capitolu iii bazele teoriei
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Cursuri electrotehnicaTRANSCRIPT
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Cap. 3. BAZELE TEORIEI CIRCUITELOR ELECTRICE
§ 3.1. TOPOLOGIE si MARIMI
3.1.1. TOPOLOGIA CIRCUITELOR
• latura: i = const.; (l);• latura: i = const.; (l);
• nod: (n);
• ochi (bucla, ciclu): (o);
• numarul de ochiuri independente:
o = l – n + 1
Retea electrica (a) si schema sa topologica (b)
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3.1.2. CLASIFICAREA CIRCUITELOR
a) natura elementelor de circuit:-- liniare;
- neliniare..
b) regimul de functionare:
-- stationar (c.c.);
- cuasistationar (c.a.);
- nestationar (variabil).
c) legatura cu exteriorul:-- izolatete;;
-- neizolate: neizolate:
-- dipol (latura);
- cuadripol;
frecventa f = 0
c) legatura cu exteriorul:-- neizolate: neizolate: - cuadripol;
- multipol..
circuite liniare in regim cuasistationar
regim permanent sinusoidal
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latura-- activa: e activa: e ≠ ≠ oo
-- pasiva: e = 0pasiva: e = 0
latura-- receptoare: preceptoare: pbb=u=ubb.i .i > 0 (> 0 (primitaprimita););
-- generatoare: pgeneratoare: pbb=u=ubb.i < 0 (.i < 0 (cedatacedata););
latura activa receptoare: latura activa receptoare: ↑i↑i ≡≡ ↑u↑ub b si ↑i si ↑i ≡≡ ↑e; u↑e; ubb + e = i.R+ e = i.R
Clasificarea laturilor: active (a, b, c); pasive ©; generatoare (a); receptoare Clasificarea laturilor: active (a, b, c); pasive ©; generatoare (a); receptoare
(b, c, d).(b, c, d).
RR
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3.1.3. MARIMI SINUSOIDATE
NNω = constant;
B = constant;
α (t) = ω·t+γ
a) producere
)tcos(BAABf γ+ω=⋅=Φ
)tsin(E)tsin(NBAdt
dNe m
f γ+ω=γ+ωω=Φ
−=
Producerea t.e.m. sinusoidate
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Marimea periodica i(t)
frecventa: f = 1/T [s-1] = [Hz]
pulsatia: ω = 2π/T = 2πf
b) caracterizare
perioada: T T [s]
Zk,)t(i)kTt(i ∈=+
00
Marimea periodica i(t)
0sau0dt)t(iT
1I
Tt
t
med
1
1
<≥= ∫+
0dt)t(iT
1I
Tt
t
21
1
≥= ∫+
valoarea medie patratica = valoare efectiva (eficace):
valoare medie:
valoarea indicata de aparatele de masura
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);sin(Ii m γω += t
)tsin(I2i γ+ω=01331 <γ−γ=ϕ
00
00
mm I0,7072
II ⋅==
2
Idtγ)t(ωsinI
T
1dt(t)i
T
1I m
T
0
22
m
T
0
2 =⋅+⋅⋅=⋅= ∫∫
],( ππ−∈γ
],( ππ−∈ϕ•defazaj: φ12 = -φ21 = ;
Faze initiale si defazaje
•faza: [rad]
•faza initiala:
φ = 0: marimi in faza;
φ = π: marimi in opozitie de faza.
01221 >γ−γ=ϕ
01331 <γ−γ=ϕ
γtω +⋅
γ-γ 21
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Problema 3.1.
Caracterizati sinusoida tensiunii de la priza monofazata, daca valoarea instantanee
are expresia u(t) = √2·220 sin(100π·t) [V].
γ)tsin(ωU2u(t) +⋅⋅=
U = 220[V];
Um= √2·220 =311[V].
ω = 100π = 314 [rad/s];
f = ω/2π =50 [Hz];
T = 1/f = 0,02 [s]
Tema 3.1:
1. Calculati frecventa si pulsatia unei marimi sinusoidale daca T = 20[ms].
2. Scrieti ecuatia si reprezentati sinusoida tensiunii u(t), daca T = 2[ms] si valoarea
maxima (de virf) Um = 331[V] este atinsa la t = 0,2[ms].
