# Response of a Series RLC Circuit

#### p75213

Joined May 24, 2011
70
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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#### Vahe

Joined Mar 3, 2011
75
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

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• p75213