Mastering Thevenin’s Theorem: Practical Circuit Analysis

We have seen the schematics of different circuits in the previous articles. When it comes to circuit solving then some predefined theorems are used. Just because at a higher level circuits become too complex, and we can not solve them just by looking at them. That's why to solve these complex systems, theorems are developed. One of the most popular ones is Thevenin’s Theorem. It allows us to replace any complex linear network with a single voltage source (called Thevenin voltage, Vth) in series with a single resistance (called Thevenin resistance, Rth). This equivalent circuit is advantageous when you need to study the behavior of a circuit under different load conditions. Today in this blog, we will see how to solve the network using Thevenin’s theorem.

What Thevenin's Theorem States:

Any linear two-terminal network of voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source (Vth) in series with a resistance (Rth).

• Vth (Thevenin Voltage): The open-circuit voltage across the terminals A and B.
• Rth (Thevenin Resistance): The resistance seen at the terminals with all independent sources turned off (voltage sources shorted, current sources opened). If dependent sources are present, a test source method is required.

Steps to Analyze an Electric Circuit using Thevenin's Theorem

1. Open the load resistor(RL).

2. Calculate/measure the open circuit voltage. This is the Thevenin Voltage (Vth). It is computed at the end of load by removing it.

3. Open all current sources and short all voltage sources.

4. Calculate /measure the open-circuit resistance. This is the Thevenin Resistance (Rth).

5. Redraw the circuit using the measured open circuit voltage (Vth) in Step (2) as the voltage source and the measured open circuit resistance (Rth) from Step (4) as the series resistance. Connect the load resistor removed in Step 1. This forms the equivalent Thevenin circuit of the original linear network or complex circuit shown in the figure above.

6. Find the Total current or load current flowing through the load resistor by using Ohm’s Law:

Examples of Thevenin's Theorem:

Let’s understand these by three practical examples, in which we solve the circuits using Thevenin theorem.

Example 1: Calculate the current (IL) through the load resistance.

Step 1: Remove the Load ( 10 Ω) as shown below in the image and find Vth.

When the 10 Ω load is removed, we’re left with two branches only:

• Branch 1: 10 V source in series with 4 Ω.
• Branch 2: 12 V source in series with 6 Ω.

Now the circuit has two sources opposing each other and connected in series through their resistors.

That means the loop is:

10V  −  12V  +  I(4)  +  I(6)=0

So, Vth = VAB

= 6I - 12

= 6(0.2) - 12

= 1.2 - 12

= -10.8V  [B is negative Polarity and A is positive Polarity]

Step 2: To find Rth, open all current sources and short all voltage sources. From the above circuit, open all current sources and then short the voltage sources. The equivalent circuit is:

So, Rth=2.4 Ω

Step 3: Connect the load as removed in Step 1. This forms the equivalent Thevenin circuit of the original linear network or complex circuit, as shown below:

Step 4: Calculate Load Current(IL)

So, IL = 0.871 A.

Example 2: Calculate Load Current(IL) through the 0.1 Ω resistor.

Step 1: Remove the Load ( 0.1 Ω) as shown below in the image and find Vth.

VTh = VAB

= -5 × 0.5 + 2 × 0.4

= -1.7V [B -ve and A +ve]

Step 2: To find Rth, open all current sources in the above circuit. The equivalent circuit is:

Rth = RAB

= 0.5 + 0.4 = 0.9 Ω

Step 3: Connect the load as removed in Step 1. This forms the equivalent Thevenin circuit of the original linear network or complex circuit, as shown below:

Step 4: Calculate Load Current(IL)

So, IL= 1.7 A.

Example 3: Determine Thevenin's equivalent across the terminals A and B.

Step 1: For a clear understanding, let us redraw the given circuit as shown below:

Step 2: To find VTh: For each loop, there are two resistors in series. Each loop is a simple series divider.

Left loop has the 20 V source driving 8Ω and 12Ω in series:

With I1​ flowing down through the 12Ω: VA = I1 × 12Ω= 1 × 12= 12 V. (So node A is +12 V relative to the bottom reference.)

Right loop has the 10V source driving 9Ω and 1Ω in series:

With I2 flowing down through the 1Ω: VB= I2 × 1Ω= 1 × 1= 1 V.

=12 V - 1 V  = 11 V.

Step 3: To find Rth: Short all voltage sources in the above circuit.

Rth= RA + RB

= 4.8Ω + 0.9Ω =5.7 Ω

Step 4: Thevenin's equivalent circuit is shown below:

From these examples, it becomes clear that Thevenin’s theorem can be used in circuits that can not be solved with a simpler approach. Whether the network has only resistors and a voltage source, or a mix of sources, or even dependent sources, the same process applies to all.

Conclusion:

Using Thevenin's theorem simplifies additive circuits, voltage dividers, and combined sources. The process remains the same whether or not you have dependent elements. Once you become familiar with the steps, you should intuitively replace large parts of a network with refined "clean" Thevenin equivalents. Any additional analysis will be faster and more intuitive. For engineers or students, mastering Thevenin's theorem is about acquiring a practical skill for real-world engineers, not an academic exercise.