3. Demonstrati ca valoarea medie a marimilor sinusoidale este nula.
4. Scrieti expresia sinusoidei curentului electric, daca ampermetrul indica 2[A] iar
frecventmetrul 100[Hz]. Alegeti faza initiala a curentului sinusoidal egala cu un
sfert de perioada.
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1j −=
γ=↔γ+ω= jeII)tsin(I2i
IImIReI 22 +=
c) reprezentare in complex
valoarea efectiva valoarea efectiva → modul→ modul
faza initiata faza initiata → argument→ argument
fazor: segment orientat atasat numarului complex I
) jsinγI(cosγeII jγ +=⋅=
Ie2Imi tjω=
)exp(j II γ=
+1
I
) jsinγI(cosγeII jγ +=⋅=
IRe
IImarctgγ =
Planul complex (planul Gauss).
+j
0
I
Problema 3.2.
π/3)tsin(ω4,24u(t) +⋅⋅=
2U4,24 =
j2,61,5)2
3j
2
13()
3
πjsin
3
π3(cose
2
4,24U 3
πj
+=+=+=⋅=
j2,61,5U +=
32,61,5U 22 =+=
π/3[rad]1,05[rad]601,5
2,6arctg
IRe
IImarctgγ 0 =====
γ
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d) operaţii cu mărimi sinusoidale reprezentate in complex
ADUNAREA
)tsin(I2)tsin(I2)tsin(I2 2211 γ+ω+γ+ω≡γ+ω
2211
2211
cosγIcosγI
sinγIsinγIarctgγ
⋅+⋅⋅+⋅
=
)γcos(γI2IIII 2121
2
2
2
1 −⋅⋅++=
i = i1 + i2::
2122112211
jγ II)sinγIsinγj(IcosγIcosγIjIsinγIcosγIeI +=+++=+==
γ=γ+γ
γ=γ+γ
sinIsinIsinI
cosIcosIcosI
2211
2211
Adunarea marimilor sinusoidale
reprezentate in complex
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AMPLIFICAREA CU UN SCALAR
IλeλI γ)tsin(ωI2λiλ jγ ⋅=⋅⇔+⋅⋅⋅=⋅
DERIVAREA
IjωeωI )2
πγtsin(ωIω2 γ)tcos(ωωI2
dt
di π/2)j(γ ⋅=⋅⇔++⋅⋅⋅=+⋅⋅⋅= +
ππ
INTEGRAREA
jω
Ie
ω
I )
2
πγtsin(ω
ω
I2 γ)tcos(ω
ω
1I2dti π/2)j(γ =⇔−+⋅=+⋅−=⋅ −∫
j1j0) 2
πsinj
2
π(cose /2j =⋅+=⋅+=π
jj
1
e
1e
/2j
/2j- −=== ππ
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00
Operatii cu marimi sinusoidale Operatii cu marimi sinusoidale
reprezentate in complexreprezentate in complex
derivare = amplificare cu scalarul ω & rotire cu π/2 [rad] = 900
integrare = amplificare cu scalarul 1/ω & rotire cu -π/2[rad] = -900
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Avantajul operatiilor cu fazori:
transforma ecuatiile integro - diferentiale (functie de timp) in ecuatii algebrice (cu
numere complexe). R L C
u(t)
i(t)
uR uL uC
(t);u(t)u(t)uu(t) CLR ++=
i(t);R(t)uR ⋅= ;dt
diL(t)u L = ;dti
C
1(t)uC ∫ ⋅=
dtiC
1
dt
diLi(t)Ru(t) ∫ ⋅++⋅=
)]1
Lj(ω[R II1
jILjωIRU −⋅+=−⋅⋅+⋅=
Circuit RLC serie
)1
Lj(ωRZU
−⋅+==)]Cω
1Lj(ω[R II
Cω
1jILjωIRU
⋅−⋅+=
⋅−⋅⋅+⋅= )
Cω
1Lj(ωRZ
I
U
⋅−⋅+==
+j
0 φ<0
-j/ωC
jωL
Z
R
Semiplanul impedantei Z
+1
+j
0
Iφ U
UL=jωCI
UR=RI
Diagrama fazoriala
Inmultire cu fazorul I = I·ejγi
(amplificare cu I si rotire cu γi)
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Problema 3.3.
