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Background

In our course, we have covered everything related to RC-RL-RLC circuit, however, all questions were based on the fact that you either have parallel or series connection.

I've not taken any course related to differential equations yet, however, I learned a little bit about it.

I have not taken Laplace, instead, mech-analysis, nodal analysis, and so on. Regarding what we have covered related to RC, RL, and RLC, we have taken the cases when we have natural or step response for these circuits, but they give us the solutions of the differential equations related to these circuits, and we just use the solutions to find the unknown voltages and current. For instance, the general equation that can describe the behavior of RL/C circuits is as follows $$\mathrm{x(t)} = \mathrm{x(\infty)} + \mathrm{x(t_0)}-\mathrm{x(\infty)} \cdot e^{\left(\textstyle -\frac{(t-t_0)}{q}\right)} \text{, where } \mathrm{q} \text{ is the time constant} $$

They teach us the cases in which eventually you can simplify your circuits to have one of four known forms, i.e., one resistor and one capacitor/ inductor, or parallel/ series combination between one resistor, one capacitor, and one inductor.


Problem

In the circuit below, my instructor challenged me to come up with a solution.

enter image description here

Given that \$\mathrm{V_1(0)} = 3 \,\mathrm{V}\$ and \$\mathrm{V_2(0)} = 1 \,\mathrm{V}\$.

How to calculate \$\mathrm{V_2(t)}\$ for \$\mathrm{t} \ge 0\$ ?


Approach

I have attempted to write KVL & KCL and then try to find \$\mathrm{V_1}\$ & \$\mathrm{V_2}\$.

The equations that I know are as follows:

\$\mathrm{I_c(t)} = C \frac{d\mathrm{V}}{dt} \implies \mathrm{V} = \frac{1}{C}\int\mathrm{I_c(t)}\,\mathrm{dt} + \mathrm{V(t_0)}\$

The transient response starts when the voltage is applied at \$\mathrm{t} = 0\$.
What I have tried is \$\frac{d\mathrm{V_2}}{dt} + \mathrm{V_2}-\mathrm{V_1} = 0\$ and I am stuck here.

Unfortunately, I have spent more than a day without any noticeable progress in the question.
So, I seek any hints or tips that can put me on the right path.
Thank you in advance.

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  • \$\begingroup\$ If you haven't any coursework on diff-eq, this means no class on integrals, no class Laplace, and... well... Um. Why are you being asked "to come up with a solution?" Have you been studying without taking any courses? (Worked for me. Might for you.) Perhaps you could explain what you do know, since you say you've covered RC, RL and RLC. \$\endgroup\$ – jonk Apr 5 at 2:40
  • \$\begingroup\$ I have not taken Laplace, instead, mech-analysis, nodal analysis, and so on. Regarding what we have covered related to RC, RL, and RLC, we have taken the cases when we have natural or step response for these circuits, but they give us the solutions of the differential equations related to these circuits, and we just use the solutions to find the unknown voltages and current. For instance, the general equation that can describe the behavior of RL/C circuits is as follows x(t) = x(infinity) +(x(t0)-x(infinity)*e^-(t-t0)/q, q is time constant \$\endgroup\$ – ABDELRAHMAN SAID ABDELRAHMAN Apr 5 at 3:04
  • \$\begingroup\$ So, they teach us the cases in which eventually you can simplify your circuits to have one of four known forms, i.e., one resistor and one capacitor/ inductor, or parallel/ series combination between one resistor, one capacitor, and one inductor.. \$\endgroup\$ – ABDELRAHMAN SAID ABDELRAHMAN Apr 5 at 3:08
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    \$\begingroup\$ (Replacing deleted comment, since I can't edit, and I got my math backwards): I'm going to start by coaching you on the circuits part. Get that done, and then I, or someone else, will coach you on the differential equations part. The voltage on a capacitor is \$i=C\frac{dv}{dt}\$ -- so it looks a lot like the resistor equation, except with a \$\frac{dv}{dt}\$ in there. Then, remember from calculus that \$\frac{d}{dt}(v_1+v_2)=\frac{d v_1}{dt} + \frac{d v_2}{dt}\$. You may have to stretch your brain a bit, but do you think you can write the mesh or node equations for the circuit? \$\endgroup\$ – TimWescott Apr 5 at 15:07
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    \$\begingroup\$ You should end up with a system of 1st-order differential equations, which will need to be turned into a 2nd-order differential equation and solved, or solved as is. It's quite a stretch to ask someone to solve those without having taken a diff-eq class, but I suspect it can be done. \$\endgroup\$ – TimWescott Apr 5 at 15:09
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Perhaps this may help:


$$\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_2}+C_1\cdot\frac{\text{d}\,V_1}{\text{d}\,t}&=\frac{V_s}{R_1}+\frac{V_2}{R_2}\tag{1a}\\\\ \frac{V_2}{R_2}+C_2\cdot\frac{\text{d}\,V_2}{\text{d}\,t}&=\frac{V_1}{R_2}\tag{2a} \end{align*}$$


