Thevenin's theorem PPT, Network analysis

PPT of Thevenin's theorem , explanation of thevenin's theorem with examples and limitations.

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, V th , in series with a single resistance, R th connected across the load“.   V th is equal to open circuit voltage at the terminals . R th is equivalent of inp ut resistance when the independent sources in the linear circuit are turned off.  

6. Circuit Schematic Of Thevenin’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 .  R l  → load resistance across AB .    

8. T o find  v th  :   T he load resistance  R l is removed. the current i in the circuit is = E/R 1 +R 2   The voltage across AB = thevenin’s  voltage ,V th                   V th  = IR 2   =>  V th   = ER 2 /R 1 +R 2    

9. T o find  R th :     T he load resistance is removed. The cell is disconnected and the wires are short as shown.   T he effective resistance across AB = Thevenin’s resistance R th . R th = R3+R1R2/R1+R2 [ R1 is parallel to R2 and this combination in series with R3 ]

10. I f the cell has internal resistance r ,     then V th  = ER 2 /R 1 +R 2 +r and  R th  = R 3 +(R 1 +r)R 2 /R 1 +r+R 2  

11. Find V th , R th and the load current I L flowing through and load voltage across the load resistor by using Thevenin’ theorem .

12. Step 1. O pen the 5kΩ load resistor 

13. Step 2. C alculate the open circuit voltage. T his is the  thevenin voltage ( v th ) Now the circuit became an open circuit . N ow 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. T his 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. S o 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.

14. Step 3 O pen current sources and short voltage sources.

15. Step 4. C alculate the open circuit resistance. This is the thevenin resistance ( R th ). R emove 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. R th  = 8kΩ +  [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)] R th   = 8kΩ + 3kΩ R th  = 11kΩ

16. step 5. C onnect the R th in series with voltage source V th and re-connect the load resistor. T his is the thevenin’s equivalent circuit.

17. Step 6. Now, applying ohm’s law I L  = V th   / ( R th  + R L ) I L   = 12V / (11kΩ + 5kΩ) = 12/16kΩ I L  = 0.75mA   V L  = I L  x R L V L  = 0.75mA x 5kΩ V L = 3.75V

18. In our day-to-day life, whenev e r 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

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