Fasores - Eletricidade e Energia UFJF

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Fasores
Pedro Machado de Almeida
Departamento de Energia Ele´trica
Universidade Federal de Juiz de Fora
Juiz de Fora, MG, 36.036-900 Brasil
e-mail: pedro.machado@engenharia.ufjf.br
Novembro 2015
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 1 / 36
Contextualizac¸a˜o
As tenso˜es geradas pelos geradores s´ıncronos em sistemas de poteˆncia
e pela maioria dos geradores dispon´ıveis comercialmente, possuem uma
forma de onda essencialmente senoidal;
Fontes senoidais sa˜o comuns tanto em sistemas de poteˆncia como em
sistemas eletroˆnicos;
Embora entradas senoidais possam ser analisadas atrave´s de equac¸o˜es
diferenciais, torna-se importante desenvolver novas te´cnicas que redu-
zam o esforc¸o computacional para sua ana´lise.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 2 / 36
Contextualizac¸a˜o
As tenso˜es geradas pelos geradores s´ıncronos em sistemas de poteˆncia
e pela maioria dos geradores dispon´ıveis comercialmente, possuem uma
forma de onda essencialmente senoidal;
Fontes senoidais sa˜o comuns tanto em sistemas de poteˆncia como em
sistemas eletroˆnicos;
Embora entradas senoidais possam ser analisadas atrave´s de equac¸o˜es
diferenciais, torna-se importante desenvolver novas te´cnicas que redu-
zam o esforc¸o computacional para sua ana´lise.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 2 / 36
Contextualizac¸a˜o
As tenso˜es geradas pelos geradores s´ıncronos em sistemas de poteˆncia
e pela maioria dos geradores dispon´ıveis comercialmente, possuem uma
forma de onda essencialmente senoidal;
Fontes senoidais sa˜o comuns tanto em sistemas de poteˆncia como em
sistemas eletroˆnicos;
Embora entradas senoidais possam ser analisadas atrave´s de equac¸o˜es
diferenciais, torna-se importante desenvolver novas te´cnicas que redu-
zam o esforc¸o computacional para sua ana´lise.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 2 / 36
Ana´lise senoidal em estado permanente
A ana´lise cla´ssica de circuitos ele´tricos requer a soluc¸a˜o de equac¸o˜es
diferenciais;
Resposta natural decai a zero ao longo do tempo =⇒ se anula quanto
t tende ao infinito;
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Utilizac¸a˜o da func¸a˜o exponencial complexa e jωt , ao inve´s da func¸a˜o
senoidal.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 36
Ana´lise senoidal em estado permanente
A ana´lise cla´ssica de circuitos ele´tricos requer a soluc¸a˜o de equac¸o˜es
diferenciais;
Resposta natural decai a zero ao longo do tempo =⇒ se anula quanto
t tende ao infinito;
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Utilizac¸a˜o da func¸a˜o exponencial complexa e jωt , ao inve´s da func¸a˜o
senoidal.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 36
Ana´lise senoidal em estado permanente
A ana´lise cla´ssica de circuitos ele´tricos requer a soluc¸a˜o de equac¸o˜es
diferenciais;
Resposta natural decai a zero ao longo do tempo =⇒ se anula quanto
t tende ao infinito;
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Utilizac¸a˜o da func¸a˜o exponencial complexa e jωt , ao inve´s da func¸a˜o
senoidal.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 36
Ana´lise senoidal em estado permanente
A ana´lise cla´ssica de circuitos ele´tricos requer a soluc¸a˜o de equac¸o˜es
diferenciais;
Resposta natural decai a zero ao longo do tempo =⇒ se anula quanto
t tende ao infinito;
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Utilizac¸a˜o da func¸a˜o exponencial complexa e jωt , ao inve´s da func¸a˜o
senoidal.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 36
Ana´lise senoidal em estado permanente
A ana´lise cla´ssica de circuitos ele´tricos requer a soluc¸a˜o de equac¸o˜es
diferenciais;
Resposta natural decai a zero ao longo do tempo =⇒ se anula quanto
t tende ao infinito;
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Utilizac¸a˜o da func¸a˜o exponencial complexa e jωt , ao inve´s da func¸a˜o
senoidal.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 36
Ana´lise senoidal em estado permanente
A Utilizac¸a˜o da func¸a˜o exponencial complexa resulta em uma simpli-
ficac¸a˜o dos procedimentos de ana´lise em estado permanente, transfor-
mando as equac¸o˜es diferenciais em equac¸o˜es alge´bricas complexas.
Outro aspecto muito importante, que vem forc¸ar a necessidade de es-
tudar a resposta em estado permanente senoidal, adve´m do fato de que
uma func¸a˜o perio´dica qualquer pode ser expressa, atrave´s da se´rie de
Fourier, como a soma de va´rias componentes senoidais de frequeˆncias
diferentes.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 4 / 36
Ana´lise senoidal em estado permanente
A Utilizac¸a˜o da func¸a˜o exponencial complexa resulta em uma simpli-
ficac¸a˜o dos procedimentos de ana´lise em estado permanente, transfor-
mando as equac¸o˜es diferenciais em equac¸o˜es alge´bricas complexas.
Outro aspecto muito importante, que vem forc¸ar a necessidade de es-
tudar a resposta em estado permanente senoidal, adve´m do fato de que
uma func¸a˜o perio´dica qualquer pode ser expressa, atrave´s da se´rie de
Fourier, como a soma de va´rias componentes senoidais de frequeˆncias
diferentes.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 4 / 36
Considerac¸o˜es iniciais
O Teorema de Euler fornece uma relac¸a˜o entre func¸o˜es senoidais e
exponenciais;
Possibilitando, a substituic¸a˜o das func¸o˜es senoidais pelas exponenciais
complexas correspondentes;
O conceito de fasor permite a conversa˜o das equac¸o˜es diferencias line-
ares em equac¸o˜es alge´bricas lineares envolvendo nu´meros complexos;
A utilizac¸a˜o de fasores resulta nos conceitos de impedaˆncia e admi-
taˆncia, ana´logos a` resisteˆncia e condutaˆncia em circuitos puramente
resistivos;
Os me´todos de ana´lise e teoremas desenvolvidos para circuitos resistivos
podem ser facilmente modificados/adaptados para a ana´lise em estado
permanente de circuitos com resistores, indutores e capacitores.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 36
Considerac¸o˜es iniciais
O Teorema de Euler fornece uma relac¸a˜o entre func¸o˜es senoidais e
exponenciais;
Possibilitando, a substituic¸a˜o das func¸o˜es senoidais pelas exponenciais
complexas correspondentes;
O conceito de fasor permite a conversa˜o das equac¸o˜es diferencias line-
ares em equac¸o˜es alge´bricas lineares envolvendo nu´meros complexos;
A utilizac¸a˜o de fasores resulta nos conceitos de impedaˆncia e admi-
taˆncia, ana´logos a` resisteˆncia e condutaˆncia em circuitos puramente
resistivos;
Os me´todos de ana´lise e teoremas desenvolvidos para circuitos resistivos
podem ser facilmente modificados/adaptados para a ana´lise em estado
permanente de circuitos com resistores, indutores e capacitores.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 36
Considerac¸o˜es iniciais
O Teorema de Euler fornece uma relac¸a˜o entre func¸o˜es senoidais e
exponenciais;
Possibilitando, a substituic¸a˜o das func¸o˜es senoidais pelas exponenciais
complexas correspondentes;
O conceito de fasor permite a conversa˜o das equac¸o˜es diferencias line-
