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THEVENIN'S THEOREM PPT

Thevenin's theorem PPT with explanation , limitations and examples.<br>Network analysis lab<br>

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THEVENIN'S THEOREM PPT

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  1. Faculty Of Engineering and Technology Jamia Millia Islamia

  2. (EE – 332)

  3. Prepared By:- Aman Malik ( 19BEE010 ) Asma Khan ( 19BEE011 ) Bashar Imam ( 19BEE012 )

  4. Theveni’s theorem

  5. Statement It states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage, Vth, in series with a single resistance, Rthconnected across the load“. Vth is equal to open circuit voltage at the terminals. Rth is equivalent of input resistance when the independent sources in the linear circuit are turned off.

  6. Circuit SchematicOfThevenin’s Theorem

  7. Let us consider a network or a circuit as shown. let E be the emf of the cell having its internal resistance r = 0 . Rl → load resistance across AB .

  8. To find vth :The load resistance Rlis removed. the current i in the circuit is = E/R1+R2The voltage across AB = thevenin’s voltage ,Vth              Vth = IR2  =>  Vth= ER2/R1+R2

  9. To find Rth:The load resistance is removed. The cell is disconnected and the wires are short as shown.The effective resistance across AB = Thevenin’s resistance Rth .Rth= R3+R1R2/R1+R2 [ R1 is parallel to R2 and this combination in series with R3 ]

  10. If the cell has internal resistance r ,then Vth = ER2/R1+R2+r and Rth = R3+(R1+r)R2/R1+r+R2

  11. Find Vth, Rth and the load current IL flowing through and load voltage across the load resistor by using Thevenin’ theorem.

  12. Step 1.Open the 5kΩ load resistor 

  13. Step 2.Calculate the open circuit voltage. This is the thevenin voltage (vth) Now the circuit became an open circuit . Now 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. Also the 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 8kΩresistor as well as 4kΩ resistor. Therefore 12V will appear across the ab terminals.

  14. Step 3Open current sources and short voltage sources.

  15. Step 4.Calculatethe open circuit resistance. This is the thevenin resistance (Rth).Remove the 48v dc source to zero as equivalent.Since, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and  12k Ω resistor. Rth = 8kΩ +  [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]Rth= 8kΩ + 3kΩRth = 11kΩ

  16. step 5.Connect the Rth in series with voltage source Vth and re-connect the load resistor. This is the thevenin’s equivalent circuit.

  17. Step 6. Now, applying ohm’s law IL = Vth/ (Rth + RL) IL= 12V / (11kΩ + 5kΩ) = 12/16kΩ IL = 0.75mA VL = IL x RL VL = 0.75mA x 5kΩ VL= 3.75V

  18. In our day-to-day life, whenever we overload a voltage source we observe a dip in voltage. This is basically an application of thevenins theorem, in the most observable form.It also have applications in all types of circuit designs at theoretical level starting from basic electrical circuits, Automation, control system design, PLC to VLSI in microprocessor. 

  19. The Thevenin equivalent has an equivalent I–V characteristic only from the point of view of the load.Many circuits are only linear over a certain range of values, thus the Thevenin equivalent is valid only within this linear range

  20. Quora Slideserve Electrical4U

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