CIRCUITI IN CORRENTE ALTERNATAIn questo capitolo indicheremo in grassetto variabili a valori complessi e con Re() la parte reale di un numero complesso.

La motivazione matematica per l’uso di quantita complesse nello studio dei circuiti in corrente alternatasta nella grande semplificazione che l’uso della formula di Eulero consente nella trattazione di espressionicontenenti funzioni trigonometriche.Cio porta all’introduzione dell’impedenza complessa di un elemento circuitale, una quantita fisica di cuianalizzeremo il significato.

1 Il circuito RLC e la notazione complessa

Consideriamo il circuito RLC in serie

V0cos(ωt)

L C

R

L’equazione del circuito si scrive:

V = LI + RI +1

CI (1)

L’equazione omogenea associata e dello stesso tipo di quella risolta nel capitolo sulle correnti continue, ele soluzioni per I hanno in funzione del tempo lo stesso andamento di quelle trovate per Q. Ora dovremocercare una soluzione particolare della equazione completa.Dalla teoria delle equazioni differenziali lineari sappiamo che tale soluzione ha la forma

I = I0cos(ωt + ϕ) (2)

dove I0 e ϕ sono costanti che dipendono da V0, L, R, C ed ω. Notate che la soluzione generaledella omogenea, quali che siano le condizioni iniziali, decresce esponenzialmente nel tempo; a tempisufficientemente grandi la soluzione sara dunque all’incirca uguale alla (2). Diciamo che il circuito e nel

• regime transitorio, nel periodo iniziale in cui i termini esponenziali non sono trascurabili rispettoalla (2);

• regime stazionario nel periodo successivo, in cui la corrente e uguale all’incirca a (2).

Le costanti d’integrazione stanno solo nella soluzione generale della omogenea, quindi le condizioni inizialihanno effetti solo nel regime transitorio, mentre nel regime stazionario l’andamento della corrente edeterminato solo dalle costanti del circuito.Il metodo delle impedenze complesse si applica al regime stazionario e d’ora in avanti ci occuperemo solodi questo.Poiche cercheremo le fasi di tensioni e corrente nel circuito relativamente alla fase di V (t), assegneremofase zero a V (t):

V = V0cos(ωt) = Re(

Veiωt)

; V = V0 (3)

Cerchiamo la soluzione particolare nella forma:

I = I0cos(ωt + ϕ) = Re(

Ieiωt)

; I = I0eiϕ (4)

1

Ora il compito e trovare I0 e ϕ. Procederemo sostituendo le espressioni complesse della tensione e dellacorrente:

Ieiωt ; Veiωt (5)

nella equazione (1):

iωVeiωt =

(

−ω2LI + iωRI +1

CI

)

eiωt (6)

questa uguaglianza deve valere per ogni t; quindi:

iωV = −ω2LI + iωRI +1

CI (7)

V =

(

iωL + R +1

iωC

)

I ≡ ZI ; Z = R + i

(

ωL − 1

ωC

)

(8)

Otteniamo dunque un’equazione algebrica per I; la risolviamo e calcoliamo la corrente fisica reale:

I(t) = Re(

Ieiωt)

= Re

(

V

Zeiωt

)

= Re

( |V||Z| e

i(ωt−ϕZ)

)

=|V||Z| cos(ωt − ϕZ) (9)

ϕZ e la fase di Z:

ϕZ = arctanωL − 1

ωC

R(10)

Se avessimo introdotto anche la fase di V (t), ϕZ sarebbe la fase di Z meno quella di V.L’ampiezza della corrente e dunque data da:

I0 =|V||Z| =

V0√

R2 +(

ωL − 1ωC

)2(11)

Provate ad ottenere gli stessi risultati senza utilizzare le espressioni complesse.Dalla (10) deduciamo che se nel circuito abbiamo:

• Solo R: ϕZ = 0 ⇒ corrente e tensione sono in fase.

• Solo L: ϕZ = π2⇒ I = I0cos(ωt − π

2) ⇒ la corrente ’segue’ la tensione di 90o.