ares em equac¸o˜es alge´bricas lineares envolvendo nu´meros complexos;
A utilizac¸a˜o de fasores resulta nos conceitos de impedaˆncia e admi-
taˆncia, ana´logos a` resisteˆncia e condutaˆncia em circuitos puramente
resistivos;
Os me´todos de ana´lise e teoremas desenvolvidos para circuitos resistivos
podem ser facilmente modificados/adaptados para a ana´lise em estado
permanente de circuitos com resistores, indutores e capacitores.Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 36
Considerac¸o˜es iniciais
O Teorema de Euler fornece uma relac¸a˜o entre func¸o˜es senoidais e
exponenciais;
Possibilitando, a substituic¸a˜o das func¸o˜es senoidais pelas exponenciais
complexas correspondentes;
O conceito de fasor permite a conversa˜o das equac¸o˜es diferencias line-
ares em equac¸o˜es alge´bricas lineares envolvendo nu´meros complexos;
A utilizac¸a˜o de fasores resulta nos conceitos de impedaˆncia e admi-
taˆncia, ana´logos a` resisteˆncia e condutaˆncia em circuitos puramente
resistivos;
Os me´todos de ana´lise e teoremas desenvolvidos para circuitos resistivos
podem ser facilmente modificados/adaptados para a ana´lise em estado
permanente de circuitos com resistores, indutores e capacitores.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 36
Considerac¸o˜es iniciais
O Teorema de Euler fornece uma relac¸a˜o entre func¸o˜es senoidais e
exponenciais;
Possibilitando, a substituic¸a˜o das func¸o˜es senoidais pelas exponenciais
complexas correspondentes;
O conceito de fasor permite a conversa˜o das equac¸o˜es diferencias line-
ares em equac¸o˜es alge´bricas lineares envolvendo nu´meros complexos;
A utilizac¸a˜o de fasores resulta nos conceitos de impedaˆncia e admi-
taˆncia, ana´logos a` resisteˆncia e condutaˆncia em circuitos puramente
resistivos;
Os me´todos de ana´lise e teoremas desenvolvidos para circuitos resistivos
podem ser facilmente modificados/adaptados para a ana´lise em estado
permanente de circuitos com resistores, indutores e capacitores.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 36
Considerac¸o˜es iniciais
Os treˆs paraˆmetros que descrevem um sinal senoidal sa˜o amplitude,
aˆngulo de fase e frequeˆncia;
Quando um circuito linear e´ excitado por uma ou mais fontes senoidais,
todas de mesma frequeˆncia ω, as amplitudes e os aˆngulos de fase
das correntes e tenso˜es variam de ramo para ramo;
Contudo, todas as tenso˜es e correntes esta˜o nesta frequeˆncia ω;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 6 / 36
Considerac¸o˜es iniciais
Os treˆs paraˆmetros que descrevem um sinal senoidal sa˜o amplitude,
aˆngulo de fase e frequeˆncia;
Quando um circuito linear e´ excitado por uma ou mais fontes senoidais,
todas de mesma frequeˆncia ω, as amplitudes e os aˆngulos de fase
das correntes e tenso˜es variam de ramo para ramo;
Contudo, todas as tenso˜es e correntes esta˜o nesta frequeˆncia ω;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 6 / 36
Considerac¸o˜es iniciais
Os treˆs paraˆmetros que descrevem um sinal senoidal sa˜o amplitude,
aˆngulo de fase e frequeˆncia;
Quando um circuito linear e´ excitado por uma ou mais fontes senoidais,
todas de mesma frequeˆncia ω, as amplitudes e os aˆngulos de fase
das correntes e tenso˜es variam de ramo para ramo;
Contudo, todas as tenso˜es e correntes esta˜o nesta frequeˆncia ω;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 6 / 36
Vetor espacial
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 7 / 36
Conceito de Fasor
Seja um vetor de comprimento Fm , num aˆngulo Φ em t = 0, girando
a uma velocidade angular ω.
Im
Re
Fm
Φ
ωt
Fm cos(ωt + Φ)
ω
Figura 1: Conceito de Fasor
O vetor pode ser visto como a represen-
tac¸a˜o gra´fica de um nu´mero complexo de
mo´dulo Fm e aˆngulo Φ, ou seja, Fme
jΦ.
O vetor assume novos aˆngulos (ωt + Φ)
a` medida que gira, sendo descrito por
Fme
j (ωt+Φ) = Fme
jωte jΦ
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 8 / 36
Conceito de Fasor
Seja
Fm = Fme
jΦ (1)
Enta˜o a func¸a˜o Fme
jωt descreve o vetor em qualquer instante de tempo
t .
A projec¸a˜o horizontal deste vetor resulta
<[Fme jωt ] = Fm cos(ωt + Φ) (2)
Assim, a func¸a˜o fisicamente realiza´vel Fm cos(ωt + Φ) pode ser ima-
ginada como sendo a parte real da func¸a˜o complexa Fme
jωt , descrita
por um vetor girante de mo´dulo Fm e aˆngulo de fase inicial Φ.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 9 / 36
Conceito de Fasor
Seja
Fm = Fme
jΦ (1)
Enta˜o a func¸a˜o Fme
jωt descreve o vetor em qualquer instante de tempo
t .
A projec¸a˜o horizontal deste vetor resulta
<[Fme jωt ] = Fm cos(ωt + Φ) (2)
Assim, a func¸a˜o fisicamente realiza´vel Fm cos(ωt + Φ) pode ser ima-
ginada como sendo a parte real da func¸a˜o complexa Fme
jωt , descrita
por um vetor girante de mo´dulo Fm e aˆngulo de fase inicial Φ.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 9 / 36
Conceito de Fasor
Seja
Fm = Fme
jΦ (1)
Enta˜o a func¸a˜o Fme
jωt descreve o vetor em qualquer instante de tempo
t .
A projec¸a˜o horizontal deste vetor resulta
<[Fme jωt ] = Fm cos(ωt + Φ) (2)
Assim, a func¸a˜o fisicamente realiza´vel Fm cos(ωt + Φ) pode ser ima-
ginada como sendo a parte real da func¸a˜o complexa Fme
jωt , descrita
por um vetor girante de mo´dulo Fm e aˆngulo de fase inicial Φ.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 9 / 36
Conceito de Fasor
Seja
Fm = Fme
jΦ (1)
Enta˜o a func¸a˜o Fme
jωt descreve o vetor em qualquer instante de tempo
t .
A projec¸a˜o horizontal deste vetor resulta
<[Fme jωt ] = Fm cos(ωt + Φ) (2)
Assim, a func¸a˜o fisicamente realiza´vel Fm cos(ωt + Φ) pode ser ima-
ginada como sendo a parte real da func¸a˜o complexa Fme
jωt , descrita
por um vetor girante de mo´dulo Fm e aˆngulo de fase inicial Φ.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 9 / 36
Conceito de Fasor
O coeficiente complexo Fm e´ chamado de fasor.
Embora todos os fasores sejam nu´meros complexos, nem todos os nu´-
meros complexos sa˜o fasores;
O fasor descreve a func¸a˜o complexa completamente, a menos do valor
da frequeˆncia ω que deve ser especificada em separado.
A convenc¸a˜o usual na ana´lise de circuitos e´ expressar a func¸a˜o senoidal
como parte real da func¸a˜o exponencial complexa;
Assim, todas as tenso˜es e correntes em estado permanente podem ser
determinadas como projec¸o˜es no eixo real de seus fasores girantes em
sentido anti-hota´rio;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 36
Conceito de Fasor
O coeficiente complexo Fm e´ chamado de fasor.
Embora todos os fasores sejam nu´meros complexos, nem todos os nu´-
meros complexos sa˜o fasores;
O fasor descreve a func¸a˜o complexa completamente, a menos do valor
da frequeˆncia ω que deve ser especificada em separado.
A convenc¸a˜o usual na ana´lise de circuitos e´ expressar a func¸a˜o senoidal
como parte real da func¸a˜o exponencial complexa;
Assim, todas as tenso˜es e correntes em estado permanente podem ser
determinadas como projec¸o˜es no eixo real de seus fasores girantes em
sentido anti-hota´rio;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 36
Conceito de Fasor
O coeficiente complexo Fm e´ chamado de fasor.
