2012/9/17

1

Basic Laws

•Ohm’s Law (resistors)•Nodes, Branches, and Loops•Kirchhoff’s Laws•Series Resistors and Voltage Division•Parallel Resistors and Current Division•Wye-Delta Transformations•Applications

Ohm’s Law•Resistance R is represented by

•Ohm’s Law:

AR

Rv+

_

i

1 = 1 V/A

Cross-sectionarea A

Meterialresistivity

ohm

Riv

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2

Resistors

0Riv == R = 0v = 0

+

_

i

R = v

+

_

i = 0

0Rv

limiR

==∞→

Variable resistor Potentiometer (pot)

Open circuitShort circuit

Nonlinear Resistors

i

v

Slope = R

v

i

Slope = R(i) or R(v)

Linear resistor Nonlinear resistor

•Examples: lightbulb, diodes•All practical resistors may exhibit certain

nonlinear behavior.

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Conductance and PowerDissipation

•Conductance G is represented by

vi

RG

11 S = 1 = 1 A/V

siemens mho

Gi

Gvivp

Rv

Riivp

vGi

22

22

===

===

=

positive R : power absorption (+)

negative R: power generation (-)

Nodes, Branches, & Loops•Branch: a single element (R,

C, L, v, i)

•Node: a point of connectionbetween branches (a, b, c)

•Loop: a closed path in acircuit (abca, bcb, etc)–An independent loop contains

at least one branch which isnot included in other indep.loops.

–Independent loops result inindependent sets of equations.

+_

a

c

b

+_

c

ba

redrawn

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ContinuedElements in parallelElements in series

•Elements in series–(10V, 5)

• Elements in parallel–(2, 3, 2A)

•Neither–((5/10V), (2/3/2A))

10V

5

2 3 2A+_

Kirchhoff’s Laws•Introduced in 1847 by German physicist G. R.

Kirchhoff (1824-1887).

•Based on conversation of charge and energy.

•Two laws are included,Kirchhoff’s current law (KCL) andKirchhoff’s votage law (KVL).

•Combined withOhm’s law, we have apowerful set of tools for analyzing resistivecircuits.

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5

KCL

i1i2

in

0211

nn

N

niiii

•Assumptions–The law of conservation of charge–The algebraic sum of charges within a system

cannot change.

•Statement–The algebraic sum of currents entering a node

(or a closed boundary) is zero.

Proof of KCL

proved)(KCLanyfor0)()(

anyfor0)(it.onstoredbetoallowednotisCharge

object.physicalanotisnodeA

)()(

)()(1

ttidt

tdqttq

dttitq

titi

TT

T

TT

n

N

nT

i1i2

in

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Example 1

i1

i3i2

i4

i5

leaving,entering,

52431

54321 0)-()-(

TT

T

ii

iiiii

iiiiii

Example 2

321

312

IIII

IIII

T

T

I1 I2 I3

ITIT

321 IIIIS

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Case with A Closed Boundary

cancelled.arecurrentsbranchInternal

0

0

0

1111

baab

nbn

mam

jbj

iai

iiii

ii

ii

a

Treat the surfaceas a big node

leavingentering ii

b

ia1

ib1

KVL

01

m

M

mv

•Assumption–The principle of conservation of energy

•Statement–The algebraic sum of all voltage drops (or rises)

around a closed path (or loop) is zero.

v1+ _ v2+ _ vm+ _

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Proof of KVL

proved)(KVLanyfor0)(0)(

anyfor0)()()()()(

anyforconstant)(givesenergyofonconservatitheofprincipleThe

)()()(

)()(

1

1

ttvti

ttitvtitvdt

tdwttw

dttitvtw

tvtv

T

Tm

M

m

T

T

TT

m

M

mT

v1+ _ v2+ _ vM+ _

i

Example 1

41532

54321 0

vvvvv

vvvvvvRT

v4v1

v5

+_ +_

+_

v2+ _ v3+ _

Sum of voltage drops = Sum of voltage rises

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Example 2

321

321 0

VVVV

VVVV

ab

ab

V3

V2

V1

Vab

+_

+_

+_

+

_

a

b

Vab+_

+

_

a

b321 VVVVS

Example 3Q: Find v1 and v2.Sol:

V12,V8A4205

03220(2),Eq.into(1)Eq.ngSubstituti

(2)020givesKVLApplying

(1)3,2,lawsOhm'From

21

21

21

vvii

ii

vv

iviv

v1+ _

v2

+

_

20V

2

3+_ i

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Example 4Q: Find currents and voltages.Sol:

