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KCL and First order circuit theory

  1. Nov 15, 2015 #1
    My question applies to the case when the switch opens. By applying KCL in order to get a first order diff equation, the following problem arises when I choose different current directions (which shouldn't happen because KCL says the current direction doesn't matter because it will be fixed after all is said and done.

    So the first one (NODE A) has

    v/(RC) + dv/dt = 0 whose solution is $$Ke^{^{\frac{-t}{RC}}}$$

    while the next case (NODE B) has
    v/(RC) - dv/dt = 0 whose solution is $$Ke^{^{\frac{t}{RC}}}$$

    I know the correct form is the first one, with a -t for the argument but doesn't a KCL "ignore" current direction and should then produce the solution, irregardless of direction?

    Am I missing something?
     

    Attached Files:

  2. jcsd
  3. Nov 16, 2015 #2

    Svein

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    Yes. When you have only a capacitor and a resistor in the circuit, the capacitor voltage drives a current through the resistor. This current reduces the capacitor charge.
     
  4. Nov 16, 2015 #3
    r I understand your point. Because the form tells me I have a decaying voltage, I need to have a negative argument to e. I need to apologize for not making my question very clear. I know the first form is correct (Node A) but is the second form equally valid but incorrect? And because they are equally valid, the only reason to choose the method of Node A vs. the method of Node b is because the argument of e should be negative?
     
  5. Nov 17, 2015 #4

    Svein

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    Your current arrows are correct in Node B (and wrong in node A). Your equations are shaky.

    Consider: With no external source, the current through R is sourced by the voltage across the capacitor. So in a time interval Δt, the current I removes a charge Q = I⋅Δt from the capacitor. Combine that with I = V/R and Q=C⋅V.
     
  6. Nov 17, 2015 #5
    Can you please explain what are the correct equations for node A and B?
     
  7. Nov 17, 2015 #6

    Svein

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    I thought I spelled it out quite clearly (and if I just show you the formula, a moderator may delete it). That aside:
    [tex] I=\frac{V}{R}=-\frac{dQ}{dt}[/tex] since current leaving the capacitor reduces the charge in the capacitor. Now insert the capacitor formula (Q=C⋅V): [tex]I=\frac{V}{R}=-\frac{dQ}{dt}=-C\frac{dV}{dt} [/tex] and you have your equation for V.
     
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