• Solo C: ϕZ = −π2⇒ I = I0cos(ωt + π

2) ⇒ la corrente ’precede’ la tensione di 90o.

• Solo L e C: la corrente precede o segue la tensione di 90o a seconda del valore di ωL relativamentea quello di 1

ωC.

Inoltre:

• Per ω = 1√LC

tensione e corrente sono in fase qualunque sia il valore di R.

Dalla (11) vediamo che, al variare di ω, la corrente ha un massimo (risonanza) per:

ω =1√LC

≡ ω0 (12)

Come abbiamo gia visto, alla risonanza la fase si annulla.Introducendo il fattore di merito alla risonanza Q0 (adimensionale):

Q0 = ω0L

R=

1

R

√

L

C(13)

riscriviamo I0:

I0 =V0

R

√

1 + Q20

(

ωω0

− ω0

ω

)2(14)

Come potete vedere nelle figure che seguono, a parita di L e di C, quindi di ω0, il circuito e tanto piu

selettivo alla risonanza quanto piu e grande Q0, cioe piu piccola e R rispetto a√

LC

. Cosa succede perR = 0 ?.

2

0

V0

R

I0(ω)

ω

············································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································Q0 = 1

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ω0

−π

2

0

π

2

ϕ(ω)

ω

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·

·

·

·

·

·

·

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2 L’impedenza complessa

La (7) e la (8) ci mostrano che, introducendo le impedenze complesse dei tre elementi circuitali fonda-mentali:

ZR = R ; ZL = iωL ; ZC =1

iωC(15)

ed utilizzando la notazione complessa possiamo scrivere una generalizzazione della legge di Ohm tra leampiezze complesse in cui la resistenza e sostituita dall’impedenza di ciascun elemento:

V = Z · I (16)

Notate che in questa equazione compaiono solo le ampiezze indipendenti dal tempo: la dipendenza daltempo compare nel fattore eiωt che abbiamo ’semplificato’.Con le stesse equazioni abbiamo gia visto che le impedenze in serie si sommano come le resistenze,possiamo facilmente predire che cio e vero anche per le impedenze in parallelo. Verifichiamolo in unsemplice circuito:

V

I(t)

CL IL(t) IC(t)

Per le correnti nei due elementi e per la corrente totale abbiamo:

IL =V

ZL

(17)

IC =V

ZC

(18)

I = IL + IC =

(

1

ZL

+1

ZC

)

V =V

Ztot

; Ztot =ZLZC

ZL + ZC

=LC

i(

ωL − 1ωC

) (19)

I =i(

ωL− 1ωC

)

LC

V (20)

I(t) = I0cos(ωt + ϕ) (21)

3

I0 = V0

∣

∣ωL − 1ωC

∣

∣

LC

; ϕ = arctan

ωL−1

ωCLC

0= ±π

2a seconda del segno di ωL − 1

ωC(22)

La fase di ZL e π2, quindi IL = V

|ZL|e−i π

2 e ritroviamo il fatto che per l’induttanza la corrente e sfasatarispetto alla tensione applicata di −π

2; analogamente IC e sfasata rispetto a V di π

2, e IL e IC sono

sfasate tra loro di π, quindi sono in opposizione di fase. Notate che per ωL = 1ωC

, I(t) e identicamentenulla; la ragione e che in questo caso IL(t) ed IC(t) hanno ampiezza uguale; essendo in opposizione difase, la loro somma e identicamente nulla.Tutto cio puo essere rappresentato nel piano complesso nel seguente modo:

Im

ReV

IC

IL

(rappresentiamo nello stesso piano quantita di dimensioni diverse, ma ci serve solo per mostrare le fasirelative).Nel caso in cui in serie all’induttanza ed al condensatore ci fossero due resistenze RL ed RC rispettivamenteavremmo la seguente rappresentazione:

