Thevenin’s theorem allows us to reduce complex electrical circuits to a simple series containing voltage source and a series resistor.
Statement of Thevenin Theorem: Any linear bilateral network can be reduced to a series circuit containing one voltage source and a series resistor
Thevenin theorem is useful for circuit analysis when we have a complex circuit with variable load and we want to do circuit analysis for the circuit.
Steps to obtain Thevenin Equivalent Circuits
- Identify load and remove the load from the circuit
- Label the leftover terminals
- Replace all sources (current source by open and voltage source by short)
- Find the equivalent resistance at terminals (This is Thevenin equivalent resistance)
- Again connect all sources
- Find the voltage at open circuit terminals. This voltage value will be the value of series voltage source
- Connect Thevenin voltage source in series with the Thevenin resistance
- Perform desired operations for solving the circuit
An Example
Find the voltage across variable load resistor for value of resistance from 2 to 10 ohms.
![](https://electricalandelectronicsengineering.com/wp-content/uploads/2019/06/thevenin-theorem-example-circuit-300x148.jpg)
Step 1: Identify load and remove resistor. Here the load is variable resistor.
Step 2: Labeling the leftover terminals. Here we’ll use x and y.
![](https://electricalandelectronicsengineering.com/wp-content/uploads/2019/06/step-2-t-t-1-300x173.jpg)
Step 3: Now replace current source (2A) by open and 12 V source by a short
Step 4: Find the equivalent resistance at terminals x and y
R (eq) = (10 ohms + 10 ohms) || (5 ohms) = 4 ohms
Also learn (How to solve resistors in series and parallel)
Step 5: Again connect all sources in the circuit
Step 6: To find voltage at open circuits terminals we need to apply Superposition principle (Since this circuit contains multiple sources)
Firstly remove current source and find voltage at output terminals
![](https://electricalandelectronicsengineering.com/wp-content/uploads/2019/06/step-6-t-t-1-300x136.jpg)
R(1), R(2) and R(3) are in series. Using voltage divider rule:
V(xy1) = (5/25) * 12 = 2.4 volt … (1)
Now consider the current source and find the voltage across xy terminals due to current source
![](https://electricalandelectronicsengineering.com/wp-content/uploads/2019/06/step-6-t-t-2-300x225.jpg)
Using current divider rule:
I = 0.8 A
From Ohm’s law:
V = IR = 0.8 A * 5 ohms = 4 volts … (2)
From equations 1 and 2:
overall voltage = 2.4 + 4 = 6.4 volts
Step 7: To obtain Thevenin equivalent:
![](https://electricalandelectronicsengineering.com/wp-content/uploads/2019/06/final-thevenin-equivalent-circuit-300x175.jpg)
Step 8: Using voltage divider we can find the behavior of voltage:
- For R(load)=2Ω: V=(2 Ω / 12 Ω ) * 6.4 = 1.0 V
- For R(load)=2Ω: V=(3 Ω / 12 Ω ) * 6.4 = 1.6 V
- For R(load)=2Ω: V=(4 Ω / 12 Ω ) * 6.4 = 2.1 V
- For R(load)=2Ω: V=(5 Ω / 12 Ω ) * 6.4 = 2.6 V
- For R(load)=2Ω: V=(6 Ω / 12 Ω ) * 6.4 = 3.2 V
- For R(load)=2Ω: V=(7 Ω / 12 Ω ) * 6.4 = 3.7 V
- For R(load)=2Ω: V=(8 Ω / 12 Ω ) * 6.4 = 4.2 V
- For R(load)=2Ω: V=(9 Ω / 12 Ω ) * 6.4 = 4.8 V
- For R(load)=2Ω: V=(10 Ω / 12 Ω ) * 6.4 = 5.3 V