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

2. 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

3. Source Transformation
vs
vs = is R or is =
R

4. Source Transformation
Vs = Rs I s
Vs
Is =
Rs

5. 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.

6. Example
 Use
source transformation to find vo in the
circuit.

7. Example

8. Example
we use current division to get
and
2
i=
(2) = 0.4A
2+8
vo = 8i = 8(0.4) = 3.2V

9. 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.

10. Steps to apply superposition principle
1.
Turn off all independent sources except one source. Find
the output (voltage or current) due to that active source
using nodal or mesh analysis.


1.
2.

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.

11. Superposition - Problem
2kΩ
4mA
12V
– +
2mA
1kΩ
I0
2kΩ

12. 2mA Source Contribution
2kΩ
2mA
1kΩ
I’0
I’0 = -4/3 mA
2kΩ

13. 4mA Source Contribution
2kΩ
4mA
1kΩ
I’’0
I’’0 = 0
2kΩ

14. 12V Source Contribution
12V
2kΩ
– +
1kΩ
I’’’0
I’’’0 = -4 mA
2kΩ

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