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Thevenin’s Theorem. Step by Step Procedure with Solved Example

Thevenin’s Theorem in DC Circuit Analysis

Step by Step Procedure with Examples.

A French Engineer, M.L Thevenin , made one of these quantum leaps in 1893. Thevenin’s Theorem is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks because we can solve complex linear circuits and networks especially electronic networks easily and quickly.
 
Thevenin’s Theorem may be stated below:
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.

Steps to Analyze Electric Circuit through Thevenin’s Theorem

  1. Open the load resistor.
  2. Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH).
  3. Open current sources and short voltage sources.
  4. Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (RTH).
  5. 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.
  6. 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:

Example:

Find VTH, RTH and the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.

Click image to enlargeThevenin's Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

Solution:-
 
Step 1.

Open the 5kΩ load resistor (Fig 2).

Click image to enlargeThevenin’s Theorem 
 
Step 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 

Click image to enlargeThevenin's Theorem may be stated below:

Step 3.

Open current sources and short voltage sources as shown below. Fig (4)

Click image to enlargeThevenin's Theorem: Simple explanation

Step 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

Click image to enlargeThevenin's Theorem solved examples

Step 5.

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

Thevenin's equivalent circuit
Thevenin’s equivalent circuit

Step 6.

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

IL= 0.75mA

And

VL = ILx RL

VL = 0.75mA x 5kΩ

VL= 3.75V

Click image to enlargethevenin's theorem examples thevenin's theorem solved problems

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.

You may also read: 

 

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99 Comments

      1. Surely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the cu

      2. Surely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the current and voltage to change oops just figured it out the long way and the 5k gets 3.75 out of the 9.75 volts across that part of the circuit your way of showing thevenin is better thank you

    1. If the voltage source is more than one you calculate two components of Nortons current that is I’ and I”, add them to get the total Nortons current and repeat the procedure explained above.

    1. I = V / R …<br />= 48V / (12kΩ + 4kΩ )<br />= 3mA <br /><br />So 3mA Current will flow 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<br /><br />

    2. when the load terminals are open then all the current passes through only first circuit and the current is from v=i/r can be calculated

  1. i am proud that u can make this subject such intresting..<br /><br />hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

  2. hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

  3. Thanks for your very informative post. I had a very bad going with this topic, but now I can assure keep my hands dirty on complex circuits related to this phenomenon by using your idea.

  4. it is very easy to understand the concept,but if more than one voltage sources are present then how will be the solution and procedure

    1. If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.
      i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.

      In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

    1. Yes…. It can be used on both AC and DC Circuits,,, We will post about Thevinin’s theorem about AC Circuits in coming days… Thanks

  5. sir I have a confusion when you calculate Rth because I dont think that Rth=8+4||12. But It should be Rth=12+4||8. Am I wrong Sir ?, if so then please guide me.
    Thanks
    Ankit

    1. Hi Ankit,

      As mentioned above that 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)

  6. have we already been provided with 3Am or you have solved it? if you have solved it how have you done it?? a ain’t sure if I have followed you on that one

    1. It is not provided.
      In Step 2, You may see that this is an Open Circuit, I.e. Current will not flow in the 8kΩ.
      So the circuit of 48V, 4kΩ and 12kΩ becomes a series circuit.
      By using Ohms Law, (I = V/R), We found the value of current as 3mA.
      Thanks

    1. Yes Dear…. If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.
      i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.

      In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.
      Thanks.
      let me know if you want to know more… Thanks

  7. if I find current across 5 ohm resister there is a greater value then the total load current that is 0.75 can anyone answer

  8. If there are a voltage or current source between A & B terminals or in the place of load resistor, Then how can i solve this????????

  9. Very good explanation. I tend to solve complex circuits using Kirchoff’s laws…Are there particular cases when one method is favourably used instead of the other? ( I have not read all posts…maybe you have already gone into this)

    1. 3mA current will flows in both 12kΩ and 4kΩ resistors as it is a series circuit because current will not flow in the 8kΩ resistor as it is open (Fig 2)

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