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RLC circuit phasor form

  1. Sep 17, 2011 #1
    Consider the circuit show below, where [tex] V1 = V_{peak}sin(\omega t - 36^\circ) [/tex]

    http://img195.imageshack.us/img195/9687/rlc.png [Broken]

    I wish to find an expression for I in the time domain and in phasor form and then sketch the magnitude/phase vs frequency plots. I already managed to get a solution but it seems way too complex to be able to sketch the above mentioned plots from it. So I'm just going to list my solution and perhaps someone will be kind enough to suggest other ways to tackle this problem (if there are any).

    to find i(t):
    [tex] V1(t) = i(t)R_1 + i(t)\bigg[2 \pi fL || \frac{1}{2 \pi fC} || R_2\bigg] [/tex]

    But [itex] f = \frac{\omega}{2 \pi} [/itex]

    Hence, [itex] V1(t) = i(t)R_1 + i(t)\bigg[\omega L || \frac{1}{\omega C} || R_2\bigg] = i(t)\bigg[R_1 + \frac{\omega LR_2}{R_2 + \omega^2 LCR_2 + \omega L}\bigg] [/itex]

    After substituting for V1 and simplifying, I ended up with:

    [tex] i(t) = \frac{\omega^2 LCR_2 + \omega L + R_2}{\omega^2 LCR_1R_2 + \omega L(R_1+R_2) + R_1R_2}V_{peak} sin(\omega t - 36^\circ) [/tex]

    Which seems ok.

    Now, for the phasor part:
    [tex] v(j \omega) = i(j \omega)R_1 + i(j \omega)\bigg[j\omega L || \frac{1}{j\omega C} || R_2\bigg] = i(j \omega)R_1 + i(j\omega)\bigg[\frac{1}{\frac{1}{j \omega L} + j \omega C + \frac{1}{R_2}}\bigg] = i(j\omega)\bigg(R_1 + \frac{j\omega LR_2}{R_2 - \omega^2 LCR_2 + j\omega L}\bigg) = i(j \omega)\bigg(\frac{R_1R_2 + j \omega LR_1 - \omega^2 LCR_1R_2 + j\omega LR_2}{R_2 + j \omega L - \omega^2 LCR_2}\bigg)[/tex]

    Hence, [tex] i(j \omega) = \frac{R_2(1 - \omega^2 LC) + j \omega L}{R_1R_2 + j \omega L (R_1+R_2) - \omega^2 LCR_1R_2}v(j \omega) [/tex]

    And now I proceded to finding [itex] |\bar{I}| [/itex] and [itex] \phi [/itex]

    Note that [itex] V1 = V_{peak}sin(\omega t - 36^\circ)= V_{peak} \angle{-36^\circ} [/itex]

    [tex] \bar{I} = \frac{\sqrt{R_2^2(1 - \omega^2 LC)^2 + (\omega L)^2}}{\sqrt{R_1^2R_2^2(1 - LC \omega^2)^2 + (\omega L)^2(R_1 + R_2)^2}} V_{peak} [/tex]
    and
    [tex] \phi = tan^{-1}\frac{\omega L}{R_2(1 - \omega^2 LC)} - 36^\circ - tan^{-1}\frac{\omega L(R_1+R_2)}{R_1R_2 - \omega^2 LCR_1R_2} [/tex]

    How would one go about sketching this without the help of any software package? There must be another way to solve this problem which is way simpler. Any ideas? Any help will be greatly appreciated
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 17, 2011 #2

    NascentOxygen

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    In your expression for current, set w to a low value (e.g., w=0) to allow you to determine the current at low frequencies. Set w to a high value (e.g., w -> infinity) to determine the current at high frequencies. The current will reach a maximum or minimum where the circuit is resonant, i.e., close to w = 1/ sqrt (LC).

    Looking at your circuit, when w is low the inductor appears as a short-circuit, so the load is basically R1 to ground. When w is high, the capacitor appears as a short-circuit, so again the load is basically R1 to ground. Your expression for i should support these figures. (It's an easy check.)
     
  4. Sep 18, 2011 #3
    That is honestly what I did for the current magnitude. However, when it came to the phase I just couldn't sketch the whole thing. I simulated the circuit and the phase is just too complicated!
     
  5. Sep 20, 2011 #4

    rude man

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    Unfortunately, the first equations are incorrect. You can't compute the impedance of a resistive and reactive component in parallel by pretending they are both purely resistive or reactive. You have to use complex numbers thruout.

    Look at your expression for i(t). The phase of i with respect to V is independent of w! You know that's not right.

    BTW when you see parallel components it's much easier to use admittance, susceptance and conductance than their impedance duals. So for example I can readily write
    Y = G2 + jwC + 1/jwL for the parallel network, where G2 = 1/R2. Then Z = 1/Y and i = V/(Z + R1)

    (BTW giving you V at a lagging angle of 36 degrees makes little sense. I would ignore it until the very end, then with your i(t) expression you can just subtract 36 deg for the phase angle. Which I think you've already done. That number has no meaning in the context of your schematic unless there is in addition some external time reference defining phase = 0.)

    When you started using complex numbers that part looks OK though I haven't checked your math. If done right your complex expression for current is what you use for getting i(t): compute i(jw) = i1 + j i2, then
    i(t) = √[i1^2 + i2^2]sin(wt + φ), φ = arc tan i2/i1. Minus 36 deg of course.

    The frequency response graph is made complicated by the second-order LC network. As another poster suggested, the parallel circuit looks like infinite impedance at w^2 = 1/LC so your graph of |i|will start flat at i = V/R1, droop down to i = V/(R1 + R2) at
    w = 1/√LC, then pick up & flatten again to V/R1 at high frequencies. I have to confess I forgot the rules on constructing the asymptotes for a resonant network, but you should be able to dig them up somewhere. They involve the damping coefficient ζ and "Q" of the i/V transfer function, both of which are just functions of your component values. Hopefully some kind and informed soul will help us?

    Graph log |i| on the y axis and log w or log f on the x axis.

    For the phase, graph phase in degrees on the y axis and log w or log f on the x axis. Keep track of the signs of i1 and i2. Example: arc tan (-1/2) is not the same angle as arc tan (1/-2.) All logs are base 10.
     
  6. Sep 20, 2011 #5

    rude man

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    Forget the edit. Should be OK as is.
     
    Last edited: Sep 21, 2011
  7. Sep 21, 2011 #6

    NascentOxygen

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    I think you should indicate more clearly exactly what the reader should "forget"? Are you saying the reader should disregard your entire post where you pointed out [apparently] glaring errors in his analysis? If so, perhaps it might be possible to request an administrator delete your posts in this thread.
     
  8. Sep 21, 2011 #7

    rude man

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    What I meant was that my post of 4:40 a.m. on 9/21/11 is OK. Sorry.
     
  9. Sep 21, 2011 #8

    NascentOxygen

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    So you stand by your post stating
    In that case I'm now confused. The OP's post appears not to have been edited subsequent to your criticism, yet in it he provides a formula for phase that involves both w and w2, just as one would expect.
     
  10. Sep 22, 2011 #9

    rude man

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    nascent - yes, his expression for i(t) contains w and w^2 but that is the amplitude, not the phase. His phase is a constant 36 degrees! Phase will obviously vary with w since there are reactive components.

    And if we disregard the 36 degrees, which as I say is a nonsensical spec to begin with, then he has 0 phase shift between V and i thruout 0 < w < ∞ !
     
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