schedule…

Discussion #5 – Ohm’s LawECEN 301 1

Date Day ClassNo.

Title Chapters HWDue date

LabDue date

Exam

17 Sept Wed 5 Ohm’s Law 2.6

18 Sept Thu LAB 1

19 Sept Fri Recitation HW 2

20 Sept Sat

21 Sept Sun

22 Sept Mon 6 Things Practical 2.6 – 2.8

LAB 2

23 Sept Tue

24 Sept Wed 7 Network Analysis 3.1 – 3.2

Schedule…


Divided We FallMatthew 12:25-26 25 And Jesus knew their thoughts, and said unto them,

Every kingdom divided against itself is brought to desolation; and every city or house divided against itself shall not stand:

Nephi 7:2 2 And the people were divided one against another; and

they did separate one from another into tribes, every man according to his family and his kindred and friends; and thus they did destroy the government of the land.


Lecture 5 – Resistance & Ohm’s Law


Open and Short Circuits

Short circuit: a circuit element across which the voltage is zero regardless of the current flowing through itResistance approaches zeroFlow of current is unimpededex: an ideal wire

• In reality there is a small resistance i

R = 0v = 0v

+

–



Undesirable short circuit – accidental connection between two nodes that are meant to be at different voltages. The resulting excessive current causes: overheating, fire, or explosion



Open circuit: a circuit element through which zero current flows regardless of the voltage applied to itResistance approaches infinityNo current flowsex: a break in a circuit

• At sufficiently high voltages arcing occurs i

R ∞i = 0v

+

–

Discussion #7 – Node and Mesh Methods

ECEN 301 7


i2+v2

–

vs+_

+v3

–

+ v1 –

R2

R1

R3

i3i1

Example: What happens to R2 and R3 if R1 is shorted? Vs = 2V, R1 = 2Ω, R2 = R3 = 4Ω, i1 = 500mA

R2 and R3 = ¼ W rating, R1 = ½ W rating


ECEN 301 8




i2+v2

–

vs+_

+v3

–

+ v1 –

R2

R1

R3

i3i1

V

mA

Riv

1

2500111

V

vvv

vvvs

1

0

122

21

mA

R

vi

2504

12

22


ECEN 301 9


i2+v2

–

vs+_

+v3

–

+ v1 –

R2

R1

R3

i3i1

W

mW

VmA

viP

2

1

500

1500111



W

mW

VmA

viP

4

1

250

1250222


ECEN 301 10


i2+v2

–

vs+_

+v3

–R2

R3

i3i1



V

vv

vv

s

s

2

0

2

2

mA

R

vi

5004

22

22

W

VmA

viP

1

2500222

BAD! R2 and R3 are damaged


Series Resistors Series Rule: two or more circuit elements are said to

be in series if the current from one element exclusively flows into the next element. From KCL: all series elements have the same current

N

nnEQ RR

1

R1

∙ ∙ ∙ ∙ ∙ ∙R2 R3 Rn RN

REQ


Series Resistors Demonstration of series rule: apply KVL and

Ohm’s law on the circuit

i

VR

Ri

RRRi

RiRiRi

vvvV

vvvV

sEQ

EQ

s

s

)(

)()()(

0

321

321

321

321

Vs

+_

+ v1 –

R1

– v3 +

R3

R2

+v2

–

1.5V

i


Series Resistors Voltage divider: the voltage across each resistor in a

series circuit is directly proportional to the ratio of its resistance to the total series resistance of the circuit

NB: the ratio will always be <= 1

EQ

n

R

R

sEQ

n

sNn

nn

VR

R

VRRRRR

Rv

321


Series Resistors Voltage divider: the voltage across each resistor in

a series circuit is directly proportional to the ratio of its resistance to the total series resistance of the circuit

sEQ

VR

Rv 1

1 Vs

+_

+ v1 –

R1

– v3 +

R3

R2

+v2

–

1.5V

i

sEQ

VR

Rv 2

2 sEQ

VR

Rv 3

3


Series Resistors

Example1: Determine v3

Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω

Vs

+_

– v1 +

R1

+ v3 –

R3

R2

+v2

–

i


Series Resistors

Example1: Determine v3

Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω

V

V

V

s

1

324

8

R

R v

:Divider oltage Using

EQ

33

Vs

+_

– v1 +

R1

+ v3 –

R3

R2

+v2

–

i

24

8610

:Rule Series Using

321 RRRREQ


Parallel Resistors Parallel Rule: two or more circuit elements are said to

be in parallel if the elements share the same terminalsFrom KVL: the elements will have the same voltage

