# Basic Laws [§â€¸‡®¹ˆ¨Œ‡¼ˆ] - National Chiao Tung Basic Laws ¢â‚¬¢Ohm¢â‚¬â„¢s Law (resistors) ¢â‚¬¢Nodes, Branches,

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• 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:

A R

 

Rv +

_

i

1 = 1 V/A

Cross-section area A

Meterial resistivity  

ohm

Riv 

• 2012/9/17

2

Resistors

0Riv == R = 0v = 0

+

_

i

R = v

+

_

i = 0

0 R v

limi R

== ∞→

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.

• 2012/9/17

3

Conductance and Power Dissipation

•Conductance G is represented by

v i

R G 

1 1 S = 1 = 1 A/V

siemens mho

G i

Gvivp

R v

Riivp

vGi

2 2

2 2

===

===

=

positive R : power absorption (+)

negative R: power generation (-)

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

C, L, v, i)

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

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

at 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

• 2012/9/17

4

Continued Elements 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 a powerful set of tools for analyzing resistive circuits.

• 2012/9/17

5

KCL

i1 i2

in 0211  nn

N

n iiii 

•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

tti dt

tdq ttq

dttitq

titi

T T

T

TT

n

N

nT



 



 

i1 i2

in

• 2012/9/17

6

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

IT IT

321 IIIIS 

• 2012/9/17

7

Case with A Closed Boundary

cancelled.arecurrentsbranchInternal

0

0

0

1111











baab

n bn

m am

j bj

i ai

iiii

ii

ii

a

Treat the surface as a big node

leavingentering ii 

b

ia1 ib1

KVL

0 1

  m

M

m v

•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+ _

• 2012/9/17

8

Proof of KVL

 

proved)(KVLanyfor0)( 0)(

anyfor0)()()()( )(

anyforconstant)( givesenergyofonconservatitheofprincipleThe

)()()(

)()(

1

1

ttv ti

ttitvtitv dt

tdw ttw

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

• 2012/9/17

9

Example 2

321

321 0

VVVV

VVVV

ab

ab

 

V3

V2

V1

Vab

+_

+_

+_

+

_

a

b

Vab +_

+

_

a

b 321 VVVVS 

Example 3 Q: Find v1 and v2. Sol:

V12,V8 A4 205

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

(2)020 givesKVLApplying

(1)3,2 ,lawsOhm'From

21

21

21

  







vv i i

ii

vv

iviv

v1+ _

v2 +

_

20V

2

3+_ i

• 2012/9/17

10

Example 4 Q: Find currents and voltages. Sol:

  (3)8330 03830

030 1,looptoKVLAppying

(2)0 givesKCL,nodeAt

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

21

21

21

321

332211

ii

ii

vv

iii a

iviviv

 







V6V,6V,24

A1A,3A2 gives(2)Eq.(5),Eq.&(3)Eq.By

(5)236 (1),Eq.By

(4)0 2,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,A1 053035

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

(2)0235 givesKVLApplying

(1)5,10 ,lawsOhm'From

 





o

oxx

ox

vi ii

vvv

iviv

i

• 2012/9/17

11

Series Resistors

 

(5)

,Let

(4)or

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

(2)0 KVL,Applying

(1), ,lawsOhm'By

21eq

eq

21

2121

21

2211

RRR

iRv RR

v i

RRivvv

vvv

iRviRv



 





 v1+ _

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

Voltage Division

v R R

v RR

R iRv

v R R

v RR

R iRv

eq

2

21

2 22

eq

1

21

1 11

 



 

 v1+

_

v

R1

+_

i

v2+ _

R2a

b

v +_

i

v+ _

Reqa

b

• 2012/9/17

12

Continued

v R R

v RRR

R v

GGGG

RRRR

eq

n

N21

n n

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

2 2

1 1

2211

RR RR

R

RRR

R v

RR v

R v

R v

i

iii a

R v

i R v

i

RiRiv

eq  



 

  

 





 i a

b

R1+_ R2v

i1 i2

i a

b

Req or Geq+_v v

• 2012/9/17

13

Current Division

i GG

G RR iR

R v

i

i GG

G RR iR

R v

i

RR RiR

iRv

21

2

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