Embora todos os fasores sejam nu´meros complexos, nem todos os nu´-
meros complexos sa˜o fasores;
O fasor descreve a func¸a˜o complexa completamente, a menos do valor
da frequeˆncia ω que deve ser especificada em separado.
A convenc¸a˜o usual na ana´lise de circuitos e´ expressar a func¸a˜o senoidal
como parte real da func¸a˜o exponencial complexa;
Assim, todas as tenso˜es e correntes em estado permanente podem ser
determinadas como projec¸o˜es no eixo real de seus fasores girantes em
sentido anti-hota´rio;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 36
Conceito de Fasor
O coeficiente complexo Fm e´ chamado de fasor.
Embora todos os fasores sejam nu´meros complexos, nem todos os nu´-
meros complexos sa˜o fasores;
O fasor descreve a func¸a˜o complexa completamente, a menos do valor
da frequeˆncia ω que deve ser especificada em separado.
A convenc¸a˜o usual na ana´lise de circuitos e´ expressar a func¸a˜osenoidal
como parte real da func¸a˜o exponencial complexa;
Assim, todas as tenso˜es e correntes em estado permanente podem ser
determinadas como projec¸o˜es no eixo real de seus fasores girantes em
sentido anti-hota´rio;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 36
Conceito de Fasor
O coeficiente complexo Fm e´ chamado de fasor.
Embora todos os fasores sejam nu´meros complexos, nem todos os nu´-
meros complexos sa˜o fasores;
O fasor descreve a func¸a˜o complexa completamente, a menos do valor
da frequeˆncia ω que deve ser especificada em separado.
A convenc¸a˜o usual na ana´lise de circuitos e´ expressar a func¸a˜o senoidal
como parte real da func¸a˜o exponencial complexa;
Assim, todas as tenso˜es e correntes em estado permanente podem ser
determinadas como projec¸o˜es no eixo real de seus fasores girantes em
sentido anti-hota´rio;
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 36
Conceito de Fasor
O fasor e´ um nu´mero complexo, representado por um ponto sobre a
varia´vel (ou em negrito) que independe do tempo;
A conversa˜o entre a func¸a˜o senoidal e a representac¸a˜o fasorial, ou vice-
versa, pode ser descrito da seguinte forma:
Fm cos(ωt + Φ)⇐⇒ Fm = Fme jΦ (3)
Desta forma, os fasores sa˜o u´teis na soma e na subtrac¸a˜o de func¸o˜es
senoidais de mesma frequeˆncia, bastando, para isso, converteˆ-las em
fasores, obter o fasor resultante e voltar para o dom´ınio do tempo.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 11 / 36
Conceito de Fasor
O fasor e´ um nu´mero complexo, representado por um ponto sobre a
varia´vel (ou em negrito) que independe do tempo;
A conversa˜o entre a func¸a˜o senoidal e a representac¸a˜o fasorial, ou vice-
versa, pode ser descrito da seguinte forma:
Fm cos(ωt + Φ)⇐⇒ Fm = Fme jΦ (3)
Desta forma, os fasores sa˜o u´teis na soma e na subtrac¸a˜o de func¸o˜es
senoidais de mesma frequeˆncia, bastando, para isso, converteˆ-las em
fasores, obter o fasor resultante e voltar para o dom´ınio do tempo.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 11 / 36
Conceito de Fasor
O fasor e´ um nu´mero complexo, representado por um ponto sobre a
varia´vel (ou em negrito) que independe do tempo;
A conversa˜o entre a func¸a˜o senoidal e a representac¸a˜o fasorial, ou vice-
versa, pode ser descrito da seguinte forma:
Fm cos(ωt + Φ)⇐⇒ Fm = Fme jΦ (3)
Desta forma, os fasores sa˜o u´teis na soma e na subtrac¸a˜o de func¸o˜es
senoidais de mesma frequeˆncia, bastando, para isso, converteˆ-las em
fasores, obter o fasor resultante e voltar para o dom´ınio do tempo.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 11 / 36
Fasor
Dada a senoide v(t) = Vm cos(ωt + φ), pode–se escrever
v(t) = Vm cos(ωt + φ) = <(Vme j (ωt+φ)) = <(Vme jφe jωt)
Portanto
v(t) = <(Ve jωt) (4)
Em que
V = Vme
jφ = Vm φ (5)
V e´ a representac¸a˜o fasorial da senoide v(t) para t = 0
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 12 / 36
Fasor
Definic¸a˜o
Fasor e´ a representac¸a˜o complexa da magnitude e fase de um sinal
senoidal.
v(t) = Vm cos(ωt + φ) ⇐⇒(
Representac¸a˜o no
dom´ınio do tempo
) V = Vm φ(
Representac¸a˜o no
dom´ınio da frequeˆncia
)
(6)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 13 / 36
Fasor
Considere os seguintes fasores
V = Vm φ e I = Im −θ
A representac¸a˜o gra´fica ou diagrama fasorial pode ser vista na figura
abaixo
362 PART 2 AC Circuits
Rotation at v rad ⁄s
at t = to
at t = 0
f
Vm
Im
Re 0 to t
Vm
v(t) = Re(Ve jvt )
(a) (b)
Figure 9.7 Representation of Vejωt : (a) sinor rotating counterclockwise, (b) its
projection on the real axis, as a function of time.
Equation (9.23) states that to obtain the sinusoid corresponding to
a given phasor V, multiply the phasor by the time factor ejωt and take
the real part. As a complex quantity, a phasor may be expressed in
rectangular form, polar form, or exponential form. Since a phasor has
magnitude and phase (“direction”), it behaves as a vector and is printed
in boldface. For example, phasors V = Vm φ and I = Im − θ are
graphically represented in Fig. 9.8. Such a graphical representation of
phasors is known as a phasor diagram.
We use lightface italic letters such as z to repre-
sent complex numbers but boldface letters such
as V to represent phasors, because phasors are
vectorlike quantities.
Lagging direction
Leading direction
Real axis
Imaginary axis
Vm
Im
v
v
V
I
–u
f
Figure 9.8 A phasor diagram showing V = Vm φ and I = Im − θ .
Equations (9.21) through (9.23) reveal that to get the phasor corre-
sponding to a sinusoid, we first express the sinusoid in the cosine form
so that the sinusoid can be written as the real part of a complex number.
Then we take out the time factor ejωt , andwhatever is left is the pha-
Figura 2: Diagrama fasorial.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 14 / 36
Fasor
Transformac¸a˜o senoide–fasor
Tabela 1: Transformac¸a˜o senoide–fasor.
Dom´ınio do tempo Dom´ınio da frequeˆncia
v(t) = Vm cos(ωt + φ) V = Vm φ
v(t) = Vm sen(ωt + φ) V = Vm φ− 90
i(t) = Im cos(ωt + θ) I = Im θ
i(t) = Im sen(ωt + θ) I = Im θ − 90
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 15 / 36
Fasor
Exemplo 1
Represente as formas de onda senoidais a seguir como fasores no
plano complexo.
v1(t) = 3 cos(ωt + 40
◦) e v2(t) = 4 cos(ωt − 20◦)
Soluc¸a˜o
Os fasores correspondentes sa˜o
V1 = 3 40◦
V2 = 4 −20◦
Figura 3: Representac¸a˜o dos fasores
no plano complexo.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 16 / 36
Fasor
Exemplo 1
Represente as formas de onda senoidais a seguir como fasores no
plano complexo.
v1(t) = 3 cos(ωt + 40
◦) e v2(t) = 4 cos(ωt − 20◦)
Soluc¸a˜o
Os fasores correspondentes sa˜o
V1 = 3 40◦
V2 = 4 −20◦
Section 5.2 Phasors 227
Figure 5.5 A sinusoid can be
represented as the real part of a vector
rotating counterclockwise in the
complex plane.
