1. Example
Find Norton equivalent circuit.

2. IN=12/~0 → A

3. Is the Norton’s equivalent circuit.

4. Example
Calculate Norton’s current through the circuit.
Inject a 1A current source into the port and define Vx across 200.

5. For node 1
1=(V1/100)+(V1-Vx)/50
100=V1+2V1-2Vx
3V1-2Vx=100

6. For node 2 -0.1V1=(Vx/200)+(Vx-V1)/50 -20V1=Vx+4Vx-4V1 5Vx+16V1=0 V1 2

7. Solving simultaneously
10Vx+32V1=0
-10Vx+15V1=500
47V1=500
V1=10.64V
Also Rth=Rn=V1/1A=10.64.
Since no independent source is involved in the circuit, hence IN=0

8. Example
We want to calculate the voltage V0.By
Norton’s theorem and we have a condition
that we cant use node or loop methods .
We will follow these four steps.

9. First Step:
We have removed RL to calculate IN.

10. We cant use loop or node methods, so lets
Second Step: IN
We cant use loop or node methods, so lets
use super position method to calculate IN

11. Only current source is acting.
Due to short circuit all current will follow
through the short circuit so

12. IN1
IN1 = 2mA

13. Only voltage source is acting.
4k is in parallel with 2k resistor which in
return in series with 2k resistor .So
total resistance.

14. IN2
R = (4k||2k) +2k
=8/6 k k
=3.33 k

15. IN2 So
IN2 = 6/3.33k
= 1.80mA

16. IN So
IN = IN1 + IN2
=2mA mA
=3.80mA

17. Third Step:
4k is in parallel with 2k.The combined
effect of these is in series with 2k.

18. RN
4k||2k + 2k = k
=3.33k
=RN

19. Fourth Step:
I0 = ( 3.80m)(3.33) x 1/9.33k
= 1.356
By ohm’s Law we have
V0 = 6 x = volts

20. Example
We want to calculate the current Io through 6k resistor using Norton’s theorem.

21. First Step
Removing RL and short circuiting open terminals to calculate IN.

22. Second Step
From loop 2, we can write
I2=2mA

23. IN I3 I1 I2
For loop 1 4k(I1-IN)+2k(I1-I2)-6=0 4kI1-4kIN+2kI1-2kI2-6=0 6kI1-4kIN-2kI2=6

24. IN I3 I1 I2
Putting the value of I2
6kI1-4kIN-2k(2m)=6
6kI1-4kIN=10

25. For Loop 3 4k(IN-I2)+4k(IN-I1)=0 4kIN-4kI2+4kIN-4kI1=0 -4kI1+8kIN=8 I1

26. Solving equations for loops 1 and 3 12kI1-8kIN=20 -12kI1+24kIN=24
16kIN=44=IN=2.75 mA

27. Third Step: Calculate RN
2kII4k+4k=((2kx4k)/(2k+4k))+4k
=5.33k

28. Fourth Step
Io= (2.75m)(5.33k)/(6+5.33k)
=1.29mA.

29. Linearity Principle In this technique we assume the unknown quantity and analyze the circuit in reverse manner. Until we reach the source which is producing all voltages or currents and calculate its value. Now by comparing this value with the original value of the source we calculate the exact unknown quantity of the circuit.

30. Example
We want to calculate Voby linearity principle. Let us assume that Vo is 1 Volt.

31. Our assumption is Vo=1V
Therefore, V2=1V
Hence I2=1/2k=0.5mA
Also, V4k=(0.5m)(4k)=2V

32. Now, V1=V4k+V2= 2+1= 3V Then, I1=3/3k=1ma Also, Is=I2+I1=0.5m+1m=1.5mAVo
Now, V1=V4k+V2= 2+1= 3V
Then, I1=3/3k=1ma
Also, Is=I2+I1=0.5m+1m=1.5mA

33. V2k=Is(2k)=2k(1.5m)=3V Vs=V2k+V1=3+3=6VVo
V2k=Is(2k)=2k(1.5m)=3V
Vs=V2k+V1=3+3=6V
When Vo is 1, source voltage is 6V, but original source voltage is 12 V, hence output voltage will be 2 Volts.