Basic Laws

Instructor: Chia-Ming Tsai

Electronics Engineering

National Chiao Tung University

Hsinchu, Taiwan, R.O.C.

Contents

• 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

Aρ

i

vR

==

Rv+

_

i

1 = 1 V/A

Cross-sectionarea A

Meterialresistivity

ohm

Resistors

0Riv == R = 0v = 0

+

_

i

R = v

+

_

i = 0

0R

vlimiR

==∞→

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 resistors exhibit nonlinear behavior.

Conductance and Power Dissipation

• Conductance G is represented by

v

i

R

1G == 1 S = 1 = 1 A/V

siemens mho

G

iGvivp

R

vRiivp

vGi

22

22

===

===

=

A positive R results in power absorption.

A negative R results in power generation.

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

C, L, v, i)

• Node: a point of connection between braches (a, b, c)

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

t least one branch which is not included in other indep. loops.

– Independent loops result in independent sets of equations.

+_

a

c

b

+_

c

ba

redrawn

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).

• Combined with Ohm’s law, we have a powerful set of tools for analyzing circuits.

• Two laws included, Kirchhoff’s current law (KCL) and Kirchhoff’s votage law (KVL)

Kirchhoff’s Current Law (KCL)

i1

i2

in

01

n

N

ni

• 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

(KCL) any for 0)()(

any for 0)( Thus

)()(

)()(1

ttidt

tdq

ttq

dttitq

titi

TT

T

TT

n

N

nT

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

Case with A Closed Boundary

surface closed theleaving

surface closed theentering

i

i

Treat the surfaceas a node

Kirchhoff’s Voltage Law (KVL)

01

m

M

mv

• Statement– The algebraic sum of all voltages

around a closed path (or loop) is zero.

v1+ _ v2+ _ vm+ _

Example 1

41532

54321

0

vvvvv

vvvvvvRT

v4v1

v5

+_ +_

+_

v2+ _ v3+ _

Sum of voltage drops = Sum of voltage rises

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:

V 12 ,V 8

A 4

205

03220

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

(2) 020

gives KVL Applying

(1) 3 ,2

,law sOhm' From

21

21

21

vv

i

i

ii

vv

iviv

v1+ _

v2

+

_

20V

2

3+_ i

Example 4Q: Find currents and voltages.

Sol:

(3) 8330

03830

030

1, loop toKVL Appying

(2) 0

gives KCL , nodeAt

(1) 6 ,3 ,8

,law sOhm'By

21

21

21

321

332211

ii

ii

vv

iii

a

iviviv

V 6 V, 6 V, 24

A 1 A, 3A 2

gives (2) Eq. (5), Eq. & (3) Eq.By

(5) 236

(1), Eq.By

(4) 0

2, loop toKVL Appying

321

312

2323

2332

vvv

iii

iiii

vvvv

v1+ _

30V

8

3+_

i1

6+

_v3

i3

i2

Loop 1 Loop 2

a

+

_v2

b

Series Resistors

(5)

, Let

(4) or

(3)

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

(2) 0

KVL, Applying

(1) ,

,law sOhm'By

21eq

eq

21

2121

21

2211

RRR

iRv

RR

vi

RRivvv

vvv

iRviRv

v1+ _

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

Voltage Division

vR

Rv

RR

RiRv

vR

Rv

RR

RiRv

eq

2

21

222

eq

1

21

111

v1+ _

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

Continued

vR

Rv

RRR

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)

, nodeat KCL Applying

(1) ,or

,law sOhm'By

21

21

21eq

eq2121

21

22

11

2211

RR

RRR

RRR

R

v

RRv

R

v

R

vi

iii

a

R

vi

R

vi

RiRiv

eq

i a

b

R1+_ R2v

i1 i2

i a

b

Req or Geq +_v v

Current Division

iGG

G

RR

iR

R

vi

iGG

G

RR

iR

R

vi

RR

RiRiRv

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

iG

Gi

GGG

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

iG

Gi

GGG

Gi

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

vR

Rv

RRR

Rv

GGGG

RRRR

eq

n

N21

nn

N21eq

N21eq

1111

Brief Summary

i a

b

R1+_ R2v

i1 i2

RN

iNv1+ _v

R1

+_

i

v2+ _

R2a

b

vN+ _

RN

Example

Req

6 35

8

2

4 1

Req

26

8

2

4

Req 2.48

4

Req 14.4

How to solve the bridge network?

R1

+_vS

R2 R3

R4

R5R6

• Resistors are neither in series nor in parallel.

• Can be simplified by using 3-terminal equivalent networks.

Wye (Y)-Delta () Transformations

R3

R1 R2

1

2

3

4

R3

R1 R2

3

4

1

2

Rb

Rc

1

2

3

4

RaRb

Rc

1

2

3

4

Ra

Y T

to Y Conversion

cba

bac

cab

RRRRRRR

RRRRRRR

RRRRRRR

//)()Y(

//)()Y(

//)()Y(

323434

211313

311212

cba

ba

cba

ac

cba

cb

RRR

RRR

RRR

RRR

RRR

RRR

3

2

1

R3

R1 R2

3

4

1

2

Y

Rb

Rc

1

2

3

4

Ra

Y- Transformations

3

133221

2

133221

1

133221

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

c

b

a

cba

ba

cba

ac

cba

cb

RRR

RRR

RRR

RRR

RRR

RRR

3

2

1

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 N

Vvvv 0

N21 ...

Applications: DC Meters

Parameters: IFS and Rm

Continued

RR V R

RR A

R

Continued

elementmeter

elementmeter

VV

II

elementmeter

elementmeter

VV

II

Voltmeters

Single-range

Multiple-range

Ammeters

Single-range

Multiple-range