Vm Vm
Real
u
v
Imaginary
v(t)
v(t)
t
Figure 5.6 Because the vectors rotate
counterclockwise, V1 leads V2 by 60
(or, equivalently, V2 lags V1 by 60 ).
V1
V2
3
40*
20*
4
The phasor diagram is shown in Figure 5.6. Notice that the angle betweenV1 andV2
is 60 . Because the complex vectors rotate counterclockwise, we say thatV1 leadsV2
by 60 . (An alternative way to state the phase relationship is to state thatV2 lagsV1
by 60 .)
Wehave seen that the voltages versus timecanbeobtainedby tracing the real part To determine phase
relationships between
sinusoids from their plots
versus time, nd the shortest
time interval tp between
positive peaks of the two
waveforms. Then, the phase
angle is = (tp/T)× 360 . If
the peak of v1(t) occurs rst,
we say that v1(t) leads v2(t) or
that v2(t) lags v1(t).
of the rotating vectors. The plots of v1(t) and v2(t) versus t are shown in Figure 5.7.
Notice that v1(t) reaches its peak 60 earlier than v2(t). This is the meaning of the
statement that v1(t) leads v2(t) by 60 .
Exercise 5.5 Consider the voltages given by
v1(t) = 10 cos( t 30 )
v2(t) = 10 cos( t + 30 )
v3(t) = 10 sin( t + 45 )
State the phase relationship between each pair of the voltages. (Hint:Find the phasor
for each voltage and draw the phasor diagram.)
Figura 3: Representac¸a˜o dos fasores
no plano complexo.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 16 / 36
Fasor
Exemplo 2
Dados i1(t) = 4 cos(ωt + 30
◦) e i2(t) = 5 sen(ωt − 20◦), calcule
i(t) = i1(t) + i2(t)
Soluc¸a˜o
Reescrevendo as correntes na forma fasorial
I1 = 4 30◦ e I2 = 5 −110◦
A soma e´ dada por
I = I1 + I2 = 4 30◦ + 5 −110◦
I = 3,464 + j2− 1,71− j4,698 = 1,754− j2,698 = 3,218 −56,97◦ A
Voltando para o dom´ınio do tempo
i(t) = 3,218 cos(ωt − 56,97◦) A
PedroMachado de Almeida (UFJF) ENE077 Novembro 2015 17 / 36
Fasor
Exemplo 2
Dados i1(t) = 4 cos(ωt + 30
◦) e i2(t) = 5 sen(ωt − 20◦), calcule
i(t) = i1(t) + i2(t)
Soluc¸a˜o
Reescrevendo as correntes na forma fasorial
I1 = 4 30◦ e I2 = 5 −110◦
A soma e´ dada por
I = I1 + I2 = 4 30◦ + 5 −110◦
I = 3,464 + j2− 1,71− j4,698 = 1,754− j2,698 = 3,218 −56,97◦ A
Voltando para o dom´ınio do tempo
i(t) = 3,218 cos(ωt − 56,97◦) A
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 17 / 36
Fasor
Considere um sinal senoidal da forma
v(t) = <(Ve jωt) = Vm cos(ωt + φ)
A derivada no tempo e´ dada por
dv(t)
dt
= −ωVm sen(ωt + φ) = ωVm cos(ωt + φ+ 90◦)
dv(t)
dt
= <(ωVme jωte jφe j90◦)
Portanto
dv(t)
dt
= <(jωVe jωt) (7)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 18 / 36
Fasor
Dom´ınio do tempo Dom´ınio da frequeˆncia
dv(t)
dt
⇐⇒ jωV∫
v(t)dt ⇐⇒ V
jω
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 19 / 36
Impedaˆncia complexa
Definic¸a˜o
A impedaˆncia Z de um circuito e´ dada pela raza˜o entre o fasor de
tensa˜o e o fasor de corrente, medido em Ω
Z =
V
I
(8)
A impedaˆncia e´ uma func¸a˜o da frequeˆncia ω
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 20 / 36
Impedaˆncia complexa
Resistor
368 PART 2 AC Circuits
showing that the voltage-current relation for the resistor in the phasor
domain continues to be Ohm’s law, as in the time domain. Figure 9.9
illustrates the voltage-current relations of a resistor. We should note from
Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor
diagram in Fig. 9.10.
(a)
i
v
+
−
R
v = iR
(b)
I
V
+
−
R
V = IR
Figure 9.9 Voltage-current
relations for a resistor in the:
(a) time domain, (b) frequency
domain.
I
f
V
0 Re
Im
Figure 9.10 Phasor diagram for the
resistor.
For the inductor L, assume the current through it is i =
Im cos(ωt + φ). The voltage across the inductor is
v = Ldi
dt
= −ωLIm sin(ωt + φ) (9.32)
Recall from Eq. (9.10) that − sinA = cos(A + 90◦). We can write the
voltage as
v = ωLIm cos(ωt + φ + 90◦) (9.33)
which transforms to the phasor
V = ωLImej(φ+90◦) = ωLImejφej90◦ = ωLIm φej90◦ (9.34)
But Im φ = I, and from Eq. (9.19), ej90◦ = j . Thus,
V = jωLI (9.35)
showing that the voltage has a magnitude ofωLIm and a phase of φ+90◦.
The voltage and current are 90◦ out of phase. Specifically, the current
lags the voltage by 90◦. Figure 9.11 shows the voltage-current relations
for the inductor. Figure 9.12 shows the phasor diagram.
Although it is equally correct to say that the in-
ductor voltage leads the current by 90◦, con-
vention gives the current phase relative to the
voltage.
i
v
+
−
L
v = L didt
(a)
I
V
+
−
L
V = jvLI 
(b)
Figure 9.11 Voltage-current
relations for an inductor in the:
(a) time domain, (b) frequency
domain.
For the capacitor C, assume the voltage across it is v =
Vm cos(ωt + φ). The current through the capacitor is
i = C dv
dt
(9.36)
By following the same steps as we took for the inductor or by applying
Eq. (9.27) on Eq. (9.36), we obtain
I = jωCV �⇒ V = I
jωC
(9.37)
showing that the current and voltage are 90◦ out of phase. To be specific,
the current leads the voltage by 90◦. Figure 9.13 shows the voltage-current
v
Re
Im
V
I
0
f
Figure 9.12 Phasor diagram for the
inductor; I lags V.
i
v
+
−
C
(a)
i = C dvdt
I
V
+
−
C
(b)
I = jvCV 
Figure 9.13 Voltage-current
relations for a capacitor in the:
(a) time domain, (b) frequency
domain.
Figura 4: Resistor.
Considere que a seguinte corrente flui atrave´s
do resistor
iR(t) = Im cos(ωt + θ) (9)
Sabendo que
vR(t) = RiR(t) = RIm cos(ωt + θ) (10)
Na forma fasorial
VR = RIm θ = RIR (11)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 21 / 36
Impedaˆncia complexa
Resistor
368 PART 2 AC Circuits
showing that the voltage-current relation for the resistor in the phasor
domain continues to be Ohm’s law, as in the time domain. Figure 9.9
illustrates the voltage-current relations of a resistor. We should note from
Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor
diagram in Fig. 9.10.
(a)
i
v
+
−
R
v = iR
(b)
I
V
+
−
R
V = IR
Figure 9.9 Voltage-current
relations for a resistor in the:
(a) time domain, (b) frequency
domain.
I
f
V
0 Re
Im
Figure 9.10 Phasor diagram for the
resistor.
