1. Example
We want to calculate the current I0. By
Thevenin’s theorem. We will follow the
same four steps.

2. First Step:
Removing RL to calculate Vth.

3. Second Step:
Now we will calculate the voltage
Vth.
Vx = (4 x12)/(4+8)
=4 volts =VA

4. Now VB = (8 x4)/16 =2 volts So VAB= 4 -2 = 2 volts = Vth Vth + 2Vx +

5. Third Step:
Now to find Isc.
For node 1
(V )/8k + V1/4k +V1/4k +
(V1 -2Vx)/12k=0

6. Vx = V1 (V1 - 12)/8k + V1/4k +V1/4k + (V1 -2V1)/12k=0I3
Isc I1 + I5 I4
2Vx + Vx - I2 -
Vx = V1
(V )/8k + V1/4k +V1/4k
+ (V1 -2V1)/12k=0
3V V1 +6V1 +2V1 -4V1 =0
V1 =36/13

7. I1 = (12 – V1 )/8k =(12 – 36/13)/8k =15/13 mA =1.15mA V1 I3 Isc I1 +
2Vx + Vx - I2 -
I1 = (12 – V1 )/8k
=(12 – 36/13)/8k
=15/13 mA
=1.15mA

8. I2 = (36/13)/4k =9/13 =0.692mA I4= (36/13)/4 =9/13mA = 0.692mA V1 I3
Isc I1 + I5 I4
2Vx + Vx - I2 -
I2 = (36/13)/4k
=9/13
=0.692mA
I4= (36/13)/4
=9/13mA = 0.692mA

9. I5 = (36/13 – 2(36/13))/12k =-3/13mA =0.23mA V1 I3 Isc I1 + I5 I4 2Vx

10. I3 =Isc = I5 + I4 =0.692 – 0.23 =0.46mA V1 I3 Isc I1 + I5 I4 2Vx + Vx- I2 -
I3 =Isc = I5 + I4
=0.692 – 0.23
=0.46mA

11. V1 I3
Isc I1 + I5 I4
2Vx + Vx - I2 - So
Rth = Vth/Isc
=2/0.46m
=4.33 k

12. Fourth Step: So
I0 = 2/ (2k k)
=0.315mA

13. Example
We want to calculate the voltage V0.By
Thevenin’s theorem.

14. First Step:
We removed RL to calculate Vth.

15. Second Step:
Vx = 12 + V1
For node 1
V1/2k + 2Vx/1k +Vx/2k= 0
V1 + 4Vx + Vx =0

16. Putting the value of V1 Vx – 12 +4Vx +Vx =0 6Vx =12 Vx =2volts = Vth
2Vx/1k
Vth Vx -
Putting the value of V1
Vx – 12 +4Vx +Vx =0
6Vx =12
Vx =2volts = Vth

17. Third Step:
here
Vx = V1 +12
V1/2k + 2Vx/1k + Vx/2k + Vx/2k =0
Vx – 12 +4Vx + Vx + Vx =0
7Vx =12

18. Isc = 12/7 x1/2 = 6/7 mA Rth = Vth/ Isc = 2/(6/7)m =7/3 k = 2.33k V1Vx +
2Vx/1k
Isc Vx -
Isc = 12/7 x1/2
= 6/7 mA
Rth = Vth/ Isc
= 2/(6/7)m
=7/3 k = 2.33k

19. Fourth Step:
V0 = 2 x 2k/2k k
= 0.92volts

20. NORTON’s THEOREM
Working with Norton’s theorem is same as working with Thevenin’s theorem .
We will follow the same four steps.
But in second step instead of finding Vth we will find INor by short circuiting the open terminals of the circuit.

21. NORTON’s THEOREM
In fourth step our circuit will consist of a current source of value equal to INor and the
Norton resistance will become parallel with RL.

22. Example
We want to calculate the voltage V0.By
Norton’s theorem. We will follow these
four steps.

23. First Step:
Replacing RL with a short circuit to find
IN.

24. Second Step:
We want to calculate IN.
KVL for loop1
6kI1 +3k(I1 –I2) =0

25. 9kI1 -3kI2 =18 KVL for loop 2 2kI2 + 12 + 3k(I2 – I1 ) =0IN
9kI1 -3kI2 =18
KVL for loop 2
2kI k(I2 – I1 ) =0
-3kI1 +5kI2 = -12

26. Solving equations for loop1 and loop2 I2 = -1.5mA So IN = -1.5mA

27. Third Step:
To calculate RN we have short circuit all
voltage sources.
Now 3k is in parallel with 6k and 2k is in
series with them.

28. 6k||3k + 2k = (6k x 3k)/(6k+ 3k) +2k =2k +2k =4k =RN

29. Fourth Step:
Current through RL= (-1.5m x4k)/8k
=- 0.75mA SO
V0 = (-0.75m)(4k)
= -3volts

30. Example
We want to calculate the voltage V0.By
Norton’s theorem. We will follow these
four steps.

31. First Step:
Replacing RL with a short circuit to find
IN.

32. Second Step:
Isc =I1 + I2
For node 1
V1/4k + (V1 – 2)/3k + V1/6k +4m=0
3V1 +4V1 – 8 +2V1 +48 =0
9V1+40=0 or V1=4.44V

33. For node 2
V2/2k + V2/8k -4 =0
4V2 +V2 – 32 =0
5kV2 =32
V2 =32/5 = 6.4V

34. I2 V2 V1 I1 IN
Now
I1 = V2/8k
=6.4/8k
= 0.8 mA

35. I2 = V1/4k =4.44/4k =1.11mA So IN = I1+ I2 = (1.11 + 0.8)m I2 V2 V1 I1

36. I2 V2 V1 I1 IN
IN = 1.91mA

37. Third Step:
For RN
3k|| 6k = 2k
2k is in series with 4k =2k + 4k
=6k

38. 8k is in series with 2k = 8k + 2k =10k 10k ||6k = 10 x6/16 =60/16 RN
8k is in series with 2k = 8k + 2k
=10k
10k ||6k = 10 x6/16
=60/16
=3.75k

39. Fourth Step:
To calculate V0
I0 = 1.91m x 3.75k x 1/(4k+3.75k)
= 0.92mA
V0 = 4k x 0.92m
=3.68 Volts