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# Thevenin Theorem

## Thevenin Theorem

According to Thevenin theorem an active circuit between two load terminals can be considered as an individual voltage source. This source’s voltage would be open circuit voltage across terminals, and the source’s internal impedance is the equivalent circuit impedance across terminals.

Let’s take a simple example of a resistive active circuit as shown below to understand Thevenin Theorem.

We will first disconnect the load from the circuit here and then evaluate the voltage across the circuit terminals. If we imagine that this whole circuit is a voltage source, this open circuit voltage across the terminals will be the source voltage. Also this open circuit voltage is known as Thevenin Voltage.

Now we measure the resistance between the terminals.

By replacing the individual sources with their internal resistance, this can be performed mathematically. We can do this in the position of an ideal source of voltage by replacing the individual source of voltage with a short circuit.

The circuit’s measured or calculated equivalent resistance across terminals is called the equivalent resistance of Thevenin.

The entire active circuit or network is a Thevenin Voltage  source’s voltage that is attached to Thevenin Resistance in series.

## Thevenin Theorem on a circuit containing current source

Here current source is first transformed to equal voltage source, then the open circuit voltage or thevenin voltage and circuit resistance can be easily calculated and the corresponding Thevenin voltage source can be drawn.

The same theorem may also apply to an active circuit where we have to deal with impedance rather than resistance.

We can better understand the theorem after going through the following examples. By implementing the Theorem of Thevenin on the following circuit, find out the current of the load

Now we have to follow the steps one by one.

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

Step 2: View back from the open terminal A and B to the open circuited network. Calculate circuit equivalent resistance, i.e. RTh.

Now we calculate internal resistance of the network.

we get the value of RTh = 5ohm. Of Thevenin’s equivalent circuit.

Now we have to find out the value of VTh.

Step 3: Draw the circuit as before, but keep the load resistance out from the terminal A and B.

Step 4: The individual loops are to be found. Apply KVL (the voltage law of Kirchhoff) and find out the current of the loop.

You’ve got two single circuit loops. Mark the clockwise arrow loops as the current flowing direction.

Start applying KVL in the first loop now.

By applying KVL in the second loop, we will get

By solving these two equations we get value of I1 = 1.041A and I2 = 1.25 A.

So the actual direction of the currents is marked in this figure.

Step 5: Start by selecting any branch route from Terminal A to B. Calculate the complete voltage you faced during your journey. This is the VTh voltage.

Assume that this VTh is linked to the terminal A and B.

Start your journey via any branch from Terminal A to Terminal B.

Let’s begin the journey as per Red Color’s labeled route.

Now by applying KVL again, we can get

There is no current through 2 ohm resistor its just connected to terminal A.

So Thevenin’s Voltage VTh is 17.5V.

You can check this VTh value by selecting another route from Terminal A to B in the circuit. Let’s choose a different route as shown below.

Now applying KVL, we will get

[there is no current through 2ohm resistor so just connected to terminal A]

Step 6: Draw the equivalent circuit of Thevenin with the RTh and VTh calculation value. Connect RL to the terminal of AB. Use KVL again to figure out the present load IL or placed the value of VTh, RTh and RL directly in the formula.

Now connect RL = 10 ohm across terminal A and B.

Again here  apply KVL and get