Thevenin’s Theorem in DC Circuit Analysis
Step by Step Procedure with Examples.
Any Linear Electric Network or complex circuit with current and voltage sources can be replaced by an equivalent circuit containing of a single independent voltage source VTH and a Series Resistance RTH.
- You may Also Read: Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)
Steps to Analyze Electric Circuit through 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.
- Now find the Total current flowing through load resistor by using the Ohm’s Law: IT = VTH/ (RTH + RL).
Solved Example by Thevenin’s Theorem:
Find VTH, RTH and the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.
Open the 5kΩ load resistor (Fig 2).
Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH). Fig (3).
We have already removed the load resistor from 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 because current will not flow in the 8kΩ resistor as it is open.
So 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 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. So,
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 = 11kΩ
Connect the RTHin 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
Click image to enlarge
Now apply the last step i.e Ohm’s law . Calculate the total load current & load voltage as shown in fig 6.
IL = VTH / (RTH + RL)
= 12V / (11kΩ + 5kΩ) → = 12/16kΩ
VL = ILx RL
VL = 0.75mA x 5kΩ
Now compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current for different load resistors by Thevenin’s Theorem? Yes and only yes.
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