Part V References and Appendix

209

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Appendix A The Questionnaires applied to the students A.1

Questionnaire 1 - Catalonia

This is the first questionnaire, that was applied to the students in Catalonia. Refer to Study 1 for the results.

221

222

A. Questionnaires

NOMBRE: 1.- ¿Cuál es tu sentir hacia la asignatura de Análisis de Circuitos I? a) Interesante b) Aburrida f) Importante g) Fácil ¿por qué?

c) Incomprensible h) Complicada

d) Indiferente i) Necesaria

e) Intrascendente

2.- ¿Crees que es necesaria esta asignatura para lo que desarrollarás en el ámbito laboral? ¿qué crees que te aportará?

3.- Menciona un tema (de Análisis de Circuitos I) que consideres más difícil de asimilar y ¿cuál crees que sea la causa?

4.- ¿Qué tema (de Análisis de Circuitos I) es el que más recuerdas? 5.- ¿Crees que es necesario ser muy hábil en matemáticas para comprender el Análisis de Circuitos Eléctricos? ¿por qué?

En las siguientes preguntas puedes indicar mas una opción. 6.- Al resolver los problemas tu interés estuvo centrado en: a) Llegar al resultado correcto aunque no siempre supe interpretarlo. b) Saber que método matemático debo utilizar o que teorema aplicar. c) Entender el problema para darme una idea de lo que tengo que obtener. d) Resolverlo rápidamente con la calculadora. e) ¿otra?

1

A.1 Questionnaire 1 - Catalonia

223

7.- Tu actitud durante la clase, al resolver problemas, fue:

a)

c)

No entiendo...

b)

d)

Es muy sencillo!!

e) Explícame! ¿a qué se refiere el profesor?

Mañana es el partido del Barça...

No me queda claro todo pero después lo reviso con calma

¿otra?

8.- Señala con cual de las siguientes opciones te identificabas al resolver problemas de la asignatura:

a)

b)

Entiendo los problemas y, a veces, sin necesidad de métodos o teoremas llego al resultado

No entiendo el fenómeno físico

c)

¡No es posible! Fallé por errores matemáticos

d)

De tantos métodos matemáticos ya no sé cual usar

e)

Sin entender mucho el fenómeno físico, con los métodos matemáticos siempre llego al resultado.

f) ¿Otra?

2

224

A. Questionnaires

EN CADA PREGUNTA SELECCIONA UNA RESPUESTA. 9.- Si una resistencia da una lectura infinita es que existe..... a) cortocircuito b) circuito abierto c) corriente continua

d) corriente alterna

10.- Si la tensión entre los extremos de una resistencia se mantiene constante , la corriente de la resistencia es ...... a su valor ohmico. a) igual b) directamente proporcional

c) inversamente proporcional

d)independiente

11.- El valor ohmico total (R t) de varias resistencias conectadas en serie es ..... que la resistencia de cualquiera de estas resistencias consideradas individualmente. a) menor b) igual c) proporcional d) mayor 12.- En un circuito serie la corriente ...... en todos los puntos. a) es diferente

b) se distribuye

c) se incrementa

d) es la misma

13.- “El Teorema de ...... establece que: en cualquier red lineal excitada con diferentes fuentes, la respuesta a través de cualquier elemento es determinada mediante la suma algebraica de las respuestas parciales producidas por cada una de las fuentes actuando por separado cuando las demás están nulificadas”. a) Norton b)Thevenin c) Superposición d)Sustitución INDICA SI ES FALSO O VERDADERO EN CADA ENUNCIADO. 14.- “Las tensiones entre los extremos de cada rama de una red en paralelo no son iguales” (F/V) 15.-“Resistencias de 22Ω, 33Ω y 47Ω están conectadas en paralelo. Su resistencia total será menos que 22Ω” (F/V) 16.-“La corriente puede existir solamente en un circuito cerrado” (F/V) DESARROLLA LO QUE SE TE PIDE.

17.-Muestra como combinar cuatro resistores de 100Ω para obtener una resistencia equivalente de 60Ω

100Ω 100Ω 100Ω

60Ω

100Ω

3

A.2 Questionnaire 2 - Catalonia

A.2

225

Questionnaire 2 - Catalonia

This is the second questionnaire, that was applied to the students in Catalonia. Refer to Study 1 for the results.

226

A. Questionnaires

NOMBRE 1.

Encuentra la corriente en el inductor y justifica tu respuesta:

4Ω

i

1H ¼F 6Ω

2. Observa el siguiente problema y contesta lo que se te pide: A un condensador de 5µ f se le aplica un voltaje igual a un escalón unitario. Encontrar la corriente. Solución: La corriente del condensador está dada por:

I =C

dV a)Explica, con tus propias palabras lo que representa la expresión (1): dt ... ( 1 )

en donde: C = 5µf = 5x10-6 farads V= U(t) volts

1

A.2 Questionnaire 2 - Catalonia

227

Sustituyendo en ( 1 ):

I = 5 x10− 6

dU amps. ... (2) dt

Se define el “Impulso Unitario” denotado δ(t), como una función de área unitaria tal que:

⎡U (t ) − U (t − σ ) ⎤ dU ⎥ = dt ...(3) σ ⎦ ⎣

δ (t ) = lim ⎢ σ →0

b) Explica ¿qué representa la expresión (3)?

Sustituyendo (3) en (2):

I = 5 x10−6 δ (t ) amps. Es decir, la corriente es una función de la forma:

I

5x10-6 δ(t) t

Por ello se dice que un condensador se comporta como un “circuito abierto” en el estado permanente ante voltajes aplicados constantes

c)Ahora explica con tus propias palabras una justificación del resultado anterior (¿a qué se refiere?):

2

228

A. Questionnaires

3. Menciona un ejemplo (de la vida real) donde tu observes que se aplique el análisis de los circuitos eléctricos:

4. Explica con tus propias palabras el concepto de electricidad:

5. Explica ¿Cuándo se produce electricidad?

6. ¿En donde observas que se utiliza la corriente alterna?

7. Explica lo que entiendes por caída de tensión

8. Si la siguiente gráfica representa la característica VL-IL de una fuente de corriente de laboratorio ¿cuánto vale la resistencia interna RS y la corriente de cortocircuito?

IL 3 2

VL 5 3

A.3 Questionnaire 3 - Mexico

A.3

229

Questionnaire 3 - Mexico

This is the third questionnaire, that was applied to the students in Mexico. Refer to Studies 2 and 3 for the results.

230

A. Questionnaires

CUESTIONARIO Nombre: 1.- Dibuja como va la corriente para cada bombilla del circuito de la figura. Ordena las bombillas según su luminosidad (las bombillas son idénticas).

B1

B2 B3

B4

B5

2.-En el siguiente circuito todas las bombillas son del mismo tipo. Completa las intensidades I1, I2, Ig en los cables: I1 I2

I=12A Ig

3.- ¿Cómo defines la Diferencia de Potencial o voltaje?

A.3 Questionnaire 3 - Mexico

231

4.-En los circuitos de la siguiente figura R1>R2>R3. Razona para cada circuito cuál es la resistencia que disipa más potencia. R1 R2 R3 a)

R1

R2

R3

b)

5.- Explique ¿Qué diferencia existe entre Corriente Continua y Corriente Alterna?

t

t

232

A. Questionnaires

6.- Observa el siguiente problema y contesta lo que se te pide: A un condensador de 5µ f se le aplica un voltaje igual a un escalón unitario. Encontrar la corriente.

5µf

V, I Solución: La corriente del condensador está dada por:

a)Este modelo matemático (1) ¿qué representa físicamente?: I =C

dV dt ... ( 1 )

en donde: C = 5µf = 5x10-6 farads V= U(t) volts

Sustituyendo en ( 1 ):

I = 5 x10− 6

dU amps. ... (2) dt

b)¿Cuánto vale la derivada de un Escalón Unitario? ¿por qué?

Se define el “Impulso Unitario” denotado δ(t), como una función de área unitaria tal que:

⎡U (t ) − U (t − σ ) ⎤ dU ⎥ = dt ...(3) σ ⎦ ⎣

δ (t ) = lim ⎢ σ →0

A.3 Questionnaire 3 - Mexico

233

siendo una función que vale _______________ para todo tiempo distinto de cero y que en t=0 tiene un valor muy grande.

c)¿Cómo se grafica el Escalón Unitario (con respecto al tiempo)? ¿por qué? n

-t

t -n

Sustituyendo (3) en (2):

I = 5 x10−6 δ (t ) amps. Es decir, la corriente es una función de la forma:

I

5x10-6 δ(t) t

d)¿Por qué se dice que el condensador se comporta como circuito abierto en el estado permanente ante voltajes aplicados constantes?

