1. THEVENIN’S THEOREM In solving any problem by Thevenin’s theorem. We will follow the steps as stated below (1) First of all we will remove the load resistor( The resistor we want to calculate the unknown quantity that is voltage, current or power) from the circuit leaving behind an open circuit.

2. THEVENIN’S THEOREM (2) Calculating the V th at open terminals of the open circuit by any method. (3) Calculating the R th (The total resistance in the Thevenin’s circuit) by open circuiting all current sources and short circuiting all voltage sources.

3. THEVENIN’S THEOREM After calculating V th and R th, re-inserting the load resistance R L in the circuit in series with R th and considering the V th as a battery in series with these two resistances.

4. Example We want to calculate V 0 by using Thevenin’s theorem.we will follow the steps given earlier.

5. First step: Removing R L Here R L is 6k resistor at which we want to calculate the voltage V o.

6. 1k is in series with 2k so Second step: calculating V th

7. 3k is in parallel with 2mA source, so by source transformation

8. Now the combined effect of these two source will be 9 volts.

9. So V th = 9 volts.

10. Third step: calculating R th R th = 3k

11. Fourth step: Calculating the unknown quantity. V 0 = (6k/9k) x9 = 6volts.

12. Example We want to calculate V 0 by using Thevenin’s theorem.we will follow the steps given earlier.

13. First step: Removing R L Here R L is 8k resistor at which we want to calculate the voltage V o.

14. Second step: calculating V th Now we will follow the superposition method to calculate V th.

15. Only voltage source is acting No current in ‘CA’ branch so voltage drop across 4k and 2k resistor is zero so V CD = V th1

16. Applying voltage division rule V CD =V th1 =V 6k =(12 x6)/(3+6) =8 volts

17. Only current source is acting 3k||6k = (3k x6k)/(3k+6k) = 2k 2k is in series with 2k =4k

18. Current through open circuit is zero so V th2 = 4k x2m =8 volts.

19. So V th =V th1 +V th2 = 8 +8 = 16 volts.

20. Third step: calculating R th 3k||6k = (3k x6k)/9k =2k

21. R th = 2k+2k+4k =8k

22. Fourth step: calculating unknown quantity. V 0 = (8k/16k) x16 =8 volts.

23. Example We want to calculate V 0 by using Thevenin’s theorem.

24. First step: Removing R L Here R L is 6k resistor at which we want to calculate the voltage V o.

25. Second step: calculating V th Apply KVL to the circuit

26. V th I2I2 I1I1 Here I 2 = 2mA For loop 1 4kI 1 +2k( I 1 –I 2 ) =6

27. V th I2I2 I1I1 4kI 1 +2kI 1 – 2kI 2 =6 6kI 1 -2kI 2 =6 Putting value of I 2 6kI 1 – 4 =6

28. V th I2I2 I1I1 I 1 = 10/6k =1.6mA

29. V th I2I2 I1I1 Now voltage across 4k resistor V 4k = 1.6m x4k=20/3 =6.66volts

30. V th I2I2 I1I1 Voltage across 2k resistor V 2k = 2m x2k =4volts

31. V th I2I2 I1I1 So V th = V4k +V2k =20/3 + 4 = 6.66 +4 =10.66 volts.

32. Third step: calculating R th 4k||2k = 4k x2k/6k =8/6 k =4/3 k

33. They are in series so Rth=1.33k +2k=10/3 =3.33k

34. Fourth step: Calculating unknown quantity. V 0 =32/3 x6 x 1/(10/3+6 ) = 64 x3/10 +18 =192/28 =48/7 volts.