This article we will discuss about circuit simplification to thevenin equivalent circuit when a problem circuit might have both independent and dependent sources. Previously we have showed circuit simplification techniques to **Thevenin** or **Norton** equivalent circuit when the problem circuit has only independent sources.

## Dependent and Independent sources

Dependent source can be any voltage source or current source which depends on any other sources in the circuit. For example if any source denoted by $\frac{V_{in}}{4000},$ which means this is a dependent source. Because it depends on other voltage source, $V_{in}$. On the other hand dependent sources are individual sources and can supply power to circuit without depending on other sources. In above example $V_{in}$ is independent source. Learn more about **sources** in the circuit.

## Circuit Simplification | Thevenin Equivalent

Suppose we have a problem circuit given below. The circuit has both independent, and dependent sources. Find the thevenin equivalent of the circuit.

Above this circuit $v_{x}$ becomes the open circuit voltage $v_{oc}$ after cutting the load out. In this situation, as no current passes through to the **3k** resistor, therefore we also can remove that from the circuit. and we have left with the circuit below right.

Suppose, **current*** i flowing through the circuit to the clockwise direction from the 4V voltage source*. And our loop is also traveling the circuit clockwise direction. Now, considering the close loop applying KVL around the loop we get,

$$4-2i-V_{oc}=0$$

For 2k resistor, loop traveling the same direction as current flow, therefore voltage will drop across 2k resistor. Then traveling $V_{oc}$ from high to low, therefore $V_{oc}$ will drop (negative). Finally, loop travel the 4V resistor from low to high, therefore, voltage will rise (positive voltage).

$$\text{Or, }4+2(\frac{V_{oc}}{4k})-V_{oc}=0, \text{ for }i=-\frac{V_{oc}}{4k}$$

$$\text{Or, }V_{oc}-\frac{V_{oc}}{2}=4$$

Finally,

$$V_{oc}=8V=V_{th}$$

Now we will find R-thevenin $(R_{th})$. Here the circuit has both dependent and independent sources.So, if the circuit has independent and dependent sources, then $R_{th}$=R-norton $(R_{N})$ =$\frac{V_{oc}}{i_{sc}}$. Fo find $R_{N},$ we have to find short circuit current $i_{sc}$ of the circuit below:

So, $i_{sc}=\frac{4}{2+3},$ or $i_{sc}=0.8mA$. Therefore,

$$R_{th}=R_{N}=\frac{V_{oc}}{i_{sc}}=10k$$

Now we have thevenin equivalent voltage, V-thevenin $(V_{th}),$ and thevenin equivalent resistance, R-thevenin $(R_{th})$. THerefore we can now easily build the thevenin equivalent circuit from it. So, our thevenin equivalent of the problem circuit is like this:

//This the end of discussion on circuit simplification to thevenin circuit, when the problem circuit might have both dependent and independent sources. Read more articles on electrical circuits, **here>> **