234

A. Questionnaires

7.- Encuentra las diferencias de potencial que se indican:

R1

C

A

3V

6V

3V

R3

R5

0V

12V

R4

12V

VP

C

VC

R2 B

D

3V VCD= VAB= VP = VC =

8.-Diga si son verdaderas o falsas las siguientes proposiciones: a) Para que haya diferencia de potencial es necesario que haya corriente. ( V/F) ¿por qué?

b) Para que haya corriente es necesario que haya diferencia de potencial. ( V/F) ¿por qué?

A.3 Questionnaire 3 - Mexico

c) El voltaje es el impulso de la corriente eléctrica (V / F) d)El voltaje es la fuerza de la corriente eléctrica (V / F) e)La corriente eléctrica es energía (V / F) f)El voltaje es energía (V / F) g)El voltaje es la causa de la corriente eléctrica (V / F) h)La corriente eléctrica es causa del voltaje (V / F) i)El voltaje se puede percibir tocando ligeramente un alambre (V / F) j)La corriente eléctrica se puede percibir tocando ligeramente un alambre (V / F) k)Para hacer funcionar un dispositivo eléctrico tu necesitas una corriente eléctrica V / F) l)Para hacer funcionar un dispositivo eléctrico tu necesitas un voltaje (V / F) m)La corriente eléctrica fluye (V / F) n)El voltaje fluye (V / F) o)El voltaje y la corriente eléctrica siempre ocurren juntos (V / F) p)El voltaje puede ocurrir sin la corriente eléctrica (V / F) q)La corriente eléctrica puede ocurrir sin el voltaje (V / F)

235

236

A. Questionnaires

9.-Para cada uno de los circuitos indica cuales bombillas no encienden y ¿por qué?

2

a)

1

2

b) 1

c) 1

3 2

d)

A.3 Questionnaire 3 - Mexico

237

10.-¿Cuál es el circuito equivalente de Norton del circuito de la figura?

1Ω

4Ω

+ 2V

3Ω

1A

2Ω

-

11.-Mencione un ejemplo práctico donde se utiliza Corriente Continua y ¿a tu ejemplo se le pude suministrar Corriente Alterna? ¿por qué?

238

A. Questionnaires

12.- En cada uno de los circuitos dibujados se tienen 3 bombillas. L1, L2, L3 expresan la luminosidad de las bombillas 1, 2 y 3. Señale las afirmaciones verdaderas y explique por qué.

1 a) L1>L2>L3 b) L1L3 b) L1=L2=L3 c) L1=L2>L3 d) L1=L2 0 U =⎨ ⎩0V , t ≤ 0 Sätt in i (1): t

I = 10

2

∫ 20U (t )dt = 2000t[ A]

b) Vad representerar detta uttryck fysikaliskt?

0

Funktionen får detta grafiska utseende: c) Beskriv vad figuren visar? I

t

252

A. Questionnaires

d) På vilket sätt liknar alltså kretsen med spolen och en spänningskälla en kortslutnen krets, och vad skiljer den från den kortslutna kretsen?

9.-Betrakta nedanstående kretsproblem och den givna lösningen.

Bestäm Vc (t) för t ≥ 0. 1/2H a t=0 Vf=12V

+ _

i

+ R=1/5Ω

3F

Vc _

Figur 1

Lösning: Betrakta systemet initialt som anslutet källa, där allt hunnit svänga in sig.

till

en

12V-

Resulterande figur vid tiden t=0 blir då figuren nedan, där i(0)=60A och Vc(0)=12V a

a)Hur tolkar du värdena på i(0) och Vc(0) fysikaliskt?

i 12V

+ _

R=1/5Ω

Vc

b) Hur uppstår dessa värden? c) Vad händer i kondensatorn spolen och motståndet?

Figur 2

A.4 Questionnaire 3 - Sweden

253

Vi behöver ställa upp två differentialekvationer av första ordningen: Kirchoffs spänningslag ger i figur (1):

L

di + V c = 0 ......(1) dt

d) Hur lyder Kirchoffs spänningslag?

Kirchoffs strömlag ger i nod a: ic +

Vc −i = 0 R

Anta att ic=C(dVc/dt): C

dVc Vc + − i = 0 .......(2) dt R

Vi Laplace-transformerar ekvationerna 1 och 2 och får: L[sI (s) − I (0)] + Vc (s) = 0

e) Vad betyder bokstaven "s"?

och

f) Varför Laplace-transformerar vi?

C[sVc (s) −Vc (0)] +

Vc (s) − I (s) = 0 R

g) Vilken fysikalisk betydelse har dessa ekvationer?

Sätt in värdena på L, C, R, i(0) och Vc (0):

[sI(s) − 60] + VC (s) = 0 3[sVc (s) −12] + 5Vc (s) − I(s) = 0 1 2

Lös alltså ekvationssystemet: sI ( s) + 2Vc ( s) = 60 − I ( s ) + (3s + 5)Vc ( s ) = 36 g)Lös detta på egen hand för att erhålla I(t) och Vc(t)...

254

A.5

A. Questionnaires

Questionnaire 3 - Catalonia

This is the third questionnaire, that was applied to the students in Catalonia. Refer to Studies 2 and 3 for the results.

A.5 Questionnaire 3 - Catalonia

255

QÜESTIONARI Nom: 1.- Llegeix atentament el següent problema i contesta el que se’t demana: A un condensador de 5µ f se li aplica un voltatge igual a un escaló unitari. Trobeu el corrent.

5µf

U, I Solució: El corrent del condensador ve donada per:

a)Aquest model matemàtic (1) ¿què representa físicament?: I =C

dU ... ( 1 ) dt

on: C = 5µf = 5x10-6 F

⎧1, t > 0 U =⎨ ⎩0, t ≤ 0 Substituint a ( 1 ):

I = 5 x10− 6

dU A. ... (2) dt

b)¿Quant val la derivada d’un Escaló Unitari? ¿per què?

256

A. Questionnaires

Es defineix l’ “Impuls Unitari” denominat δ(t), com una funció d’àrea unitària tal que:

⎡U (t ) − U (t − σ ) ⎤ dU ⎥ = dt ...(3) σ ⎣ ⎦

δ (t ) = lim ⎢ σ →0

essent una funció que val _______________ per tot instant de temps diferent de zero i que a t=0 té un valor molt gran.

c)¿Com es representa gràficament l’Escaló Unitari (respecte del temps)? ¿per què? n

-t

t -n

Substituint (3) a (2):

I = 5 x10−6 δ (t ) A. És a dir, el corrent és una funció de la forma:

I

5x10-6 δ(t) t

d)¿Per què es diu que el condensador es comporta com a circuit obert en l’estat permanent amb voltatges aplicats constants?

A.5 Questionnaire 3 - Catalonia

257

2.- Troba les diferències de potencial que s’indiquen:

R1

C

A

3V

6V

3V

R3

R5

0V

12V

R4

12V

VP

C

VC

R2 B

D

3V VCD= VAB= VP = VC =

3.-Digues si són verdaderes o falses les següents proposicions i digues perque: a) Perquè hi hagi diferència de potencial és necessari que hi hagi corrent. (V/F)

b) Perquè hi hagi corrent és necessari que hi hagi diferència de potencial. (V/F)

c) El voltatge és l’impuls del corrent elèctric (V / F)

258

A. Questionnaires

d) El voltatge és la força del corrent elèctric (V / F)

e) El corrent elèctric és energia (V / F)

f) El voltatge és energia (V / F)

g) El voltatge és la causa del corrent elèctric (V / F)

h) El corrent elèctric és causa del voltatge (V / F)

i) El voltatge es pot percebre tocant lleugerament un fil elèctric (V / F)

j) El corrent elèctric es pot percebre tocant lleugerament un fil elèctric (V / F)

k) Si vols fer funcionar un dispositiu elèctric necessites un corrent elèctric (V/F)

l) Si vols fer funcionar un dispositiu elèctric necessites un voltatge (V / F)

m) El corrent elèctric flueix (V / F)

n) El voltatge flueix (V / F)

A.5 Questionnaire 3 - Catalonia

o) El voltatge i el corrent elèctric sempre esdevenen junts (V / F)

p) El voltatge pot donar-se sense el corrent elèctric (V / F)

q) El corrent elèctric pot esdevenir sense el voltatge (V / F)

4.-Dóna un exemple pràctic on es faci servir Corrent Continu i ¿al teu exemple se li pot subministrar Corrent Altern? ¿per què?

5.-Dóna un exemple pràctic on es faci servir Corrent Altern i ¿al teu exemple se li pot subministrar Corrent Continu? ¿per què?

259

260

A. Questionnaires

6..-Per cadas un dels circuits indica quines són les bombetes que no s’encenen i ¿per què?

