# 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.
$\begin{array}{l}
{v_c} = \frac{1}{c}\int_0^t i \,dt \\
\to 30 - 10{e^{ - 20t}} + 30{e^{ - 10t}} = \frac{1}{1000c}\int_0^t {40{e^{ - 20t}} - 60{e^{ - 10t}}\,dt} \\
\end{array}$

It comes close to 200 micro F but not equal.

I am not sure how to get L

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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}}
$

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,
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.