# How to construct a differential equation from this RLC circuit?

I'm trying to figure out how to construct a differential equation for the natural response of the following circuit and I am having trouble. The capacitor is initially charged with a voltage $v_0$ and I want to solve for $v(t)$, the voltage across the capacitor.

simulate this circuit – Schematic created using CircuitLab

Here is my attempt so far:

By Kirchhoff's laws, $v_L=v_2$, $v=v_1+v_L$, and $i_1+i_2+i_3=0$. The currents and voltages are defined so that $i_1=-\frac{v_1}{R_1}=C\frac{dv}{dt}$, $i_2=\frac{1}{L}\int{v_Ldt}$, and $i_3=\frac{v_2}{R_2}=\frac{v_L}{R_2}$. $$C\frac{dv}{dt}+\frac{1}{L}\int{v_Ldt}+\frac{v_L}{R_2}=0$$ And I cannot come up with an equation the exclusively solves for $v$. Is there even an existing solution to this problem?

• You need to have an initial value for the inductor current as well as the capacitor voltage. – The Photon Mar 31 '18 at 5:18
• It's self-explanatory. The inductor acts like an open circuit at t=0, by the time the capacitor is connected to the circuit. The capacitor is initially charged with $v_0$ as stated above and cannot change abruptly. But, how can this help me form the correct differential equation? – mjtsquared Mar 31 '18 at 5:25
• nowhwere did you say the capacitor was connected to the circuit at t=0, you just stated it's charge at that time. To solve this kind of circuit you need to know the initial conditions of each energy storage element --- both capacitors and inductors. – The Photon Mar 31 '18 at 5:30
• I know that initial conditions are required to solve the differential equation. That's not what I wanted to do; I wanted to construct the correct differential equation, if any. – mjtsquared Mar 31 '18 at 5:44

You just need to add one more equation

$$v_1 = -i_1 R_1$$ $$v_1 = -CR_1\frac{dv}{dt}$$

Thus $$v_L = v - v_1 = v + C R_1 \frac{dv}{dt}$$

Now you can substitute in your integro-differential equation to get an equation with just one variable

$$C\frac{dv}{dt} + \frac{1}{L}\int_{-\infty}^t (v + CR_1\frac{dv}{dt})dt +\frac{v+CR_1\frac{dv}{dt}}{R_2} = 0$$

From here you can rearrange terms and take derivatives until you reach a differential equation in one variable, that can be solved by the usual methods (recognizing a known form, Laplace transform, etc.)

• If I had more patience for Mathjax I'd rearrange the 2nd i-d equation to push the dv/dt back into the integral until the next step where it would fall out again properly. But it's late here so hopefully this is clear enough. – The Photon Mar 31 '18 at 5:50
• I see, but the coefficient to the integral is $\frac{1}{L}$ though. – mjtsquared Mar 31 '18 at 6:04
• I got almost that, but (C R1 / L) not (L C R1) and also v/L and not L v. Oh, well. – jonk Mar 31 '18 at 6:54
• @jonk, probably related to the transcription error that mjt already pointed out. – The Photon Mar 31 '18 at 16:32

Just so you have my approach to consider, as well. (I know you've already selected an answer.) Here's the redrawn schematic which I prefer:

simulate this circuit – Schematic created using CircuitLab

I apply nodal analysis and get these two equations from the two nodes:

\begin{align*} \frac{V_\text{C}}{R_1}+C\frac{\text{d} V_\text{C}}{\text{d} t}&= \frac{V_\text{L}}{R_1}\label{n1}\tag{node V_\text{C}}\\\\ \frac{V_\text{L}}{R_1}+\frac{V_\text{L}}{R_2}+\frac{1}{L}\int V_\text{L}\:\text{d} t&=\frac{V_\text{C}}{R_1}\label{n2}\tag{node V_\text{L}} \end{align*}

Just solve the $\ref{n1}$ equation for $V_\text{L}$:

\begin{align*} V_\text{L}&=V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\label{n3}\tag{solved for V_\text{L}} \end{align*}

, and then substitute that into the above for $\ref{n2}$:

\begin{align*} \frac{V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}}{R_1}+\frac{V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}}{R_2}+\frac{1}{L}\int \left[V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\right]\:\text{d} t&=\frac{V_\text{C}}{R_1} \\\\ \frac{V_\text{C}}{R_1}+\frac{V_\text{C}}{R_2}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{1}{L}\left[\int V_\text{C}\:\text{d} t+R_1\:C\int\text{d} V_\text{C}\right]&=\frac{V_\text{C}}{R_1}\\\\ \frac{V_\text{C}}{R_2}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{1}{L}\left[\int V_\text{C}\:\text{d} t+R_1\:C\int\text{d} V_\text{C}\right]&=0 \end{align*}

, now take everything with respect to the derivative of time:

\begin{align*} \frac{1}{R_2}\frac{\text{d} V_\text{C}}{\text{d} t}+C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d}^2 V_\text{C}}{\text{d} t^2}+\frac{1}{L}\left[V_\text{C}+R_1\:C\frac{\text{d} V_\text{C}}{\text{d} t}\right]&=0\\\\ C\left(1+\frac{R_1}{R_2}\right)\frac{\text{d}^2 V_\text{C}}{\text{d} t^2}+\left(\frac{1}{R_2}+\frac{R_1\:C}{L}\right)\frac{\text{d} V_\text{C}}{\text{d} t}+\frac{V_\text{C}}{L}&=0\\\\ \end{align*}

You can easily put that into standard form and solve using the usual 2nd order diff-eq approach or else use Laplace.