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RL Circuit ODE

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the RL circuit shown in the figure. Assume that the current ##i(t)## has reached a steady state with the switch at position ##A##. At time ##t = 0##, the switch is moved from position ##A## to position ##B##.

    http://imgur.com/dRIOrp0
    If I use the image button, nothing shows. Why is that?

    Find the differential equation relating ##i(t)## and ##v_2## for ##t > 0^-##. Specify the initial condition (that is, the value of ##i(0^-)##for this differential equation in terms of ##v_1##).

    3. The attempt at a solution

    I have come up with
    $$
    v_2(t) = Ri + L\frac{di}{dt} = i + \frac{di}{dt}
    $$
    1. Is this correct?
    2. How do I determine the initial condition?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 12, 2014 #2

    Simon Bridge

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    t=0- is t=0 just before the switch is thrown.
    you are told this is the steady-state situation for a particular circuit - so what is that circuit and what is the steady state?

    at t=0+ the circuit has changed.

    Note: are V1 and V2 supplied by AC or DC voltage sources?
     
  4. Mar 12, 2014 #3
    Those are battery symbols.
     
  5. Mar 12, 2014 #4

    Simon Bridge

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    OK - so battries are DC right?
    So ##v_1(t)=V_1## and ##v_2(t)=V_2## ... so with the switch in it's first position, and the circuit in equilibrium, what is the current?
     
  6. Mar 12, 2014 #5
    With the circuit is position 1, we have ##V_1 = v_1(t) = i + \frac{di}{dt}##, and in position 2, ##V_2 = v_2(t) = i + \frac{di}{dt}##.

    Is that what you are asking. Also, it is DC.
     
  7. Mar 12, 2014 #6

    Simon Bridge

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    Close - focus on the equation for V1 - that situation has existed for a very long time ... long enough for the condition to be at equilibrium. It's a steady-state. What does it mean for a circuit to be in a steady state?

    In a steady-state DC circuit - what does an inductor do?

    If you don't know, then you'll have to solve the equation for V1 - for the condition that it has been switched on at ##t=-\infty## from a state of 0V, then use that to work out i(0). It's a lot of work - it is really easier just to use the physics.
     
    Last edited: Mar 12, 2014
  8. Mar 12, 2014 #7
    Can you explain what steady state means in a DC circuit? I looked it up and it didn't help. I only ask for an explanation since I am not versed in EE. I have a BS in Math and working on a MS in Mech E so this is a little foreign to me.

    Edit:

    If let ##V_1 = 0##, then ##i(t) = Ce^{-t}##. As ##t\to -\infty##, our solution blows up.
     
  9. Mar 13, 2014 #8

    Simon Bridge

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    Well you've managed to look at it three ways ... so I'll cover them from hardest to easiest:
    But ##V_1\neq 0## you are told that ##V_1>0##, and has been like that for all time before t=0.

    Reverse the situation - switching the circuit on an infinite time ago and looking for the situation now is the same as switching it on now and waiting for an infinite time.

    if you went from ##v(0^-)=0## to ##v(0^+)=V_1## at ##t=0##, then what is the current when ##t=+\infty## ?


    It means the state of the circuit is steady" i.e. "unchanging".

    The current and the voltage are not changing.
    If the current does not change - then what is di/dt?
    Rewrite the equation for V1, taking this into account.

    Shortcut:
    Reactive components only come into their own when there is some change in the potential.

    In the DC steady-state case,
    ... an inductor is the same as a length of straight wire and a capacitor is the same as a gap in the circuit.
    ... so redraw the circuit to take this into account: this is the DC equivalent circuit.
    ... evaluate the new equivalent circuit.

    In this problem - at t=0 you apply a step potential difference: the inductor reacts to the sharp change in voltage and the circuit takes a while to settle to a new steady state again.
     
  10. Mar 13, 2014 #9
    So if ##V_1\neq 0## but we are at a steady state, we can assume ##V_1## is some arbitrary constant?

    If the current isn't changing, then ##\frac{di}{dt} = 0##.

    Therefore, we have ##V_1 = i(t)##, correct?

    Then at ##t = -\infty##, we have ##V_1##.
     
  11. Mar 13, 2014 #10

    Simon Bridge

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    ##V_1## is not arbitrary! ##V_1## is the voltage of the first battery.

    Need to be a bit careful here - you have ##R=1\Omega## right?
    The steady state current before t=0 would be ##I=V_1/R## ... because the equivalent circuit is just a battery and a resistor.

    Since R=1 Ohm, you get ##I=V_1## ... if ##V_1## is in volts, then ##I## will be in Amperes.
    Since that is the current just before t=0 was I, then ##i(0^-)=V_1## ... and that is your initial condition.

    Now you can solve the equations for V2.
     
  12. Mar 13, 2014 #11
    Ok so now I understand that piece. Next it says:

    Using the properties of the unilateral Laplace transform, determine and plot the current ##i(t)## for ##v_1 = 4V## and ##v_2 = 0V##. Using your answer, argue that the current ##i(t)## may be expressed as a sum of the circuit's zero-state response and zero-input response.

    If ##v_1 = 4V##, we have the unit step function.
    $$
    4V\cdot\mathcal{U}(-t) =
    \begin{cases}
    4V, & \text{if } t < 0\\
    0, & \text{otherwise}
    \end{cases}
    $$
    If ##V_2 = 0V##, then our DE is
    $$
    0 = i + \frac{di}{dt}\Rightarrow Ce^{-t} = i(t)
    $$
    I am not sure if I have enough information to find ##C##.
    The solution would then be
    $$
    i(t) =
    \begin{cases}
    4V, & \text{if } t < 0\\
    Ce^{-t}, & \text{if } t\geq 0
    \end{cases}
    $$
    At ##t = 0##, is it customary in EE that we don't have a discontinuity? If that is the case, I can find ##C##. Otherwise, we have a plot of 4V for ##t < 0## and a dying exponential for ##t \geq 0##.
     
  13. Mar 14, 2014 #12

    Simon Bridge

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    Except ##V_2\neq 0\text{V}## - it is the voltage of the second battery.
    Why put a dead battery in the circuit?

    If you switch form V1 to V2 at t=0, then the voltage waveform is:
    ##v(t)=V_1+(V_2-V_1)\text{h}(t)## ... where ##\text{h}(t)## is the heaviside step function.

    If ##V_2 > V_1## then it is a step up.

    ... what about the initial condition?
     
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