Thevenin’s Theorem in DC Circuit Analysis
A French engineer, M.L Thevenin, made one of these quantum leaps in 1893. Thevenin’s Theorem (also known as Helmholtz–Thévenin Theorem) is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks. This theorem is useful to quickly and easily solve complex linear circuits and networks, especially electric circuits and electronic networks.
Thevenin’s Theorem may be stated below:
Any linear electric network or a complex circuit with current and voltage sources can be replaced by an equivalent circuit containing a single independent voltage source VTH and a Series Resistance RTH.
- VTH = Thevenin’s Voltage
- RTH = Thevenin’s Resistance
Steps to Analyze an Electric Circuit using Thevenin’s Theorem
- Open the load resistor.
- Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH).
- Open current sources and short voltage sources.
- Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (RTH).
- Now, redraw the circuit with measured open circuit Voltage (VTH) in Step (2) as voltage source and measured open circuit resistance (RTH) in step (4) as a series resistance and connect the load resistor which we had removed in Step (1). This is the equivalent Thevenin circuit of that linear electric network or complex circuit which had to be simplified and analyzed by Thevenin’s Theorem. You have done it.
- Now find the Total current flowing through the load resistor by using the Ohm’s Law: IT = VTH / (RTH + RL).
Solved Example by Thevenin’s Theorem:
Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH). Fig (3).
We have already removed the load resistor in figure 1, so the circuit became an open circuit as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since 3mA current flows in both 12kΩ and 4kΩ resistors as this is a series circuit and current will not flow in the 8kΩ resistor as it is open.
This way, 12V (3mA x 4kΩ) will appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is an open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage i.e. 12V will appear across the 8kΩ resistor as well as 4kΩ resistor. Therefore 12V will appear across the AB terminals. i.e,
VTH = 12V
Open current sources and short voltage sources as shown below. Fig (4)
Calculate / measure the open circuit resistance. This is the Thevenin Resistance (RTH)
We have removed the 48V DC source to zero as equivalent i.e. 48V DC source has been replaced with a short in step 3 (as shown in figure 3). We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:
8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)
RTH = 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]
RTH = 8kΩ + 3kΩ
RTH = 11kΩ
Connect the RTH in series with Voltage Source VTH and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin’s equivalent circuit.
Now apply the last step i.e Ohm’s law . Calculate the total load current and load voltage as shown in fig 6.
IL = VTH / (RTH + RL)
IL = 12V / (11kΩ + 5kΩ) → = 12/16kΩ
IL = 0.75mA
VL = IL x RL
VL = 0.75mA x 5kΩ
Now compare this simple circuit with the original circuit shown in figure 1. Do you see how much easier it will be to measure and calculate the load current in complex circuit and network for different load resistors by Thevenin’s Theorem? Yes and only yes.
Good to know: Both Thevenin’s and Norton’s theorems can be applied to both AC and DC circuits containing difference components such as resistors, inductors and capacitors etc. Keep in mind that the Thevenin’s voltage “VTH” in AC circuit is expressed in complex number (polar form) whereas, the Thevenin’s resistance “RTH” is stated in rectangular form.
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