Ce valoare indica ampermetrul, daca i1(t) = 3√2 sin(314t+π/3)[A]
si i2(t) = 5,65 sin(100πt+π/6)[A]. Scrieti expresia sinusoidei
curentului ce parcurge ampermetrul.
Rezolvare
• indentificind sinusoidele: I1= 3[A]; 1 = π/3[rad] si I2= 5,65/√2 = 4[A]; 2 = π/6[rad];
• fazorii curentilor:
γ γ
j2,6[A];1,5)2
3j
2
13()
3
πjsin
3
π3(cose3I 3
πj
1 +=+=+=⋅=
j2[A];3,46)2
1j
2
34()
6
πjsin
6
π4(cose4I 6
πj
2 +=+=+=⋅=
A
i1
i2
i
•prima teorema a lui Kirchhoff: i1 +i2 – i = 0; →
• i(t) = 6,76·√2 sin(314t+0,74)[A].
j2[A];3,46)2
j2
4()6
jsin6
4(cose4I2 +=+=+=⋅=
;III 0;III 2121 =+⇒=−+
6,76[A]4,64,96I 22 =+=0430,74[rad]
4,96
4,6arctgγ ===
[A]e6,76e6,76j4,64,96III0j43j0,74
21 ⋅=⋅=+=+=
Tema 3.2.
1. Determinati, cu ajutorul fazorilor, valoarea efectiva, faza initiala a curentului i = i1– i2 si
scrieti expresia valorii instantanee i(t), daca i1(t)= √2I1sin(ωt+π/3),
i2(t)= √2I2sin(ωt+2π/3) si I1 = 2I2 = 3[A].
2. Deduceti forma canonica in sinus a t.e.m. autoinduse in bobina cu inductivitatea L=3[mH]
parcursa de curentul i1(t) = 3√2 sin(314t+π/3)[A].
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dipol pasiv
ek=0u
i
p
3.1.4. CARACTERIZAREA DIPOLULUI PASIV
• Caracterizare = cunoasterea valorii parametrilor electrici echivalenti;
• Parametri electrici pot fi determinati experimental prin incercari electrice
(mers in gol si scurtcircuit de proba).
uju eUU)tsin(U2u
γ=↔γ+ω=
iji eII)tsin(I2i
γ=↔γ+ω=
u
Dipol electric pasiv (receptor)
excitatie:
raspuns:
parametri dipolului:
?
- grupe de cite 2 parametri reali;
- parametri complexi.
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a) Impedanţa Z şi defazaj φ
ππ−∈γ−γ=ϕ
Ω≥=
]rad[,2
,2
][,0I
UZ
iu
)tsin(Z
U2i u ϕ−γ+ω=
φ>0 – dipol cu caracter inductiv; φ=0
– dipol cu caracter rezistiv; φ<0 –
dipol cu caracter capacitiv.
b) Rezistenta R şi reactanta X
Ω<≥ϕ=
Ω≥ϕ=
][,0sau0sinZX
][,0cosZR
=ϕ
+=
R
Xarctg
XRZ 22
)R
Xarctgtsin(
XR
U2i u
22−γ+ω
+=
X>0 – dipol cu caracter inductiv; X=0
– dipol cu caracter rezistiv; X<0 –
dipol cu caracter capacitiv.
b) Rezistenta R şi reactanta X
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Z
UI =ϕj)γj(γ
jγ
jγ
eZeI
U
eI
eU
I
UZ iu
i
u
⋅=⋅=⋅⋅
== −
Z = Z·ejφ = Z(cosφ +j·sinφ) = R + j X
c) Impedanţa complexa Z Nu este fazor ci operator complex
I → i(t)
Planul complex al fazorilor tensiune si curent (a) si semiplanele parametrilor impedanta
(b) si admitanta (c).
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c) Admitanta Y si defazaj φ
−∈−=
≥==
[rad],2
π,
2
πγγ
[S]0,Z
1
U
IY
iuϕ)tsin(UY2i u ϕ−γ+ω=
d) Conductanta G şi susceptanta Bd) Conductanta G şi susceptanta B
<≥ϕ=
≥ϕ=
]S[,0sau0sinYB
]S[,0cosYG
−γ+ω+=G
BarctgtsinBGU2i u
22
=ϕ
+=
G
Barctg
BGY 22
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jBGsinjYcosYYeeZ
1
Z
1
U
IY j
j−=ϕ−ϕ===== ϕ−
ϕ
YUI =
c) Admitanta complexa Y Nu este fazor ci operator complex
I → i(t)
Planul complex al fazorilor tensiune si curent (a) si semiplanele parametrilor impedanta
(b) si admitanta (c).
