So, i've been asked to get the differential equations (or model) for this circuit:

RC circuit

V1=15 u(t)V, R1=10 ohm, R2 = 20 ohm, C1=10uF, C2=20uF

I'm just a beginner, and don't know if the differential equations i made are actually ok (I think they are not). The model i came with is this one. What I came with D.E Then i would go with laplace, but it's very difficult to simplify and then come with the inverse laplace, so I guess it´s wrong.

I also tried simulating it on matlab, but i'm not sure if it is okay (again, just a beginner), and I got this: Matlab results

it doesn't seem very convincing, but i think that's the solution. I'm being asked to graph the voltage in relation to time and the current in relation to time. If someone could help me telling me how would the equations look and if what I'm doing is ok, or the best way to approach this type of exercises to get with an easier model, i would appreciate it.

  • \$\begingroup\$ " graph the voltage in relation to time and the current in relation to time"; which voltage and which current? Also which differential equations? There are several possible ones. \$\endgroup\$
    – RussellH
    Commented Feb 26 at 3:10

2 Answers 2



Assigning \$v_1\$ to the node between \$R_1\$ and \$C_1\$ and assigning \$v_2\$ to the remaining node (with \$R_2\$ and \$C_2\$) the KCL says:

$$\begin{align*} \frac1{R_1}v_1+C_1\frac{\text{d}}{\text{d}t}v_1 &= \frac1{R_1}v_{in}+C_1\frac{\text{d}}{\text{d}t}v_2 \\\\ \frac1{R_2}v_2+C_2\frac{\text{d}}{\text{d}t}v_2+C_1\frac{\text{d}}{\text{d}t}v_2&=C_1\frac{\text{d}}{\text{d}t}v_1 \end{align*}$$

You can solve that with Cramer's Rule., easy enough. Either for \$v_1\$ or \$v_2\$ (or both.) If you need hand-holding on that process, say so.

Using Cramer's by hand (no computer), and for \$v_2\$, I find:

$$\begin{align*} \left[\left(\frac{\text{d}}{\text{d}t}\right)^2+\left(\frac1{R_1\, C_1}+\frac1{R_1\, C_2}+\frac1{R_2\, C_2}\right)\frac{\text{d}}{\text{d}t}+\frac1{R_1\, R_2\, C_1\, C_2}\right]v_2&=\frac1{R_1\, C_2}\frac{\text{d}}{\text{d}t}15\:\text{V} \cdot u(t) \\\\ &= 0 \end{align*}$$

(Obviously, the derivative on the right side goes to zero with constant voltage.)

That's homogeneous. And no Laplace required!

general solution

The above may be quickly seen as \$\frac{\text{d}^2}{\text{d}t^2}+a\frac{\text{d}}{\text{d}t}+b\$ where \$a=\frac1{R_1\, C_1}+\frac1{R_1\, C_2}+\frac1{R_2\, C_2}\$ and \$b=\frac1{R_1\, R_2\, C_1\, C_2}\$.

We'd prefer it in better standard form of \$\left(\frac{\text{d}}{\text{d}t}-\alpha\right)^2+\beta^2\$. So I find \$\alpha=-\frac12 a=-8750\$ and \$\beta=\frac12\sqrt{4b-a^2}= 7180.70330817254\:j\$. (Just using a hand-calculator. No need for a computer.)

Then, taking note that \$\beta\$ is imaginary, just use the standard corresponding solution form but here selecting \$\cosh\$ and \$\sinh\$ for the solution rather than \$\cos\$ and \$\sin\$:

$$\begin{align*} v_2 &=\exp\left(\alpha\,t\right)\cdot\left[A_1\cdot \cosh\left(\frac{\beta}{j}\,t\right)+A_2\cdot \sinh\left(\frac{\beta}{j}\,t\right)\right] \end{align*}$$

specific solution

From the above, we know that at \$t=0\$ it must be that \$A_1=0\$. That simplifies things a lot. Then from the rate of change (derivative) we know that \$A_2=\frac{\frac{15\:\text{V}}{R_1\,C_2}\,j}{\beta}=10.4446593573419\$.

So the final answer:

$$v_2\approx 10.4446593573419\cdot\exp\left(-8750\,t\right)\cdot\sinh\left(7180.70330817254\,t\right)$$

You can expand the \$\sinh\$ and combine a few things, if you want. But that should get you there.


Now, at this point, I'll finally invoke LTspice:

enter image description here

I've included a behavioral voltage source that uses the above equation so that the circuit simulation can be compared with the mathematical answer.

They are the same.


Your equations look fine.

The reason you don't see any charging curves is because you chose to plot on a time scale which is way larger than the time it would take those capacitors to charge, and their voltages to settle.

As a first estimate, the time constant of the capacitances and resistances (their product) is tens of ohms × tens of microfarads = hundreds of microseconds.

The plotted graphs are 10s wide, so the entire charging process is over in less than a pixel's width!

Plot over 1ms, and set your step input voltage source to step at 10μs.


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