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Solving a parallel LC circuit with DC current

  1. Oct 18, 2015 #1
    Take the following circuit (capacitor initially not charged) :


    I am trying to solve for the current everywhere as a function of time. I set up the following equations (are they right ?) :


    Then, any way I try it, I end up with some messy integrals. Even using wolfram alpha to get those integrals solutions, I can't find out I_0 (current in the bottom wire with the battery), I_C (current in the middle wire with the capacitor), and I_L (current in the upper wire with the inductor) as functions of time.

    I suspect it shouldn't be that messy, I must miss something somewhere. Can anyone solve this for me?

    I understand that :

    - at first (t = 0, battery just plugged) the current must be 0 in the inductor (upper wire) and equal to VR in the middle and lower wires
    - after a long time (t = infinity), current must be 0 in the capacitor (middle wire) and equal to VR in the inductor (upper wire) and in the resistance (lower wire)

    What I am trying to understand is what happens in between t = 0 and t = infinity.
  2. jcsd
  3. Oct 18, 2015 #2


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    Do you know about Laplace transforms?

    The impedance in the upper wire will be: sL.
    In the middle wire: 1/sC.
    In the lower wire: R.

    Now, sketch a model of the circuit and solve the transfer functions as for the currents, when the battery is connected.
    The currents through upper and middle wire will oscillate.

    The input function when you switch on the the battery will be: V/s. ( step function ).
    Last edited: Oct 18, 2015
  4. Oct 18, 2015 #3


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    It would help if you drew in the currents and labeled the potential drops across each element.

    If you differentiate (1) and use (4), you get
    $$\frac{dI_0}{dt} = \frac{1}{RC} I_C.$$ Then use (3) to eliminate ##I_C## to get
    $$\frac{dI_0}{dt} = \frac{1}{RC} I_0 - \frac{1}{RC} I_L.$$ Rewriting (2) slightly gives
    $$\frac{dI_L}{dt} = \frac VL - \frac RL I_0.$$ You can solve this system using matrix methods. Alternately, you can solve the bottom equation for ##I_0## and then substitute the result into the second equation to get a second-order differential equation for ##I_L##.
  5. Oct 19, 2015 #4
    I don't, sadly. Is it really necessary here?

    Hmm, that looks like what I know, if you refer to s as the driving frequency. However, the voltage source here is DC, not AC.

    Are you sure? This is DC current. One thing seems pretty sure, at t = infinity, nothing oscillates. Do you mean there are oscillations before that time? Why? My guess would have been that the current trough the capacitor decreases continuously while it increases in the inductor.

    Thanks. Sadly I don't know yet how to solve 2nd order differential equations. Do the matrix methods you refer need linear algebra? I don't know those as well. But nice to get the right equation to solve!
  6. Oct 19, 2015 #5


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    Then forget about it.
    But anyway:
    S is not a driving frequency. Say you have a time function: f(t). If you Laplace transform it to f(s). Then:

    L(df(t)/dt) = s*f(s)


    L( f(t)dt ) = f(s)/s
    Yes I am. A LC-connenction will oscillate by any disturbance, such as beeing connected to a battery.
    You are right because also the voltage across the LC will oscillate, thus giving some losses in the resistor. But if you disconnect the resistor just after the oscillation has started, the remaining LC-circuit will in principle oscillate forever, because no resistance is included in the coil.
    No, the energy will by turns be stored in L ( E = ½*L*I2 ) or in C ( E = ½*V2*C ). The current/voltage will overshoot all the time.
  7. Nov 2, 2015 #6
    Thank you for the answers, and sorry for the delay.

    Ok. I kind of instinctively, and incorrectly, thought it was different with DC.

    Ok. It's really hard to picture. With simple LC or RLC circuits in series, it's easier to understand. Is there some intuitive way to understand this ?

    Some of my main interrogations are :

    When does the oscillation starts ? Right after current starts to flow, inductance prevents current in the upper part, so does the oscillation really starts just after current is prevented, right when some of it starts to flow through the inductor ?
    How does the battery reacts to this oscillation ? Does it recharge/discharge at each oscillation, while always discharging a little more at each cycle due to heat dissipated by R ?
  8. Nov 3, 2015 #7


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    With an oscillating voltage across C and L there will be that same voltage across R + the DC source. So, yes, this means an alternating current through R and through the voltage source.
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