EE40 Lec04 Mesh Analysis Th i d N t E i l t Thevenin and Norton ...

EE40 Lec04

Mesh AnalysisTh i d N t E i l tThevenin and Norton Equivalent

Superposition Principle

Prof. Nathan Cheung09/08/200909/08/2009

Slide 1EE40 Fall 2009 Prof. Cheung

Reading: Hambley Chapter 2

Mesh-Current Method

1. Identify all meshes and assign each of them an unknown mesh current. For convenience, define the mesh currents to be clockwise in direction.

2. Apply Kirchoff’s voltage law (KVL) to each mesh.

3 S l th lt t i lt ti t


3. Solve the resultant simultaneous equations to determine the mesh currents.

Mesh-Current Method Example

V0 = 18 VV0 18 V

R1 = 6 ohms

R2 = R3 = 2 ohms2 3

R4 = 4 ohms

R5 = R6 = 4 ohms

( ) ( )( ) ( ) 0315212110 =−+−++− IIRIIRIRV 21 =I A

( ) ( ) 032423122 =−++− IIRIRIIR( ) ( ) 036234135 =+−+− IRIIRIIR

12 =I A13 =I A


( ) ( ) 036234135 ++ IRIIRIIR

Mesh Analysis: Supermesh

Definition: Two adjoining meshes that share a current source

Problem: We cannot write KVL for meshes a and b because there is no

Definition: Two adjoining meshes that share a current source constitute a supermesh.

way to express the voltage drop across the current source in terms of the mesh currents.

Solution: Define a “supermesh” – a mesh which avoids the branch


Solution: Define a supermesh a mesh which avoids the branch containing the current source. Apply KVL for this supermesh.

Supermesh Example


Mesh Analysis w/ Dependent Sources

• Treat as independent source in organizing and writing mesh equationsand writing mesh equations, but i l d th ti th tinclude another equation that expresses the relationship of the dependent source


Dependent Source Example

Outside supermesh KVL : -20+ 4i1 + 6i2+ 2i2 =0

i2 - i1 = vx / 4vx = 2i2


Solving, we get i1 =1A, i2 = 2A

Summary of Techniques for Circuit Analysis

• Node Analysis– Node voltage is the unknown– Solve for KCL– Floating voltage source using super node

M h A l i• Mesh Analysis– Loop current is the unknown

S l f KVL– Solve for KVL– Current source using super mesh


Equivalent Circuit Concept• A network of voltage sources, current sources,

and resistors can be replaced by an i l t i it hi h h id ti l t i lequivalent circuit which has identical terminal

properties (i-v characteristics) without affecting the operation of the rest of the circuitthe operation of the rest of the circuit.

network A

iAnetwork B

iB+vA

network Aof

sourcesand

≡+vB

network Bof

sourcesand_

resistors_

resistors

i (v ) = i (v )


iA(vA) = iB(vB)

Thévenin Equivalent Circuit• Any* linear 2-terminal (1-port) network of indep. voltage

sources, indep. current sources, and linear resistors can be replaced by an equivalent circuit consisting of anbe replaced by an equivalent circuit consisting of an independent voltage source in series with a resistorwithout affecting the operation of the rest of the circuit.

N V V d R R i H bl 4th Edi i

RThévenin equivalent circuit

Note: VTh =Vt and RTh=Rt in Hambley 4th Edition

networkof +

RTh

R

iL+

v

a

R

iL+

v

a

sourcesand

resistors

≡ –+

VThRLvL–

b

RLvL–

b


b b“load” resistor



Example: voc and isc method


Example: voc and isc method (cont)


Example: voc and isc method (cont)



Equivalent Resistance Method

• Cannot use if with dependent sources


Norton Equivalent Circuit• Any* linear 2-terminal (1-port) network of indep. voltage

sources, indep. current sources, and linear resistors can be replaced by an equivalent circuit consisting of anbe replaced by an equivalent circuit consisting of an independent current source in parallel with a resistorwithout affecting the operation of the rest of the circuit.

Norton equivalent circuita a

networkof

sourcesand

≡RL

iL+

vL RL

iL+

vLiN RNand

resistors –

b

–

b


Finding IN and RN

• We can derive the Norton equivalent circuit from a Thévenin equivalent circuit simply by making a

t f tisource transformation: aRTh a

RLRN

iL

iN

+

vL–+

RL

iL+

vLvTh–

b

–

b

scTh

ThN

sc

ocThN ; i

Rvi

ivRR ====


External-Source Method

• Can contain dependent sources

( ) 042668 211 =+−++− xIIII

( ) 04624 +++ IIIII ( ) 04624 2221 =++−+− IIIII x

I1 = 8 AI2 = 2 AV = VVTh = Vab

= 4 I2


= 8 V

External-Source Method

• Can contain dependent sourcesMesh equations:

( ) 0426 '2

'1

'1 =+−+ xIIII

Mesh equations:

( ) 04 '' =+− VII

( ) ( ) 04624 '3

'2

'2

'2

'1 =−++−+− IIIIII x

( ) 04 23 =+ exVII

Find RTh:

Iex = – I3’ = – 17/3 VexDeactivate voltage source

Apply external voltageΩ==

36exVR


Ω==17ex

Th IR

Source Transformation


Source Transformation Example


Linear Circuit

• A linear circuit is one constructed only ofA linear circuit is one constructed only of linear elements (linear resistors, and linear capacitors and inductors, linear dependent p psources) and independent sources.

• Linear means I-V characteristic ofLinear means I V characteristic of elements/sources are straight lines when plotted


p

Principle of Superposition

• In any linear circuit containing multiple i d d t th tindependent sources, the current or voltage at any point in the network may be calculated as the algebraic sum of thecalculated as the algebraic sum of the partial responses (i.e. individual contributions of each source actingcontributions of each source acting alone)


Superposition Example

Find i


Superposition Example (cont)

First, zero current source to find i due to voltage source



Then, zero voltage source and find i due to current source


Superposition Example (cont)

Summing two partial responses:Summing two partial responses:



Find i


Superposition with Dependent Source Example(see Hambley pp 100-103)


Superposition with Dependent Source Example (cont)

Substitute dependent source as an independent source I.Set 1A source =0



Set I source =0





Wheatstone Bridge

• Originally used as an ohmmeter:

Advantage of a wheatstone bridge?


Measurement is independent of V0.

Wheatstone Bridge

• Wheatstone bridges are good for sensing applications:


EE40 Lec04 Mesh Analysis Th i d N t E i l t Thevenin and Norton ...

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