Electrical Circuit Simplification | Thevenin Equivalent | Dependent Sources

# Circuit Simplification | Thevenin Equivalent | Dependent Sources

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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>>

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