2

a)

1

2

b) 1

c) 1

3 2

d)

A.5 Questionnaire 3 - Catalonia

261

7.-Determina els valors del Corrent elèctrica total (It), el Voltatge total (Vt) i la Resistència total (Rt) del següent circuit:

R1=10Ω VR1= 20V

It=

R2=5Ω VR2=

Vt= Rt=

R3=15Ω VR3=

262

A. Questionnaires

8.-Llegeix atentament el següent problema i contesta el que es demana.

A un inductor de 10 mH se li aplica un voltatge de 20 U (t) v. El corrent inicial a l’inductor és nul·la. Troba el corrent a l’inductor. 10 mH

V, I Solució: El corrent en la bobina ve donada per : t

I = Γ ∫ Vdt = Γ ∫ Vdt + I 0 ...(1)

a)Aquest model matemàtic (1) ¿què representa físicament?:

0

On: 1 Γ= 1 = = 10 2 yrnehs L 10 x10 − 3 I0 = 0 V = 20U (t )volts

Substituint a (1): t

I = 10 2 ∫ 20U (t )dt = 2000tU (t ) A 0

b) ¿què representa físicament aquest resultat?

A.5 Questionnaire 3 - Catalonia

263

Aquesta funció és de la forma: c)Interpreta físicament la gràfica dibuixada. I

t

d)¿Per què es diu que un inductor es comporta com un curtcircuit a l’estat permanent per a voltatges aplicats constants?

264

A. Questionnaires

9.-Observa el siguiente problema y contesta lo que se pide.

Determinar Vc (t) para t ≥ 0. 1/2Hy a t=0 Vf=12V

i

+ _

+ R=1/5Ω

Vc

3f _

figura1 Solución: Se determinan las condiciones iniciales considerando que el circuito lleva largo tiempo con la fuente de 12V conectada. Como se muestra en la siguiente figura resulta evidente que i(0)=60ª y Vc(0)=12V a a)¿Cómo interpretas (con tus conocimientos de física) los valores obtenidos en i(0) y Vc (0)?

i 12V

+ _

R=1/5Ω

Vc b)¿Por qué o de donde surgieron estos valores?

Figura 2

c) ¿Qué ocurre en el inductor, en el condensador (capacitor) y en el resistor?

A.5 Questionnaire 3 - Catalonia

265

Hacen falta dos ecuaciones diferenciales de primer orden en función de Vc e i para t≥=0. La ecuación de la LKV para la malla con i de la figura (1) es: d) A partir de la Teoría de Circuitos ¿cómo se llega a esa expresión? L

di + Vc = 0 ......(1) dt

La ecuación para la corriente del condensador (capacitor) ic en el nodo “a” es: ic +

Puesto que ic=C(dVc/dt): C

Vc −i = 0 R

dVc Vc + − i = 0 .......(2) dt R

Con las transformadas de Laplace de las ecuaciones (1) y (2) se obtiene: L[sI ( s ) − I (0)] + Vc ( s ) = 0

e)¿Qué significado tiene el termino “(s)”?

y

f) Por qué se hace esta sustitución y de dónde surge?

V (s) C [sVc ( s ) − Vc (0)] + c − I (s) = 0 R

g)¿Qué significado físico tienen estas identidades?

Sustituyendo los valores de L, C, R, i(0) y Vc (0) se obtiene: 1 [sI ( s) − 60] + Vc ( s) = 0 2 3[sVc ( s ) − 12] + 5Vc ( s ) − I ( s ) = 0

266

A. Questionnaires

Reordenando para aplicar la regla de Cramer: sI ( s) + 2Vc ( s) = 60

− I ( s) + (3s + 5)Vc ( s ) = 36

g)A partir de estas ecuaciones termina de resolver el problema para obtener el resultado que se pide...

Appendix B Tables of answers for Questionnaire 3

267

268

B. Tables of answers for Questionnaire 3

(a): Current is necessary to get Potential Difference (p): Voltage can occur without Current

No Answer

a\p True 8 25 False 21 1 True & False 0 0 No Answer 3 4 32 30

True & False

False

True

Mexico

0 0 0 0 0

1 1 0 2 4

34 23 0 9 66

Relation between questions (a) and (p), where the following was the highest frequency in each country 1. Mexico (a): True and (p): False

No Answer

a\p True 3 7 False 10 3 True & False 0 0 No Answer 0 0 13 10

2. Sweden

True & False

False

True

Sweden

0 0 0 0 0

1 0 0 1 2

(a): False and (p): True 3. Catalonia (a): False and (p): True 11 13 0 1 25

False

True & False

No Answer

a\p True False True & False No Answer

True

Catalonia

2 9 1 0 12

1 4 0 0 5

0 0 0 0 0

1 0 0 0 1

4 13 1 0 18

Table B.1: Relations between questions (a) and (p)

269 (p): Voltage can occur without Current (h): The Current is the cause of electric Voltage

No Answer

p\h True 14 17 False 20 8 True & False 0 0 No Answer 0 1 34 26

True & False

False

True

Mexico

1 0 0 0 1

1 2 0 2 5

33 30 0 3 66

Relation between questions (p) and (h), where the following was the highest frequency in each country 1. Mexico (p): False and (h): True

Sweden

True

False

True & False

No Answer

2 5 0 0 7

8 4 0 1 13

0 0 0 0 0

3 2 0 0 5

True

False

True & False

No Answer

p\h True False True & False No Answer

2. Sweden

5 1 0 0 6

6 4 0 0 11

0 0 0 0 0

0 0 0 1 1

(p): True and (h): False 3. Catalonia (p): True and (h): False 13 11 0 1 25

Catalonia

p\h True False True & False No Answer

11 5 0 1 18

Table B.2: Relations between questions (p) and (h)

270

B. Tables of answers for Questionnaire 3

(a): Current is necessary to get Potential Difference (h): The Current is the cause of electric Voltage

No Answer

a\h True 21 12 False 10 11 True & False 0 0 No Answer 3 3 34 26

True & False

False

True

Mexico

0 1 0 0 1

1 0 0 4 5

34 22 0 10 66

Relation between questions (a) and (h), where the following was the highest frequency in each country 1. Mexico (a): True and (h): True

Sweden

False

True & False

No Answer

0 0 0 0 0

0 2 0 0 2

False

True & False

No Answer

0 5 1 0 6

3 8 0 0 11

0 0 0 0 0

1 0 0 0 1

True

4 5 3 8 0 0 3 0 10 13

True

a\h True False True & False No Answer

2. Sweden (a): True and (h): False 3. Catalonia (a): False and (h): False 9 13 0 3 25

Catalonia

a\h True False True & False No Answer

4 13 1 0 18

Table B.3: Relations between questions (a) and (h)

271 (a): Current is necessary to get Potential Difference (o): Voltage and Current occur together

No Answer

a\o True 23 8 False 1 21 True & False 0 0 No Answer 4 3 28 32

True & False

False

True

Mexico

0 0 0 0 0

3 0 0 3 6

34 22 0 10 66

Relation between questions (a) and (o), where the following was the highest frequency in each country 1. Mexico (a): True and (o): True

Sweden

False

True & False

No Answer

7 3 3 11 0 0 0 0 10 14

0 0 0 0 0

1 0 0 0 1

True

a\o True False True & False No Answer

2. Sweden (a): False and (o): False 3. Catalonia (a): False and (o): True 11 14 0 0 25

and also (a): False and (o): False

False

True & False

No Answer

a\o True False True & False No Answer

True

Catalonia

2 5 1 0 8

1 5 0 0 6

0 0 0 0 0

1 3 0 0 4

4 13 1 0 18

Table B.4: Relations between questions (a) and (o)

272

B. Tables of answers for Questionnaire 3

(o): Voltage and Current occur together (p): Voltage can occur without Current

No Answer

o\p True 6 22 False 24 8 True & False 0 0 No Answer 1 2 31 32

True & False

False

True

Mexico

0 0 0 0 0

0 0 0 3 3

28 32 0 6 66

Relation between questions (o) and (p), where the following was the highest frequency in each country 1. Mexico (o): False and (p): True

No Answer

o\p True 0 10 False 14 0 True & False 0 0 No Answer 0 0 14 10

2. Sweden

True & False

False

True

Sweden

0 0 0 0 0

0 0 0 1 1

(o): False and (p): True 3. Catalonia (o): True and (p): True 10 14 0 1 25

and also (o): True and (p): False

False

True & False

No Answer

o\p True False True & False No Answer

True

Catalonia

3 7 0 2 12

3 1 0 1 5

0 0 0 0 0

0 0 0 1 1

6 8 0 4 18

Table B.5: Relations between questions (o) and (p)

273 (h): The Current is the cause of electric Voltage (o): Voltage and Current occur together

