Thevenin Theorem This theorem is very conceptual. If we think deeply about an electrical circuit , we can visualize the statements mad...

## Thevenin Theorem

This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in

**Thevenin theorem**. Suppose we have to calculate the current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The current through the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch was connected and the internal resistance of the source is nothing but the equivalent resistance of the circuit looking back into the terminals where, the branch was connected. So the**Thevenin theorem**can be stated as follows,
An active bilateral linear network containing energy sources (generators) and impedances can be replaced by an equivalent circuit containing a voltage source (E

_{Th}or V_{Th}) in series with an impedance (Z_{Th}), where the E_{Th}or V_{Th}is the open circuit voltage between terminals of the network and Z_{Th}is the impedance measured between the terminals of this network with all energy sources eliminated (not eliminating their impedances).
In other words

1.When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is

2.The equivalent resistance of the circuit network looking back into the terminals, is

3.If we replace the rest of the circuit network by a single voltage source, then the voltage of the source would be

1.When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is

**Thevenin equivalent voltage**and,2.The equivalent resistance of the circuit network looking back into the terminals, is

**Thevenin equivalent resistance**.3.If we replace the rest of the circuit network by a single voltage source, then the voltage of the source would be

**Thevenin equivalent voltage**and internal resistance of the voltage source would be**Thevenin equivalent resistance**which would be connected in series with the source as shown in the figure below.
The load current is calculated as

Where, Z

_{L}is the Load impedance and Z_{Th}is the internal impedance of the circuit as viewed back into the open circuited network from terminal A & B with all voltage sources replaced by their internal impedances (if any) current sources with infinite impedance.

**Some Important Terms to be kept in mind****1.**Bilateral network is a network which does not change its characteristics due to change the direction of its voltage and current sources. Bilateral network elements are R, L, and C.

**2**.Linear Network is a network in which its parameters (like resistor, capacitor and inductor) never change their magnitude with respect to the variation of current or voltage or both.

**3**.Unilateral network always change its characteristics with respect to the change of direction of voltage and current. Example: vacuum diode, silicon diode, crystal detectors etc

To make Thevenin theorem easy to understand, we have shown the circuit below,

Here two resistors R

_{1}and R_{2}are connected in series and this series combination is connected across one voltage source of emf E with internal resistance R_{i}as shown. One resistive branch of R_{L}is connected across the resistance R_{2}as shown. Now we have to calculate the current through R_{L}.
First, we have to remove the resistor R

_{L}from the terminals A and B.
Second, we have to calculate the open circuit voltage or Thevenin equivalent voltage V

_{T}across the terminals A and B.
The current through resistance R

_{2},
Hence voltage appears across the terminals A and B i.e.

Third, for applying Thevenin theorem, we have to determine the Thevenin equivalent electrical resistance of the circuit, and for that; first we have to replace the voltage source from the circuit, leaving behind only its internal resistance R

_{i}. Now view the circuit inwards from the open terminals A and B. It is found the circuits now consist of two parallel paths - one consisting of resistance R_{2}only and the other consisting of resistance R_{1}and R_{i}in series.
Thus the Thevenin equivalent resistance R

_{T}is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance R_{L}is connected across terminals A and B, the network behaves as a source of voltage V_{T}and internal resistance R_{T}and this is called Thevenin equivalent circuit. The current through R_{L}is given as,## Thevenin Equivalent Circuit

### Application of Thevenin’s Theorem in DC Network

If you apply Thevenin’s theorem in DC system, then Z

_{Th}can be considered as R_{Th}. Whatever may be the circuit, we have to make that one as per the circuit shown below.
Load resistance R

_{L}is inserted to find out the Load current, following the figure below.**Procedure to Solve a Problem by Applying Thevenin’s Theorem**

Some important steps are to be followed carefully.

**Step 1**: Draw the circuit by removing load resistance, shortening voltage sources and opening the current sources from the circuit. Name the Load terminals with A & B.**Step 2**: View back into the open circuited network i.e. from the open terminal A & B. Calculate equivalent resistance of the circuit, i.e. R_{Th}.**Step 3**: Draw the circuit as previous but keeping the Load Resistance removed from A & B terminal.**Step 4**: Find the individual Loops. Apply KVL (Kirchhoff’s Voltage Law) and find out loop current.**Step 5**: Start journey from terminal A to B by choosing any path of branches. Calculate total Voltage that you have faced during journey. This voltage is V_{Th}.**Step 6**: Draw the Thevenin’s equivalent circuit with the value of calculate R_{Th}and V_{Th}. Connect R_{L}across AB terminal. Again apply KVL to find out the load current I_{L}directly put the value of V_{Th}, R_{Th}and R_{L}in the formula
Some important conventions to solve a circuit by applying Thevenin’s Theorem

When you travel along a branch against the direction of the current flowing, take the voltage for the resistance R as positive voltage drop or + IR.