Im

Re

IC = VRC+ i

ωC

R2

C+ 1

ω2C2

IL = VRL−iωL

R2

L+ω2L2

Come vediamo, l’impedenza complessa contiene l’informazione, per un singolo elemento o per un circuito,sulla risposta in ampiezza e fase alla tensione applicata; e per questo che l’impedenza complessa e compostada due quantita (parte reale e parte immaginaria) indipendenti.Consideriamo ora un circuito, complesso quanto vogliamo, di elementi in serie e parallelo; per calcolare lacorrente nel regime stazionario non sara necessario scrivere l’equazione differenziale del circuito: basteracalcolare l’impedenza totale del circuito applicando le regole per la combinazione delle impedenze in seriee parallelo e poi applicare la legge di Ohm generalizzata.Il metodo che qui abbiamo chiamato ’delle impedenze complesse’ viene anche detto metodo simbolico.

4

3 La funzione di trasferimento

Pensiamo ora che al posto della differenza di potenziale fornita dal generatore al nostro circuito RLC visia un segnale periodico qualsiasi di periodo T , ad esempio il segnale proveniente da un’antenna radio.Quale sara il segnale, cioe la differenza di potenziale, ai capi della resistenza ?. Useremo i termini disegnale di ingresso (o di input) e di segnale di uscita (o di output).

Vin(t)

L C

Vout(t)R

Il segnale di input puo essere sviluppato in serie di Fourier:

Vin(t) = 12a0 +

∞∑

n=1[ancos(nωt) + bnsin(nωt)]

= 12a0 +

∞∑

n=1Ancos(nωt + ϕn)

ω = 2πT

; an = 2T

T2∫

−T2

Vin(t)cos(nωt) dt ; bn = 2T

T2∫

−T2

Vin(t)sin(nωt) dt

(23)

nei paragrafi precedenti abbiamo imparato a calcolare la risposta del circuito, in ampiezza e fase, perqualunque frequenza di input; per l’ampiezza abbiamo dalla (14):

V0out = RI0out =R

RC

V0in√

1 + Q20

(

ωω0

− ω0

ω

)2(24)

dove abbiamo tenuto conto del fatto che nella (14) R e la resistenza totale del circuito, che comprendela resistenza della bobina dell’induttanza e quella dell’elemento circuitale resistivo (R in figura); abbiamorinominato RC tale resistenza totale.

Quindi RC = R + RL; perche non sommiamo anche la resistenza interna di un eventuale generatore che fornisca Vin ?.

Nella (24) invece R e la resistenza del solo elemento circuitale resistivo ai cui capi preleviamo il segnaledi uscita. A questo punto definiamo la funzione di trasferimento T (ω) del circuito:

T (ω) =V0out(ω)

V0in(ω)=

R

RC

1√

1 + Q20

(

ωω0

− ω0

ω

)2(25)

Lo sfasamento del segnale di uscita rispetto a quello di ingresso e dato dalla (10) cambiata di segno.Riscriviamolo utilizzando Q0:

δϕ(ω) = arctan

[

Q0

(

ω0

ω− ω

ω0

)]

(26)

(ricordiamo che per una resistenza non c’e sfasamento tra corrente e tensione).Utilizziamo ora T (ω) e δϕ(ω) nella (23) per calcolare il segnale di uscita:

Vout(t) =

∞∑

n=1

AnT (nω)cos(nωt + ϕn + δϕ(nω)) (27)

(il livello continuo 12a0 non viene trasmesso all’uscita).

5

Senza addentrarci nella trattazione di questa serie, possiamo dire in buona sostanza che T (ω) e δϕ(nω) cidicono quali frequenze presenti nel segnale di ingresso saranno trasmesse dal circuito all’uscita, con qualeattenuazione e sfasamento; per riassumere il tutto possiamo definire una banda passante del circuito, icui estremi, ad esempio, sono le pulsazioni per le quali l’attenuazione si riduce di un fattore 1√

2rispetto

all’attenuazione alla risonanza.

In laboratorio potete provare ad utilizzare l’onda quadra in input al circuito risonante, per frequenze prossime alla risonanza o lontane da questa;

osservate all’oscillografo il segnale sulla resistenza e date un’interpretazione qualitativa del suo andamento.

