RLC Circuit qualitative explanation! any1?

Discussion in 'General Electronics Chat' started by sudar_dhoni, Nov 9, 2009.

  1. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    can some one explain me the qualitative aspects of RLC circuit
    i not am satisfied with the vector diagrams i want the physical meaning behind
    say for instance in ac with inductor when current varies induced electric field is set up which opposes the current so there is a voltage drop
    similarly in capacitor it stores charges and they repel the incoming current and also for resistor
    like this i want a physical explanation of RLC series circuit what happens there and also about resonance and q factor

    can u explain in terms of voltage drops in inductor capacitor and resistor
    see the source supplies energy to current or charges,these charges then somehow or the other they have to return to lower energy state so they drop their energy (energy is same as voltage -enegy possesed by unit charge) so they drop their voltage.All i have said till now is in
    my doubt is that how does this happen in RLC circuit

    plz dont explain in terms of vector diagrams as it is not convincing
    i want the physical meaning behind
  2. KL7AJ

    Senior Member

    Nov 4, 2008

    Greetings, Sudar:

    I deal with this all the time when I explain antennas at my amateur radio club meetings....most of these folks are not electrical engineers, so I have to be very "mechanical" about it there, too!

    In the circuit you show, at RESONANCE the circuit is reduced to a simple resistor. In the Inductor the current lags the voltage by 90 degrees; however in the capacitor, the current LEADS the voltage by 90 degrees. The result is that the TOTAL phase shift is perfectly cancelled out (at precisely the resonant frequency) The L and C reactances are precisely cancelled leaving only the resistor to account for.

    For this simple circuit, Q is very simple to account for. It is simply XL/R. (Or Xc/R). XL/R is usually a bit more practical to use, because in the real world its the inductor that places the limit on the Q...because of the wire resistance. All things being equal, an inductor will have greater Q for a greater inductance. Why? As you wind a coil, the resistance goes up proportionally to the lengh of the wire...but the inductance goes up as a SQUARE of the number of turns!

    Anyway...that's a little off the topic. The real matter is this: At resonance, the current is determined ONLY by the resistance and the applied voltage, by Ohm's Law.

    What happens OFF resonance is highly dependent on the total Q of the circuit. If the Q is very high, the current will drop off very quickly as you move away from resonant frequency. If the Q is low (either by having a high R value, or a low L/C ratio) the current will taper off more slowly as you tune away from the center resonant frequency.

    Here's another thing to look at. If you look at the VOLTAGE at the junction of the L and the C, you will find that it will be GREATER than the supply voltage....by a factor precisely equal to the value of Q! This is a very neat and simple experiment to set up and look at with an oscilloscope, or even a good A.C. voltmeter!

    Another thing to keep in mind .....and sometimes diffictult to believe....is that even though the NET phase change in the circuit can be complex as you change frequency, the phase change through the capacitor alone is ALWAYS 90 degress, through the inductor alone, ALWAYS 90 degrees, and through the resistor alone, ALWAYS 0 degrees!

    Things are a bit more complicated, as well, if you look at PARALLEL circuits, where you have two different Q values, the LOADED Q and the UNLOADED Q. The UNLOADED Q is determined by the resistance IN SERIES with the L and C loop, while the LOADED Q is determined by the resistance ACROSS, or in PARALLEL with the L and C. This becomes very important when working with transmitter tank circuits and oscillators.

    Hope I've shed some light!

  3. KL7AJ

    Senior Member

    Nov 4, 2008
    If you imagine a series resonant circuit as being a well-lubricated pendulum, you can model an RLC circuit fairly well, too.

    Pendulum swings to the left, energy stored in the inductor
    Pendulum swings to the right, energy stored in the capacitor.
    Velocity of pendulum....analogous to current amplitude
    Direction of pendulum swing...direction of current flow....swinging left, current flowing into inductor...swinging right, current flowing into capacitor
    Friction of pendulum bearing: R
    Height of pendulum at any time=voltage

    L/C ratio= mass of pendulum
    LC product (period)= length of pendulum

    Q= Mass of pendulum/friction of pivot bearing

    You wll see that maximum current flow (pendulum velocity) occurs at the BOTTOM of the swing, where the stored energy (voltage) is minimum

    Now, you have to be a LITTLE careful about applying this analogy too literally. First, GRAVITY is a FIXED potential, you don't really have an equivalent to "A.C. gravity" When you nudge a pendulum, you're applying an A.C. LATERAL force, which is then translated to a VERTICAL potential, as a result of the pivot. There's no direct electrical equivalent to a pivot.

