## Thevenin Theorem

Thevenin Theorem tells that an active circuit between two load terminals can be considered as an individual voltage source. The voltage of this source would be open circuit voltage across the terminals and the internal impedance of the source is the equivalent impedance of the circuit across the terminals.

To understand Thevenin Theorem let us take a simple example of a resistive active circuit as shown below.Here first we will disconnect the load from the circuit and then measure the voltage across the terminals of the circuit. This open circuit voltage across the terminals will be the source voltage if we think this entire circuit is a voltage source. This open circuit voltage is also known as Thevenin Voltage.

Now we will measure the resistance between the terminals.

Mathematically it can be done by replacing the individual sources with their internal resistance. In case of ideal voltage source, we can do that by replacing the individual voltage source by a short circuit. The measured or calculated equivalent resistance of the circuit across the terminals is called Thevenin Equivalent resistance.

The entire active circuit or network is a voltage source of Thevenin Voltage with Thevenin resistance connected in series with it.

### Thevenin Theorem on a circuit containing current source

Here current source is first converted to equivalent voltage source then we can easily calculate the open circuit voltage or Thevenin Voltage and resistance of the circuit and can draw the equivalent Thevenin Voltage source.

There is nothing extra to discuss Thevenin Theorem as it so simple.

The same theorem can also be applicable for ac active circuit where we have to deal with impedance instead of resistance.

After going through the following examples we can better understand the theorem further.

By applying Thevenin’s Theorem on the circuit below, find out the load current

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 and B.

Step 2: View back into the open circuited network i.e. from the open terminal A and 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 and 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 and 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 and 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 and B terminal and remove the voltage 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.

Ω

In the Thevenin theorem problem section, while applying KVL, the current direction on both loop is taken clockwise. But we know the direction of the current in the second loop is in the opposite direction. The same has been used while indicating current flow in the next section where current values are written. The current values of loop1 and loop 2 are added together. But while formulating equations but KVL this current I1 and I2 were subtracted.

Why did you use the reverse current direction in the second loop?

Would love to help you out Mebin – do you mind linking to the image you’re referencing? It will make it much easier to formulate a proper response.