Response of a Series RLC Circuit

Thread Starter


Joined 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.
{v_c} = \frac{1}{c}\int_0^t i \,dt \\
\to 30 - 10{e^{ - 20t}} + 30{e^{ - 10t}} = \frac{1}{1000c}\int_0^t {40{e^{ - 20t}} - 60{e^{ - 10t}}\,dt} \\
It comes close to 200 micro F but not equal.

I am not sure how to get L


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Joined 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}}

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}

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,
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