For the inductor L, assume the current through it is i =
Im cos(ωt + φ). The voltage across the inductor is
v = Ldi
dt
= −ωLIm sin(ωt + φ) (9.32)
Recall from Eq. (9.10) that − sinA = cos(A + 90◦). We can write the
voltage as
v = ωLIm cos(ωt + φ + 90◦) (9.33)
which transforms to the phasor
V = ωLImej(φ+90◦) = ωLImejφej90◦ = ωLIm φej90◦ (9.34)
But Im φ = I, and from Eq. (9.19), ej90◦ = j . Thus,
V = jωLI (9.35)
showing that the voltage has a magnitude ofωLIm and a phase of φ+90◦.
The voltage and current are 90◦ out of phase. Specifically, the current
lags the voltage by 90◦. Figure 9.11 shows the voltage-current relations
for the inductor. Figure 9.12 shows the phasor diagram.
Although it is equally correct to say that the in-
ductor voltage leads the current by 90◦, con-
vention gives the current phase relative to the
voltage.
i
v
+
−
L
v = L didt
(a)
I
V
+
−
L
V = jvLI 
(b)
Figure 9.11 Voltage-current
relations for an inductor in the:
(a) time domain, (b) frequency
domain.
For the capacitor C, assume the voltage across it is v =
Vm cos(ωt + φ). The current through the capacitor is
i = C dv
dt
(9.36)
By following the same steps as we took for the inductor or by applying
Eq. (9.27) on Eq. (9.36), we obtain
I = jωCV �⇒ V = I
jωC
(9.37)
showing that the current and voltage are 90◦ out of phase. To be specific,
the current leads the voltage by 90◦. Figure 9.13 shows the voltage-current
v
Re
Im
V
I
0
f
Figure 9.12 Phasor diagram for the
inductor; I lags V.
i
v
+
−
C
(a)
i = C dvdt
I
V
+
−
C
(b)
I = jvCV 
Figure 9.13 Voltage-current
relations for a capacitor in the:
(a) time domain, (b) frequency
domain.
Figura 4: Resistor.
Considere que a seguinte corrente flui atrave´s
do resistor
iR(t) = Im cos(ωt + θ) (9)
Sabendo que
vR(t) = RiR(t) = RIm cos(ωt + θ) (10)
Na forma fasorial
VR = RIm θ = RIR (11)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 21 / 36
Impedaˆncia complexa
Resistor232 Chapter 5 Steady-State Sinusoidal Analysis
VR
IR
(a) Phasor diagram (b) Current and voltage versus time
vt
vR(t)
iR(t)
u
u
Figure 5.10 For a pure resistance, current and voltage are in phase.
VL = 100 0+
(a) Exercise 5.6 (0.25 H inductance) (b) Exercise 5.7 (100 mF capacitance)
IL = 2 90+
VC = 100 0+
IC = 2 90+
(c) Exercise 5.8 (50 * resistance)
VR = 100 0+
IR = 2 0+
Figure 5.11 Answers for Exercises 5.6, 5.7, and 5.8. The scale has been expanded for the currents compared with the voltages
so the current phasors can be easily seen.
Exercise 5.7 A voltage vC(t) = 100 cos(200t) is applied to a 100- F capacitance.
a. Find the impedance of the capacitance, the phasor current, and the phasor voltage.
b.Draw the phasor diagram.
Answer a. ZC = j50 = 50 90* , IC = 2 90* , VC = 100 0* ; b. the phasor
diagram is shown in Figure 5.11(b). *
Exercise 5.8 A voltage vR(t) = 100 cos(200t) is applied to a 50- resistance. a. Find
the phasor for the current and the phasor voltage. b. Draw the phasor diagram.
Answer a. IR = 2 0* , VR = 100 0* ; b. the phasor diagram is shown in
Figure 5.11(c). *
5.4 CIRCUIT ANALYSIS WITH PHASORS ANDCOMPLEX
IMPEDANCES
Kirchhoff s Laws in Phasor Form
Recall that KVL requires that the voltages sum to zero for any closed path in an
electrical network. A typical KVL equation is
v1(t)+ v2(t) v3(t) = 0 (5.50)
If thevoltages are sinusoidal, they canbe representedbyphasors.Then,Equation5.50
becomes
V1 +V2 V3 = 0 (5.51)
Thus, we can apply KVL directly to the phasors. The sum of the phasor voltages
equals zero for any closed path.
Figura 5: Tensa˜o e corrente no resistor.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 22 / 36
Impedaˆncia complexa
Indutor
If we transform this answer back to the time domain, it is evident that the
same expression for the current is obtained. We conclude that there is no
saving in time or effort when a resistive circuit is analyzed in the frequency
domain.
The Inductor
Let us now turn to the inductor. The time-domain representation is shown in
Fig. 10.14a, and the defining equation, a time-domain expression, is
v(t) = L di(t)
dt
[19]
After substituting the complex voltage equation [16] and complex current
equation [17] in Eq. [19], we have
Vme j (ωt+θ) = L ddt Ime
j (ωt+φ)
Taking the indicated derivative:
Vme j (ωt+θ) = jωL Ime j (ωt+φ)
and dividing through by e jωt :
Vme jθ = jωL Ime jφ
we obtain the desired phasor relationship
V = jωLI [20]
The time-domain differential equation [19] has become the algebraic
equation [20] in the frequency domain. The phasor relationship is indicated
in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦ and
that I must therefore lag V by 90° in an inductor.
CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS386
■ FIGURE 10.14 An inductor and its associated
voltage and current in (a) the time domain, v = L di/dt;
and (b) the frequency domain, V = jωLI.
di
dt
i
v = L 
+
–
(a)
L
I
V = j�LI
+
–
(b)
L
Apply the voltage 8/−50° V at a frequency ω � 100 rad/s to a 4 H
inductor, and determine the phasor current and the time-domain
current.
We make use of the expression we just obtained for the inductor,
I = VjωL =
8/−50◦
j100(4) = − j0.02/−50
◦ = (1/−90◦)(0.02/−50◦)
or
I = 0.02/−140◦ A
If we express this current in the time domain, it becomes
i(t) = 0.02 cos(100t − 140◦) A = 20 cos(100t − 140◦) mA
EXAMPLE 10.4
Figura 6: Indutor.
Considere que a seguinte corrente flui atrave´s
do indutor
iL(t) = Im sen(ωt + θ) (12)
Sabendo que
vL(t) = L
diL(t)
dt
(13)
Substituindo (12) em (13)
vL(t) = L
d
dt
(Im sen(ωt + θ)) = ωLIm cos(ωt + θ)
(14)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 23 / 36
Impedaˆncia complexa
Indutor
If we transform this answer back to the time domain, it is evident that the
same expression for the current is obtained. We conclude that there is no
saving in time or effort when a resistive circuit is analyzed in the frequency
domain.
The Inductor
Let us now turn to the inductor. The time-domain representation is shown in
Fig. 10.14a, and the defining equation, a time-domain expression, is
v(t) = L di(t)
dt
[19]
After substituting the complex voltage equation [16] and complex current
equation [17] in Eq. [19], we have
Vme j (ωt+θ) = L ddt Ime
j (ωt+φ)
Taking the indicated derivative:
Vme j (ωt+θ) = jωL Ime j (ωt+φ)
and dividing through by e jωt :
Vme jθ = jωL Ime jφ
we obtain the desired phasor relationship
V = jωLI [20]
The time-domain differential equation [19] has become the algebraic
equation [20] in the frequency domain. The phasor relationship is indicated
in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦ and
that I must therefore lag V by 90° in an inductor.
CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS386
■ FIGURE 10.14 An inductor and its associated
voltage and current in (a) the time domain, v = L di/dt;
and (b) the frequency domain, V = jωLI.
di
dt
i
v = L 
+
–
(a)
L
I
V = j�LI
+
–
(b)
L
Apply the voltage 8/−50° V at a frequency ω � 100 rad/s to a 4 H
inductor, and determine the phasor current and the time-domain
current.
