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Finding the Steady State Solution for a RLC circuit

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    An ideal AC voltage source generating an emf V (t) = V0 cosωt is connected in series with a resistance R, an inductance L, and a capacitance C.

    a) Find the steady-state solution for the charge, q(ω,t), which is of the form q0(ω)cos(ωt− δ(ω)).

    b) Find the steady-state current in the circuit, I(ω,t).


    2. Relevant equations
    q0(ω)cos(ωt− δ(ω)).

    3. The attempt at a solution
    From Kirchoff's voltage rule ; q = cVocosωt -cLq(double dot) - cRq(single dot)

    But I am unsure how to get from there to a steady-state solution with the things I am given. I understand that once I get "a" I will need to derive it to get "b".
     
    Last edited by a moderator: Oct 15, 2015
  2. jcsd
  3. Oct 15, 2015 #2

    Mark44

    Staff: Mentor

    I don't think the above is correct. From KVL, you get ##V_R + V_L + V_C = v(t)##, where the left side represents the voltages across the resistor, the inductor, and the capacitor. The right side represents the voltage that is applied to the circuit.

    Based on the KVL, the equation becomes
    ##Ri(t) + L\frac{di(t)}{dt} + \frac 1 C \int_{-\infty}^{\tau = t} i(\tau)d\tau = v(t)##
    If you differentiate with respect to t, you get rid of the integral, producing
    ##R\dot{i} + L\ddot{i} + \frac 1 C i = v'(t)##
    This equation can be rearranged and divided through by L, resulting in the following equation.
    ## \ddot{i} + \frac R L \dot{i} + \frac 1 {LC} i = \frac 1 L v'(t)##
    The above is a second order differential equation that can be solved for i(t), from which you can get q(t) by antidifferentiating, since dq/dt = i.
    ##- cL\ddot{q} - cR\dot{q} + cV_0\cos(\omega t) = q ##
    The question now is, do you know how to solve 2nd order differential equations?
     
  4. Oct 15, 2015 #3

    vela

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    Actually, it looks correct if you assume ##c## is the same as ##C##. Rearranging slightly yields
    $$L \ddot{q} + R\dot{q} + \frac{1}{C} q = V_0 \cos \omega t,$$ which is KVL applied to the series RLC circuit.
     
  5. Oct 16, 2015 #4
    Thanks for the replies. My equation for KVL I wrote,
    q = cVocosωt -cLq(double dot) - cRq(single dot)
    is the same as yours with some rearranging.

    I understand what the steady state represents as a non-transient state of oscillations. But, I am just stuck as to what am I suppose to do to get a steady-state solution from the equation we have above that was obtained from KVL.
     
  6. Oct 16, 2015 #5

    vela

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    As Mark asked you, do you know how to solve a second-order differential equation? That's what you need to do.
     
  7. Oct 18, 2015 #6
    I have looked online and understand the general idea of solving a second-order differential equation, but I don't know what I am trying to solve for here and how it relates to the Steady-state equation.
     
  8. Oct 18, 2015 #7

    vela

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    Yet in your previous post, you said
    In one breath, you're saying you know how to solve the problem, and in the next, you're saying you don't. It's a bit confusing as to where you're getting stuck.

    When you solve a second-order differential equation, the solution will consist of two parts: ##y_h##, the solution to the homogeneous differential equation, and ##y_p##, the particular solution, which represents the response of the system to the forcing function. In physical systems, the homogeneous solution dies off; that is, ##\lim_{t \to \infty} y_h = 0##. Its contribution doesn't persist, i.e., it's transient. The particular solution can stick around as long as the forcing function does. It represents the steady-state response of the system.

    For example, suppose you have a step-input to an RC circuit where the capacitor is initially uncharged. KVL would yield the differential equation
    $$RC \frac{dv_c}{dt} + v_c = V_0,$$ which has the solution ##v_c(t) = V_0 (1 - e^{-t/RC}).## The constant term in the solution is the steady-state solution while the exponential term, which dies off, is the transient.
     
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