No Answer

h\o True 20 14 False 7 16 True & False 0 1 No Answer 1 1 28 32

True & False

False

True

Mexico

0 0 0 0 0

0 3 0 3 6

34 26 1 5 66

Relation between questions (h) and (o), where the following was the highest frequency in each country 1. Mexico (h): True and (o): True

Sweden

True

False

True & False

No Answer

5 3 0 1 9

2 10 0 3 15

0 0 0 0 0

0 0 0 1 1

True

False

True & False

No Answer

h\o True False True & False No Answer

2. Sweden

4 2 0 0 6

0 8 0 1 9

0 0 0 0 0

2 0 0 1 3

(h): False and (o): False 3. Catalonia (h): False and (o): False 7 13 0 5 25

Catalonia

h\o True False True & False No Answer

6 10 0 2 18

Table B.6: Relations between questions (h) and (o)

274

B. Tables of answers for Questionnaire 3

(c): A Voltage impulse will cause a Current (g): The Voltage is the cause of electric Current

No Answer

c\g True 22 13 False 9 11 True & False 0 0 No Answer 3 3 34 27

True & False

False

True

Mexico

0 0 0 0 0

1 1 0 3 5

36 21 0 9 66

Relation between questions (c) and (g), where the following was the highest frequency in each country 1. Mexico (c): True and (g): True

Sweden

False

True & False

No Answer

0 0 0 0 0

2 2 0 0 4

False

True & False

No Answer

3 4 0 2 9

3 4 0 1 8

0 0 0 0 0

0 1 0 0 1

True

3 2 0 0 4

True

2. Sweden

c\g True 13 False 3 True & False 0 No Answer 0 16

(c): True and (g): True 3. Catalonia (c): False and (g): True 18 7 0 0 24

and also (c): False and (g): False

Catalonia

c\g True False True & False No Answer

6 9 0 3 18

Table B.7: Relations between questions (c) and (g)

275 (d): The Voltage is the force driving Current (g): The Voltage is the cause of electric Current

No Answer

d\g True 21 11 False 9 16 True & False 0 0 No Answer 4 0 32 27

True & False

False

True

Mexico

0 0 0 0 0

1 0 0 4 5

33 25 0 8 64

Relation between questions (d) and (g), where the following was the highest frequency in each country 1. Mexico (d): True and (g): True

Sweden

False

True & False

No Answer

0 0 0 0 0

0 1 0 1 2

False

True & False

No Answer

3 4 0 2 9

2 4 0 2 8

0 0 0 0 0

0 0 0 1 1

True

3 2 0 2 7

True

2. Sweden

d\g True 15 False 1 True & False 0 No Answer 0 16

(d): True and (g): True 3. Catalonia (d): False and (g): True 18 4 0 3 25

and also (d): False and (g): False

Catalonia

d\g True False True & False No Answer

5 8 0 5 18

Table B.8: Relations between questions (d) and (g)

276

B. Tables of answers for Questionnaire 3

(g): The Voltage is the cause of electric Current (p): Voltage can occur without Current

No Answer

g\p True 15 19 False 16 10 True & False 0 0 No Answer 1 2 32 31

True & False

False

True

Mexico

0 0 0 0 0

0 1 0 2 3

34 27 0 5 66

Relation between questions (g) and (p), where the following was the highest frequency in each country 1. Mexico (g): True and (p): False

No Answer

g\p True 10 6 False 2 3 True & False 0 0 No Answer 2 1 14 10

2. Sweden

True & False

False

True

Sweden

0 0 0 0 0

0 0 0 1 1

(g): True and (p): True 3. Catalonia (g): True and (p): True 16 5 0 4 25

False

True & False

No Answer

g\p True False True & False No Answer

True

Catalonia

7 5 0 0 12

2 3 0 0 5

0 0 0 0 0

0 0 0 1 1

9 8 0 1 18

Table B.9: Relations between questions (g) and (p)

277 (c): A Voltage impulse will cause a Current (p): Voltage can occur without Current

No Answer

c\p True 21 16 False 8 12 True & False 0 0 No Answer 2 4 31 32

True & False

False

True

Mexico

0 0 0 0 0

0 0 0 3 3

37 20 0 9 66

Relation between questions (c) and (p), where the following was the highest frequency in each country 1. Mexico (c): True and (p): True

Sweden

False

True & False

No Answer

0 0 0 0 0

0 1 0 0 1

False

True & False

No Answer

6 5 0 1 12

0 3 0 2 5

0 0 0 0 0

0 1 0 0 1

True

8 10 6 0 0 0 0 0 14 10

True

c\p True False True & False No Answer

2. Sweden (c): True and (p): False 3. Catalonia (c): True and (p): True 18 7 0 0 25

Catalonia

c\p True False True & False No Answer

6 9 0 3 18

Table B.10: Relations between questions (c) and (p)

278

B. Tables of answers for Questionnaire 3

(d): The Voltage is the force driving Current (p): Voltage can occur without Current

No Answer

d\p True 15 18 False 15 9 True & False 0 0 No Answer 2 4 32 31

True & False

False

True

Mexico

0 0 0 0 0

0 1 0 1 3

33 25 0 7 66

Relation between questions (d) and (p), where the following was the highest frequency in each country 1. Mexico (d): True and (p): False

No Answer

d\p True 10 8 False 3 2 True & False 0 0 No Answer 1 0 14 10

2. Sweden

True & False

False

True

Sweden

0 0 0 0 0

1 0 0 0 1

(d): True and (p): True 3. Catalonia (d): False and (p): True 19 5 0 1 25

False

True & False

No Answer

d\p True False True & False No Answer

True

Catalonia

4 6 0 2 12

1 2 0 2 5

0 0 0 0 0

0 0 0 1 1

5 8 0 5 18

Table B.11: Relations between questions (d) and (p)

279 (h): The Current is the cause of electric Voltage (q): Current can occur without Voltage

No Answer

h\q True 8 25 False 10 15 True & False 0 1 No Answer 0 3 18 44

True & False

False

True

Mexico

0 0 0 0 0

0 1 0 3 4

33 26 1 6 66

Relation between questions (h) and (q), where the following was the highest frequency in each country 1. Mexico (h): True and (q): False

Sweden

True

False

True & False

No Answer

1 2 0 1 4

6 11 0 3 20

0 0 0 0 0

0 0 0 1 1

True

False

True & False

No Answer

h\q True False True & False No Answer

2. Sweden

0 3 0 0 3

6 11 0 0 14

0 0 0 0 0

0 0 0 1 1

(h): False and (q): False 3. Catalonia (h): False and (q): False 7 13 0 5 25

Catalonia

h\q True False True & False No Answer

6 14 0 1 18

Table B.12: Relations between questions (h) and (q)

280

B. Tables of answers for Questionnaire 3

Appendix C Xarxes Sist` emiques Technique to analyze data, called Xarxes Sist`emiques, from the Doctoral Thesis of Marina Castells (1997) University of Barcelona, Part 3, pages 1 through 10.

281

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C. Xarxes Sist`emiques

SYSTEMIC NETWORKS Systemic networks is a method developed by Bliss and Ogborn (1979); see also for example Bliss, Monk & Ogborn, (1983); Koulaidis & Ogborn, (1988); Castells, (1997) to describe complex qualitative data. In describing such data one is faced with the problem of representing the complexity, subtlety and detail of human transactions. In doing this one strategy is to report in terms of a relatively simple category scheme and the other is to present extensive quotations from the data itself, hoping that the essentials comes through. The kind of analysis made using systemic networks falls between the two alternatives presented above. Defined categories are used, but the method attempts to elaborate those categories to the point where enough of the individual essence of data is preserved and represented. These categories are elaborated using a notation derived from systemic linguistics to show their interdependencies. Network like structures are generated in which are shown which categories belong with others, which are independent, and which are conditional on choice of others. Making a network can be regarded as an extension of the work of putting things into categories. To categorize is to label things – to put things in boxes. A network is a graphical re-presentation of the set of boxes one has chosen to use and how they are related to each other. In making a network an arbitrarily complexity can be chosen. Any category scheme, such as systemic networks, needs names for categories. Any category could be divided in plausible subdivisions. The number of categories and subcategories used depends entirely on the nature of the material being used and the researchers choices about what would make most sense. Off course subcategories can have their own subcategories. In literature about systemic networks there exist several technical terms. In this short description we will only make the distinction between system and co-selection. System is related to the system of mutually exclusive choices made within a context of alternatives. Co-selection is related to parallel aspects, aspects that are not mutually exclusive, i. e. choices which are made simultaneous. A very simple network (which does not shows the merits of the method of systemic network) could be a presentation of students where it has been chosen that the important distinctions are gender (male/female) and parents school background (elementary school/secondary school/university). Male/female here belongs to one system and elementary school/secondary school/university to another system. Gender and school background are coselections. Male Gender Female

Student

Parent

University

School

Secondary School

Background

Elementary School

283

Co-selections can be presented in a so-called contingency table. Each system are analogous to a different independent dimension that could be placed as one of the axes of a table. The network presented above corresponds to a 2 × 3 table. One cell will then represent female students with parents having elementary school background, another cell male students with parents having university background etc … In real investigations the network can be more complicated and have a larger delicacy.

