/Rsupply = 0

    Node 2: I5 + I6 + I7 + I8 = 0
    which is (V2 - V1)/Rp + C x d(V2 - V1)/dt + (V2 - V3)/Rc + C x d(V2 - V3)/dt = 0

    Node 3: I9 + I10 + I11 + I12 = 0
    which is (V3 - V2)/Rc + C x d(V3 - V2)/dt + (V3 - V4)/Rk + C x d(V3 - V4)/dt = 0

    I am not sure if this is the correct way to write the current through the capacitors.

    ps. sorry for the delay in replying, I left my laptop power cable at home, so have been without computer access.
  5. Jan 8, 2009 #4


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    Staff: Mentor

    It's hard to tell without seeing a circuit diagram, with the components labelled. Why did you change the voltage source into a current source?
  6. Jan 8, 2009 #5
    Hi, i have attatched a file containing the circuit diagram, hope it works ok.
    I changed the voltage source to a current source because this is what a book I have says to do when using nodal analysis, so you can apply KCL.

    Attached Files:

  7. Jan 8, 2009 #6


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    Staff: Mentor

    The diagram still shows a voltage source, and yes it shows 4 nodes the way it's drawn. I don't usually convert to a current source -- I just write equations for all nodes in the circuit, including the one between the ideal source and its output resistance (R1 in this drawing).

    Also, if you are dealing with phasors (a constant frequency), you can just use the equivalent impedance for the capacitors, instead of the differential equation relating voltage and current. Are you supposed to solve the full differential equations?
  8. Jan 9, 2009 #7
    Sorry, the circuit diagram I attached was the original circuit which i was asked to analyse.
    I havent been told how to analyse the circuit really. I am a maths student, and am doing a report on the basics of circuit modelling. I was asked to finish the report by analysing this circuit. I haven't covered phasors in my report yet, I was hoping the solution would be fairly simple. So is it not possible to use differential equations for the circuit?
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