Per concludere vi faccio notare che se il segnale di ingresso non fosse periodico potremmo rifare il discorsosviluppato fin qui utilizzando la trasformata di Fourier al posto della serie.

6

4 Filtri passa-alto e passa-basso

Il circuito RLC in serie e, come abbiamo visto, un circuito passa banda. Piu semplici sono i circuitipassa-alto e passa-basso che selezionano le alte e le basse frequenze rispettivamente.

A) PASSA ALTO

Nel circuito che segue

Vin(t)

C

Vout(t)R

ad un segnale di ingresso continuo corrisponde (nel regime stazionario !!) un segnale di uscita nullo;infatti la corrente e nulla, quindi e nulla la caduta di tensione ai capi della resistenza: la caduta di tensionee tutta sul condensatore. Viceversa, ad alte frequenze ( rispetto ad 1

RC) il condensatore non ha il tempo

di caricarsi completamente prima che si inverta la polarita di Vin); come sappiamo, fino a che la carica sulcondensatore resta piccola, questo si comporta, approssimativamente, come un corto circuito. Quindi lacaduta di tensione e tutta sulla resistenza.

0

1

1√

2

T (ω)

ω···················································································

·····················································································································passa − alto

········································································································································································································

passa − basso

ωL, ωH

−π

2

−π

4

0

π

4

π

2

δϕ(ω)

ω

················································································································································································································································································································································································································

ωL, ωH

Figura 1: Funzione di trasferimento e sfasamento in funzione di ω per i circuiti passa-alto e passa-basso

Verifichiamo questo comportamento calcolando la funzione di trasferimento. Indicando con ZR e Zl’impedenza della resistenza e quella totale del circuito abbiamo:

Vout =ZR

ZVin =

R

R − i 1ωC

Vin =1

1 − i 1ωRC

Vin =1

1 − iωL

ω

Vin ; ωL =1

RC(28)

Quindi la funzione di trasferimento e lo sfasamento sono dati da:

T (ω) =1

√

1 +(

ωL

ω

)2(29)

δϕ(ω) = arctanωL

ω(30)

7

ed il loro andamento e riportato nella figura (1). Notate che il valore massimo (asintotico) della funzionedi trasferimento e 1 e che per ω = ωL essa assume il valore 1√

2e lo sfasamento vale π

4.

Ripensiamo ora alla caduta di potenziale ai capi del condensatore: a basse frequenze il condensatore hail tempo di caricarsi quasi completamente e la differenza di potenziale si trasferisce quasi completamenteai suoi capi; ad alte frequenze non ha il tempo di caricarsi e la differenza di potenziale ai suoi capi restapiccola. Se preleviamo il segnale ai capi del condensatore otteniamo dunque un filtro passa-basso.

B) PASSA BASSO

Vin(t)

R

Vout(t)C

T (ω) =1

√

1 +(

ωωH

)2; δϕ(ω) = − arctan

ω

ωH

; ωH =1

RC(31)

Per finire notiamo che in tutti e tre i casi visti finora lo studio completo della funzione di filtro nonpuo prescindere dal carico che si presenta all’uscita del filtro: esso diventa un elemento del circuito e nemodifica le caratteristiche. Possiamo tuttavia determinare le condizioni sull’impedenza del carico per lequali gli effetti di tale carico sono trascurabili. Provate a scriverle.

5 Due filtri in sequenza

Consideriamo due filtri passa-basso e passa-alto in sequenza:

Vin(t)

R1

C1

C2

Vout(t)R2J1 J2

Per trovare la differenza di potenziale su R2 potremmo calcolare l’impedenza totale del circuito:

Z = ZR1+ ZC1

//(ZC2+ ZR2

) (32)

e calcolare la corrente in R1; poi dovremmo dividere questa corrente nei due rami C1 e C2 −R2 ed infinecalcolare la differenza di potenziale ai capi di R2, ogni volta utilizzando le impedenze complesse dei varirami. Poiche il procedimento e piuttosto lungo, utilizziamo invece il metodo delle correnti di maglia perle maglie indicate in figura. Il sistema si scrive:

(

ZR1+ ZC1

−ZC1

−ZC1ZR2

+ ZC1+ ZC2

)

·(

J1

J2

)

=

(

Vin

0

)

(33)

esplicitando le impedenze, la matrice dei coefficienti si scrive:

R1 + 1iωC1

− 1iωC1

− 1iωC1

R2 + 1iωC1

+ 1iωC2

(34)

8

ed il suo determinante:

D =(

R1 + 1iωC1

)(

R2 + 1iωC1

+ 1iωC2

)

+ 1ω2C2

1

= R1R2 + R11

iωC1

+ R11

iωC2

+ R21

iωC1

− 1ω2C2

1

− 1ω2C1C2

+ 1ω2C2

1

= 1iωC1

[

R1 + R2 + R1C1

C2

+ i(

ωR1R2C1 − 1ωC2

)]

(35)

Poiche siamo interessati alla corrente in R2, calcoliamo J2:

J2 =1

D

∣

∣

∣

∣

∣

∣

R1 + 1iωC1

Vin

− 1iωC1

0

∣

∣

∣

∣

∣

∣

=1

D

1

iωC1Vin =

1

R1 + R2 + R1C1

C2

+ i(

ωR1R2C1 − 1ωC2

)Vin (36)

La funzione di trasferimento e quindi data da:

T (ω) = R2

∣

∣

∣

∣

∣

∣

1

R1+R2+R1C1C2

+i

(

ωR1R2C1− 1

ωC2

)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

1R1R2

+1+R1C1R2C2

+i

(

ωR1C1− 1

ωR2C2

)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

1R1R2

+1+ω2ω1

+i

(

ω

ω1−ω2

ω

)

∣

∣

∣

∣

∣

∣

= 1√

(

R1R2

+1+ω2ω1

)2

+

(

ω

ω1−ω2

ω

)2

ω1 = 1R1C1

; ω2 = 1R2C2

(37)

La (37) ha un massimo per ω =√

ω1ω2 ed un andamento simile a quello di una risonanza (fig. 2). Qual’ela differenza sostanziale col circuito risonante ?.

0

0.33

T (ω)

ω

···············································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································

ω1 = ω2

0.083

0

T (ω)

ω

····························································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································································

ω1 ω2

Figura 2: T (ω) per i due filtri in cascata per R1 = R2 e ω2 = ω1 e ω2 = 10 · ω1 rispettivamente.

9

6 Un partitore di tensione in corrente alternata

Sonda compensata

V0cos(ωt)

C1 R1

A

C2 R2

B

Sotto quale condizione l’ampiezza della differenza di potenziale tra A e B e indipendente da

ω ?.

Le impedenze dei due paralleli:Z1 = ZC1

//ZR1=

=R1

iωC1

R1+ 1

iωC1

= R1

1+iωR1C1

Z2 = R2

1+iωR2C2

(38)

La differenza di potenziale complessa tra A e B:

VAB =Z2

Z1 + Z2V =

1

1 + Z1

Z2

V (39)

che, se:R1C1 = R2C2 (40)

si riduce a:VAB =

1

1 + R1

R2

V =R2

R1 + R2V (41)

che e la stessa relazione che vale per il partitore resistivo in corrente continua.Quando si usano strumenti di misura in corrente alternata bisogna tener conto del fatto che questidispositivi hanno una impedenza complessa di cui bisogna tener conto quando si eseguono delle misure.Il caso tipico e l’oscillografo: nella figura che segue ho indicato con A e B i terminali di ingresso di unoscillografo. Questo strumento ovviamente contiene dei circuiti, cioe resistenze, capacita e induttanzee quindi ha una impedenza di ingresso: il valore di questa impedenza e determinata dai dettagli deicircuiti che lo compongono, ma normalmente si inseriscono all’ingresso degli elementi circuitali in modoche l’impedenza di ingresso abbia valori ben definiti e compatibili con gli usi a cui lo strumento e destinato.Questa impedenza non potra essere puramente resistiva: non fosse altro per il fatto che i terminale sonoconduttori vicini tra loro e che vi sono delle piste ravvicinate sui circuiti integrati, vi sara almeno ancheuna componente capacitiva. La schematizzazione piu semplice che possiamo fare e quella della figuraseguente: R2 e C2 sono la resistenza e la capacita di ingresso.