N

EQNEQ

RRRR

RRRRRR 1111

111111

321

321

+R1

–

+R2

–

+R3

–

+Rn

–

+RN

–

+REQ

–

NB: the parallel combination of two resistors is often written: R1 || R2


Parallel Resistors Demonstration of parallel rule: apply KCL and

Ohm’s law on the circuit

i1 i2 i3

+R1

–

is+R2

–

+R3

–

+

v

–

Node a

sEQ

EQ

s

s

i

vR

Rv

RRRv

R

v

R

v

R

v

iiii

iiii

1

111

0

:a nodeat KCL

321

321

321

321


Parallel Resistors Current divider: the current in a parallel circuit divides in

proportion to the resistances of the individual parallel elements

sn

EQ

s

EQ

n

s

Nn

nn

iR

R

i

R

R

i

RRRRR

Ri

1

1

11111

1

321

NB: the ratio will always be <= 1 n

EQ

R

R


Parallel Resistors Current divider: the current in a parallel circuit

divides in proportion to the resistances of the individual parallel elements

sEQ

sEQ

iR

R

iR

Ri

1

11 /1

/1

sEQ i

R

Ri

22

sEQ i

R

Ri

33

i1 i2 i3

+R1

–

is+R2

–

+R3

–

NB: it makes sense that the smaller the resistor, the larger the amount of current that will flow.


Parallel Resistors

Example2: find i1R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A

i1 i2 i3

–R1

+

is–

R2

+

–R3

+


Parallel Resistors


13

2020

11021

201

21

101

1

1111

terminalsshare elements all :

321 RRR

REQ

NB

i1 i2 i3

–R1

+

is–

R2

+

–R3

+

A

i

R

R i s

EQ

615.065

40

4

2013101

1

1

:dividercurrent Using

11


Parallel Resistors


s

EQ

EQs

EQs

sEQ

sEQ

sEQ

s

s

i

R

Ri

R

RRRi

iR

Ri

R

Ri

R

R

iiii

iiii

/1

/1

/1

/1/1/1

/1

/1

/1

/1

/1

/1

0

:dividingcurrent with holds KCLt Verify tha

321

321

321

321

i1 i2 i3

–R1

+

is–

R2

+

–R3

+


Parallel Resistors The parallel combination of resistors is often written:

For two resistors: R1 || R2

For three resistors: R1 || R2 || R3 etc.

In each case REQ for the parallel combination must be found

213132

321

321

213132

321

321

1

1111

||||

RRRRRR

RRR

RRRRRRRRR

RRR

RRR

12

21

21

12

21

21

1

111

||

RR

RR

RRRR

RR

RR

+R1

–

+R2

–

+R3

–

+R1

–

+R2

–

+R1|| R2|| R3

–

+R1|| R2

–


Parallel Resistors Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

+R2

–

vs+_

+R3

–

+ R1 –

+

v

–



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

+R2

–

vs+_

+R3

–

+ R1 –

+

v

–

isR2 || R3

vs+_

+ R1 –

+

v

–



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

isR2 || R3

vs+_

+ R1 –

+

v

–

s

s

s

s

s

sEQ

vRRRRRR

RR

vRRRRRR

RR

RR

RR

v

RRRRRRRRRR

RR

v

RRRR

RRR

RR

vRRR

RR

vR

RRv

)(

)()(

1

1

)||(

||

||

:divider voltage Using

323121

32

323121

32

32

32

32

32312132

32

32

321

32

32

321

32

32



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

isR2 || R3

vs+_

+ R1 –

+

v

– )(210

5

)5()101010(

10

)(

:divider voltage Using

33

3

33

33

26

33

323121

32

VR

R

VRR

R

vRRRRRR

RRv s

NB: notice how R3 controls v



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

V

V

VR

Rv

kR

3

5

)(10210

105

)(210

5

:1 If

33

3

33

3

3

isR2 || R3

vs+_

+ R1 –

+

v

–


)(12

5

)(102.010

105.0

)(210

5

:1.0 If

33

3

33

3

3

V

V

VR

Rv

kR



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

0v

:0 As 3

R


2

3

v

:wordsother In 2

5v

: As

v

R

is

+R2

–

vs+_

+R3

–

+ R1 –

+

v

–



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

2

3

Rgh througoes

current no ords,other wIn

0v

:0 As

R

is

+R2

–

vs+_

+ R1 –

+

v

–



vs = 5V, R1 = 1kΩ, R2 = 1kΩ

)R through goesent curr (no

v

:wordsother In 2

5v

: As

3

2

3

v

R

is

+R2

–

vs+_

+ R1 –

+

v

–

schedule…

Documents

w rating ecen

current r1 r2r3rnrnreqecen

ohms law ecen

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ohms law5open

ohms law3lecture

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