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ANÀLISI RESPOSTES AL QÜESTIONARI

1.3.2. Anàlisi de les dades. L'anàlisi consta de tres parts: 1)Una anàlisi aprofundida de les respostes que els estudiants donen a cada pregunta. 2)Una anàlisi de les interrelacions que hi ha entre les diferents respostes a les diferents preguntes del qüestionari. 3)Una anàlisi de les relacions entre els resultats de les diverses preguntes.

1.3.2.1. Anàlisi aprofundida de les respostes que els estudiants donen a cada pregunta En aquest estudi pregunta per pregunta s'ha començat fent una anàlisi qualitativa mitjançant les xarxes sistèmiques i donant les freqüencies que corresponen a cadascuna de les categories detectades en les respostes. Tècnica de les xarxes sistèmiques A partir de la lectura de les respostes dels estudiants, per escrit o mitjançant una entrevista, a unes qüestions, s'intenta establir unes categories, però a mesura que hom s'hi endinsa una mica s'adona que algunes d'aquestes categories s'exclouen mútuament, mentre que d'altres, en canvi, poden produir-se simultàniament. A més, moltes categories tenen subcategories, i aquestes, de vegades, noves subcategories. Així doncs, quan un s'enfronta a dades reals i les intenta recollir d'alguna manera, d'adona que entre elles hi ha una complexa interestructura de relacions i que, per tant, donar tan sols unes categories o uns nombres seria simplificar i empobrir la realitat de les respostes. El mètode, que expliquen Bliss, Monk i Ogborn al seu llibre Qualitativa Data Analysis for Educational Research (1983), permet plasmar de manera simplificada, fàcil de seguir amb la vista, la complexitat de les interrelacions que hi ha entre les dades reals, podríem dir que fa com un dibuix dels diferents aspectes del conjunt de les respostes. Una xarxa seria això: un resum quasi plàstic d'unes dades que, per a ésser descrites, necessitarien pàgines i pàgines i, reduïdes a nombres, perdrien molt del seu significat. El mètode no proporciona un esquema d'anàlisi universal, l'ús del mètode és personal, s'ha de produir una xarxa que serveixi per a la proposta de la investigació concreta i no ha de tenir necessàriament una posterior aplicació. De tota manera, no es descarta que certes idees usades en l'anàlisi d'un problema, d'un qüestionari o d'una entrevista, etc.... puguin fer-se servir en d'altres. Aquest mètode i la terminologia que utilitza deriven de la "systemic linguistics" que està interessada en la descripció i representació del significat del llenguatge. Tal com indica Bliss (1979), darrere de cada paraula escrita en el context d'una frase hi ha un significat no directament expressat per les paraules. L'anàlisi sistèmica vol recollir aquest significat dels sistemes de paraules i ha desenvolupat un poderós formalisme per a representar-lo. Aquest model d'anàlisi, que inicialment s'aplicava al camp de la lingüística, s'ha anat aplicant a anàlisis no lingüístiques. D'aquesta manera es fa servir per extreure, codificar i representar informació no lingüística (pensaments, sentiments, idees...). És per aquesta raó que pot ser útil en investigacions molt diverses. De tota manera, la xarxa descriu no tant les dades com la seva interpretació de cara a un objectiu. Podríem dir que les dades són filtrades a través dels interessos, preconcepcions i percepcions pròpies del que investiga. Per això, partint d'unes mateixes dades, segons el que es proposa el que les

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analitza, es poden construir xarxes molt diferents. L'investigador en construir la xarxa, amb l'objectiu de presentar les dades d'una forma que siguin fàcils de copsar, ja les interpreta, és a dir, ja les analitza, és per això que les xarxes sistèmiques seran considerades una forma de simplificació i de presentació de les dades al mateix temps que una forma d'anàlisi qualitativa d'aquestes mateixes dades. Els aspectes de notació usats en escriure una xarxa són molt simples. Molt poques i elementals notacions són usades. Podem dir que es poden reduir a notacions de dos tipus: 1) les que són per a indicar classes de tria o selecció

2) i les que són per a indicar condicions en la tria

o per a indicar la possibilitat de repetir la tria

.

Combinant adequadament tots els símbols per a les diferents categories i subcategories, es construeix la xarxa, que en alguns casos pot ser de gran complexitat. Un cop dissenyada una determinada xarxa, aquesta permet trobar múltiples interconnexions i així establir "models" de respostes. També pot servir per a veure els efectes d'unes categories sobre unes altres. A continuació presentem les notacions usades, seguint Bliss i altres (1983). Notacions usades a les xarxes sistèmiques

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(a)Categories excloents A

B

-A x------ -B -C

C

(b)Categories amb subcategories excloents -A1 A

B

C

-A-A2

A1 A2

C1 C2

Y----- -B -C1 -C-C2

(c)Categories excloents de dos aspectes excloents -A Aspecte P

A

B

-P-B Z

Aspecte Q

a

b

-a -Q-b

(d)Categories o aspectes que poden ser simultanis -A Aspecte P

A

B

P---B Z---

Aspecte Q

a

b

-a Q---b

(e)Condicions restrictives A A

B

són C

---C B

Comparacions de resultats. Anàlisi de les mitjanes per efectes i residus L'anàlisi de mitjanes per efectes i residus (Erickson i Nosanchuck, 1983) s'ha aplicat quan s'han volgut establir comparacions, dins d'un mateix problema, entre preguntes o entre diferents grups d'estudiants (segons el nivell de formació científica de base). Aquesta anàlisi ens permet de veure quines categories són més dominants a cada pregunta o grup.

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En aquesta anàlisi, les dades originals (freqüències absolutes o relatives, generalment simples percentatges) es transformen en unes altres dades -anomenades residus- restant-los un terme d'ajustament. Residu = Dada - terme d'ajustament El terme d'ajustament prové de considerar el valor mitjà -mitjana global- i els efectes -efecte de fila i/o de columna- definits com la diferència entre el valor mitjà de la fila o de la columna, respectivament, i la mitjana global. Terme d'ajustament = mitjana global + efecte de fila + efecte de columna Una anàlisi posterior dels residus obtinguts per aquest procediment ajudarà a establir les regularitats que es donen entre les variables objecte d'anàlisi. Encara que sembli el contrari, els residus no desvirtuen les dades, perquè es recuperen fàcilment: Dada = mitjana global + efecte de fila + efecte de columna + residu. Mosteller i Tukey (1977) proposen aquesta anàlisi per analitzar taules de doble entrada i posen especial èmfasi en aquests punts: 1.L'anàlisi fa, amb cura i quantitativament, el que sovint tractem de fer qualitativament quan "mirem" la taula. 2.No només ens fa l'anàlisi - generalment la fa millor- , sinó que exposa els residus a la nostra vista -ara podem "mirar-los" per veure el que està passant-; és més, podem, i generalment ho hauríem de fer, sometre'ls a posterior anàlisi quantitativa. 3. Ara tenim més nombres a la nostra taula que no quan vàrem començar, un fenomen essencial si hem d'analitzar i preservar les dades a la vegada. En un punt donat, podem reemplaçar tots els residus o cada un dels diversos grups de residus per descripcions sumarials amb l'objecte de fer decrèixer el nombre de valors a tenir en compte, però només hauríem de fer-ho després d'haver mirat detingudament tots els residus. 4.La seva versatilitat ofereix una gran llibertat a l'hora d'escollir l'anàlisi. Les versions estàndard tenen usos importants, però podem escollir l'anàlisi que ens interessi més. Aquesta anàlisi ens ha resultat especialment més aclaridora quan els efectes de fila o de columna eren zero, en aquest cas podíem evidenciar el domini d'unes categories d'una variable fixant l'altra. A continuació, presentarem dues aplicacions completes del mètode amb unes dades tretes de la nostra anàlisi per a il.lustrar com es procedeix. Cas de dades amb efectes de fila zero: correcció de les dues preguntes a) i b) del problema 1. Presentem la taula que serveix per a comparar els percentatges de correcció d'ambdues qüestions: Correcció p.1

correcta

incorrecta

no clara

no respon

Total a) (114)

(30) 26.3 -1.8

(67) 58.8 +7

(4) 3.5 -1.3

(13) 11.4 -4

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mitj. fila efec. fila 25 0

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(34) 29.8 +1.7

(51) 44.7 -7.1

(7) 6.1 +1.3

(22) 19.3 +3.9

mitj. col. efec. col.

