1. Electrical Circuits_Lecture4
Circuit Theorems

2. Lecture 4 Objectives: To study the Superposition theorem
To study the Source Transformation techniques
Voltage to current transformation
Current to voltage transformation

3. INTRODUCTION A large complex circuits Simplify circuit analysis
Circuit Theorems
‧Thevenin’s theorem ‧ Norton theorem
‧Circuit linearity ‧ Superposition
‧source transformation ‧ max. power transfer

4. Linearity Property
A linear circuit is one whose output is linearly related (or directly proportional) to its input. v V0 I0 i
Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 5 V will give i = 1 A.

5. Superposition
The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

6. Steps to apply superposition principle
Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
Turn off voltages sources = short voltage sources; make it equal to zero voltage
Turn off current sources = open current sources; make it equal to zero current
Repeat step 1 for each of the other independent sources.
Find the total contribution by adding algebraically all the contributions due to the independent sources.
Dependent sources are left intact.

7. Example 1: use Superposition principle to find Io

8. 2mA Source Contribution
2kW
1kW
I’0
2mA
I’0 = -4/3 mA

9. 4mA Source Contribution
2kW
1kW
I’’0
4mA
I’’0 = 0

10. 12V Source Contribution
2kW
1kW
12V
I’’’0
– +
I’’’0 = -4 mA

11. Final Result I’0 = -4/3 mA I’’0 = 0 I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA

12. Example 2: find v using superposition

13. one independent source at a time, dependent source remains
KCL: i = i1 + i2
Ohm's law: i = v1 / 1 = v1
KVL: 5 = i (1 + 1) + i2(2)
KVL: 5 = i(1 + 1) + i1(2) + 2v1
10 = i(4) + (i1+i2)(2) + 2v1
10 = v1(4) + v1(2) + 2v1
v1 = 10/8 V

14. Consider the other independent source
KCL: i = i1 + i2
KVL: i(1 + 1) + i2(2) + 5 = 0 i2(2) + 5 = i1(2) + 2v2 Ohm's law: i(1) = v2 v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2 i2(2) + 5 = i1(2) + 2v2 -2v2 = (i - i2)(2) + 2v2 -2v2 = [v2 + (5+2v2)/2](2) + 2v2 -4v2 = 2v v2 -8v2 = 5 => v2 = - 5/8 V
from superposition: v = -5/8 + 10/8
v = 5/8 V

15. Source Transformation
A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa

16. Source Transformation

17. Source Transformation

18. Source Transformation
Equivalent sources can be used to simplify the analysis of some circuits.
A voltage source in series with a resistor is transformed into a current source in parallel with a resistor.
A current source in parallel with a resistor is transformed into a voltage source in series with a resistor.

19. Example 3: Use source transformation to find vo in the circuit in the following figure.

21. we use current division to get
and

22. Example5: Find vx in the circuit using source transformation

23. Applying KVL around the loop igives
(1)
Appling KVL to the loop containing only the 3V voltage source, the resistor, and vx yields
(2)

24. Substituting this into Eq.(1), we obtain
Alternatively
thus