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Problema 3.4.
Ce caracter (inductiv, capacitiv sau rezistiv) are dipolul electric caracterizat de
impedanta complexa Z = 3 + j4[Ω].
Solutie: Z = R +jX = 3 +j4; → R = 3; X = +4 >0 → caracter inductiv.
Problema 3.5.
Curent sinusoidal, de pulsatie ω = 2πf, frecventa f = 50[Hz] si valoare efectiva I =
2[mA], parcurge rezistorul de rezistenta R = 1500[Ω]. Sa se calculeze:
• valoarea instantanee a curentului: i(t)=√2·I·sin(ωt+α) = 2,83·10-3sin(314t+ α)[A];
• valoarea instantanee a tensiunii, la bornele rezistorului:
u(t)=R·i =√2·I·R·sin(ωt+α) = 4,24·sin(314t+ α)[V];u(t)=R·i =√2·I·R·sin(ωt+α) = 4,24·sin(314t+ α)[V];
• valoarea efectiva a tensiunii: U = R·I =1500·2·10-3 = 3[V].
Problema 3.6
Sa se calculeze curentul prin bobina de inductivitate L = 5[μH], alimentata la
tensiunea sinusoidala cu valoare efectiva de 3[V] si frecventa f = 20[kHz].
• valoarea instantanee a tensiunii : u(t) = √2·U·sin(ωt+α) = √2·3·sin(125,7·103t+α)[V];
• admitanta bobinei: YL= 1/ ZL= 1/ ω·L =1/(2πf·L) = 1,59[S];
• valoarea instantanee a curentului:
i(t) =√2·U·YL·sin(ωt+α - π/2) = √2·4,77·sin(125,7·103t+ α - π/2)[A].
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Tema 3.3
1. Calculati impedanta complexa, impedanta si admitanta condensatorului cu
capacitatea C = 33[nF], la frecventa f = 100[Hz]. Se da X = -1/ωC.
2. Ce caracter are dipolul electric caracterizat de admitanta Y = 3 + j4[S]. Justificati
raspunsul.
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)tsin(U2u uγ+ω=)tsin(I2i iγ+ω=
3.1.5. PUTERI ELECTRICE IN REGIM SINUSOIDAL
a) Puterea instantanee p(t)
)γtIsin(ω2)γtUsin(ω2i(t)u(t)p(t) =+⋅⋅+⋅=⋅=
dipol pasiv
ek=0u
i
p
Se dau:
Se definesc:
b) Puterea aparenta S
)γγtsin(2ωUI)γcos(γUI
)γtIsin(ω2)γtUsin(ω2i(t)u(t)p(t)
iuiu
iu
++⋅⋅−−⋅=
=+⋅⋅+⋅=⋅=
]VA[,0UYIZIUS 22 ≥===
•amplitudinea puterii instantanee;
•putere disponibila.
φ
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S=UI
0
S
Puterea instantanee p(t), activa P si aparenta S.
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]W[,0GUcosYUIRcosIZ
cosIUpdtT
1P
2222
Tt
t
1
1
≥=ϕ==ϕ=
=ϕ== ∫+
•P≥0 → absorbita;
•putere utila.
c) Puterea activa P
];[,cosS
Pkp −== ϕfactor de putere:
- factor de utilizare a puterii disponibile;
- caracterizeaza eficacitatea sistemului de distributie a energiei electrice;
- distribuitorul de energie doreste kP cit mai mare, adica kP →1.
[0,1];kP ∈
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d) Puterea reactiva Q
]VAR[,0sau0BUsinYUIXsinIZ
PSsinIUQ
2222
22
<≥=ϕ==ϕ=
=−=ϕ=
•dipol inductiv: Q > 0, absoarbe putere reactiva;
•dipol rezistiv: Q = 0;
•dipol capacitiv: Q < 0, debiteaza putere reactiva.•dipol capacitiv: Q < 0, debiteaza putere reactiva.