28.1 +3.1

51.8 +26.8

4.8 -20.2

15.4 -9.6

25 0 m =25

Per a analitzar els residus es requereix una tècnica auxiliar que sintetitza la informació proporcionada per ells eliminant les dades supèrflues. El resultat d'aplicar aquesta tècnica és un resum numèric que pot quedar reflectit, si es vol, en un diagrama de caixa o box-plot. El resum numèric consta de cinc números obtinguts en ordenar els residus. Aquests cinc núneros determinen els tipus de residus: els que són considerats anomalies, bé superiors (X) o bé inferiors (O); els que estan per sobre o per sota dels quartils, però que no són anomalies (x o o); els que són inclosos dins de l'interval interquartilar (+. si són positius i -. si són negatius). Per tant, tindrem els residus classificats i ens diran quines categories són dominants i quines deficients. I, el que és més important, ho tindrem a la "vista". En els nostres resultats destaquem només els residus dominants, molt dominants, deficients i molt deficients. La notació que usem en les nostres taules no és la mateixa que proposen Erickson i Nosanchuck i que acabem de presentar, hem canviat els símbols X i O per trames amb diferent matís de negre. Presentem a continuació com operem, a partir dels residus de la taula: Residus +7 Anomalia superior = +11.2 +3.9 qU=+2.8 +1.7 +1.3 Md = 0 -1.3 -1.7 qL=-2.8 -3.9 dq = qU- qL = 5.6 -7.1 pas = 1.5x5.6 =8.4 Anomalia inferior = -11.2

Dades 58.8 44.7 qU=37.3 29.8 26.3 19.3 11.4 6.1 3.5

qL=8.8 dq=28.5 coef dq(R)/dq/(D) = 0.2

El coeficient ens indica si el model ajusta bé. Quan el valor del coeficient és proper a 1 vol dir que el model no ajusta bé, cosa que indica que hi ha interaccions entre les dades (tindrem residus molt grans). El model que ajustaria seria llavors un model no aditiu, hauríem de transformar les dades a logaritmes, però llavors sembla que ens allunyem més de les dades i ens costa més d'interpretar els resultats. Nosaltres no hem fet la transformació de les dades a logaritmes. Presentem el sumari de les dades del damunt en un quadre:

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XU Anomalia superior: Valor superior a qU Quartil superior Md Mediana QL Quartil inferior XL Anomalia inferior XL

qL O

-10

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+11.2 +2.8 0 -2.8 -11.2 Md

_. _

-5

dq(R)/dq(D )= 0.2

qU

XU

+.

0

x

+5

+10

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Plantejament i metodologia de la primera fase de la investigació

Cas de dades amb efectes de fila i de columna: correcció de les respostes a les preguntes del segon qüestionari Taula de percentatges, efectes de fila, de columna i residuals sobre les files i les columnes Correcció

p1(I)a)

p1(I)b)

p4(I)

p1(II)

mitjana fila efecte fila

Magisteri Ciències

59.7% +2.7 -2.2

7.6% +3.9 -1

46.6% +4.6 -0.3

38.3% +8.3 +3.4

38.1 +4.9

COU

67.6% +10.6 +7.1

2.9% -0.8 -4.3

47.6% +5.6 +2.1

28.8% -1.2 -4.7

36.7 +3.5

2n.BUP

43.6% -13.4 -5.0

0.7% -3 +5.4

31.9% -10.1 -1.7

23% -7 +1.4

24.8 -8.4

mitj. col. efec. col.

57.0 +23.1

3.7 -30.2

42.0 +8.1

30.0 -0.9

M=33.2

XU Anomalia superior: Valor superior a qU Quartil superior Md Mediana QL Quartil inferior XL Anomalia inferior -12.33

-3.25

+11.9 +2.8 -0.65 -3.25 -12.33 -0.65

_ .

-10

-5

dq(R)/dq(D )= 0.2

+2.8

+11.9

+ .

0

+5

+10

Anàlisi de les interrelacions. Taules de contingència i khi-quadrat Després de categoritzar els resultats en les xarxes, és a dir, de veure de quines maneres els estudiants responen, mirem si hi ha relacions de dependència entre les diverses categories i subcategories de la xarxa. Per això, construïm les taules de contingència per a cada parell de variables categòriques que intentem veure si estan relacionades. Les taules de contingència són taules de freqüència (nombre d'estudiants que responen de tal forma). Les cel.les podrien ésser transformades en proporcions o percentatges però el que és important assenyalar és que, qualsevol que sigui la forma en què siguin presentades, les dades eren originàriament freqüències o comptes. De tota manera, dades contínues poden sovint ésser posades en forma discreta usant intèrvals sobre una escala contínua. Edat, per exemple, és una variable contínua, però si la gent es classifica en diferents grups d'edat, els intervals corresponents a aquests grups poden ser tractats com si fossin unitats discretes.

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En una taula de contingència es parla de la variable fila i de la variable columna, dels marginals fila, dels marginals columna i de les cel.les. Per saber si hi ha relació de dependència entre les variables fila i columna, fem servir la tècnica del khi-quadrat (Erickson i Nosanchuk, 1983). El khiquadrat que calculem és comparat amb un valor crític de la taula de khi-quadrat de referència. Si el valor del khi-quadrat que hem calculat és més gran que el que li correspon a la taula, voldrà dir que hi ha relació de dependència entre les dues variables. Per a poder fer servir aquesta taula ens cal saber quin nivell de significança volem usar i quants graus de llibertat tenim. Normalment serà adequat usar un nivell de significança del 5%, cosa que vol dir que les conclusions a les que arribem tenen un 95% de probabilitat d'ésser certes. El nombre de graus de llibertat per al càlcul del khi-quadrat es troba multiplicant el nombre de files menys una pel nombre de columnes menys una i correspon al nombre de cel.les que un cop fixades, si a més es coneixen els marginals, determinen el valor de les altres cel.les. En el nostre cas, les taules de contingència han estat per encreuament entre dues variables, però també és possible estudiar les interrelacions amb tres variables mitjançant la construcció de taules de contingència de tres dimensions. Per saber la força de la relació, cas que n'hi hagi, hem fet servir el coeficient per a taules 2x2 i el coeficient V de Cramer per a taules més grans. Aquests coeficients presenten la característica que el seu valor pot anar des de zero fins a 1 (força màxima). Presentem a continuació les fórmules per a calcular aquests coeficients: ¢ = (khi-quadrat)2/N, éssent N el nombre total de individus de la mostra. V de Cramer =

(khi-quadrat)2/N(S-1), on N correspon al nombre total de individus de la mostra, i

S, al nombre de columnes o de files, el que sigui menor. Amb el valor del khi-quadrat podem saber si hi ha relació de dependència entre les dues variables, però no sabem com és, hem d'examinar les dades per veure-ho i això moltes vegades és el que costa més. Forma part del que en podríem dir interpretació de les dades. Per al càlcul dels valors del khi-quadrat, dels valors esperats i dels residuals, en les taules de contingència farem servir el programa SPSS.

Parelles de categories dominants en les taules de doble entrada Un cop construïdes les taules de contingència, per a saber quines parelles de categories són dominants fem una ordenació dels residus tal com hem explicat a l'apartat de comparació de resultats quan presentàvem l'anàlisi de les mitjanes per efectes i residuals. Presentem l'anàlisi amb unes de dades d’una altra recerca. En aquest cas comptem els estudiants que responen dues preguntes del mateix problema de la mateixa manera o no (des del punt de vista de la relativitat galileana).

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Relativitat física rel.fís. b)

rel. fis. a)

Total

sí

no

no res.

sí

4 0.37 +3.63

4 4.30 -0.30

0 3.33 -3.33

no

1 3.10 -2.10

52 35.98 +16.02

14 27.92 -13.92

0 1.53 -1.53

2 17.72 -15.72

31 13.75 +17.25

33

5 -31

58 +22

45 +9

108

no res.

8 -28 67 +31

-3

mitjana cel.la = 12 df = 4 khi-quadrat = 91.58 s = 0.001 sí, relació de dependència V de Cramer = 0.65 ; Relació bastant forta

La taula expressa que respondre relativament al SR en moviment a la pregunta a) no és independent de respondre relativament al SR en moviment a la pregunta b). El coeficient V de Cramer que s'acosta a 1, ens indica que la relació és bastant forta. Els residus que destaquen significativament en positiu indiquen que el no respondre relativament al SR en moviment a les dues preguntes i el deixar les dues preguntes sense respondre es dóna significativament més que altres parelles.