10

C

V0cos(ωt)

D

C1 R1

A

C2 R2

B

cavo coassiale Cc

Supponiamo ora di volere misurare con questo strumento la differenza di potenziale ai teminali C e D.Dovremo connettere A con D e B con C; generalmente questo si fa utilizzando un cavo coassiale: duesemplici fili a distanza variabile avrebbero una capacita diversa a seconda della loro disposizione, mentrela configurazione coassiale permette di avere un capacita fissata, indicata con Cc nella figura. Un valoretipico di Cc e ∼ 100pF per metro di cavo. Completeremo il discorso sui cavi nel capitolo sulle linee ditrasmissione.Se ora connettessimo semplicemente i terminali del cavo ai punti C e D (sostituite la maglia C1 − R1

nella figura con un corto circuito) avremmo un sistema che si comporta diversamente (in termini diattenuazione e sfasamento) a seconda di ω; e per questa ragione che si inserisce la suddetta maglia conun condensatore variabile: si puo variare C1 fino a realizzare la condizione (40) con C2 sostituito dalparallelo tra Cc e C2. Se questa condizione e verificata, l’attenuazione tra segnale da misurare e segnaledi ingresso all’oscillografo e la stessa per tutte le frequenze.Per verificare la condizione di compensazione si utilizza in input un’onda quadra; perche ?.Il sistema descritto (cavo piu maglia R1C1) si chiama sonda compensata. La possibilita di variare C1

permette di adattare la sonda a diverse impedenze d’ingresso dell’oscillografo.

11

7 I coefficienti di auto e mutua induzione

Vi ricordo che il campo magnetico di un solenoide infinito costituito da n spire per unita di lunghezzaavvolto su un cilindro di permeabilita magnetica relativa µr e percorso da una corrente I e nullo all’esternodel solenoide; all’interno e invece diretto lungo l’asse del solenoide ed e uniforme. Il suo modulo vale:

B = µ0µrnI (42)

Utilizzeremo la stessa espressione come approssimazione del campo magnetico all’interno di un solenoidelungo e sottile di lunghezza ` costituito da N spire:

B ' µ0µr

N

`I (43)

se S e la sezione del cilindro, il flusso totale di B attraverso le N spire del solenoide e dato da:

φ = BNS = µ0µr

N2

`SI (44)

e la forza elettromotrice autoindotta:

E = −dφ

dt= −µ0µr

N2

`S

dI

dt≡ −L

dI

dt(45)

Il coefficiente di autoinduzione L e dunque dato da:

L = µ0µr

N2

`S (46)

Per avere un’idea degli ordini di grandezza, poniamo µr = 1 e consideriamo una bobina composta da 10spire avvolte su un cilindro di lunghezza e raggio uguali a 1 cm (Km = 10−7):

L = 4πKm

100

10−2π10−4 ' 4 · 10−6 Henry (47)

Per un materiale di permeabilita magnetica relativa µr, questo valore va moltiplicato per µr. Per lamaggior parte dei materiali µr vale circa 1, mentre per i materiali ferromagnetici:

µr ' 102 ÷ 105 (48)

Bisogna tuttavia tener presente che per questi materiali si presenta un effetto di saturazione, per cui pergrandi valori della corrente non vale piu la relazione di proporzionalita tra campo magnetico e corrente.Consideriamo ora due spire percorse dalle correnti I1 ed I2 rispettivamente. Il flusso attraverso la primaspira sara dato dalla somma di due termini, uno dovuto al campo magnetico generato da I1, l’altro aquello generato da I2; analogamente per la seconda spira:

φ1 = L1I1 + L12I2 ; φ2 = L21I1 + L2I2 (49)

L1 ed L2 sono i coefficienti di autoinduzione delle due spire e sono entrambi positivi, mentre per L12 edL21 si puo dimostrare che:

L12 = L21 = M ; 0 ≤ |M | ≤√

L1L2 (50)

M e positivo o negativo a seconda del verso (orario o antiorario) scelto come positivo per le correnti nelledue spire: e positivo se i due versi sono concordi, negativo se sono discordi.Quanto al valore assoluto di M , esso sara ovviamente molto piccolo se le due spire sono molto distantitra loro, e vicino o uguale a

√L1L2 se le due spire sono vicine o sovrapposte. Consideriamo ad esempio

due solenoidi lunghi e stretti costituiti da N1 ed N2 spire avvolte sullo stesso cilindro di lunghezza ` edaventi la stessa sezione S; indicando con B1 il campo magnetico del primo solenoide, il flusso indotto dalprimo nel secondo sara dato da:

B1N2S = µ0µrn1I1N2S = µ0µr

N1

`N2SI1 (51)

quindi

M = µ0µr

N1N2

`S =

√

L1L2 (52)

che e appunto il valore massimo previsto dalla (50).

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8 Osservazioni sperimentali sull’induzione elettromagnetica

In questo paragrafo faremo qualche osservazione qualitativa sul fenomeno dell’induzione elettromagnetica,preliminare ai calcoli sul trasformatore presentati nel paragrafo successivo.Il circuito utilizzato e il seguente:

R

CH1

CH2

R vale circa 5 Ω, le due bobine, entrambe composte da circa 20 spire, sono avvolte su un supportotoroidale di plexiglas di ∼ 1 cm di raggio e ∼ 0.25 cm2 di sezione. CH1 e CH2 sono i due canalidell’oscilloscopio.

Notate la posizione della massa nel circuito primario: a quel ramo del circuito vanno collegate la massa del generatore e quella di CH1, e questa

e l’unica disposizione possibile.

CH1 misura la differenza di potenziale ai capi di R, che e proporzionale alla corrente che scorre nelprimario.

L’oscillografo misura solo differenze di potenziale: per misurare una corrente, inseriamo una resistenza R piccola rispetto a tutte le altre presenti

nel circuito: la differenza di potenziale ai suoi capi sara proporzionale alla corrente.

CH2 misura invece la differenza di potenziale ai capi del secondario. Quando la frequenza del segnalefornito dal generatore e di 1 MHz osserviamo i seguenti segnali:

Figura 3: ν = 1 MHz. Sinistra: CH1 in alto, CH2 in basso. Destra: CH2 vs CH1.

Possiamo notare che il segnale osservato sul secondario e proporzionale alla derivata della corrente nelprimario cambiata di segno. Per un segnale sinusoidale questo comporta uno sfasamento di π

2di CH2

rispetto a CH1.

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La differenza di potenziale ai capi della bobina del primario (non mostrata in figura) e in fase con quellasul secondario e di ampiezza all’incirca uguale. Siamo nella situazione di un trasformatore ideale: ilrapporto tra le ampiezze e uguale a quello tra il numero di spire nei due avvolgimenti. Questo si verificaperche a questa frequenza le impedenze dei due avvolgimenti sono molto maggiori rispetto a tutte le altreimpedenze presenti nel circuito.Ben diversa la situazione a 10 kHz (notate la scala verticale di CH2). A questa frequenza il generatoreci permette di utilizzare anche segnali triangolari e ad onda quadra, e di verificare ancora una volta che sulsecondario abbiamo un segnale proporzionale alla derivata della corrente nel primario cambiata di segno.Notate l’ampiezza del segnale sul secondario nel caso dell’onda quadra ( ∼ 10 V ); questo e il caso in cuila variazione della corrente nel primario e piu rapida:

Figura 4: ν = 10 kHz, onda sinusoidale e triangolare sul primario.

Figura 5: ν = 10 kHz, onda quadra sul primario. A destra: dettaglio intorno al fronte di salita delsegnale sul primario.

