How can I find the Thevenin Equivalent in a RC circuit with dependent current source?

Question

In the picture below, at time $t=0$ the switch closes and the voltage of the capacitor is 10 volt.

I am asked to find the Thevenin equivalent right from A,B terminals.But there is a dependent current source in the circuit.How can I proceed ? Any help ?

Answer 1

May as well draw out the schematic and number the parts (something you should actually do to aide communication when posting a question):

^{simulate this circuit – Schematic created using CircuitLab}

To make what I'm about to say a little easier, I'm going to assign a ground to one of the wires (\$V_{\text{B}}=0\:\text{V}\$.) The other node is \$V_{\text{A}}\$ and is its value is to be understood to be in reference to that ground:

^{simulate this circuit}

The above two schematics are identical for all analysis purposes.

I'll do the math in a moment. But this is a good moment to just think about things.

What is the current in \$R_2\$? It's just a current source that depends on the applied voltage, \$V_{\text{A}}\$, by the factor \$\frac1{R_2}\$. But \$I_1\$ is tied between the same nodes (wires), isn't it? True, \$I_1\$ is \$7\times\$ larger. But that just means \$7\times\$ lower resistance. Not so? In short, isn't \$I_1\$ also a current source depending on the same voltage source, except that its current is \$7\times\$ more than in \$R_2\$?

A voltage controlled current source, controlled by the applied voltage across it, is just the same thing as a resistor. By definition, a resistor has the voltage dependent current of \$I_R=\frac{V_R}{R}=f\left(V_R\right)=\frac1{R}\cdot V_R\$. (A voltage dependent current source with the applied voltage being the controlling voltage.)

Anyway, that's a bunch of words to try and get you to think in a different mode for a moment. Let's go back to the math, now.

\$I_\Delta=f\left(V_{\text{A}}\right)=\frac1{R_2}\cdot V_{\text{A}}\$

and,

\$I_1=7\cdot I_\Delta=7\cdot f\left(V_{\text{A}}\right)=\frac7{R_2}\cdot V_{\text{A}}=\frac{V_\text{A}}{\left[\frac{R_2}{7}\right]}\$

So this would behave exactly the same as if we made up a new resistor, \$R_3=\frac{R_2}{7}\$. Not so? Then:

\$I_1=g\left(V_{\text{A}}\right)=7\cdot f\left(V_\text{A}\right)=\frac1{R_3}\cdot V_{\text{A}}\$

Yeah. That's a resistor behavior.

And now you can compute \$R_3\$'s value, too.

Given three resistors in parallel like this, I'm sure you can work out the equivalent circuit using only one resistor to replace them.

The KCL for just the right side would have been:

$$\begin{align*} \frac{V_\text{A}}{R_1}+\frac{V_\text{A}}{R_2}+7\cdot I_\Delta&=0 \\\\ \frac{V_\text{A}}{R_1}+\frac{V_\text{A}}{R_2}+7\cdot\frac{V_\text{A}}{R_1}&=0 \\\\ \frac{V_\text{A}}{R_1}+\frac{V_\text{A}}{R_2}+\frac{V_\text{A}}{\frac{R_1}7}&=0 \end{align*}$$

How can it be otherwise that \$\frac{R_1}7\$ isn't just another resistance?

Answer 2

Try to apply a test voltage (\$V_X\$) source across A and B terminals. And next, solve for \$I_X\$ (voltage source current).
And from there you can calculate the \$R_{TH} = \frac{V_X}{I_X} = 10k\Omega||\frac{20k\Omega}{7 + 1}\$

Take a look at this example:

From KCL we have

$$I_X = \frac{V_X}{20k\Omega} + 7 \frac{V_X}{20k\Omega}$$ $$\frac{I_X}{V_X} = \frac{1}{20k\Omega} + \frac{7}{20k\Omega} = \frac{7 +1}{20k\Omega} $$ $$\frac{I_X}{V_X} = \frac{7 +1}{20k\Omega} $$

And the Thevenin resistance is equal to

$$R_{TH} = \frac{V_X}{I_X} = \frac{20k\Omega}{7 + 1} = 2.5k\Omega$$