INTRODUCTION

A largecomplex circuits

A largecomplex circuits

Simplifycircuit analysis

Simplifycircuit analysis

Circuit TheoremsCircuit Theorems

‧Thevenin’s theorem Norton theorem‧‧Circuit linearity Superposition‧‧source transformation max. power transfer‧

‧Thevenin’s theorem Norton theorem‧‧Circuit linearity Superposition‧‧source transformation max. power transfer‧

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

Source Transformation

Rv

iRiv ssss or

Source Transformation

sss IRV s

ss R

VI

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.

Example

Use source transformation to find vo in the circuit.

Example

Example

we use current division to get

and

A4.0)2(82

2

i

V2.3)4.0(88 ivo

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.

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. 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

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to the independent sources.

Dependent sources are left intact.

Superposition - Problem

2k1k

2k12V

I0

2mA

4mA

– +

2mA Source Contribution

2k1k

2k

I’0

2mA

I’0 = -4/3 mA

4mA Source Contribution

2k1k

2k

I’’0

4mA

I’’0 = 0

12V Source Contribution

2k1k

2k12V

I’’’0

– +

I’’’0 = -4 mA

Final Result

I’0 = -4/3 mAI’’0 = 0I’’’0 = -4 mA

I0 = I’0+ I’’0+ I’’’0 = -16/3 mA