1.3.2.2. Comparacions i interrelacions entre problemes diferents Per a fer comparacions entre problemes, podem fer una anàlisi global comparant entre si les xarxes sistèmiques respectives (quines categories tenim en un problema que no són a l'altre, quines categories tenim a tots els problemes analitzats....). També podem fer una anàlisi més detallada comparant percentatge de les categories específiques o globals trobades als diversos problemes aplicant l'anàlisi de les mitjanes per efectes i residus (Erickson i Nosanchuck, 1983) ja comentada abans. La construcció de taules de contingència entre variables categòriques de parelles de problemes ens permetrà adonar-nos de possibles inconsistències i de les tendències coincidents entre problemes. Amb tot això podrem adonar-nos de quines variables o factors influeixen en els models de raonament dels estudiants i quines són les tendències més majoritàries i més deficients.

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Appendix D Diagram of Relations

295

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D. Diagram of Relations

ÅSIKTSBEROENDE RELATIONSDIAGRAM (Perspective -dependent Relational Diagram) ÅSIKTSBEROENDE RELATIONSDIAGRAM is a way to analyze and summarize data from the transcriptions of interviews developed by author of this thesis. It consist in three spheres that each of them represents an specific aspect of interest in the study. The way that the spheres are drawn represent the links or how that aspects are related each other.

Aspect 1

Aspect 2

? Aspect 3

ÅSIKTSBEROENDE RELATIONSDIAGRAM

The drawn notation used in Åsiktsberoende Relationsdiagram are very simples and elementary : ●

The three spheres represent different aspects that can be called with a name that the research decided as more convenient according to the study, this name has to express the general idea of the aspect in study. Each sphere contains the most important information of the interviewee transcription.

Sphere

The “link” symbol is used to indicate relation between 2 aspects (2 spheres). Not always this symbol appear. It will depend the answer.

“link”

It is very important how the question is formulated to get the information of interest. There is not a strict rule for this. It will depend of the ability and creativity of the interviewer. Then when we have the answers, the next step is to transcribe all the information received, included small details that in the beginning do not look interesting. We will illustrate with some examples the use of this tool: Task: “We want to know the relation among mathematics, physics and technology aspects with the Laplace transform topic”. Example A: Transcription of interviewee “A”:

297

“I think that is very difficult to divided in 3 aspects. I think that are all connected and you cannot you divided very clear I meant, you need 1st to understand what the Laplace transform is to be able to use it; you most then understand…I cannot see very much the physical aspect of the Laplace transform, honestly, I meant, I cannot see very much the difference between physical and technological aspect at the least from my point of view I mean the Laplace transform is the transform so it is a mathematical definition with all the proprieties and then it makes it possible to change, to study some problems in Laplace domain instead the time domain that it can be very much more difficult for us because the transfer function becomes much easier to study the proprieties of the system items of Laplace. I cannot see very much for me the difference between physic and technology? could be(program)? One that I have understood how you can use in some… then of course you can use also any program, I can think only Matlab for me because is one that I use more then I can use Matlab but is only way of computation. Matlab makes only faster computation that you don’t have to do by hand but steel one is to understand what is behind, in my opinion, other wise the risk is that you can give a wrong interpretation of the results if you just do not have clear what the Laplace transform represents. This is so there are…. A mathematical definition more understand why and how they can use and then of course you can also using in the program that is the easier step.” Then we mark the most important points from the speech. 1.

I think that is very difficult to divided in 3 aspects.

2.

I cannot see very much the difference between physical and technological aspect.

3.

The Laplace transform /.../ is a mathematical definition.

4.

I can use Matlab but is only way of computation

Example B: Transcription of interviewee “B”: “Ok. For mathematical aspect in the Laplace transform you can always do a lot of mathematics that are really details and not really useful for engineering and you take up especially all this parts using distributions which you can do a lot of things. Is not that useful for engineering part, when we use Laplace transform we like simplify the Laplace transform so we don’t… we purpose skip this details,

298

D. Diagram of Relations

mathematical details. For the physics, eh…for me the Physics… for me the Physics is more about differential equations than Laplace transform so for building physical models that I also have courses where I don’t use the Laplace transform at all, we only use…or only… use the differential equation as the way of describing a system it’s than if you use it in automatic control that you have to instead of having a differential equation do something with it, for example Laplace transform is. But is not necessary for all the automatic control, some part of that (automatic control) you don’t use the LT; you use the differential equation way or writing using (-state-) space instead. So for the Physic part from my perspective the LT isn’t that important. And the 3rd aspect. For me an application can be that you try to control something and if you look as try to control a system then the Laplace transform is fundamental for describing the system, almost always in the 1st step to make eh…not the simple control but the 2nd step in the control “manufacturing process” (or what to say) because the simples process then you don’t care about the system what it is? You don’t do a model of it and then you don’t have to use the LT, you can do more about logical reasoning and you don’t have to do the model of the system but then in the next step when you try to analyse in some way you have to use the LT in automatic control to control this application so in that case the (hall) subject of automatic control is application driven. 1.

For mathematical aspect /.../ you can always do a lot of mathematics that are really details and not really useful for engineering.

2.

the Physics is more about differential equations

3.

for the Physic part from my perspective the LT isn’t that important.

4.

you try to control something and if you look as try to control a system then the Laplace transform is fundamental for describing the system

In these two examples “links” were not used because the answers from informants did not justify it. As these examples show the ÅSIKTSBEROENDE RELATIONSDIAGRAM can be used as an effective way to describe the most relevant ideas from an interview.

Appendix E History of Laplace E.1

Who was Laplace

Pierre Simon de Laplace (1749-1827) was born in Beaumont-en-Auge, in Normandy, on March, 23rd 1749. He was son of a farmer. He belonged a humble family and he started his studies by economical support from some rich neighbours. He studied in the “Benedictino” college of his town and after he continued his studies of theology in the University of Caen . During his university studies, Laplace showed talent in mathematics and he published a memory about the calculus of finite differences in the bulletin edited by Lagrange. When Laplace was 18 years old, he moved to Paris and got, by his merit, the support of D’Alembert to work as a professor of mathematics in the Royal Military School of Paris. Laplace started scientific activity: he published constantly a big number of memories where he applied his mathematical abilities to understanding of relatives questions to planetary theory. In 1773 Laplace was elected mechanic adjunct of the Science Academic and in the session in 1774 was written: “This Society has hurried up to reward his works and talents, there is no one else youth able to present many important memories about divers and difficult mathematics”. Laplace continued working in this age in two principal directions: the celestial mechanic and the calculus of probability. In 1784 he was examiner of the body of artillery, in B´ezout, Laplace examined to young Napoleon Bonaparte, opening to him the military career. During the Revolution, Laplace is member of the Commission of Weigh and Measures but the Convention considered Laplace, Carles de Borda, Charles de Coulomb y Lavoisier “not worthy of confidence for their republican virtues and their distaste to the kings”, for this reason Laplace is expulsed by: the decree number III of Nivoso of second year. During the “Terror”, after the dissolution of Academies, he moved to Melun to escape of the conflictive period of France and there he wrote the “Exposition of System of the World”. When the revolution finished, Laplace work with Lagrange teaching mathematics in the Normal School of the Convention. Lagrange, as a member in the National Institute and Office of Lengths, he suggested to adopt a new calendar based in astronomic calculus. After 18 Brumario, Napoleon Bonaparte, called Laplace as minister of Interior to Laplace because was interested in to get support of culture people, but Laplace had few political aptitudes and after some 6 weeks Napoleon considered an error his assignment and gave the place to Lucien Bonaparte. In Santa Elena isle, Napoleon wrote in his memories about Laplace: 299

300

E. History of Laplace “Geometrician of first rank, Laplace, early showed himself as mediocre administrator; from his first work we recognized that we committed a mistake. Laplace seized no question under his real point of view: he looked for subtleties everywhere, had only problematic ideas, and carried finally the spirit of ’infinitesimal’ to the administration.”

In 1803, Laplace became a president and then canceller of the “Conservator Senate”. Although Laplace received honours from Napoleon – Count of Empire, official of the Honour Legion – he noticed that the Empire will disappear and offered services to the “Borbones” and he joined to the rehabilitation of the monarchy. In 1816 is called member of the French Academy of Language. In 1806 Laplace acquired propriety in Arcueli, near of his college and friend Berthollet. There born the “Society of Arcueli”. Laplace and Berthollet gathered chemists and physicist together as Chaptal, Th´enard, Gay-Lussac, Dulong, most of them were disciples of Laplace. As a result of those meetings where written three volumes of memories of this Society where Laplace presented important approaches about mathematical physic. In this period, Laplace supported the work of his disciples and he was generous omitting (in some cases) his name in documents to give credit just to his collaborators. Laplace would have had good reputation just realizing his scientific work; but he coveted, especially, social reputation. When Laplace reached social position and aristocracy, he kept distance from his relatives and from all those that have helped him in his youth to realize his studies. Also it was known his ability and facility to change his politic position according the occasion. His contemporaneous reproach his vanity that didn’t let recognize the work of other sciences that he considered his rivals. Laplace were not a pure mathematician as Lagrange, who did not have the qualities of personality, neither the elegant clarity, neither the perfection of his works. Laplace was considered as a teacher surrounded by his disciples like Biot and Poisson. Laplace wrote also: “Exposition of the system of the world” (1796), “Celestial Mechanics” (1798-1825), “Analytical Theory of Probabilities” (1812) and more memories about the study of movement, the figure of stars, Theory of gasses and capillarity, electromagnetic laws, etc. Laplace was interested in natural philosophy and he considered mathematical analysis as a means and not as an end and he tried to build mathematical theories and perfection them to explain the mysteries of the celestial mechanic and apply the theory of probabilities to the civil life.