We make use of the expression we just obtained for the inductor,
I = VjωL =
8/−50◦
j100(4) = − j0.02/−50
◦ = (1/−90◦)(0.02/−50◦)
or
I = 0.02/−140◦ A
If we express this current in the time domain, it becomes
i(t) = 0.02 cos(100t − 140◦) A = 20 cos(100t − 140◦) mA
EXAMPLE 10.4
Figura 6: Indutor.
Considere que a seguinte corrente flui atrave´s
do indutor
iL(t) = Im sen(ωt + θ) (12)
Sabendo que
vL(t) = L
diL(t)
dt
(13)
Substituindo (12) em (13)
vL(t) = L
d
dt
(Im sen(ωt + θ)) = ωLIm cos(ωt + θ)
(14)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 23 / 36
Impedaˆncia complexa
Indutor
If we transform this answer back to the time domain, it is evident that the
same expression for the current is obtained. We conclude that there is no
saving in time or effort when a resistive circuit is analyzed in the frequency
domain.
The Inductor
Let us now turn to the inductor. The time-domain representation is shown in
Fig. 10.14a, and the defining equation, a time-domain expression, is
v(t) = L di(t)
dt
[19]
After substituting the complex voltage equation [16] and complex current
equation [17] in Eq. [19], we have
Vme j (ωt+θ) = L ddt Ime
j (ωt+φ)
Taking the indicated derivative:
Vme j (ωt+θ) = jωL Ime j (ωt+φ)
and dividing through by e jωt :
Vme jθ = jωL Ime jφ
we obtain the desired phasor relationship
V = jωLI [20]
The time-domain differential equation [19] has become the algebraic
equation [20] in the frequency domain. The phasor relationship is indicated
in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦ and
that I must therefore lag V by 90° in an inductor.
CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS386
■ FIGURE 10.14 An inductor and its associated
voltage and current in (a) the time domain, v = L di/dt;
and (b) the frequency domain, V = jωLI.
di
dt
i
v = L 
+
–
(a)
L
I
V = j�LI
+
–
(b)
L
Apply the voltage 8/−50° V at a frequency ω � 100 rad/s to a 4 H
inductor, and determine the phasor current and the time-domain
current.
We make use of the expression we just obtained for the inductor,
I = VjωL =
8/−50◦
j100(4) = − j0.02/−50
◦ = (1/−90◦)(0.02/−50◦)
or
I = 0.02/−140◦ A
If we express this current in the time domain, it becomes
i(t) = 0.02 cos(100t − 140◦) A = 20 cos(100t − 140◦) mA
EXAMPLE 10.4
Figura 7: Indutor.
Em termos de fasores
IL = Im θ − 90◦ (15)
VL = ωLIm θ (16)
No indutor a tensa˜o esta´ adiantada da
corrente em 90◦
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 24 / 36
Impedaˆncia complexa
Indutor
If we transform this answer back to the time domain, it is evident that the
same expression for the current is obtained. We conclude that there is no
saving in time or effort when a resistive circuit is analyzed in the frequency
domain.
The Inductor
Let us now turn to the inductor. The time-domain representation is shown in
Fig. 10.14a, and the defining equation, a time-domain expression, is
v(t) = L di(t)
dt
[19]
After substituting the complex voltage equation [16] and complex current
equation [17] in Eq. [19], we have
Vme j (ωt+θ) = L ddt Ime
j (ωt+φ)
Taking the indicated derivative:
Vme j (ωt+θ) = jωL Ime j (ωt+φ)
and dividing through by e jωt :
Vme jθ = jωL Ime jφ
we obtain the desired phasor relationship
V = jωLI [20]
The time-domain differential equation [19] has become the algebraic
equation [20] in the frequency domain. The phasor relationship is indicated
in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦ and
that I must therefore lag V by 90° in an inductor.
CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS386
■ FIGURE 10.14 An inductorand its associated
voltage and current in (a) the time domain, v = L di/dt;
and (b) the frequency domain, V = jωLI.
di
dt
i
v = L 
+
–
(a)
L
I
V = j�LI
+
–
(b)
L
Apply the voltage 8/−50° V at a frequency ω � 100 rad/s to a 4 H
inductor, and determine the phasor current and the time-domain
current.
We make use of the expression we just obtained for the inductor,
I = VjωL =
8/−50◦
j100(4) = − j0.02/−50
◦ = (1/−90◦)(0.02/−50◦)
or
I = 0.02/−140◦ A
If we express this current in the time domain, it becomes
i(t) = 0.02 cos(100t − 140◦) A = 20 cos(100t − 140◦) mA
EXAMPLE 10.4
Figura 8: Indutor.
Impedaˆncia
ZL =
VL
IL
=
ωLIm θ
Im θ − 90◦
Ou
ZL =
VL
IL
= ωL 90◦
Portanto
ZL = jωL (17)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 25 / 36
Impedaˆncia complexa
Indutor
If we transform this answer back to the time domain, it is evident that the
same expression for the current is obtained. We conclude that there is no
saving in time or effort when a resistive circuit is analyzed in the frequency
domain.
The Inductor
Let us now turn to the inductor. The time-domain representation is shown in
Fig. 10.14a, and the defining equation, a time-domain expression, is
v(t) = L di(t)
dt
[19]
After substituting the complex voltage equation [16] and complex current
equation [17] in Eq. [19], we have
Vme j (ωt+θ) = L ddt Ime
j (ωt+φ)
Taking the indicated derivative:
Vme j (ωt+θ) = jωL Ime j (ωt+φ)
and dividing through by e jωt :
Vme jθ = jωL Ime jφ
we obtain the desired phasor relationship
V = jωLI [20]
The time-domain differential equation [19] has become the algebraic
equation [20] in the frequency domain. The phasor relationship is indicated
in Fig. 10.14b. Note that the angle of the factor jωL is exactly +90◦ and
that I must therefore lag V by 90° in an inductor.
CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS386
■ FIGURE 10.14 An inductor and its associated
voltage and current in (a) the time domain, v = L di/dt;
and (b) the frequency domain, V = jωLI.
di
dt
i
v = L 
+
–
(a)
L
I
V = j�LI
+
–
(b)
L
Apply the voltage 8/−50° V at a frequency ω � 100 rad/s to a 4 H
inductor, and determine the phasor current and the time-domain
current.
We make use of the expression we just obtained for the inductor,
I = VjωL =
8/−50◦
j100(4) = − j0.02/−50
◦ = (1/−90◦)(0.02/−50◦)
or
I = 0.02/−140◦ A
If we express this current in the time domain, it becomes
i(t) = 0.02 cos(100t − 140◦) A = 20 cos(100t − 140◦) mA
EXAMPLE 10.4
Figura 8: Indutor.
Impedaˆncia
ZL =
VL
IL
=
ωLIm θ
Im θ − 90◦
Ou
ZL =
VL
IL
= ωL 90◦
Portanto
ZL = jωL (17)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 25 / 36
Impedaˆncia complexa
Indutor Section 5.3 Complex Impedances 229
VL = VM u
IL = IM u 90°
(a) Phasor diagram (b) Current and voltage versus time
vt
vL(t)
iL(t)
90°
 2p
u
Figure 5.8 Current lags voltage by 90 in a pure inductance.
The phasor diagram of the current and voltage is shown in Figure 5.8(a). The corre-
sponding waveforms of current and voltage are shown in Figure 5.8(b). Notice that
the current lags the voltage by 90 for a pure inductance.
Current lags voltage by 90
for a pure inductance.
