# What is Thevenin theorem

Many electronic circuits contain a combination of batteries, resistors and make it very complicated. So simplifying these complex circuits we need Thevenin's Theorem. This theorem states that it is possible to simplify any linear circuits, to an equivalent circuit with just a single voltage source and impedance in series with the load, no matter how complex they are.

## Thevenin theorem statements

According to this theorem, any two-terminal linear network containing energy sources and impedances can be replaced by an equivalent circuit consisting of a voltage source (VTH) in series with an impedance (RTH), where (VTH) is the open-circuit voltage between the terminals of the network and (RTHis the impedance measured between the terminals with all the energy sources replaced by their internal impedances.

### Thevenin's equivalent circuit

To show the Thevenin's equivalent circuit we consider a circuit with a complicated passive network driven by an energy source (Vs). The network contains three resistors (R1, R2, and R3) and they are connected with a load (RL). This circuit will be replaced by an equivalent circuit with a voltage source (VTH) called Thevenin's voltage and impedance (RTH) called Thevenin's impedance.

To calculate the Thevenin's voltage at first remove the load. When the load has removed the voltage across AB is equal to the voltage across the resistor (R2). So the Thevenin's voltage is

$\small&space;V_{TH}=IR_{2}=\frac{V_{s}R_{2}}{R_{1}+R_{2}}\left&space;[&space;\therefore&space;I=&space;\frac{V_{s}}{R_{1}+R_{2}}\right&space;]$

Where I = The flow of current through the circuit when the load is removed.

Now to calculate the Thevenin's impedance at first replace the energy sources by their internal impedance and the load (RL) also disconnected.
Note: If the internal impedance of the energy sources is given then it will be added to the resistor network.

Here the internal impedance is zero so the  Thevenin's impedance is

$\small&space;R_{TH}=R_{3}+R_{1}\parallel&space;R_{2}$

$\small&space;R_{TH}=R_{3}+\frac{R_{1}R_{2}}{R_{1}+R_{2}}$

Therefore the Thevenin's equivalent circuit for the above circuit is

Here the load current for this equivalent circuit is

### Steps to follow for solving Thevenin's Theorem

Step 1 :
Step 2 :
Remove the load and calculate the open-circuit voltage (VTH).
Step 3 :
To calculate Thevenin's impedance (RTH), replace the sources by their internal impedance.
Step 4 :
Construct the Thevenin's equivalent circuit by connecting (VTH) in series with (RTH).

### Thevenin's Theorem solved problems

Example 1: Calculate the current through the resistor of resistance 6 Ω.

Solution :

Here the load (RL) = 6 Ω

To calculate (VTH) :

Now remove the load. When the load is removed the open-circuit voltage is the same as that of the voltage across the resistor of resistance 4 Ω.

∴ The current in the circuit is

∴ The Thevenin's voltage is

To calculate (RTH) :

After replacing the source by their internal impedance the Thevenin's impedance is

$\small&space;R_{TH}=\left&space;[&space;2+2\parallel&space;4&space;\right&space;]&space;=\left&space;[&space;\frac{2\times&space;4}{2+4}+2&space;\right&space;]&space;=\left&space;[&space;\frac{4}{3}+2&space;\right&space;]&space;=\frac{10}{3}\Omega&space;=3.34\Omega$

Thevenin's equivalent circuit :

∴ The current through the load,

$\small&space;I_{L}=\frac{4}{6+3.34}=\frac{4}{9.34}=0.43Amp$

Example 2: Calculate the Thevenin's voltage and Thevenin's resistance.

Solution :

To calculate (VTH) :
Here the open-circuit voltage is the same as that of the voltage across the resistor of resistance 7 Ω.

∴ The current in the circuit is

∴ The Thevenin's voltage is

$\small&space;V_{TH}=1\times&space;7=7V$

To calculate (RTH) :

After replacing the source by their internal impedance the Thevenin's impedance is

Example 3: Calculate the current through the load resistance (RL) = 5 Ω.

Solution :

Here the load (RL) = 5 Ω

To calculate (VTH) :

Now remove the load. When the load is removed the open-circuit voltage is the same as that of the voltage across the resistor of resistance 10 Ω.

Here the current through the first loop is

$\small&space;I_{1}=\frac{V_{S}}{R_{1}}=\frac{20}{\frac{80}{11}}=\frac{20\times&space;11}{80}=\frac{11}{4}=2.75Amp$

Where

$\small&space;R_{1}=\left&space;[&space;\left&space;(&space;8+10&space;\right&space;)&space;\parallel&space;4\right&space;]+4=\frac{18\times&space;4}{18+4}+4=\frac{36}{11}+4=\frac{80}{11}\Omega$

And the current through the second loop is

$\small&space;I_{2}=\frac{4\times&space;I_{1}}{R_{2}}=\frac{4\times&space;2.75}{22}=\frac{11}{22}=\frac{1}{2}=0.5Amp$

Where

∴ The Thevenin's voltage is

$\small&space;V_{TH}=0.5\times&space;10=5V$

To calculate (RTH) :

After replacing the source by their internal impedance the Thevenin's impedance is

$\small&space;R_{TH}=\left&space;[&space;\left&space;(&space;4\parallel&space;4&space;\right&space;)&space;+8\right&space;]\parallel&space;10=\left&space;\{&space;\frac{4\times&space;4}{4+4}&space;+8\right&space;\}\parallel&space;10=\left&space;(&space;2+8&space;\right&space;)\parallel&space;10=\frac{10\times&space;10}{10+10}=5\Omega$

Thevenin's equivalent circuit :

∴ The current through the load,

$\small&space;I_{L}=\frac{5}{5+5}=\frac{1}{2}=0.5Amp$

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