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

Thevenin’s Theorem in DC Circuit Analysis

In 1893, a French engineer, M. L. Thévenin, made one of these quantum leaps to analyze electric circuits. Thevenin’s Theorem (also known as the Helmholtz-Thévenin Theorem) is not, by itself, a circuit analysis tool, but rather the foundation for a highly useful method of simplifying active circuits and complex networks. This theorem is particularly valuable for quickly and efficiently solving complex linear circuits and electrical 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

Related Post: Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

Steps to Analyze an Electric Circuit using Thevenin’s Theorem

  1. Open the load resistor.
  2. Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH).
  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 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 original linear network or complex circuit.
  6. Find the Total current flowing through the load resistor by using the Ohm’s Law: IT = VTH / (RTH + RL).

By following these simple steps, complex networks and circuits can be effectively analyzed and simplified using Thevenin’s Theorem.

Related Post: SUPERMESH Circuit Analysis | Step by Step with Solved Example

Solved Example using Thevenin’s Theorem:

Example:

Find VTH, RTH, the load current IL flowing through the load resistor, and the load voltage across it in Fig. 1 using Thevenin’s Theorem.Thevenin's Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

Solution:-

STEP 1.

Open the 5kΩ load resistor (Fig 2).Thevenin’s Theorem

STEP 2.

Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH) as shown in Fig (3).

We have already removed the load resistor from Fig. 1, so the circuit becomes an open circuit, as shown in Fig. 2. Now we need to calculate the Thevenin voltage.

Since a current of 3mA of current flows through both the 12kΩ and 4kΩ resistors (they are in series), no current flows through the 8kΩ resistor because it is an open circuit.

The voltage drop across the 4kΩ resistor based on V = I × R is:

3mA × 4kΩ = 12V

We also know that the 8kΩ resistor is in parallel with the 4kΩ resistor. In parallel branches, the voltage is the same, so the 8kΩ resistor will also have 12V across it. Therefore, 12V appears across terminals A and B, i.e.:

VTH = 12V

Thevenin's Theorem may be stated below:

STEP 3.

Open all current sources and short all voltage sources, as shown in Fig. 4.

Thevenin'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 by replacing it with a short circuit in Step 3 (as shown in Fig. 3). We can see that the 8kΩ resistor is in series with the parallel combination of the 4kΩ resistor and the 12kΩ resistor, i.e.:

8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)

RTH = 8kΩ +  [(4kΩ × 12kΩ) / (4kΩ + 12kΩ)]

RTH = 8kΩ + 3kΩ

RTH = 11kΩ

Thevenin's Theorem solved examples

STEP 5.

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 is the Thevenin’s equivalent circuit.

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 and load voltage as shown in fig 6.

IL = VTH / (RTH + RL)

IL = 12V / (11kΩ + 5kΩ) → = 12/16kΩ

IL = 0.75mA

And

VL = IL × RL

VL = 0.75mA × 5kΩ

VL= 3.75V

thevenin's theorem examples thevenin's theorem solved problems

Now compare this simple circuit with the original circuit shown in Figure 1. Do you see how much easier it becomes to measure and calculate the load current in a complex circuit or network for different load resistors using 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 various components such as resistors, inductors, and capacitors. Keep in mind that the Thevenin voltage (VTH) in an AC circuit is expressed in complex form (polar notation), whereas the Thevenin resistance (RTH) is given in rectangular form.

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110 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 />

      1. If the rounded off value of 3mA is used to find Vth you get 12V.
        If the less rounded off value of 3.273mA is used we’d get 13V.

    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.

  5. can i get more relevant problems using thevinin’s theorem & may i know that thevinin’s theorem can use for AC supplied circuits.

    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

  6. 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)

  7. 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

  8. what will be total R in the circuit if 4k shud the load resistance in revolving towards the source

    1. You may just change the value of the resistance and solve again same as above…. No Difficulties…. :)

    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

  9. 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

  10. 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????????

  11. VERY SIMPLY AND THOROUGHLY EXPLAINED THE CONCEPT OF THEVENINS THEOREM. PLEASE INCLUDE THE NOTES OF TWO POLE NETWORKS ALSO

  12. 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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