LC Resonance Tank Circuit

Discussion in 'General Electronics Chat' started by peter_morley, Jul 22, 2011.

  1. peter_morley

    Thread Starter Member

    Mar 12, 2011
    Under the volume II AC section of the e-book their is a chapter devoted to the electric pendulum which is an LC circuit that is exposed to a DC source then cut off. I am confused about this sentence.

    "If either the capacitor or inductor starts out in a charged state, the two components will exchange energy between them, back and forth, creating their own AC voltage and current cycles."

    I did some rough calculations on paper and it didn't make sense to me. When the LC circuit gets exposed a DC voltage the capacitor should have the same voltage drop as the source and no current going through it eventually. While the inductor should have zero voltage drop across it and all the current going through it eventually. I am making these conclusions based on the fact that the circuit is in steady state with DC voltage applied to it. Well when the DC voltage is popped off we are left with just an inductor and capacitor in series now. So if an inductor resists changes in current shouldn't the instantaneous current going through the inductor stay at its max current when the DC voltage is cut off? The example says the current going through the inductor is at 0 when it starts oscillating.

    So my question is in order for their to be oscillation in a tank circuit should only one component (L or C) be charged in order for oscillation to occur? And if both are charged then oscillation can't happen?

    Thanks in advance!
  2. SgtWookie


    Jul 17, 2007
    If either or both components were charged, there would still be oscillations.

    Capacitors and resistors are basically reciprocal functions.
    Capacitors resist changes in voltage.
    Inductors resist changes in current.
    Voltage is electrical pressure.
    Amperage is electrical current flow.

    If you place a voltage across an inductor that has no current flowing through it, the current flow through the inductor will increase.
    If you place a current across a capacitor that has zero volts potential across its' leads, the voltage across the capacitor will increase.

    So, let's say you have a capacitor with no voltage across its' terminals, and an inductor with zero current flowing through it, connected together.

    Then you connect a voltage across the capacitor for an instant to give it a charge. At time zero, current is not flowing through the inductor, and now the capacitor has a charge.

    The voltage pressure in the capacitor starts current flowing in the inductor. Even as the voltage level in the capacitor drops, the current flow through the inductor continues to build until the voltage across the capacitor reaches zero.

    At that point, you have maximum current flow in the inductor, and no voltage on the cap.

    The current flow through the inductor charges the capacitor in reverse polarity to what it was originally. If both the inductor and capacitor are "ideal" with no parasitic properties, the capacitor will eventually charge to the same level it started off with, but opposite in polarity; and at that point the current through the inductor will be zero; the building voltage in the capacitor started the slowing of the current, and eventually caused it to reverse direction.

    The current flow in the capacitor and in the inductor are equal, but 180º out of phase with each other.
    Last edited: Jul 22, 2011
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  3. peter_morley

    Thread Starter Member

    Mar 12, 2011
    Thanks for the explanation but still a time interval of an "instant" doesn't make sense to me. Do you mean an instant in that the inductor does not have enough time to reach max current? I feel like I am not understanding how an inductor works properly. An inductor resists against current change by dropping a voltage in repulsion to the current. But again I would think the inductor would act as a short even if a dc voltage was attached for even an "instant".
  4. praondevou

    AAC Fanatic!

    Jul 9, 2011
    R represents wire/coil resistance. Voltage is max. at the moment where the coil is connected. then the voltage drops, and the current increases to the point where it's limited only by the resistance R and the internal resistance of the voltage source.

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  5. peter_morley

    Thread Starter Member

    Mar 12, 2011
    Ok so you are saying an instant is less than the time it takes for the current going through the inductor to be maxed out? I'm getting confused probably just thinking too hard.
  6. Wendy


    Mar 24, 2008
    Instant in this case is a fixed unit of time. A L C circuit follows the same math as a pendulum, where gravity and inertia create a back and forth action. Inertia and gravitational potential swap back and forth for a pendulum.

    The actions of capacitors and inductors react much the same way, with the capacitor giving the inductor it's charge, then the inductor returning it. Back and forth, with resistance eating the energy bit by bit until there is nothing there. But while it is happening, the frequency (as with the pendulum) is fixed.
  7. Adjuster

    Late Member

    Dec 26, 2010
    There is no room for doubt that LC circuits can resonate. This has been observed in practice innumerable times, at least back to the late 19th Century. It is also the basis of very many functional systems, like the tuning circuits used extensively in radio. In addition, it has been described by mathematics which can give good predictions of the practical results.

    Although it is physically impossible to charge a capacitor instantaneously, it is very common in calculations and simulations to assume certain states of charge on capacitors (or of currents in inductors) at some point in time. When this is done at t=0, it is sometimes referred to as setting initial conditions. We don't worry about how these states were achieved, we just follow the progress of the circuit conditions from that time onwards.

    I think it may also be worth pointing out that one statement you have made is not correct:
    The fact that an inductor resists current change means that it CANNOT behave as a "short" instantaneously: if a voltage is applied suddenly, the current through the inductor can only build up gradually.
  8. peter_morley

    Thread Starter Member

    Mar 12, 2011
    Adjustter you are right but I think what I was trying to say was that neither can the capacitor be fully charged up instantaneously. I wanted to discuss the transient response of when the voltage is initially applied. The voltage across the capacitor is increasing at its max initially so the current through the capacitor is i = c*dv/dt which means the current is at its max initially. The current through the inductor is increasing at its max initially so the voltage across the inductor is v = L*di/dt which means the voltage is at its max initially. Voltage across the cap eventually stops changing thus 0 current through cap. As does the current through the inductor stops changing thus 0 voltage drop across inductor.

    Wow I have a headache now but what I understand from Adjuster is that for this circuit we are given an initial condition. This condition describes a capacitor being charged up and then being attached to an inductor which is not charged. If I am right in this assumption could this circuit be described as taking a cap and just charging it up and then removing it from the circuit. Then taking that capacitor and hooking it up in series with an inductor to complete a circuit? I need to understand this its blowing my mind how I can't figure this out.