# Thevenin Theorem: The Thevenin Equivalent Circuit

According to the Thevenin Theorem, a complex two-terminal circuit can be replaced with a fixed resistor, R­­th, in series with a variable resistor, which varies frequently, R­­­L, and a voltage source, V­th, the voltage across the variable resistor.

In circuit analysis, when a single resistor changes, the whole circuit parameters (voltage and current) changes. When a circuit has some variable resistor, it becomes frustrating to calculate over and over. Practically, it happens more often, consider your home outlet where you plug in fan, iron and phone charger according to your requirements. In 1883 M. Leon Thevenin provided the handy technique to avoid this problem and save time.

Norton Theorem can also reduce a complex circuit to a three elements circuit. But Norton Theorem replaces the source with a current source and Norton Resistor in parallel with the load resistor.

## Thevenin Resistance and Voltage :

Thevenin theorem is very a very handy tool in simplification of large electrical circuit analysis. A complex circuit may be replaced by an independent voltage source and a two resistor; first load resistor which changes frequently and Thevenin Resistance, the remaining circuit resistance, both in series. And a voltage source, $V_{th}$, the voltage across the load resistor when the circuit is open, which is called Thevenin Voltage. So, the technique is quite easy, just find Thevenin resistor and voltage and you are done. To illustrate I will explain it through example.

## Example:

Suppose we have the following circuit, with load resistor which changes frequently that’s why its shown as a variable resistor.

### Solution :

#### Find Thevenin Resistance:

To find the Resistance, remove the load resistor, short the independent voltage source and open current sources. The circuit will look like this:

Now we have to find out the Resistance, R$_{th}$, from right-hand side (as the arrow points in the diagram).

$R_{th}=\frac{4 \times 12}{4+12} +1$
$R_{th}=4 \Omega$

### Find Thevenin Voltage:

Now, to find the Voltage, Vth, we remove the load resistance R­L and find the voltage at that terminals.

First, make changes to the circuit and redraw as follow:

Now in the circuit above, we have to find the Vth, the thevenin voltage, using any technique we want. Here, I am applying mesh analysis to the circuit, it gives me the following equation:

$-32+4 I_{1}+2 ( I_{1}-I_{2})=0$

$I_{2}= -2$

Solving for I$_{1}$, we get I$_{1}$ = 0.5A and V$_{th}$:

$V_{th}=12 ( I_{1}-I_{2} )=12 ( 0.5-2)$

$V_{th}=30v$

Notice that I have ignored the 1Ω resistor in finding Vth because no current will through it.

So far we have calculated the Thevenin Resistance and Voltage, we will now plug it into the following circuit. In order to calculate the current, the supply voltage is divided by the equivalent resistance, keeping Ohm’s law in mind.

Using Voltage Divider Rule, the formula for current $I_{L}$ is:

$I_{L}= \frac{V_{th}}{R_{th}+R_{L}}= \frac{30}{4+R_{L}}$

Suppose if R$_{L }$= 6Ω, the current I$_{L}$ will be:

$I_{L}= \frac{30}{4+6}= 3A$

Suppose if $R_{L }$= 16Ω, the current $I_{L}$ will be:

$I_{L}= \frac{30}{4+16}= 1.5A$

Suppose if R$_{L }$= 36Ω, the current I$_{L}$ will be:

$I_{L}= \frac{30}{4+36}= 0.75A$

## The Theorem with dependent sources :

It is quite easy to solve circuit using Thevenin theorem when it has only independent source in it. The fact that the output of dependent source varies with circuit parameters, so they are treated differently. To find Resistance and Voltage (Rth and Vth) in case of dependent sources, we use two approaches as explained below.

1. We apply a voltage source, V­o, at terminals of the load resistor, $R_{L}$ and determine the resulting current, Io, Then Rth = Vo / I­o.
2. Alternatively, we may insert a current source, Io, at terminals load resistor, R­L, and find the terminal voltage, V­o, then Rth = Vo / I­o.

Either of the approach assumes some voltage or current at the terminal of the load resistor, R­­L, and will lead to same results. It often occurs R­th takes negative values, it implies that the circuit is supplying power.

## Conclusion:

• A complex circuit can be converted into a single voltage source circuit by using the theorem
• Thevenin theorem is used where a single resistor is considered which varies frequently
• Thevenin resistor and voltage source should be connected in series with the load resistor
• All the voltage sources are short circuited while all the current sources are open circuited to find the Resistance Rth
• Original circuit is replaced with Rth, Vth and the load resistor
• The technique is used for linear electrical circuit with number of voltage and current sources
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