electrical engineering: principles and applications · pdf fileelectrical engineering:...

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.

Lec

ture

3

Circu

it L

aws,

Voltag

e &

Curren

t D

ivid

ers


KIR

CH

HO

FF

’S C

UR

RE

NT

LA

W

•The net current entering a node is zero.

•Alternatively, the sum of the currents

entering a node equals the sum of the

currents leaving a node.


KIR

CH

HO

FF

’S V

OL

TA

GE

LA

W

The algebraic sum of the voltages equals

zero for any closed path (loop) in an

electrical circuit.


Usi

ng K

VL

, K

CL

, and O

hm

’s

Law

to S

olv

e a C

ircu

it


A 3 5

V 15

=Ω

=yi

yx

xi

ii

=+

5.0

A 2=

xi


V 20

10

==

xx

iv

15

+=

xs

vV

V 35

=sVA 2

=xi


Chapter 2

Resistive Circuits

1.Solve circuits (i.e., find currents and

voltages of interest) by combining

resistances in series and parallel.

2. Apply the voltage-division and current-

division principles.

3. Solve circuits by the node-voltage

technique.


4. Solve circuits by the mesh-current

technique.

5. Find Thévenin and Norton equivalents and

apply source transform

ations.

6. Apply the superposition principle.

7. Draw the circuit diagram and state the

principles of operation for the W

heatstone

bridge.


eqiR

RR

Ri

iRiR

iRv

vv

v=

++

=+

+=

++

=)

(3

21

32

13

21


eqRv

RR

Rv

Rv

Rv

Rvi

ii

i=

++

=+

+=

++

=3

21

32

13

21

11

1


163

62

61

21

31

611

11

1

43

2

=+

+=

++

=+

+=

RR

RReq


1Ω

Ω=

Ω+

=3

11R

Rtotal


21

63

61

62

61

311

==

+=

+=

eqR

Ω=

0.2

eqR


1.5Ω

Ω=

Ω+

Ω=

10

28

eqR


Ω=

→Ω

=+

=5

102

101

101

1eq

eq

RR


Cir

cuit

Analy

sis

usi

ng

Ser

ies/

Para

llel

Equiv

ale

nts

1.

Beg

in b

y loca

ting a

com

bin

atio

n o

f re

sistan

ces th

at a

re in ser

ies or par

alle

l.

Often

the

pla

ce to sta

rt is fa

rthes

t from

the

sourc

e.

2.

Red

raw

the

circ

uit w

ith the

equiv

alen

t re

sistan

ce for th

e co

mbin

atio

n found in

step

1.


3. Rep

eat step

s 1 a

nd 2

until th

e ci

rcuit is

reduce

d a

s fa

r as

poss

ible

. O

ften

(but not

alw

ays)

we

end u

p w

ith a

sin

gle

sourc

e an

d

a single

res

ista

nce

.

4. Solv

e fo

r th

e cu

rren

ts a

nd v

oltag

es in the

final

equiv

alen

t ci

rcuit.


()(

)V

Ai

Rv

AV

Rvi

AV

Rvi

30

310

16060

23060

11

1

323

222

=Ω

==

=Ω

==

=Ω

==


Voltage Division

tota

l

32

1

11

1v

RR

R

Ri

Rv

++

==

tota

l

32

1

22

2v

RR

R

Ri

Rv

++

==

Of th

e to

tal voltag

e, the

frac

tion that

appea

rs a

cross

a

giv

en res

ista

nce

in a

ser

ies ci

rcuit is th

e ra

tio o

f th

e giv

en

resi

stan

ce to the

tota

l se

ries

res

ista

nce

.


Application of the Voltage-

Division Principle

V5.

1

15

6000

2000

1000

1000

1000

tota

l

43

21

11

=

×+

++

=

++

+=

vR

RR

R

Rv


total

total

eqeq

iR

R

RR

iR

vR

R

RR

R2

1

21

21

21

+=

=→

+=


Current Division

tota

l

21

2

1

1i

RR

R

Rvi

+=

=

tota

l

21

1

2

2i

RR

R

Rvi

+=

=

For tw

o res

ista

nce

s in

par

alle

l, the

frac

tion o

f th

e to

tal

curren

t flow

ing in a

res

ista

nce

is th

e ra

tio o

f th

e oth

er

resi

stan

ce to the

sum

of th

e tw

o res

ista

nce

s.


()

Ai

RR

Ri

AR

R

vi

s

x

ss

417

.0

25

.1

60

303

0

25

.1

20

60100

32

23

1

=+

=+

=

=+

=+

=

Find i3


Find i1


Application of the Current-

Division Principle

A10

15

20

10

20

eq1

eq

1=

+=

+=

siR

R

Ri

Ω=

+×=

+=

20

60

30

60

30

32

32

eqR

R

RR

R


Use

the

voltag

e div

isio

n p

rinci

ple

to

find the

unknow

n v

oltag

es


()

()

()

()

VV

Vv

RR

RR

Rv

VV

Vv

RR

RR

Rv

VV

Vv

RR

RR

Rv

VV

Vv

RR

RR

Rv

ssss

60

2

120

120

30

15

10

5

30

4

30

4

120

120

30

15

10

5

15

20

6

120

120

30

15

10

5

10

10

12

120

120

30

15

10

5

5

43

21

4

43

21

33

43

21

22

43

21

11

==

++

+=

++

+=

==

++

+=

++

+=

==

++

+=

++

+=

==

++

+=

++

+=


Ω=

+=

92

.2

51

71

1eqR

()

()(

)

()

()(

)

()

()(

)V

VV

vR

RR

Rv

VV

Vv

RR

R

Rv

VV

Vv

RR

R

Rv

seq

seqeq

seq

06

.8

92

.9

20

420

492

.2

3

4

89

.5

92

.9

20

92

.2

20

492

.2

3

92

.2

05

.6

92

.9

20

320

492

.2

3

3

41

44

41

2

41

11

==

++

=+

+=

==

++

=+

+=

==

++

=+

+=


Use

the

curren

t div

isio

n p

rinci

ple

to

find the

curren

ts


() ()

Ai

RR

R

RR

i

Ai

RR

R

Ri

tt

23

15

20

10

30

13

15

20

10

15

32

1

21

3

32

1

31

=+

+=

++

+=

=+

+=

++

=


5

101

101

1=

+=

eqR

() ()

()

Ai

RR

Ri

Ai

RR

Ri

Ai

RR

Ri

seq

eq

seq

eq

seq

eq

13

10

5

5

13

10

5

5

13

10

5

5

13

12

11

=+

=+

=

=+

=+

=

=+

=+

=


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isto

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