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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.

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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 $R=\frac{V_{in}}{I_{in}}$. – Mostafa Jun 7 '13 at 18:26
Would 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

2 Answers 2

up vote 1 down vote accepted

Without dependent sources, zeroing all the independent sources leaves a resistor network so that the Thevenin resistance can be calculated directly.

However, dependent sources typically alter the Thevenin resistance so those can't be zeroed.

One technique is to calculate the open circuit voltage and then place a wire across the nodes and calculate the short circuit current. The ratio of the open circuit voltage to the short circuit current gives the Thevenin resistance.

Another technique is to zero the independent sources and then place a test source across the nodes.

For example, if you place a 1A current source across the nodes and calculate the resulting voltage, the Thevenin resistance in Ohms is just the value of the calculated voltage.

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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.

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