# Thevenin equivalent of a circuit

Ok so I can't post the image of my question and my attempt at it,

so I would like to ask, if anyone knows the steps to simplifying a circuit with dependent current sources, independent voltage sources and resistors, into a Thevenin equivalent circuit.

I tried to use Kirchoff's current law, to find voltage at nodes and then using that I calculated the value of the current flow, but I am not able to find out how to find the Thevenin equivalent resistance of the circuit,

which is the very last step I need to take to get the Thevenin equivalent of the circuit I'm trying to simplify.

• The simplest way is to open-circuit all independent current sources and short-circuit all independent voltage sources, then connect a voltage source $V_{in}$ to the input and find the current $I_{in}$ flowing through the circuit. Thevenin equivalent resistance will be $R=\frac{V_{in}}{I_{in}}$. – Mo_ Jun 7 '13 at 18:26
• Would electronics.stackexchange.com be a better home for this question? – Qmechanic Jun 7 '13 at 19:17
• @Qmechanic oh thanks! i was wondering why i couldnt find a stackexchange for engineering questions – user1343502 Jun 7 '13 at 20:41

One thing you can do to get it all at once is to attach a resistor of variable resistance $R$ over the load, and calculate the current going through it as a function of $R$. The expression for current will come in the form $\frac{A}{B+R}$ for some $A$ and $B$. Looking at the equivalent expression in the Thevenin equivalent circuit, we see that $A=V_{th}$ and $B=R_{th}$. Alternatively, the voltage will be $\frac{V_{th} R}{R_{th}+R}$ and you can figure it out with this. We can see that the short-circuit and open circuit methods come from the special cases $R=0$ and $R=\infty$, but keeping $R$ variable lets you calculate both of them by solving the circuit once.