    (Well, maybe there IS, but it probably requires something like an Op-Amp gyrator or the like!)

    Hope this helps even more!

  4. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    thanks for ur reply
    but what i am expecting is that
    consider first half cyle current flows in one direction and is slowly increasing
    the induced electric field is set up in the inductor which opposes this current(self induced emf)so there will be a voltage drop there
    now in resistor again there will be a voltage drop
    in capacitor it would have deposited charges on the plate which would oppose it now(but actually in capacitor during the start of the cylce there wont be a voltage drop as capacitor would be uncharged)

    that means there wont be a voltage drop initially in capacitor AM I RIGHT?

    ok now consider second half i can explain only what will happen in resistor
    what will happen in inductor which now has no induced electric field unless current flows across and capacitor which is now fully charged
    also the self induced emf is due to induced electric field which is due to changing magnetic field which is due to changing current
    since no current is flowing how can the induced electric field be maximum
    i have these sort of doubts
    i dont want the mathematical expressions and all i require a qualitative physical meaning behind it could some1 explain my doubts like how i expect
  5. steinar96

    Active Member

    Apr 18, 2009
    Well, the thing is it's hard to understand this without mathematical expressions and graphs displaying the voltages over the components.
    The circuit is governed by a second order differential equation. Which means that the rate of change of both the capacitor and inductor contribute to the proplem. And trying to visualize the rate of change of many components all at once and over time.... is quite difficult.
    It's pretty darn hard thing to do to predict the voltage values by simply looking at the circuit.


    check that link out if you want to see a graphical view of what is happening in terms of instantanious quantities.
    Last edited: Nov 11, 2009
  6. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    i already know this site
    i want in detail explanation like i have mentioned above
    because i can explan LR RC only L only C and only R but cant RLC in a physical manner not involving mathematics as it is not convincing
    i want the physical meaning like what i have mentioned previously
  7. Ratch

    New Member

    Mar 20, 2007

    Yes, I can. But first, I know you do not know written English very well. Therefore you should practice capitalizing the first word of each sentence and ending it with a period.

    Electric fields are not associated with inductors unless stray capacitances are involved. Inductors produce magnetic fields.

    A capacitor does not store charge. It stores energy. No matter how much charge accumulates on one plate of a capacitor, an equal amount depletes on the opposite plate for a net charge of zero. The capacitor becomes charge imbalanced, which produces a voltage that opposes the energizing voltage.

    For AC, the inductor and capacitor store and release energy twice during each cycle. During resonance, the storage of the inductive energy coincides with the release of the capacitive energy and vice versa. It is as simple as that. The Q factor is a definition you can easily look up. If you are interested in the relationships between the driving voltage or current and the voltage across the three circuit elements, then you have to accept mathematical presentations including vector analysis.

  8. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    I will improve my written english.
    Thanks for ur advice.
    I Think u r coming to the point
    but the one thing u said about inductors that they dont produce electric fields is false.The develop magnetic field which is changing, since it is changing it also produces electric field (electromagnetic field)
    leave all that.You stated that.....

    A capacitor does not store charge. It stores energy. No matter how much charge accumulates on one plate of a capacitor, an equal amount depletes on the opposite plate for a net charge of zero. The capacitor becomes charge imbalanced, which produces a voltage that opposes the energizing voltage.

    How does this happen ?
    Can u explain in terms of charge movement?Also how can the capacitor energise the current?
    You have mentioned that the capacitor produces opposing voltage or electric field within the plates but current does not flow within the plates it reaches one plate and equal amount of it comes out of the other plate.
    Then how does the voltage drop? I can understand the voltage drop in capacitors? I need to know this to learn further about RLC.