According to the Thevenin Theorem, a complex linear electrical circuit containing several electrical resistance, voltage, and current source can be simplified. The simplified circuit with a fixed resistor called Thevenin equivalent resistance R_{th}, in series with a variable load resistor, which varies frequently, R_{L}, and a Thevenin voltage, V_{th}. In other words, any one-port electrical network can be reduced to a single voltage source and a single resistor circuit. The Theorem holds for any black box linear circuit, doesn’t matter how complex the circuit is.

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.

The Thevenin Theorem also holds for AC circuit, where the circuit may contain reactive and resistive impedance load. [1]

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 Rth and Voltage Vth :

Thevenin theorem is very a very handy tool in simplification of large electrical circuit analysis. A complex original 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. I will explain the theorem with the following 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, V_{th}, we remove the load resistance R_{L} 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 V_{th}, 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 V_{th} 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 (R_{th} and V_{th}) in case of dependent sources, we use two approaches as explained below.

- We apply a voltage source, V
_{o}, at terminals of the load resistor, $R_{L}$ and determine the resulting current, I_{o}, Then R_{th }= V_{o}/ I_{o}. - Alternatively, we may insert a current source, I
_{o}, at terminals load resistor, R_{L}, and find the terminal voltage, V_{o}, then R_{th }= V_{o}/ I_{o}.

Either of the approach assumes some voltage or current at the terminal of the load resistor, R_{L}, and will lead to same results. It often occurs R_{th} 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 R
_{th} - Original circuit is replaced with R
_{th}, V_{th}and the load resistor - The technique is used for linear electrical circuit with number of voltage and current sources