When you travel along a branch in the direction of the current flowing, take the voltage drop for the resistance R as negative voltage drop or – IR.When you travel along a branch against the direction of the current flowing, take the voltage for the resistance R as positive voltage drop or + IR.

When you apply KVL consider the direction of current in a loop clock wise always whatever may be the sign of the current or actual direction of the actual current.

Get familiar with Thevenin’s Theorem by applying on a Circuit to find out Load Current

Suppose the electrical circuit is like that,

Now follow the steps one by one.

Step 1: Draw the circuit by removing load resistance, shortening voltage sources and opening the current sources (if any) from the circuit. Name the Load terminals with A & B.

Step 2: View back into the open circuited network i.e. from the open terminal A & B. Calculate equivalent resistance of the circuit, i.e. R

_{Th}.
Now calculate the internal resistance of the network.

You got the value of R

_{Th}= 5ohm. Of the Thevenin’s equivalent circuit.
So go for the next steps to find out the value of V

_{Th}.
Step 3: Draw the circuit as previous but keeping the Load Resistance removed from A & B terminal.

Step 4: Find the individual Loops. Apply KVL (Kirchhoff’s Voltage Law) and find out loop current.

You got two individual two loops from the circuit. Mark the loops with clock wise arrow as the direction of the current flowing.

Now start to apply KVL in the first loop.

[as you are in the loop 1, consider I

_{1}> I_{2}, through 6 ohm resistor (I_{1}– I_{2}) current will flow in]
By applying KVL in the second loop, we get

[As you are in the loop 2, consider I

_{2}> I_{1}, through 6 ohm resistor (I_{2}– I_{1}) current will flow in]
By solving two equations we get the value of I

_{1}= 1.041A and I_{2}= 1.25 A.
So the actual direction of the currents is marked in the figure below.

Step 5: Start journey from terminal A to B by choosing any path of branches. Calculate total Voltage that you have faced during journey. This voltage is V

_{Th}.
Assume this V

_{Th}connected across A and B terminal.
From terminal A start your journey along any branch to reach terminal B.

Let us start journey as per marked path by Red Color.
Now by applying KVL again, we can write that
[There is no current through 2 ohm resistor just connected to terminal A]
So Thevenin’s Voltage V_{Th}is 17.5V. You can verify this value of V

_{Th}choosing another path in the circuit from terminal A to B. Let us choose another path as per drawing below. Now applying KVL, we get [there is no current through 2ohm resistor just connected to terminal A] Step 6: Draw the Thevenin’s equivalent circuit with the value of calculate R

_{Th}and V

_{Th}. Connect R

_{L}across AB terminal. Again apply KVL to find out the load current I

_{L}or directly put the value of V

_{Th}, R

_{Th}and R

_{L}in the formula Now connect R

_{L}= 10 ohm across A and B terminal. Again apply KVL here and get

## Application of Thevenin’s Theorem in AC System

To calculate V_{Th}and Z

_{Th}we have to follow the steps which are followed in DC system to solve a problem. But one thing extra i.e. phase angle consideration as it AC system. Let’s start to solve a problem and get familiar with the steps again. Suppose the circuit is like Here terminal A and B are load terminals. Step 1: Draw the circuit by removing load resistance, shortening voltage sources and opening the current sources from the circuit. Step 2: View back into the open circuited network i.e. from the open terminal A & B. Calculate equivalent resistance of the circuit, i.e Z

_{Th}Step 3: Draw the circuit as previous but keeping the Load Resistance removed from A & B terminal. Step 4: Find the individual Loops. Apply KVL (Kirchhoff’s Voltage Law) and find out loop current. For the loop 1, For the loop 2, Solving them we get, Step 5: Start journey from terminal A to B by choosing any path of branches. Calculate total Voltage that you have faced during journey. This voltage is V

_{Th}. Choose a path as per red marked way on the figure below. A special Problem considering mutual induction in the circuit Without any mutual induction application of Thevenin’s theorem is easily applicable to find out V

_{Th}and Z

_{Th}. But when there the effect of mutual induction in the circuit then a special approach to be considered along with the general steps. This type of circuit is given below. First draw this circuit as per presence of mutual inductance. Now follow the steps one by one. But a problem will arise to find out Z

_{Th}. But short cut and easy method is that connect a voltage source of 1 volt across A & B terminal and remove the volage sources from the circuit. Now appy KVL in each loop. Hence calculate the value Of I

_{3}only. Now the value of Z

_{Th}= -1∠0

^{o}/I

_{3}ohm. But to find out V

_{Th}only you need to calculate the value of I

_{3}from the given circuit below. Now calculate the value of I

_{2}. V

_{Th}= I

_{2}.4 volt. So you get the data of the equvalent circuit of Thevenin.

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