1. Circuit Analysis III
Section 06

2. Linearity and Superposition

3. Basically, a mathematical equation is said to be linear
Linearity
Basically, a mathematical equation is said to be linear
if the following properties hold.
homogenity
additivity 1

4. Basic Electric Circuits
Linearity :
Many of us were brought-up to think that if plotting
an equation yields a straight line, then the equation
is linear. From the following illustrations we have; 5

5. Using Linearity The procedure can be made entirely algorithmic
1. Give to Vo any arbitrary value (e.g., V’out =1 )
2. Compute the resulting source value and call it V’_s
3. Use linearity.
This is a nice little tool
for special problems.
Normally when there is
only one source and
when in our judgement
solving the problem
backwards is actually
easier
Assume that the answer is known. Can we
Compute the input in a very easy way ?!!
If Vout is given then V1 can be computed
using an inverse voltage divider.
… And Vs using a second voltage divider
Solve now for the variable Vout

6. Basic Electric Circuits
Linearity : Example: Given the circuit shown
in Figure. Use the concept of linearity to find
the current I0.
Figure 9.1: Circuit for Example 9.1.
Assume I0 = 1 A. Work back to find that this gives
VS = 45 V. But since VS = 90 V this means the true
I0 = 2 A. 9

7. Basic Electric Circuits
Linearity : Example 9.2: In the circuit shown
below it is known that I0 = 4 A when IS = 6 A. Find
I0 when IS = 18 A.
Figure 9.2: Circuit for Example 9.2
Since I0 = 4A  IS = 6A
and, I0 = X  Is = 18A
So, I0= 12A 10

8. Source Superposition This technique is a direct application of
linearity.
It is normally useful when the circuit has only
a few sources.

9. + = Circuit with voltage source set to zero (SHORT CIRCUITED)
SOURCE SUPERPOSITION =
Circuit with current
source set to zero(OPEN) +
Due to the linearity of the models we must have
Principle of Source Superposition
The approach will be useful if solving the two circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient

10. SOURCE SUPERPOSITION
The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they'll do with all power sources in effect.
To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).

11. look at to the following circuit and apply Superposition Theorem :
Since we have two sources of power in this circuit, we will have to calculate two sets of values one for the circuit with only the 28 volt battery in effect. and one for the circuit with only the 7 volt battery in effect: .

12. Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:
Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:

13. Applying these superimposed voltage figures to the circuit, the end result looks something like this:
Currents add up algebraically as well:

14. Basic Electric Circuits
Superposition: Example Given the circuit below.
Find the current I by using superposition.
Figure 9.5: Circuit for Example 9.4.
First, deactivate the source IS and find I in the 6  resistor.
Second, deactivate the source VS and find I in the 6  resistor.
Sum the two currents for the total current. 19

15. Basic Electric Circuits
Superposition: Example Given the circuit below.
Find the current I by using superposition.
IVs = 3 A 20

16. Basic Electric Circuits
Superposition: Example Given the circuit below.
Find the current I by using superposition.
Total current I: I = IS + Ivs = 5 A 21

17. Thevenin's theorm

18. Thevenin and Norton Equivalent Circuits
M. Leon Thévenin ( ), published his famous theorem in 1883. + _ v i Rs V0
Fig.2.17 (a) Thevenin equivalent circuit

19. Thevenin & Norton Equivalent Circuits
Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.
A series combination of Thevenin equivalent voltage source V0 and Thevenin equivalent resistance Rs

20. Thévenin Equivalent Circuit
Thévenin’s Theorem: A resistive circuit can be represented
by one voltage source and one resistor:
Resistive Circuit
Thévenin Equivalent Circuit
VTh
RTh

22. Calculating the equivalent Thevenin source voltage and series resistance is actually quite easy. First, the chosen load resistor is removed from the original circuit, replaced with a break (open circuit):
Next, the voltage between the two points where the load resistor used to be attached is determined. 

23. To find the Thevenin series resistance for our equivalent circuit, we need to take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure the resistance from one load terminal to the other:

24. Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.
Steps to follow for Thevenin's Theorem:
(1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
(2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
(3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
(4) Analyze voltage and current for the load resistor following the rules for series circuits.

25. Thévenin’s Theorem – example 1
Replace everything except the load resistor R with its Thévenin equivalent

26. Example 1 – Get RTH

27. LEARNING BY DOING
“PART B”

28. Source Transformations
The Thévenin and Norton equivalent circuits both represent the same circuit
They have the same voltage-current characteristics

29. Using Source Transformations in Circuit Analysis
Any voltage source in series with a resistance can be modeled as a current source in parallel with the same resistance and vice-versa

30. EXAMPLE: SOLVE BY SOURCE TRANSFORMATION
The equivalent current source will have the
value 12V/3k
The 3k and the 6k resistors now are in parallel
and can be combined
In between the terminals we connect a current
source and a resistance in parallel
In between the terminals we connect a voltage
source in series with the resistor
The equivalent source has value 4mA*2k
The 2k and the 2k resistor become connected
in series and can be combined
After the transformation the sources can be combined
The equivalent current source has value 8V/4k
and the combined current source has value 4mA
Options at this point
1. Do another source transformation and get
a single loop circuit
2. Use current divider to compute I_0 and then
compute V_0 using Ohm’s law

31. HomeWork

33. HomeWork
Q. Find I3 using superposition theorm? I3

34. Source Transformation – Homework
Use source transformations to determine the voltage v

35. Homework Using thevenin therorm find the equivielent circuit Vth and Rth ? Note Vth = VOC