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22*jj***
22jj**
U)jBG(UYeUeUYUYU
I)jXR(IZeIeIZIIZIUS
uu
ii
+====
=+=====γ−γ
γ−γ
d) Puterea complexa S
jQPjSsinScoseSeIUeIUeIUS j)γj(γjγjγ*iuiu +=+===== −− ϕϕϕ
Nu este fazor ci operator complex
Semiplanul puterii complexe S.
0
>0
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Problema 3.7.
Un motor asincron monofazat este alimentat la tensiunea de 220[V], frecventa
50[Hz]. Motorul este receptor inductiv, care poate fi reprezentat prin impedanta
complexa Z = R+jX = 42+j26[Ω]. Calculati puterile electrice absorbite de motor.Rezolvare
]; [ 49,42642XRZ 2222 =+=+=
4,45[A];49,6
220
Z
UI ===
516[var];4,4526IXQ 833[W];4,4542IRP 2222 =⋅=⋅==⋅=⋅=833P
Ω
0,85.980
833
S
Pcosk 980[VA];4,45220IUS P =====⋅=⋅= ϕ
Tema 3.4.
1. Aceeasi problema dar rezolvata in complex. Calculati si inductivitatea motorului.
2. Rezistorul cu R = 3[Ω], parcurs de un curent sinusoidal, disipa P = 675[W]. Care
este valoarea maxima a curentului si valoarea efectiva a tensiunii la bornele
rezistorului?
3. Ce caracter (rezistiv, inductiv sau capacitiv) are dipolul care absoarbe atit putere
activa cit si putere reactiva. Justificati raspunsul.
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Problema 3.8.
O bobina reala cu rezistenta R = 6[Ω] si inductivitatea L = 15[mH] este alimentata cu
tensiunea u(t) = √2·24·sin(314t)[V]. Calculati parametrii electrici ai bobinei si puterile
electrice absorbite de aceasta.
Rezolvare:
• Valoarea efectiva, pulsatia, frecventa si faza initiala a tensiunii sint:
U = 24[V]; ω = 100π = 314[rad/s]; f = ω/2π = 50[Hz]; γu = 0.
• Reactanta: X=ωL= 100π·15·10-3 = 4,71[Ω], permite calculul impedantei si defajazului:
Z = (R2+X2)1/2 = 7,63[Ω]; φ = arctgX/R = arctg4,71/6 = 0,66[rad] = 380.
• Valoarea efectiva, faza initiala si expresia curentului absorbit:
I = U/Z = 24/7,63 = 3,13[A]; γi= γu- φ = -0,66[rad]; i(t) = √2·3,14·sin(314t-0,66)[A];
• Puterile activa si reactiva consumate de bobina:
P = R·I2 = U·Icosφ = 59,3[W]; Q = X·I2 = U·Isinφ = 46,6[var].
• Utilizind simbolurile complexe ale marimilor si parametrilor:
U = U·ejγu = 24[V]; I = I·ejγi = 3,14·e-j0,66 = 2,47-j1,94[A];
Z = Zejφ = R+jX = 6+j4,71[Ω];
S = Z·I2 = U·I* = 24(2,47+j1,94) = 59,3 + j46,6[VA].
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Tema 3.5
1. Calculati impedanta si admitanta unui condensator de capacitate C =47[nF] la
frecventa f=100[kHz]. Repetati calculele pentru bobina cu inductivitatea
L=3,3[μH]. Calculati valorile maxime ale tensiunii la bornele acestor elemente
ideale de circuit parcurse de curentul sinusoidal de frecventa f si valoare efectiva I
= 5[mA].
2. La ce tensiune se poate alimenta un rezistor cu caracteristicile: P = 1[W] si R
=10[kΩ].
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§ 3.2. ECUATIILE CIRCUITELOR ELECTRICE
3.2.1. TEOREMA LUI JOUBERT - latura
3.2.2. TEOREMELE LUI KIRCHHOFF – nod; ochi
3.2.3. TEOREMA CONSERVARII PUTERILOR - circuit3.2.3. TEOREMA CONSERVARII PUTERILOR - circuit
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3.2.1. FORMA COMPLEXA A LEGII LUI OHM
Teorema lui Joubert; legea conductiei electrice, in regim
cuasistatianar, pentru o latura de circuit electric.