Equation 5.41 can be written in the form
VL = ( L 90* )× Im 90* (5.42)
Using Equation 5.40 to substitute into Equation 5.42, we nd that
VL = ( L 90* )× IL (5.43)
which can also be written as
VL = j L× IL (5.44)
We refer to the term j L = L 90* as the impedance of the inductance and denote
it as ZL. Thus, we have
ZL = j L = L 90* (5.45)
and
VL = ZLIL (5.46)
Thus, the phasor voltage is equal to the impedance times the phasor current. Equation 5.46 shows that
phasor voltage and phasor
current for an inductance are
related in a manner analogous
to Ohm s law.
This is Ohm s law in phasor form. However, for an inductance, the impedance is an
imaginary number, whereas resistance is a real number. (Impedances that are pure
imaginary are also called reactances.)
Capacitance
In a similar faashion for a capacitance, we can show that if the current and voltage
are sinusoidal, the phasors are related by
VC = ZCIC (5.47)
in which the impedance of the capacitance is
ZC = j
1
C
=
1
j C
=
1
C
90* (5.48)
Figura 9: Tensa˜o e corrente no indutor.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 26 / 36
Impedaˆncia complexa
Exemplo 3
Uma tensa˜o vL = 100 cos(200t) e´ aplicada a uma indutaˆncia de
0,25 H .Determine:
a) A impedaˆncia;
b) Os fasores de tensa˜o e corrente;
c) Desenhe o diagrama fasorial.
Respostas
a) ZL = j50 Ω
b) IL = 2 −90◦; VL = 100 0◦.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 27 / 36
Impedaˆncia complexa
Capacitor
SECTION 10.4 THE PHASOR 387
The Capacitor
The final element to consider is the capacitor. The time-domain current-
voltage relationship is
i(t) = C dv(t)
dt
The equivalent expression in the frequency domain is obtained once more
by letting v(t) and i(t) be the complex quantities of Eqs. [16] and [17], tak-
ing the indicated derivative, suppressing e jωt , and recognizing the phasors V
and I. Doing this, we find
I = jωCV [21]
Thus, I leads V by 90° in a capacitor. This, of course, does not mean that a
current response is present one-quarter of a period earlier than the voltage
that caused it! We are studying steady-state response, and we find that the
current maximum is caused by the increasing voltage that occurs 90° earlier
than the voltage maximum.
The time-domain and frequency-domain representations are compared
in Fig. 10.15a and b. We have now obtained the V-I relationships for the
three passive elements. These results are summarized in Table 10.1, where
the time-domain v-i expressions and the frequency-domain V-I relation-
ships are shown in adjacent columns for the three circuit elements. All the
phasor equations are algebraic. Each is also linear, and the equations relat-
ing to inductance and capacitance bear a great similarity to Ohm’s law. In
fact, we will indeed use them as we use Ohm’s law.
■ FIGURE 10.15 (a) The time-domain and (b) the
frequency-domain relationships between capacitor
current and voltage.
i = C
v
+
–
(a)
C
dv
dt I = j�CV
V
+
–
(b)
C
Ri
v+ –
Li
v+ –
Ci
v+ –
RI
V+ –
V+ –
V+ –
I
I
j�L
1/j�C
TABLE 10.1 Comparison of Time-Domain and Frequency-Domain 
● Voltage-Current Expressions
Time Domain Frequency Domain
v = Ri V = RI
v = L di
dt
V = jωLI
v = 1
C
∫
i dt V = 1jωC I
Kirchhoff’s Laws Using Phasors
Kirchhoff’s voltage law in the time domain is
v1(t) + v2(t) + · · · + vN (t) = 0
We now use Euler’s identity to replace each real voltage vi by a complex
voltage having the same real part, suppress e jωt throughout, and obtain
V1 + V2 + · · · + VN = 0
Thus, we see that Kirchhoff’s voltage law applies to phasor voltages just as
it did in the time domain. Kirchhoff’s current law can be shown to hold for
phasor currents by a similar argument.
Figura 10: Capacitor.
Utilizando a mesma metodologia
aplicada ao indutor e sabendo que:
ic(t) = C
dvc(t)
dt
(18)
Obte´m-se
ZC =
Vc
Ic
=
1
ωC
−90◦
Portanto
ZC = −j 1
ωC
=
1
jωC
(19)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 28 / 36
Impedaˆncia complexa
Capacitor
SECTION 10.4 THE PHASOR 387
The Capacitor
The final element to consider is the capacitor. The time-domain current-
voltage relationship is
i(t) = C dv(t)
dt
The equivalent expression in the frequency domain is obtained once more
by letting v(t) and i(t) be the complex quantities of Eqs. [16] and [17], tak-
ing the indicated derivative,suppressing e jωt , and recognizing the phasors V
and I. Doing this, we find
I = jωCV [21]
Thus, I leads V by 90° in a capacitor. This, of course, does not mean that a
current response is present one-quarter of a period earlier than the voltage
that caused it! We are studying steady-state response, and we find that the
current maximum is caused by the increasing voltage that occurs 90° earlier
than the voltage maximum.
The time-domain and frequency-domain representations are compared
in Fig. 10.15a and b. We have now obtained the V-I relationships for the
three passive elements. These results are summarized in Table 10.1, where
the time-domain v-i expressions and the frequency-domain V-I relation-
ships are shown in adjacent columns for the three circuit elements. All the
phasor equations are algebraic. Each is also linear, and the equations relat-
ing to inductance and capacitance bear a great similarity to Ohm’s law. In
fact, we will indeed use them as we use Ohm’s law.
■ FIGURE 10.15 (a) The time-domain and (b) the
frequency-domain relationships between capacitor
current and voltage.
i = C
v
+
–
(a)
C
dv
dt I = j�CV
V
+
–
(b)
C
Ri
v+ –
Li
v+ –
Ci
v+ –
RI
V+ –
V+ –
V+ –
I
I
j�L
1/j�C
TABLE 10.1 Comparison of Time-Domain and Frequency-Domain 
● Voltage-Current Expressions
Time Domain Frequency Domain
v = Ri V = RI
v = L di
dt
V = jωLI
v = 1
C
∫
i dt V = 1jωC I
Kirchhoff’s Laws Using Phasors
Kirchhoff’s voltage law in the time domain is
v1(t) + v2(t) + · · · + vN (t) = 0
We now use Euler’s identity to replace each real voltage vi by a complex
voltage having the same real part, suppress e jωt throughout, and obtain
V1 + V2 + · · · + VN = 0
Thus, we see that Kirchhoff’s voltage law applies to phasor voltages just as
it did in the time domain. Kirchhoff’s current law can be shown to hold for
phasor currents by a similar argument.
Figura 10: Capacitor.
Utilizando a mesma metodologia
aplicada ao indutor e sabendo que:
ic(t) = C
dvc(t)
dt
(18)
Obte´m-se
ZC =
Vc
Ic
=
1
ωC
−90◦
Portanto
ZC = −j 1
ωC
=
1
jωC
(19)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 28 / 36
Impedaˆncia complexa
Capacitor
SECTION 10.4 THE PHASOR 387
The Capacitor
The final element to consider is the capacitor. The time-domain current-
voltage relationship is
i(t) = C dv(t)
dt
The equivalent expression in the frequency domain is obtained once more
by letting v(t) and i(t) be the complex quantities of Eqs. [16] and [17], tak-
ing the indicated derivative, suppressing e jωt , and recognizing the phasors V
and I. Doing this, we find
I = jωCV [21]
Thus, I leads V by 90° in a capacitor. This, of course, does not mean that a
current response is present one-quarter of a period earlier than the voltage
that caused it! We are studying steady-state response, and we find that the
current maximum is caused by the increasing voltage that occurs 90° earlier
than the voltage maximum.
The time-domain and frequency-domain representations are compared
in Fig. 10.15a and b. We have now obtained the V-I relationships for the
three passive elements. These results are summarized in Table 10.1, where
the time-domain v-i expressions and the frequency-domain V-I relation-
ships are shown in adjacent columns for the three circuit elements. All the
phasor equations are algebraic. Each is also linear, and the equations relat-
ing to inductance and capacitance bear a great similarity to Ohm’s law. In
fact, we will indeed use them as we use Ohm’s law.
