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I have a question about a parallel series RLC circuit; the capacitor is parallel to the {inductor + resistor}. The capacitor is charged at an initial voltage $U_{C,0}$ and the inductor has initially no energy.

I cant include schematics yet, but it would look like that: parallel circuit with

  • on the left a capacitor (voltage Uc across it, the + top, the - bottom)
  • on the right side a resistor (voltage $U_R$ with the + top, - bottom) in series with a inductor (voltage $U_L$ with + top, - bottom)
           ---------------
     |     |             |   |
     |     |             R   |  Ur
     |     |             |   V
 Uc  |     C             |   -
     |     |             |   |
     |     |             L   |  Ul
     V     |             |   V
           ---------------

I wonder what the differential equation is.

Indeed, I believe I get a wrong result...

$U_c=U_r+U_L$ $<=> \frac{d²i}{dt²} + \frac{R}{L} \frac{di}{dt} - \frac{1}{LC} = 0$

And here is a problem since delta is always positive... It should behave like a LC circuit but damped over time.

Therefore, I am lead to believe that my convention for $U_c$ is the problem (for me, the + of the capacitor is like that since it behaves like a voltage source because of its initial charge... maybe this is the problem?).

Any help would be very welcome

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I believe I get a wrong result...

$$ \frac{d²i}{dt²} + \frac{R}{L} \frac{di}{dt} - \frac{1}{LC} = 0$$

I think besides being wrong or right, a result should also be useful. If you arrived at a differential equation for $i$, but have boundary value for $U_C(t)$, that is $U_C(0) = U_{C,0}$, then I wonder if this differential equation is the result you are looking for.

I am lead to believe that my convention for $U_C$ is the problem

The convention is arbitrary as long as you stick to it.

Let's add some currents to the schematic:

           ---------------
     |     |             |   |
     |     V  Ic         R   |  Ur
     |     |             |   V
 Uc  |     C        Irc  V   -
     |     |             |   |
     |     |             L   |  Ul
     V     |             |   V
           ---------------

The indices in the schematic are lower case for readability while in the formulas they are upper case.

As you said

$$U_C = U_R + U_L$$

additionally

$$I_C = C\dot U_C$$ $$U_L = L\dot I_{RL}$$ $$U_R = RI_{RL}$$

from the circuit it looks like $$I_C = -I_{RL}$$

which gives $$U_L = L\dot I_{RL}= L\dot I_C$$ $$U_R = RI_C$$

and with $I_C = C\dot U_C$ and $\dot I_C = C\ddot U_C$, respectively

$$U_L = L\dot I_C = LC\ddot U_C$$ $$U_R = RI_C = RC\dot U_C $$

you end up with a differential equation for $U_C$ which is what you are looking for (I guess)

$$0= LC\ddot U_C + RC\dot U_C - U_C$$


clumsy me taking over, trying to solve this

$$0= LC\lambda^2 + RC\lambda - 1$$

$$0= \lambda^2 + \frac{R}{L}\lambda - \frac{1}{LC}$$ $$\lambda_{1,2}=\frac{-RC\pm\sqrt{(RC)^2-4}}{2LC}$$

with the general solution for $U_C$ $$U_C(t) = a_1e^{\lambda_1 t} + a_2e^{\lambda_2 t}$$

interpreting the different kinds of possible $\lambda_{1,2}$

  • $(RC)^2 = 4$, yielding only one solution from the root $$\lambda_{1,2}=\lambda=\frac{-RC\pm\sqrt{(RC)^2-4}}{2LC}=\frac{-2\pm\sqrt{0}}{2LC}=-\frac{1}{2LC}$$ R,C and L are all > 0, which is why I can conclude that $RC=2$, not $\pm2$. I also say that $\lambda <0$ for the same reason. The general solution $$U_C(t) = ae^{\lambda t}$$ with the boundary value yields $$U_C(0) = U_{C,0}= ae^{\lambda 0} = a$$ hence $$U_C(t) = U_{C,0}e^{\lambda t}$$ With $\lambda <0$ the function decays over time or is damped over time as you put it. So it actually does what you expect it to do. Good?
  • if $(RC)^2>4$ then $\lambda_{1,2} \in \mathbb{R}$, which is the same as the general solution $$U_C(t) = a_1e^{\lambda_1 t} + a_2e^{\lambda_2 t}$$ With two variables $a_1$ and $a_2$, the single boundary value is not enough to determine the two. But I say that $\lambda_{1,2} <0$, because $RC>\sqrt{(RC)^2-4}$. The same reasoning as in the first case applies: $U_C(t)$ is damped over time. Ok?
  • the last case is $(RC)^2<4$ so that $\lambda_{1,2} \in \mathbb{C}$ $$\lambda_{1,2}=\alpha \pm\beta i$$ with the solution for $U_C(t)$ $$U_C(t)=e^{\alpha t}\sin (\beta t) + e^{\alpha t}\cos (\beta t)$$ Only the real part of $\lambda_{1,2}$ influences the damping of the function. If split into real and imaginary part $\lambda_{1,2}$ looks like this $$\lambda_{1,2}=\underbrace{{\frac{-RC}{2LC}}}_{\alpha}\pm\underbrace{\frac{\sqrt{(RC)^2-4}}{2LC}}_{\beta i}$$ Given that $\alpha <0$, the function $U_C(t)$ is also damped in this case. It does some fancy wobbling, but it's still damped.

The conclusion is that no matter how the values are, $U_C(t)$ is always damped.

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  • $\begingroup$ Thanks a lot, that really helped. I came to realize thanks to your comment that I did not use the same current convention, therefore I had my problem. $\endgroup$ – alexanzi Aug 28 '15 at 8:25
  • $\begingroup$ that also solved the damping thing, thanks $\endgroup$ – alexanzi Aug 28 '15 at 8:26
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Your differential equation is formally right but has a small error. For $U_c$ you have $U_c = \frac{q}{C}$. Now if the current $i$ flows out from the capacitor you will have the relationship for the charge: $q(t) = q(t=0) - \int_0^t i(t')dt'$ with $q(t=0) = CU_{C,0}$. The capacitor is discharging with time while the current $i(t)$ flows through the LR sector of the circuit.

After differentiating by time you will get the right result.

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  • $\begingroup$ I'm not sure to understand you correctly. If I differenciate your equation I get $i = \frac{dq}{dt}$ ?? $U_C=U_R+U_L$ \\ $ => \frac{dU_C}{dt}=\frac{dU_R}{dt}+L\frac{d^2 i}{dt}$\\ $ <=>\frac{i}{C}=Ri+L\frac{d^2 i}{dt}$\\ $<=> \frac{d^2 i}{dt^2} + 2\zeta\omega_{0}\cdot \frac{di}{dt} + \omega_{0}^2\cdot i=0$ Here I still get the -1/LC .. $\endgroup$ – alexanzi Aug 27 '15 at 14:57
  • $\begingroup$ after correct equation manipulations you should arrive at a term with positive $\frac{1}{LC}$. $\endgroup$ – kryomaxim Aug 27 '15 at 15:26

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