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9 Il trasformatore

Consideriamo ora due solenoidi accoppiati; collegando un generatore ai capi del primo (circuito primario)ed un carico di impedenza Z ai capi del secondo (circuito secondario) otterremo ai capi del carico unadifferenza di potenziale la cui ampiezza puo essere regolata variando il numero di spire dei due solenoidi.In particolari condizioni il rapporto tra le ampiezze della differenza di potenziale ai capi del carico e diquella fornita dal generatore dipende solo dal rapporto tra il numero di spire dei due solenoidi.

-

V0cos(ωt)

+Rp Rs

+

Z

-

Jp Js

Scegliamo per le due maglie i versi di percorrenza indicati in figura; con questa scelta e con M positivo, unincremento della corrente nel primario produce una corrente negativa nel secondario, cosa che ritroveremonelle equazioni che seguono. Le equazioni per le due maglie sono le seguenti:

V = Jp (Rp + iωLp) + JsiωM (53)

0 = Jp (iωM) + Js (Z + iωLs + Rs) (54)

ricavando Jp dalla seconda e sostituendo nella prima:

Jp = −Js

Z + iωLs + Rs

iωM(55)

V = Js

(

−Z + iωLs + Rs

iωM(Rp + iωLp) + iωM

)

= −Js

(Rp + iωLp) (Z + iωLs + Rs) + ω2M2

iωM(56)

Js = −ViωM

(Rp + iωLp) (Z + iωLs + Rs) + ω2M2(57)

Indicando con Vout la differenza di potenziale ai capi del carico e tenendo con conto dei versi positiviscelti, abbiamo:

Vout = −ZJs = ViωMZ

(Rp + iωLp) (Z + iωLs + Rs) + ω2M2(58)

Che e la relazione tra l’entrata e l’uscita del circuito. Possiamo semplificare qualcosa se assegnamo a M

il suo valore massimo√

LpLs:

Vout = Viω

√

LpLsZ

(Rp + iωLp) (Z + Rs) + iωLsRp

(59)

La differenza di potenziale a circuito aperto si puo calcolare ponendo Z reale e facendolo tendere ainfinito:

Vout ca = Viω

√

LpLs

(Rp + iωLp)= V

√

Ls

Lp

1

1 +Rp

iωLp

(60)

e per Rp ωLp:

Vout ca ' V

√

Ls

Lp

= VNs

Np

; Rp ωLp (61)

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che e il risultato anticipato all’inizio (l’ultima uguaglianza vale se le due bobine hanno la stessa lunghezzae la stessa sezione).Ponendo poi Z = 0 nella (57) possiamo ricavare la corrente di corto circuito nel secondario:

Js cc = −ViωM

(Rp + iωLp) Rs + iωLsRp

(62)

e calcolare quindi l’impedenza equivalente di Thevenin (il segno meno viene dalla scelta dei versi positiviper la tensione e la corrente):

ZTh = −Vout ca

Js cc

=(Rp + iωLp) Rs + iωLsRp

Rp + iωLp

= Rs + Rp

iωLs

Rp + iωLp

' Rs + Rp

N2s

N2p

(63)

(l’ultima uguaglianza vale per Rp ωLp).Abbiamo dunque caratterizzato completamente il circuito di Thevenin equivalente, sul carico, al trasfor-matore.Infine, utilizzando le equazioni (53) e (55) possiamo scrivere:

V = Jp

(

Rp + iωLp +ω2M2

Z + iωLs + Rs

)

= Jp

(

Rp +iωLp (Z + Rs)

Z + iωLs + Rs

)

' Jp

(

Rp +N2

s

N2p

(Z + Rs)

)

(64)

(l’ultima uguaglianza e ottenuta ponendo Z reale e trascurando Rs+Z rispetto a ωLs). Questa equazioneci mostra come il carico sul secondario viene ’visto’ dal generatore sul primario.

Il dispositivo permette sia di aumentare che di diminuire l’ampiezza della tensione all’uscita rispetto all‘entrata; ma, se fossimo interessati solo ad

una diminuzione, anche un semplice partitore resistivo (vedi capitolo sulle correnti continue) realizzerebbe lo scopo. Perche non accontentarci di

questo ?.

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