Response of a Series RLC Circuit

Discussion in 'Homework Help' started by p75213, Nov 29, 2011.

  1. p75213

    Thread Starter Member

    May 24, 2011
    39
    0
    Can anybody help with the attached exercise?
    The answers are R=750 ohms, C=200 micro F, L=25H

    To get R I tried Vc(t)/Il(t) but couldn't get 750 ohms.

    To find C I have used the following equation.
    \begin{array}{l}<br />
 {v_c} = \frac{1}{c}\int_0^t i \,dt \\ <br />
  \to 30 - 10{e^{ - 20t}} + 30{e^{ - 10t}} = \frac{1}{1000c}\int_0^t {40{e^{ - 20t}} - 60{e^{ - 10t}}\,dt}  \\ <br />
 \end{array}
    It comes close to 200 micro F but not equal.

    I am not sure how to get L
     
    Last edited: Nov 29, 2011
  2. Vahe

    Member

    Mar 3, 2011
    75
    9
    p75213,

    to calculate the capacitance, consider the following argument.

    Depending on the direction of the inductor current i_L (t) and the capacitor voltage v_C(t). We have the following relationship

     i_L(t) = \pm C \frac{dv_C(t)}{dt}

    The \pm is necessary since the problem does not tell us if the inductor current is directed into the + or the - side of the capacitor voltage. Either way, this is fine. Now, evaluate the inductor current and capacitor voltage at a convenient time value, for example at t=0. Now the above equation becomes,

     i_L(0) = \pm C \frac{dv_C(0)}{dt} \Rightarrow C = \pm \frac{i_L(0)}{\frac{dv_C(0)}{dt}}<br />

    We can find i_L(0) pretty easily. Just utilize the equations given in the problem statement with t=0.

     i_L(0) = 40 e^{0} - 60 e^{0} = -20 \text{mA}

    Now calculate the derivative of v_C(t)

    \frac{dv_C(t)}{dt} = 200 e^{-20t} - 300 e^{10t} \text{V/s}

    Now calculate \frac{dv_C(0)}{dt}
    \frac{dv_C(0)}{dt} = 200 e^{0} - 300 e^{0} \text{V/s}= -100 \text{V/s}

    Now we can calculate the numerical value of the capacitor
     C = \pm \frac{i_L(0)}{\frac{dv_C(0)}{dt}} = \pm \frac{-20 \text{mA}}{-100 \text{V/s}} = \pm 200 \mu \text{F}<br />

    Of course, the only the positive value makes sense.

    You could get the resistance and inductance values using the characteristic equation for the circuit. The characteristic equation for this circuit is  s^2 + (R/L) s + 1/(LC) = 0 . You can tell from the response that the natural frequencies (solutions of the characteristic equation for s)) of the circuit are at s=-10, -20. These are the factors multiplying t on the exponentials. This means that the characteristic equation must be of the form (s+10)(s+20)=0 in order to yield the correct natural frequencies. If we multiply out the polynomial we get

    (s+10)(s+20)=s^2 + 30s + 200 = 0

    But this has to equal the characteristic equation as well; therefore, we have

    s^2 + (R/L) s + 1/(LC)=s^2 + 30s + 200

    Now we match the coefficients; therefore R/L = 30 and 1/(LC)=200. Now we solve for R and L knowing that the capacitor C=200\mu\text{F}.

    \frac{1}{LC}=200 \Rightarrow L = \frac{1}{200C} = \frac{1}{(200)(200 \mu\text{F})} = 25 \text{H}

     \frac{R}{L} = 30 \Rightarrow R = 30L = (30)(25 \text{H})=750 \Omega

    Hope it helps you out.

    Best regards,
    Vahe
     
    Last edited: Dec 3, 2011
    p75213 likes this.
  3. p75213

    Thread Starter Member

    May 24, 2011
    39
    0
    Thanks for that. I should have got C but I sure couldn't see R and L until you pointed it out.
     
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