Latura activa si receptoare.
k k+1
kkkk ZIUE ⋅=+
);dt
diL(edlE
Γ
kkk∫ −+=⋅
∫ ⋅++⋅=+t
0k
k
kkkkkk dti
C
1
dt
diLiRue
∫∫ −⋅+⋅=−+=⋅t
0kk
Γk
kkkCR udtiC
1iRuuudlE
kk
k
kkk
k
kkkkkk ZI)]
ωC
1Lj(ω[RI
Cjω
IILjωIRUE ⋅=−⋅+=
⋅+⋅⋅+⋅=+
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3.2.2. TEOREMELE LUI KIRCHHOFF
;0i(n)k
k∑∈
= 0iiii nk21 =−+−+
;0I(n)k
k∑∈
= 0IIII nk21 =−+−+
+Ik ↔ ik iese din nod
a) prima teorema - nod
b) teorema a doua - ochi Nod de circuit electric
;VVu 1kkk +−= ∑∈
=(o)k
k 0u
∑∈
=(n)k
k 0U
+Uk ↔ ↑uk ≡
↑o; +Ek ↔ ↑ek
≡ ↑o; +Ik ↔
↑ik ≡ ↑o;
kkkk ZIUE ⋅=+∑ ∑∈ ∈
=(o)k (o)k
kkk IZE
Ochi de circuit electric
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APLICATIE
∑∈
−=−=(n)k
kkkBAAB )EIZ(UU∑∈
=+(n)k
kBA 0UU
∑∈
=(n)k
k 0U
kkkk ZIUE ⋅=+
Tensiunea intre doua noduri
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3.2.3. TEOREMA CONSERVARII PUTERILOR
;0Ikj
j∑∈
= ;0Ikj
*
j∑∈
= ;0IVkj
*
jk∑∈
= ;0IVkj
*
j
n
1k
k∑∑∈=
= 0.I)VV(*
k1k
n
1k
k =− +=∑
0;IU*
k
l
1k
k =⋅∑=
0;Sl
1k
k =∑=
kkk jQPS +=
;jXRZ kkk += ∑∑ =l
2
kk
l*
kk IZIE
0Ql
1k
k =∑=
0Pl
1k
k =∑=
- se conserva atit puterile
active cit si cele reactive.
;jXRZ kkk += ∑∑== 1k
kk
1k
kk
Vk Vk+1
Latura activa si receptoare
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§ 3.3. ELEMENTE DE CIRCUIT DIPOLARE
3.3.1. ELEMENTE ACTIVE (SURSE, GENERATOARE)
element activ (e ≠ 0)
pasiv (e = 0)
rezistor
bobina
condensator
3.3.2. REZISTOARE ELECTRICE
3.3.3. BOBINE ELECTRICE
3.3.4. CONDENSATOARE ELECTRICE
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3.3.1. ELEMENTE ACTIVE (SURSE, GENERATOARE)
u =const.= e
i =const.= isc
u = e - R .iu = e - Rg.i
i = isc– u.Gg
Generator ideal de tensiune (a), ideal de curent (b) si scheme echivalente
ale generatorului real (c).
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3.3.2. REZISTOARE ELECTRICE
a) rezistor liniar: R = const. ≠ f(u,i)
IRURiu =↔=
0X =
0R
Xarctg ==ϕR
I
UZR ==
R;R =Zindependenta
de frecventa
i u +ju
u(t) i(t)
Rezistor liniar in c.a.
R
UUYIRIZjQPS
22*22 ====+= P = R·I2 = U2/R; Q
= 0.
p = u·i
u
i
R
u
i0
+1
+j
φ = 0 U
I
a b c
ωt0
i
φ = 0
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b) rezistor neliniar: R ≠ const. = f(u,i)
Rezistoare neliniare; a) -filament metalic; b) -termistor; c) -varistor; d) -dioda
semiconductoare; e) -dioda Zener; f)- dioda tunel.
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culoarea
banda 1
banda 2
banda 3
banda 4banda 5
Negru 0 0 0 x 1
Maro 1 1 1 x 10 1%
Rosu 2 2 2 x 100 2%
Portocaliu
3 3 3 x 1,000
Galbe x
Utilizare:
• producerea locala a unei caderi de tensiune;
• transformarea energiei electrice in caldura.
Tehnologie:
• rezistoare chimice;
• rezistoare bobinate;
• rezistoare cu pelicula.
