Thevenin Theorem is a popular method of solving complex circuits. 1883 The French telegraph engineer Leon Charles Thevenin Discover this theory. But 1926 The German scientist, Hermann von Helmholtz, was known at the time 1853 Discovered this same theory in the But then the Theorem has gained huge popularity as Thevenin Theorem.

Today we will talk about the Thevenin Theorem in detail. What is going on in our discussion today:

- Thevenin Theorem Statement.
- Basic Knowledge about Thevenin Theorem.
- Requirements of the Thevenin Theorem.
- Steps to apply the Thevenin Theorem.
- Circuit solution (lyrical) with the help of the Thevenin Theorem.

Table of Contents

## Thevenin’s Theorem Statement

Thevenin states in his Theorem that no matter what component is within a network circuit (Figure 1) if it is a two-terminal network and a linear bi-lateral circuit, it can be expressed by a voltage source and series resistance (Figure 2).

Any two-terminal linear networks or circuits can be released in the form of an Equivalent Network / Circuit, which will be in series with a register voltage source. This is originally called Thevenin’s vs. Circuit. A linear circuit can have an independent source, dependent source, and register.

## Basic Knowledge about Thevenin Theorem

Before solving a circuit with the Thevenin Theorem, we need to keep in mind some things, such as when calculating the voltage source and current source from the Thevenin circuit, the total resistance of the circuit is called Thevenin resistance which is R_{th} And the total voltage is called the Thevenin voltage V_{th} Is expressed by. By combining these Thevenin voltages and Thevenin resistance, the Thevenin circuit is available. It also acts as a common resistor or other device load resistor in the Thevenin circuit, which can absorb power from the circuit.

## Requirements of Thevenin Theorem

In many cases, we have to change the load resistance or some Elements from the circuit while solving the circuit. But when an element or load resistance is changed from the circuit, the whole circuit is changed. As a result, we have to solve the circuit every time, but repeatedly applying KVL, KCL, Ohm’s Law or any other formula to each circuit is often a hassle and a hassle.

So we can solve this circuit easily by using the Thevenin Theorem, transforming that circuit into a simple series circuit. Since the circuit is in series there is no difficulty in changing the load resistance or any element.

## Steps to apply the Thevenin Theorem

To solve a circuit with Thevenin Theorem, we have to follow some steps or Steps. The following is explained with:

* Step 1:* To determine the current of the resistance, open the resistance from the circuit and take it apart. (Circuit 2)

* Step 2: *Identify the loop in the circuit, the voltage source of the loop curves (KVL) The voltage from which the resistance is opened is applied (V

_{th}) Have to figure out.

* Step 3: *Shorten all voltage sources and open the current source. (R

_{th}) To be diagnosed (Circuit 4). But if the voltage source has an internal resistance, it needs to be added to the voltage source.

Thevenin Resistance, *R _{th }= (R_{1} R_{2}) / (R_{1 }+ R_{2})*

* Steps 4: *The Last of Us V

_{th}, R

_{th}ও R

_{L }It will have to calculate a Thevenin Equivalent Circuit and load current from this circuit (I

_{L}) It has to be evaluated. (Circuit 5)

Load current, * I _{L} = V_{th }/ (R_{th} + R_{L})*

## Circuit Solutions with Thevenin Theorem:

* Question: *From the following circuit, determine how much current I will flow through the R resistance of the circuit using the Thevenin theorem?

### Solution:

Since we have to calculate the current flowing through R resistance, we have to open this load and separate it.

The voltage from the circuit this time (V_{th}) Has to figure out.

Since the edge on the right side is open 4 Ω No current will flow in the resistance. So I = 0 Will be

And 50 V, 5 Ω And 20 Ω I’ll get a circuit. If we make the current of this circuit I_{1} Circuit (circuit 1.3)Then it will be of value , I_{1} = 50 / (5+20) = 50 / 25 = 2 A.

Since 2 A current is flowing in this circuit 20 Ω The voltage across it will be 2×20 V = 40 V

This time we KVL apply By doing V_{th}We have to find the value,

Applying KVL,

V_{th} – (I_{1}× R_{3})+ (I × R_{2})

⇒ V_{th} – 40 + (0×4) = 0

⇒ V_{th} – 40 + 0 = 0

.·. V_{th} = 40 V

This time the voltage source of the circuit is shortened by the open edge (R_{th}) To be diagnosed (circuit 1.5).

R_{th} = 5 ।। 20 + 4

⇒ R_{th }= {(5 x 20) / (5 + 20)} + 4

Or, R_{th }= (100 / 25) + 4

Or, R_{th }= 4 + 4

.·. R_{th }= 8 Ω

This time we V_{th} And R_{th }This will be Thevenin Equivalent Circuit,This will be Thevenin Equivalent Circuit,

So R The amount of current flowing through the resistance,

I = 40 / (8 + R)

Reference by,

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