Response of a Series RLC Circuit

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

  1. p75213

    Thread Starter Member

    May 24, 2011
    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 />
    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


    Mar 3, 2011

    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,
    Last edited: Dec 3, 2011
    p75213 likes this.
  3. p75213

    Thread Starter Member

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