$$\begin{align*} &\therefore\quad\text{the linear system of two 1st order diff-eqs is:}\\\\ \frac{\text{d}\,V_1}{\text{d}\,t}&=\frac{1}{C_1}\cdot\left(\frac{V_s-V_1}{R_1}+\frac{V_2-V_1}{R_2}\right)\\\\ &=\left[-\frac{R_1+R_2}{C_1}\right]\cdot V_1+\left[\frac{R_2}{C_1}\right]\cdot V_2+\left[V_s\cdot\frac{R_2}{C_1}\right]\tag{1b}\\\\ \frac{\text{d}\,V_2}{\text{d}\,t}&=\frac{1}{C_2}\cdot\frac{V_1-V_2}{R_2}\\\\ &=\left[\frac{1}{R_2\cdot C_2}\right]\cdot V_1 +\left[\frac{-1}{R_2\cdot C_2}\right]\cdot V_2+\left[0\right]\tag{2b}\end{align*}$$


$$\begin{align*} &\therefore\quad\text{where }\:a_{11}=-\frac{R_1+R_2}{C_1}, a_{12}=\frac{R_2}{C_1}, b_1=V_s\cdot\frac{R_2}{C_1}\\\\ &\quad\quad\quad\quad\quad\: a_{21}=-\frac{1}{R_2\cdot C_2}, a_{22}=\frac{-1}{R_2\cdot C_2}, b_2=0, \\&\quad\quad\text{then,}\\\\ \frac{\text{d}\,V_1}{\text{d}\,t}&=a_{11}\cdot V_1+a_{12}\cdot V_2+b_1\tag{1c}\\\\ \frac{\text{d}\,V_2}{\text{d}\,t}&=a_{21}\cdot V_1+a_{22}\cdot V_2+b_2\tag{2c} \end{align*}$$


You can use whatever method you like. But the substitution method and the direct "guess and verify" methods come to mind. Either way, you want the homogeneous and the steady-state parts.

For one possible example, continuing with the homogenous portion:

$$\begin{align*} \frac{\text{d}^2\,V_1}{\text{d}\,t^2}&=a_{11}\cdot \frac{\text{d}\,V_1}{\text{d}\,t}+a_{12}\cdot\frac{\text{d}\,V_2}{\text{d}\,t}\\\\&=a_{11}\cdot \frac{\text{d}\,V_1}{\text{d}\,t}+a_{12}\cdot\left(a_{21}\cdot V_1+a_{22}\cdot V_2\right)\\\\\text{from (1c) above (ignoring }b_2\text{)}, V_2&=\frac{\frac{\text{d}\,V_1}{\text{d}\,t}-a_{11}\cdot V_1}{a_{12}}, \text{then,}\\\\ \frac{\text{d}^2\,V_1}{\text{d}\,t^2}&=a_{11}\cdot \frac{\text{d}\,V_1}{\text{d}\,t}+a_{12}\cdot\left(a_{21}\cdot V_1+a_{22}\cdot \frac{\frac{\text{d}\,V_1}{\text{d}\,t}-a_{11}\cdot V_1}{a_{12}}\right)\\\\&=\left(a_{11}+a_{22}\right)\cdot\frac{\text{d}\,V_1}{\text{d}\,t}+\left(a_{11}\cdot a_{22}-a_{12}\cdot a_{21}\right)\cdot V_1 \end{align*}$$


The above is a 2nd order diff-eq, in \$V_1\$, that can be now solved with the sum of two exponential expressions. With that in hand, the rest isn't hard.

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  • \$\begingroup\$ Are the equations misaligned or just my thought? \$\endgroup\$ – Unknown123 Apr 7 at 3:00
  • \$\begingroup\$ @Unknown123 Don't know. I just edited it a little differently. Does this help in any way? \$\endgroup\$ – jonk Apr 7 at 3:19
  • \$\begingroup\$ Nevermind, probably its just my comfortable style of aligning left if there are multiple distinct equations. Also, I'm curious what is the reason that this got down-voted. \$\endgroup\$ – Unknown123 Apr 7 at 7:17
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    \$\begingroup\$ @Unknown Someone here doesn't like me writing too much. That's all. There are two or three in particular where we just don't get along about that point, though it could be others. It isn't a problem. I write accurately and if there is no specific criticism it just means I'm right and they know it and can't say anything except complain by using a down vote. I don't mind, at all. I actually kind of enjoy it in an odd way because it bothers them being powerless to control me about these policy differences or to say anything undermining what I've written. I enjoy helping and won't stop. \$\endgroup\$ – jonk Apr 7 at 10:30
  • \$\begingroup\$ @jonk thank you for doing it \$\endgroup\$ – Simon Marcoux May 10 at 20:07
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Your question is lacking some important details, so this is guesswork. Because this is schoolwork, this response is intentionally vague.

IF Vs is a DC source, then there are two possible answers, the transient voltage (when Vs is applied and the capacitors charge up) and the steady state voltage (after the caps are fully charged). The steady state solutions are pretty straightforward.

The transient solution for V1 and V2 is (I'm guessing) an equation describing the change in voltage over time. Do you know the equation for how a capacitor's voltage changes as it charges up from a constant voltage source? If yes, you can use that to describe V2 as a function of V1. Then, again, describe V1 as a function of 10 V. Then use the equation for V1 as the voltage source in the equation for V2.

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  • \$\begingroup\$ The required is the transient response when the voltage is applied at t = 0. What I have tried is as follows dv2/dt +v2-v1 = 0, and I am stuck here. Thank you for your consideration. \$\endgroup\$ – ABDELRAHMAN SAID ABDELRAHMAN Apr 5 at 3:17

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