Marimi caracteristice:Galben
4 4 4x
10,000
Verde 5 5 5x
100,0000.50%
Albastru
6 6 6 x 1060.25%
Violet 7 7 7 x 1070.10%
Gri 8 8 8 x 1080.05%
Alb 9 9 9 x 109
Auriu x 0.1 5%
Argintiu
x 0.01 10%
Codul de culori pentru marcarea rezistoarelor electronice
Categorii de rezistoare:
• fixe;
• variabile;
• neliniare
Marimi caracteristice:
• puterea disipata;
• rezistenta;
• toleranta.
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3.3.3. BOBINE ELECTRICE
a) bobina ideala (fara pierderi): L = const. ≠ f(Φ,i); R = 0.
inductivitate = inductanta: L = Φ/i; [H = Wb/A]
Rieu =+0R =
dt
diLe −=
ILjUdt
diLu ω=↔=
R =0;
X = +ωL;
φ = +π/2;
ZL= ωLLjωI
UZL ⋅==
• impedanta bobinei variaza liniar cu frecventa f;
• la f = 0 (c.c.) impedanta este nula (scurt-circuit). u(t) i(t) γu= 0
L
UjUYILjIZjQPS
22*22
ω==ω==+=
P = 0;
Q = ωL·I2 > 0; absorbita
• la f = 0 (c.c.) impedanta este nula (scurt-circuit).
ωt
φ=+π/22π
0
u
i
p = u·i
u
i
L
Φ
i0
+1
+j
φ = +π/2
U
I
a b c
Bobina electrica ideala in c.a. sinusoidal
γu= 0
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b) bobina cu pierderi (reala): L = const. ≠ f(Φ,i); R ≠ 0.
- in rezistenta proprie PR; -
prin histerezis magnetic PH; -
prin curenti turbionari PT
pierderi
U =UR+UL=R.I+jωL.I
φ = π/2- α;
P = UIcosφ = PR+PH+PT
Zs= Rs+ jωLs
Scheme echivalente serie (a) si paralel (b) ale bobinei
reale.
α = unghi de pierderi.
;I
PR
2S =
.If2π
P-IUL
2
222
S ⋅⋅=
- metoda 3 aparate:
P[W]; U[V]; I[A]
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Bobine fara miez ( cu aer):
• inductivitate perfect liniara;
• frecventa mare, care ar cauza pierderi exagerate in miez feromagbetic.• frecventa mare, care ar cauza pierderi exagerate in miez feromagbetic.
Bobine cu miez feromagnetic (inchis, sau cu intrefier):
• tola silicioasa sau otel electrotehnic: εr= 103 ÷ 105;
• intrefierul liniarizeaza caracteristica de magnerizare;
• utilizate la frecvente industriale (sute Hz): p1,0/50=1[W/kg].
Bobine cu miez ferimagnetic:
• ferite = materiale semiconductoare sinterizate: MeO Fe2O3;
• utilizate la frecvente mari si foarte mari (kHz ÷GHz).
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3.3.4. CONDENSATOARE ELECTRICEa) condensator liniar (fara pierderi): C = const. ≠ f(q,u).
capacitate: C = q/u; [F = C/V]
dt
dq i =
uC q ⋅=const. C =
UCjIdt
duC
dt
dqi ω=↔==
2;
C
1X;0R
π−=ϕ
ω−==
Cω
1j
Cjω
1
I
UZC ⋅
−=⋅
==Cω
1ZC ⋅
=
• impedanta condensatorului variaza invers proportional cu frecventa;
• la frecventa nula (c.c.) impedanta condensatorului este infinita (intrerupere de circuit).
22*2
2 CUjUYC
IjIZjQPS ω−==ω
−==+=P = 0;
Q = -ωC·U2 < 0;debitata
Condensator liniar (fara pierderi)
• la frecventa nula (c.c.) impedanta condensatorului este infinita (intrerupere de circuit).
p = u·i
u
i
C
q
u0
+1
+j
φ = -π/2
U
I
a b c
ωt
ui
u(t) i(t)
φ =-π/2
γu=0
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a) condensator real (cu pierderi): P ≠ 0; φ ≠ -π/2
- imperfectiunii dielectricului; -
polarizarii ciclice a dielectriculuipierderi datorate
I = IR+IC = U/RP+jωCP.U
φ = -π/2 + δ;
δ = unghi de pierderi.
