# How to find the time constant of a RC circuit with dependent sources

Given the following circuit, I would like to calculate the time constant $$\ \tau \$$.

I know that in an RC circuit $$\\tau=R_{eq}\cdot C \$$.

I've also found out that $$\\frac{d v_{c}}{d t}=-\frac{2-\alpha}{R C} v_{c}(t)+\frac{E(1-\alpha)}{R C}\$$, where $$\E=e(t)\$$. Let's define $$\\lambda=\frac{1}{\tau}\$$. My textbook considers $$\\lambda=-\frac{2-\alpha}{RC}\$$, so that $$\\tau=-\frac{RC}{2-\alpha}\$$.

Instinctively I would simplify the resistors so that I can find $$\R_{eq}\$$ in order to find $$\\tau\$$, but I don't know if that would be appropriate. Also why $$\\lambda=-\frac{2-\alpha}{RC}\$$?

It is easier to re-draw the circuit in a simpler representation to see if things are clearer:

Then, to determine the time constant, you turn the excitation off and replace the source by a short circuit. Redraw the circuit again in this mode:

To determine the resistance seen from the capacitor's terminals, connect a test generator $$\I_T\$$ which develops a voltage $$\V_T\$$ across its terminal. Determine $$\\frac{V_T}{I_T}\$$ and you're done:

You can see that $$\R_2\$$ is in parallel with the resistance we want. Temporarily disconnect $$\R_2\$$ to bring it back later across the intermediate result. The equation is $$\V_T=I_TR_1+kV_T\$$. Factor and reorganize to find that $$\R=(\frac{R_1}{1-k})||R_2\$$. A quick Mathcad sheet with a SPICE sim and the operating point confirms the result:

So, unless I misinterpret the controlled source symbol (a controlled voltage source I believe), the time constant is $$\\tau=C((\frac{R_1}{1-k})||R_2\$$). If both resistances are equal to $$\R\$$, the expression simplifies to $$\\tau=\frac{CR}{2-k}\$$

When you determine the natural time constants of a circuit, you always turn the excitation off and look through the energy-storing element's terminal what resistance $$\R\$$ do you see. That $$\R\$$ multiplied by the capacitor in this case forms the time constant. This is the basic approach I described in the book I wrote on fast analytical circuits techniques or FACTs.

• Why are we allowed to take out R2 and then consider it again? Thanks! Jan 21, 2020 at 22:50
• When you want to determine a resistance across two terminals and you see that other resistances are directly connected in parallel with these terminals - and provided the current in these resistors is not driving any source or used elsewhere in the circuit - then you can temporarily disconnect these resistors and bring them back in parallel with the intermediate result. No magic here, simple inspection which lets you gain a lot of time. Jan 22, 2020 at 7:35

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

Using KVL, we can write:

$$\begin{cases} \text{I}_1=\frac{\text{V}_3}{\text{R}_2}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_3}{\text{R}_1}\\ \\ \text{I}_3=\frac{\text{V}_x-\text{V}_3}{\text{R}}\\ \\ \text{V}_1=\text{V}_2-\text{V}_x \end{cases}\tag2$$

Now, using the fact that $$\\text{R}\$$ is not a resistor but a capacitor and assuming that the initial condition is equal to zero. We can set:

$$\text{R}\space\space\space\to\space\space\space\frac{1}{\text{sC}}\tag3$$

Besides that we also know that:

$$\text{V}_1=-\alpha\text{V}_3\tag4$$

• So isn't $τ=\dfrac{RC}{2α}$ the correct answer? (positive R/2 and reduced by α ) i.stack.imgur.com/T2poD.png Jan 21, 2020 at 17:00