# First order circuit with Thevenin equivalent

I am self reading linear electrical circuits and I came across to a problem that I don't know how to approach it.

The problem states :"In the following circuit with dependent current source, find $$\v_c(t)\$$ for $$\v_c(0^{-})=10V\$$ using Thevenin Theorem.

As far as I know, to the left of the two terminal A,B I have to find the Thevenin equivalent. So I can connect to the terminals a current source of 1A. But how can I proceed further ? At the end I suspect that the equivalent will be a resistor.

Can someone explain me how can I solve it ? I have a difficulty understanding the Thevenin equivalent with dependent sources.And then this is a first order linear circuit of RC (if I am right).

• $v_c(0^{-})=10V$ should change to $v_c(0^{-})=10V$ i.e. $v_c(0^{-})=10V$ (if that's what you meant to write). Commented Jul 10, 2023 at 11:35
• @Andyaka I did it thank you Commented Jul 10, 2023 at 11:37
• @HomerJaySimpson Are you looking to have someone do something like I did here or here? Or what? (No, I've not attempted a solution and don't have a comment about your intuitions about the circuit at this time.) Commented Jul 10, 2023 at 12:14
• I have just looked quickly at this and it looks as though the general solution is of the form: $\frac{\text{d} v_x}{v_x}=\frac1{2\,\cdot\, C}\:\text{d}t$, which has the obvious integral results on both sides. Initial conditions then added for the specific solution, obviously. Commented Jul 10, 2023 at 12:23

Since you didn't respond to those pages I linked in comments, I'm guessing none of it makes any sense. So I'll keep it short and simple.

You know, or should know, that $$\i_C=C\cdot\frac{\text{d}}{\text{d}t}v_x\$$. You also know that the current in the resistor is obviously $$\i_R=\frac{v_x}{R}\$$. And the dependent current source is specified, as well. So you have (subtracting the resistor current from the dependent source current to get the capacitor current):

$$i_C=0.75\mho\cdot v_x-\frac{v_x}{4\:\Omega}=0.25\:\text{F}\cdot\frac{\text{d}}{\text{d}t}v_x$$

(The resistor just reduces the magnitude of the dependent source, is all. Technically, you could just disconnect the resistor after modifying the factor of 0.75 down to 0.5 on the source.)

So, just re-arrange to get:

$$\frac{0.5\mho}{0.25\:\text{F}}\cdot \text{d}t=\frac{\text{d}v_x}{v_x}$$

The solution is $$\v_x=A\cdot\exp\left(2\cdot t\right)\$$. But you know the initial condition, so the specific solution must be:

$$v_x=10\:\text{V}\cdot\exp\left(2\:\text{Hz}\cdot t\right)$$

In one second, $$\v_x\approx 73.89\:\text{V}\$$ about which LTspice appears to agree on:

Now, if none of this makes sense then there's some serious remedial work ahead. I like the book "Fundamentals of Differential Equations" by R. Kent Nagle & Edward B. Saff & Arthur David Snider and would recommend it for self-learners as an adjacent line of inquiry.