Lecture 7 - University of California, Santa Cruz 07.pdf · Lecture 7 Norton Equivalent Circuits...

Lecture 7

Norton Equivalent Circuitsq

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Thévenin Equivalent CircuitsThévenin Equivalent Circuits



oct vV =ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

oct


Vtsc R

Vi =


tR

Thévenin EquivalentThévenin Equivalent Circuits

voc

ivRt =


sci

iI = scn iI =


Step-by-step Thévenin/Norton EquivalentThévenin/Norton-Equivalent-

Circuit Analysisy1. Perform two of these:

D t i th i it lt Va. Determine the open-circuit voltage Vt = voc.

b Determine the short-circuit current I = ib. Determine the short circuit current In = isc.

c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals.


2. Use the equation Vt = Rt In to compute the remaining value.

3. The Thévenin equivalent consists of a voltage source Vt in series with Rt .

4. The Norton equivalent consists of a current source In in parallel with Rt .


Find the Norton Equivalent

15

Rd dV l

RRv

Rvv

KCL ococx

250

015

4:

3

321=

++

−+

Vvvvv

vvRR

vdividerVoltage

ococococ

ococx

62.401525.0

25.0:32

3

=→=+−

+

=+

=


VvRRR oc 62.40

4 321→

+++


= 0vx

=Ω

=→=− 75.02015015

0

1 AViVRi

v

scsc

x

Ω===

Ω

15.662.420

VvR oc

t


Ω15.675.0 Ai

Rsc

t

++−



Vvvv ococ 63151250110==→=−+

− Vvoc 63.158

012515

==→=+



To find Rt zero the sources:

Rt 38.92515

)25)(15(Ω=

Ω+ΩΩΩ

=

AVRV

i tsc 67.1

38963.15

=Ω

==


Rt 38.9 Ω


Rt 38.9 Ω=Ain

t

67.1=ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Find the Norton Equivalentvoc

+−

vvvvvKVL xocxxoc

2302: =→=−−

Vvvvv

KCL ococxoc 3002

30102

: =→=−+−



isc+−

x

xxx

Av

C

vvvKVL−

=→=−−

20220

002:

nscscx iAii

vKCL ==→=−+ 202

1020

:



= 2Ain

Ω=== 152

30AV

iv

Rsc

oct

n


Source TransformationsSource Transformations


Source Transformations



VARiV 10)10)(1( =Ω==

ARR

iiRiR

VARiV tnt

667.01002010

10)10)(1(

11211 ==→=−++

=Ω==


RR 2111211 +


Aii

AVRV

it

tn

5

4520

=Ω

==

AAARR

RidividerCurrent

Aii

667.15105

55:

5

12

32

=+

=+

=

=+


RR 10521 ++

Transform the Norton equivalent into a Thevenin equivalent with Vt = (2A)(5Ω) = 10V, Rt = 5Ωequivalent with Vt (2A)(5Ω) 10V, Rt 5Ω

= 5ΩNote the reference di i f h

10v +−

direction for the current gives the polarity for the voltage source!


voltage source!

Source Transformations= 5Ω

10v +−

VViiVRiRiVKVL t

200205101010: 22222 =−+=−++−

AVi 33.1510

202 =

+=



AAAiiAii 667.02333.122 2112 −=−=−=→+=


Thévenin/Norton Equivalents


Maximum Power TransferMaximum Power Transfer

The load resistance that absorbs the maximum power from a two-terminalmaximum power from a two-terminal circuit is equal to the Thévenin resistanceresistance.


tLL

LL

Lt

tLLL

Lt

tL RR

dRdPR

RRV

RiPRR

VI =→=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==+

= 02

2


Find the load resistance for maximum power

)5)(20(21RRΩ=→Ω=

+=

+= 44

520)5)(20(

21

21MaxLt R

RRRRR


SUPERPOSITION PRINCIPLE

The superposition principle states that the total response is the sum of pthe responses to each of the independent sources actingindependent sources acting individually. In equation form, this is

nT rrrr +++= L21


Superposition Principle


Superposition PrincipleCurrent source

i itopen circuit

VVvRv 51552 VVvRR

v s 51510521

21 =

+=

+=


Superposition PrincipleVoltage source h t i it

Reqshort circuit

VAARRiRiv 666)333)(2()5)(10()2(21 =Ω====

VVVvvv

VAARR

iRiv

T

seqs

66.1166.65

66.6)33.3)(2(510

)2(

21

212

=+=+=

=Ω=+

=+

==


Circuit with just vs1j s1

Vs2 short circuit

Req=3.75Ω

VvR

Rv s

eqT 45.520

1075375.3

10 11 ===


R seq

T 1075.310 11 ++

Circuit with just vs2j s2

Vs1 short circuit

Req=3.33Ωeq

VvR

Rv s

eq

eqT 82.110

1533.333.3

15 12 =+

=+

=


Vvvv TTT

eq

27.782.145.521 =+=+=


VV 27720 AVVvvi TsT 27.1

1027.720

101 =

Ω−

=Ω−

=


WHEATSTONE BRIDGEThe Wheatstone bridge is used by mechanical and civil engineers to measure the resistancesand civil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildingsof machines and buildings.

R3

2 RRRRx =

1R


Lecture 7 - University of California, Santa Cruz 07.pdf · Lecture 7 Norton Equivalent Circuits...

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