(3)8330

03830

0301,looptoKVLAppying

(2)0givesKCL,nodeAt

(1)6,3,8,lawsOhm'By

21

21

21

321

332211

ii

ii

vv

iiia

iviviv

V6V,6V,24

A1A,3A2gives(2)Eq.(5),Eq.&(3)Eq.By

(5)236(1),Eq.By

(4)02,looptoKVLAppying

321

312

2323

2332

vvv

iii

iiii

vvvv

v1+ _

30V

8

3+_

i1

6+

_v3

i3

i2

Loop 1 Loop 2

a

+

_v2

b

Example 5

Q: Find vo.Sol:

V5,A1053035

(2),Eq.into(1)Eq.ngSubstituti

(2)0235givesKVLApplying

(1)5,10,lawsOhm'From

o

oxx

ox

viii

vvv

iviv

i

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Series Resistors

(5)

,Let

(4)or

(3)(2),Eq.&(1)Eq.By

(2)0KVL,Applying

(1),,lawsOhm'By

21eq

eq

21

2121

21

2211

RRR

iRvRR

vi

RRivvv

vvv

iRviRv

v1+ _

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

Voltage Division

vRR

vRR

RiRv

vRR

vRR

RiRv

eq

2

21

222

eq

1

21

111

v1+ _

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

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Continued

vRR

vRRR

Rv

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

v +_

i

v+ _

Reqa

b

v1+ _

v

R1

+_

i

v2+ _

R2a

b

vN+ _

RN

Parallel Resistors

(5)or

(4)111

(3)11

(2),Eq.&(1)Eq.By(2)

,nodeatKCLApplying

(1),or

,lawsOhm'By

21

21

21eq

eq2121

21

22

11

2211

RRRR

R

RRR

Rv

RRv

Rv

Rv

i

iiia

Rv

iRv

i

RiRiv

eq

i a

b

R1+_ R2v

i1 i2

i a

b

Req or Geq+_v v

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Current Division

iGG

GRRiR

Rv

i

iGG

GRRiR

Rv

i

RRRiR

iRv

21

2

21

1

22

21

1

21

2

11

21

21eq

i a

b

R1+_ R2v

i1 i2

i a

b

Req or Geq+_v v

Continued

iGG

iGGG

Gi

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

i a

b

Req or Geq+_v v

i a

b

R1+_ R2v

i1 i2

RN

iN

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iGG

iGGG

Gi

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

vRR

vRRR

Rv

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

Brief Summary

v1+ _v

R1

+_

i

v2+ _

R2a

b

vN+ _

RN i a

b

R1+_ R2vi1 i2

RN

iN

Example

Req

6 35

8

2

4 1

Req

26

8

2

4

Req 2.48

4

Req 14.4

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How to solve the bridge network?

•Resistors are neitherin series nor inparallel.•Can be simplified by

using 3-terminalequivalent networks.

R1

R2 R3R4

R5R6

Wye (Y)-Delta () Transformations

R3

R1 R21

2

3

4

R3

R1 R2

3

4

1

2

Rb

Rc1

2

3

4

RaRb

Rc1

2

3

4

Ra

Y T

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to Y Conversion

cba

bac

cab

RRRRRRR

RRRRRRR

RRRRRRR

//)()Y(

//)()Y(

//)()Y(same.thebemustbehaviorterminal-Two

323434

211313

311212

cba

ba

cba

ac

cba

cb

RRRRR

R

RRRRR

R

RRRRR

R

3

2

1

R3

R1 R2

3

4

1

2Y

Rb

Rc1

2

3

4

Ra

Y-Transformations

3

133221

2

133221

1

133221

RRRRRRR

R

RRRRRRR

R

RRRRRRR

R

c

b

a

cba

ba

cba

ac

cba

cb

RRRRR

R

RRRRR

R

RRRRR

R

3

2

1

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Example

Rab

12.5

15

510

30

20

a

b

Rab

12.5

15

17.570 30

a

b

35

Rab7.292

10.521

a

b

Rab9.632

a

b

Applications: Lighting Systems

0N21 ... Vvvv NV

vvv 0N21 ...

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Applications: DC Meters

Parameters:IFS: full-scale currentRm: meter resistance

Continued

RR V R

RR A

R

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Continued

elementmeter

elementmeter

VVII

elementmeter

elementmeter

VVII

Voltmeters

mnFSFS RRIV range-Single

mFSFS

mFSFS

mFSFS

RRIV

RRIV

RRIV

33

22

11

range-Multiple

2012/9/17

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Ammeters

nmnFSM RRRII :range-Single

333

222

111

:range-Multiple

RRRII

RRRII

RRRII

mFSM

mFSM

mFSM