E.2

Contributions to Mathematics

• The discussion of the general theory determinants in 1772, simultaneously with Vandermonde, • The demonstration of every equation of degree two must have at least a quadratic real factor. • The demonstration of finite solution of a difference equation of the first degree and the second order might always be obtained in the shape of a continued fraction.

E.3 The Laplace transform

301

• In the mathematical analysis he introduces the use of the potential function (1874). He demonstrates that the potential function presented by Clairaut and used by Lagrange in the field of the dynamics satisfies a differential equation in partial derivatives for which integration introduces the functions harmonic spherical, studied before for Legendre.

E.3

The Laplace transform

Transformation of a function real variable f (t), defined in the whole field of the real numbers, makes it correspond a new function L{f }, called “the Laplace transform” defined by the expression: Z∞ L{f } =

e−st f (t)dt

(E.1)

0

The function (E.1) depends on s; where s = σ + jω

(E.2)

• The demonstration of D’Alembert theorem about the forms of the roots of the algebraic equations. • He perfected the methods of integration of equations in differential partial. It was established The Laplace-Gauss law. Also is known it by the name of Gauss’s law. But of fact Laplace discovered this law in 1780 when Gauss (1777-1855) is three years old. This law is called normal Law too. His treatise has two parts: the first is “Del calculus of generates function” and the second is “General theory of probabilities”. In the first chapter of the second part of the book I, called “De la integration by approximation of differences that include high factors to big potencies”, Laplace present a method of approximation in where introduce the transformation: y = γe−r

2

(E.3)

And, in general the transformation that has his name: Z∞ f (x) =

e−xt g(t)dt

(E.4)

0

The function f is called: “the Laplace transform of the function g”. In the second chapter of this same pat is the procedure used and for first time defined integrals are used in the resolution of differential equations. The next chapter is about “approximation of different functions of big numbers” where Laplace gave an approximate value to the expression E.5 for “i” very big: Rx 1 ∆n i = s

xi−1 e−sx (e−n − 1)n dx

0

Rx 0

(E.5) xi−1 e−x dx

302

E. History of Laplace

Where ∆n means the (n-´esima) finite difference. In the chapter III of book II talk about “the laws of probability produced by define multiplication of events”, Laplace studies the Bernoulli theorem and proved that if the number of test “u” is big, the probability of reason of the possible number of events to the total number of test will situate between the limits E.6 and E.7. √ τ 2pq p− √ u √ τ 2pq p+ √ u

(E.6) (E.7)

And the value is 2 √ τ

Zτ

2

e−t dt + √

1 2 e−r 2πµpq

(E.8)

0

Where q = (1 − p) and “p” is the probability of an event in every test. The fourth chapter of the book II is the most important and difficult too, because contain the theory of “method of minimum squares” proposed by Legendre. One finds also in the chapter V, titled “application of the calculation of probabilities to the investigation of the phenomena and his reasons”, the solution to the problem of Buffon’s needle for the approximation of the number II. Laplace finds the solution of the first part of this problem and gives a solution for the second, that Buffon could not do. Besides, he gives a generalization of this problem for the case of two mutually perpendicular sets of parallel equidistant lines. If the distances are a and b, the probability of which a needle of length l (minor than a and b) falls down on one of the lines is: 2l(a + b) − l2 (E.9) πab Proposed by Thomas Bayes’s (posthumous memory in 1763), the problem of the determination of the probability of the reasons for the observed effects is considered again by Laplace In the chapter V of 2nd book (memory published in 1774), where it enunciates in a definitive way Bayes’s rule: It is x the possibility of a simple event and the possibility of a compound event that depends on the simple event of a way arbitrary, which implies that ”y” is a function known of ”x”. Let’s suppose that this compound event has been observed; then the probability of which the possibility of the simple event places between a and b is: p=

Rβ α Rl

ydx (E.10) ydx

0

Combined with the theorems of the total probability and of the compound probability, this rule allowed to Laplace to evaluate the probability of future events relying on the results obtained of the observed events. The “Laplace equation” or the Laplacian of a function is verified by the potential in Mechanics. In a memory on the theory of the attractions of the spheroids and the

E.4 Contributions in Physics and Chemistry

303

figure of the planets, included in his famous agreement of astronomy, Laplace develops the concept of potential, a function which directional derivative in every point is equal to the component of the field of intensity in the given direction. ∆V =

∂2V ∂2V ∂2V + + =0 ∂x2 ∂y 2 ∂z 2

(E.11)

Laplace was interested equally for other aspects of the mathematics. Mentioning brief the extension of the notion of singular solutions to the equations of grade superior and to the differential equations of three variables; the utilization of spherical coordinates to express the “potential equation”; the generalization of Vandermonde’s method for the development of the determinants in minor products; The demonstration for the case of six planets that move in the same direction and six typical roots are real and different; the utilization of the changeable complex functions divides to calculate integrals, happening from a real to integral to the integral complex one in order to calculate the real integral. (Laplace was thinking that this step of the real thing to the imaginary thing could be considered to be a heuristic method).

E.4

Contributions in Physics and Chemistry

In 1780 Laplace elaborated (together with Lavoisier), the first measures of heats in chemical reactions and other specific, establishing with these experiences the fundaments of thermo-chemistry. In this age, both scientists conclude that the breathing is not any more than a type than combustion. He establishes the formula of the transformations adiabatic of a gas, which he used in the expression of the speed of spread of the sound. In the field of theoretical physics he establish his theorem of capillary attraction, who accepted the idea proposed by Hauksbee in the Philosophical Transactions (1709). It explains that the phenomenon is produced by an attractive force that is “insensitive to sensible distances”. Laplace started the study about the action of a solid on a liquid and he did not develop completely the mutual action of those of liquids. Gauss did and later, Neumann completed the details. The relation that expresses the capillary pressure exercised on a liquid curled surface is known as Laplace Law and says: If the surface is spherical, then the capillary pressure behaves with the following expression: ρ=

2σ r

(E.12)

Where: σ Superficial tension, r radius of curvature and ρ capillary pressure. To emphasize his studies on the configuration of a fluid in balance submitted to a rotating movement. And he contributes to the study of the electricity and the magnetism with mathematical techniques. The Laplace laws or Laplace equation: “In all region of the field where there is no electrical charge, the potential is distributed according to a law completely independent from charges that create it”. He enunciates two fundamental laws of the electromagnetism:

304

E. History of Laplace

The first Laplace law : “In any region where there is no electrical charge, the potential change in it average value in the points of a spherical surface is equal to the potential in it center” The second Laplace law or Ampere law of the magnetic induction: “In an element of current inside of a magnetic field, the acting force is always normal to the element of current and to the field” In 1816 Laplace was the first in demonstrating why the theory of Newton vibratory movement gives an incorrect value to the speed of the sound. It is a consequence of the heat developed by the sudden compression of the air that increases it elasticity and therefore the speed of transmission of the sound is higher than Newton calculated. Laplace was admired by his talent, the unit and the extent of his conceptions. The following appointment, Poisson’s words, illustrates scientific contributions of Laplace: “Undoubtedly Laplace appeared as a man of talent in the mechanics celestial; he) gave proof of the most penetrating sagacity to discover the reasons of the phenomena; and it was as well as found the cause of the acceleration of movement of the moon and that of the big differences between Saturn and Jupiter, that Euler and Lagrange were searching fruitlessly. But he was even more in the calculation of probabilities where he demonstrated himself as a great geometer; because the numerous applications that he did about this calculation gave origin to the calculation of finite partial differences, to his method for the reduction of certain integral as series and that he called the ‘theory of the functions generative’ .... Let’s believe, so, that a topic that called the attention of similar men is worthy of ours; and let’s try, if it is possible to us, to add something what they found in such a difficult and so interesting matter.” There is not many information about his family life either his hobbies. He was married with Charlotte de Courty de Romanges in 1788 and from this union he had a daughter and a son (who was general of artillery). After 1806, the Laplace family lived in Arcueli and in March 5th of 1827 Laplace died with almost 78 years old. In conclusion, for Laplace the mathematical analysis is only a tool to solve physical problems