# Specific differential equation in RLC circuit

I have been studying differential equations in RLC circuits: specifically I am looking at

a generator with fixed EMF $=E$,
a capacitor $C$,
an inductor with inductance $L$ and internal resistance $r$,
and a separate resistor $R$

with the elementary cases accounting for
$q$ (the charge on the capacitor),
$V_c$ its voltage or
$i$ the current flowing through the circuit

For example $$\ddot q+\frac{R+r}{L}\dot q+\frac{q}{LC}=E$$

I've been trying to find such a differential equation for the compound voltage

$$V_{L,r}=V_L +V_r=ri+L\frac{di}{dt}$$

which didn't seem to satisfy the criteria for a "regular ODE": $$\fbox{\ddot V_{L,r}+\frac{R}{L}\dot V_{L,r}+\alpha V_{L,r}=\frac{\alpha r}{L}e^{-rt/L}\int e^{rt/L} \ V_{L,r} \ dt}$$ with $\alpha=\frac{-Rr}{L^2}+\frac{1}{LC}$

I started with trying to express $i$ through $V_{L,r}$ as all relevant voltages are expressed in $i$ (resistor), $q$ (capacitor) and $\frac{di}{dt}$ ($V_{L,r}$). At first through this relation by applying regular ODE properties: $V_{L,r}=ri+L\frac{di}{dt} \rightarrow \fbox{$i=\frac{1}{L}e^{-rt/L} \int e^{rt/L} \ V_{L,r} \ dt$}$, and then replaced in : $E=V_{L,r}+Ri+\frac{q}{C} \rightarrow 0=\frac{dV_{L,r}}{dt}+R\frac{di}{dt}+\frac{i}{C}$ and obtained the aforementioned DE.

Should I be using any other physical relation?

• Could you add a diagram showing the circuit - it's not clear whether you intend for all these components to be in series. – Floris Oct 3 '14 at 0:08

## 1 Answer

It is not clear to me why you want to do such a complicated thing. But if you want to follow this way, a slight easier approach is to resolve for the voltage $V_L$. The KVL for your circuit is $$\tag{1} E=V_L+(R+r)i+\frac{q}{C}$$ Now assuming zero initial conditions you have to express $i$ and $q$ in terms of $V_L$. The current $i$ is easily derived from the constitutive relation of the inductor: $$\tag{2} i(t)=\frac{1}{L}\int_0^{t}V_L(t')dt'$$ while for the (2) the charge $q(t)$ is $$\tag{3} q(t)=\int_0^{t}i(t')dt'=\frac{1}{L}\int_0^{t}dt'\int_0^{t}V_L(t'')dt''$$ substituting (2) and (3) in (1) you have the equation

$$\tag{4} E=V_L+\frac{R+r}{L}\int_0^{t}V_L(t')dt'+\frac{1}{LC}\int_0^{t}dt'\int_0^{t}V_L(t'')dt''$$

Differentiating (4) twice you get: $$\tag{5} \ddot{V_L}+\frac{R+r}{L}\dot{V_L}+\frac{1}{LC}V_L=0$$

• Thank you! Still, what led me consider this problem was that if I could obtain a DE for $V_{L,r}$ in a simple RL circuit ($V_{L,r}+\frac{L}{R+r} \frac{dV_{L,r}}{dt}=\frac{rE}{r+R}$, maybe I could obtain such a DE in a RLC one. – N.E. May 15 '14 at 6:06
• @NizarEzroura - if you believe that this answered your question, please consider "accepting" by checking the little check mark. It is how you recognize the person who took the trouble to write a useful answer... – Floris Oct 3 '14 at 0:09