Yp = 1/Rp +jωC
- metoda 3 aparate:
P[W]; U[V]; I[A]
Scheme echivalente serie (a) si paralel (b) ale
condensatorului real
;P
UR
2
P = .Uf2π
P-IUC
2
222
P ⋅⋅=
aplicatii incalzire dielectricatgδCUf2πsinδ cosδ
IUcosUI P 2C ⋅⋅⋅==⋅= ϕ
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Clasificare dupa:
• geometria armaturilor: rulate, plane, tubulare, plachete;
• tipul dielectricului: aer, hirtie, ceramica, mica, oxid metalic (Al2O3; Ta2O5);
• domeniu de utilizare: electronica, electrotehnica, inalta tensiune.
Familii:
• fixe neelectrolitice: hirtie,impregnata, mica, film termoplastic (poliester,
policarbonat, polipropilena, poliester), sticla, ceramica, ulei, gaz comprimat;
• fixe electrolitice: oxid metalic, polarizate / nepolarizate;• fixe electrolitice: oxid metalic, polarizate / nepolarizate;
• variabile / ajustabile (trimer): aer, ceramica, film plastic;
• neliniare: diode varicap, folosite in automatizari.
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tgδCUf2π P 2 ⋅⋅⋅=
d
AεεC r0 ⋅⋅
= ][W/m tgδfεε2πEV
Pp 3
0r
2 ⋅⋅⋅⋅⋅==
dEU ⋅=
- uscare materiale izolatoare (lemn, piese abrazive, textile etc.);
- lipire mase plastice;
- cuptor cu microunde.
Utilizari:
Incalzire dielectrica
Pierderi in dielectricul unui condensator (a).
Schema unei instalatii de incalzire dielectrica (b).
- cuptor cu microunde.
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Element de circuit Rezistor Bobina Condensator
Ecuatia tensiunii functie de timp u = R·i u = L·di/dt u =1/C·∫i·dt
Ecuatia tensiunii in complex UR= R·I UL= jωL·I UC= I / jωC
Impedanta complexa Z = R + jX ZR= R ZL= jωL ZC= -j / ωC
Admitanta complexa Y = G - jB YR= 1/ R YL= -j / ωL YC= jωC
Defazaj φ φR = 0 φL = +π/2 φR = – π/2
Caracterizarea elementelor ideale de circuit
Impedanta Z ZR= R ZL= ωL ZC= ωL
Admitanta Y YR= 1/ R YL= 1 / ωL YC= ωC
Rezistenta R RR= R RL= 0 RC= 0
Reactanta X XR= 0 XL= +ωL XC= -1/ωC
Conductanta G GR= 1/ R GL= 0 GC= 0
Susceptanta B BR= 0 BL= -1 / ωL BC= +ωC
Putere complexa S = P + jQ SR= R·I2 + j0 SL= 0 + jωLI2 SC= 0 –
jωCU2
Factor de putere kP = cosφ cosφ = 1 cosφ = 0 cosφ = 0
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Teme 3.6.
Problema 3.9.
Calculul parametrilor unei bobine reale.
Problema 3.10.
Care din valorile impedantelor de mai jos, caracterizeaza un condensator real si care
un condensator ideal: Z = 1 + j2[Ω], Z =1 - j2[Ω], Z = 1[Ω], Z = +j2[Ω], Z= - j2[Ω].
Justificati raspunsul.
Teme 3.6.
1. Reprezentati variatia rezistentei electrice cu frecventa pentru rezistor, bobina si
condensator.
2. Explicati comportarea bobinei si condensatorului in c.a. daca frecventa f→∞.
3. Care din valorile impedantelor de mai jos, caracterizeaza o bobina reala si care o
bobina ideala: Z=1+j2[Ω], Z= 1-j2[Ω], Z= 1[Ω], Z= +j2[Ω], Z= -j2[Ω]. Justificati
alegerea facuta.
4. Un condensator plan are caracteristicile: A=1[m2], d=5[mm] εr=2,5 si tgδ=10-3.
Care este capacitatea condensatorului şi pierderile în dielectricul său, dacă este
alimentat în c.a. cu U=1[kV] şi f=1[kHz]?