1. Example
We want to calculate the voltage Vo .We
will the same four steps.

2. First Step:
We have removed RL to calculate Vth .

3. We want to calculate Vth . Apply KVL in two
Second Step:
Vth
We want to calculate Vth . Apply KVL in two
loops to calculate the individual loop
currents.

4. Here
I1 = 2mA

5. KVL for the super mesh -6 +6kI2 +12k(I1 +I2) + 12k(I1 +I2) =0
Vth I1
KVL for the super mesh
-6 +6kI2 +12k(I1 +I2) + 12k(I1 +I2) =0
-6 +6kI2 +12kI2 + 12kI2 +24kI1 =0
30kI2 – 6 +24k(2m)=0
30KI =0

6. I2
Vth I1
I2 = -42/30mA
= -1.4mA

7. I2
Vth I1
Now
V6k = 6kI2
=6k(-1.4m)
= -8.4volts

8. I2
Vth I1
V12k = 12k(I1 +I2)
= 12k( 2m -1.4m)
=7.2volts

9. I2
Vth I1 So
Vth = V6k + V12k =
= -1.2volts.

10. Third Step:
6k is in series with 12k .The resultant of
these two is in parallel with 12k.
(6k+12k)||12k = 18k x12k/(12k+18k)
=7.2k

11. Fourth Step:
V0 = -1.2 x4k/(7.2k + 4k)
= volts.

12. Example
We want to calculate the value of RL and
the maximum power dissipation across it
by Thevenin’s Theorem.

13. First Step:
To remove RL to calculate Vth .In this case
our Rth is our RL.

14. We want to calculate Vth . Apply KVL in two
Second Step:
Vth A c
We want to calculate Vth . Apply KVL in two
loops to calculate the individual loop
currents.

15. Here
I1 = 2mA
Applying KVL to loop 2
6kI2 + 3k(I2 – I1) +3 =0

16. Vth A c I2 I1
9kI2 -3kI1 + 3 =0
9kI2 – 6 +3 =0
I2 = 1/3 mA
I2 = 0.33mA

17. Vth A c I2 I1
Now
VAB = 4k(I1)
= 4k( 2m)
= 8 volts.

18. Vth A c I2 I1
VBC = 6k(I2)
= 6k x 0.33m
= 2 volts.

19. Vth A c I2 I1 So
Vth = VAB + VBC
= 8 +2
= 10 volts.

20. Third Step:
To calculate Rth .Short circuiting the voltage
source and open circuiting the current
source.

21. 3k is in parallel with 6k and 4k is series with these two so
Vth
Rth
Rth A c I2 I1
3k is in parallel with 6k and 4k is series
with these two so
3k||6k + 4k = 3k x 6k/(3k + 6k) +4k
=2k +4k = 6k = Rth =RL

22. Fourth Step:
For maximum power dissipation
RL =Rth = 6k
PL = I2R
=(10/12k)2 (6k)

23. Rth
PL = 25/6 mW
= 4.1 mW

24. Example
We want to calculate the value of RL and
the maximum power dissipation across it
by Thevenin’s Theorem.

25. First Step:
To remove RL to calculate Vth .In this case
our Rth is our RL.

26. Second Step:
KVL for loop1
I1 – 6I2 = 0
2I1 – I2 =2

27. 6k I1
Vth I2
KVL for loop2
12I2 – 6I1 +3 =0
4I2 – 2I1 +1 =0
2I1 =4I2

28. Putting in equation for loop1 4kI2 +1 –1kI2 =2 3kI2 =1 I2 =0.33mA
Vth I2
Putting in equation for loop1
4kI2 +1 –1kI2 =2
3kI2 =1
I2 =0.33mA

29. 6k I1
Vth I2
from equation1
I1 = (I2 +2)/2
=0.33+2/2
=1.166mA

30. 6k I1
Vth I2
VR6k = 6kI1
= 6k(1.166m)
=7volts.

31. 6k I1
Vth I2
Vth = 7V +3V
=10 volts.

32. Third Step:
We want to calculate Rth
6k is in parallel with 6k .The resultant is
again parallel with third 6k resistor.

33. 6k
Rth
6k||6k = (6k x 6k)/6k + 6k
=3k
3k||6k = 6k x 3k/(6k + 3k)
=2k

34. 6k
Rth So
Rth = 2k

35. Fourth Step:
VRL = 10 x2k/4k
=5 volts
PL = V2/R
=25/2k =12.5mW

36. THEVENIN’S THEOREM and DEPENDENT SOURCES.
Working with dependent sources is different from working with independent sources while applying Thevenin’s theorem .
While calculating Rth we can simply open circuit current sources and short circuit voltage sources.

37. THEVENIN’S THEOREM and DEPENDENT SOURCES.
Because the voltage or current of the dependent sources is dependent on the voltage or current of these independent spources.

38. THEVENIN’S THEOREM and DEPENDENT SOURCES.
While calculating Rth we will short circuit the open terminals of the Thevenin circuit and will calculate the Isc and then divide Vth with Isc to calculate Rth.

39. Example
We want to calculate the voltage Vo by
Thevenin’s theorem.

40. First Step:
To remove RL to calculate Vth .

41. Voltage across 4k resistor V4k = (4k/6k) 12 =8volts
Second Step: + - + - + - + - VA
4VA VA
4VA V0
Vth
Voltage across 4k resistor
V4k = (4k/6k) 12
=8volts
Voltage across 2k resistor
V2k= (2k/6k) 12 = 4volts.

42. Vth = 8 – 4VA = 8 – 4(4) = 8 – 16 = -8 volts. + - + - + - + - VA 4VA

43. Third Step:
We will first find Isc to calculate Rth.
Here
VA = 2kI1

44. Applying KVL to loop 1 -12 +2kI1 +(I1 – Isc)4k=0
4VA - + + - VA
Isc I1
Isc
Applying KVL to loop 1
-12 +2kI1 +(I1 – Isc)4k=0
-12 +2kI1 +4kI1 – 4kIsc=0
6kI1 -4kIsc =12

45. For other loop (Isc –I1)4k +4VA = 0 4kIsc -4I1 +4(2kI1)=0
4kIsc -4I1 +8I1 =0
I1 = -Isc

46. Putting in equation for loop1 3I1 -2Isc =6 3(-Isc) -2Isc =6
4VA - + + - VA
Isc I1
Isc
Putting in equation for loop1
3I1 -2Isc =6
3(-Isc) -2Isc =6
Isc = -6/5 mA

47. 4VA - + + - VA
Isc I1
Isc So
Rth = Vth/Isc
= -8/(-6/5 m)
=6.67k

48. Fourth Step: So
V0 = (6k/6k +6.67k) x8
=6k/12.67k x8
=3.78 volts