■ FIGURE 10.15 (a) The time-domain and (b) the
frequency-domain relationships between capacitor
current and voltage.
i = C
v
+
–
(a)
C
dv
dt I = j�CV
V
+
–
(b)
C
Ri
v+ –
Li
v+ –
Ci
v+ –
RI
V+ –
V+ –
V+ –
I
I
j�L
1/j�C
TABLE 10.1 Comparison of Time-Domain and Frequency-Domain 
● Voltage-Current Expressions
Time Domain Frequency Domain
v = Ri V = RI
v = L di
dt
V = jωLI
v = 1
C
∫
i dt V = 1jωC I
Kirchhoff’s Laws Using Phasors
Kirchhoff’s voltage law in the time domain is
v1(t) + v2(t) + · · · + vN (t) = 0
We now use Euler’s identity to replace each real voltage vi by a complex
voltage having the same real part, suppress e jωt throughout, and obtain
V1 + V2 + · · · + VN = 0
Thus, we see that Kirchhoff’s voltage law applies to phasor voltages just as
it did in the time domain. Kirchhoff’s current law can be shown to hold for
phasor currents by a similar argument.
Figura 10: Capacitor.
Utilizando a mesma metodologia
aplicada ao indutor e sabendo que:
ic(t) = C
dvc(t)
dt
(18)
Obte´m-se
ZC =
Vc
Ic
=
1
ωC
−90◦
Portanto
ZC = −j 1
ωC
=
1
jωC
(19)
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 28 / 36
Impedaˆncia complexa
Capacitor
Section 5.3 Complex Impedances 231
VC = VM u
IC = IM u + 90°
(a) Phasor diagram (b) Current and voltage versus time
vt
vC(t)
iC(t)
90°
 2p
Figure 5.9 Current leads voltage by 90 in a pure capacitance.
Notice that the impedance of a capacitance is also a pure imaginary number.
Suppose that the phasor voltage is
VC = Vm *
Then, the phasor current is
IC =
VC
ZC
=
Vm *
(1/ C) 90*
= CVm + 90*
IC = Im + 90*
where Im = CVm. The phasor diagram for current and voltage in a pure capacitance Current leads voltage by 90
for a pure capacitance.is shown in Figure 5.9(a). The corresponding plots of current and voltage versus time
are shown in Figure 5.9(b). Notice that the current leads the voltage by 90 . (On
the other hand, current lags voltage for an inductance. This is easy to remember if
you know ELI the ICE man. The letter E is sometimes used to stand for electromo-
tive force, which is another term for voltage, L and C are used for inductance and
capacitance, respectively, and I is used for current.)
Resistance
For a resistance, the phasors are related by
VR = RIR (5.49)
Because resistance is a real number, the current and voltage are inphase,as illustrated Current and voltage are in
phase for a resistance.in Figure 5.10.
Exercise 5.6 A voltage vL(t) = 100 cos(200t) is applied to a 0.25-H inductance.
(Notice that = 200.) a. Find the impedance of the inductance, the phasor current,
and the phasor voltage. b. Draw the phasor diagram.
Answer a. ZL = j50 = 50 90* , IL = 2 90* , VL = 100 0* ; b. the phasor
diagram is shown in Figure 5.11(a). *
Figura 11: Tensa˜o e corrente no capacitor.
No capacitor corrente esta´ adiantada da tensa˜o em 90◦
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 29 / 36
Impedaˆncia complexa
Exemplo 4
Uma tensa˜o vc = 100 cos(200t) e´ aplicada a um capacitor de
100 uF .Determine:
a) A impedaˆncia;
b) Os fasores de tensa˜o e corrente;
c) Desenhe o diagrama fasorial.
Respostas
a) ZC = −j50 Ω
b) IL = 2 90
◦; VL = 100 0◦.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 30 / 36
Exemplo 5
A tensa˜o e a corrente em um elemento do circuito sa˜o mostradas na
figura a seguir. Determine a natureza da carga e o valor do elemento.
276 Chapter 5 Steady-State Sinusoidal Analysis
P5.27. Suppose we have two sinusoidal voltages of
the same frequency with rms values of 10V
and 7V, respectively. The phase angles are
unknown.What is the smallest rms value that
the sum of these voltages could have? The
largest? Justify your answers.
P5.28. A sinusoidal current i1(t) has a phase angle
of 60 . Furthermore, i1(t) attains its positive
peak 0.25ms earlier than current i2(t) does.
Both currents have a frequency of 500Hz.
Determine the phase angle of i2(t).
P5.29. Reduce the expression
15 sin( t 45 )+ 5 cos( t 30 )
+ 10 cos( t 120 )
to the form Vm cos( t + ).
P5.30. Suppose that v1(t) = 90 cos( t 15) and
v2(t) = 50 sin( t 60 ). Use phasors to
reduce the sum vs(t) = v1(t) + v2(t) to a
single term of the form Vm cos( t + ). State
the phase relationships between each pair of
thesephasors. (Hint: Sketchaphasordiagram
showing V1,V2, and Vs.)
P5.31. Supposewehavea circuit inwhich thevoltage
is v1(t) = 10 cos( t 30 ) V. Furthermore,
the current i1(t) has an rms value of 10A and
lags v1(t) by 40 . (The current and the volt-
age have the same frequency.) Draw a phasor
diagram and write an expression for i1(t) of
the form Im cos( t + ).
P5.32. Use MATLAB to obtain a plot of v(t) =
cos(19 t) + cos(21 t) for t ranging from 0
to 2 s. Explain why the terms in this expres-
sion cannot be combined by using phasors.
Then, considering that the two terms can be
represented as the real projection of the sum
of two vectors rotating at different speeds in
the complex plane, comment on the plot.
Section 5.3: Complex Impedances
P5.33. Write the relationship between the phasor
voltage and phasor current for an inductance.
Repeat for capacitance.
P5.34. What is the phase relationship between cur-
rent and voltage for a pure resistance? For
an inductance? For a capacitance?
*P5.35. A voltage vL(t) = 10 cos(2000 t) is applied
to a 100-mH inductance. Find the complex
impedance of the inductance. Find the phasor
voltage and current, and construct a pha-
sor diagram. Write the current as a function
of time. Sketch the voltage and current to
scale versus time. State thephase relationship
between the current and voltage.
P5.36. A certain circuit element is known to be a
pure resistance, a pure inductance, or a pure
capacitance. Determine the type and value
(in ohms, henrys, or farads) of the element if
the voltage and current for the element are
given by: a. v(t) = 100 cos(100t + 30 )V,
i(t) = 2 cos(100t + 30 )A; b. v(t) =
100 cos(400t + 30 )V, i(t) = 3 sin(400t +
30 )A; c. v(t) = 100 sin(200t + 30 )V,
i(t) = 2 cos(200t + 30 )A.
*P5.37. A voltage vC(t) = 10 cos(2000 t) is applied
to a 10- F capacitance. Find the complex
impedance of the capacitance. Find the pha-
sor voltage and current, and construct a
phasor diagram. Write the current as a func-
tion of time. Sketch the voltage and current to
scale versus time. State the phase relationship
between the current and voltage.
P5.38. a.The current and voltage for a certain circuit
element are shown in Figure P5.38(a). Deter-
mine the nature and value of the element.
b. Repeat for Figure P5.38(b).
i (mA)
v (V)
5
4
3
2
1
0
1
2
3
4
5
(a)
0 1 2 3 4 5
t
(ms)
Figure P5.38
Figura 12: Formas de onda de tensa˜o e corrente no elemento.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 31 / 36
Exemplo 6
A tensa˜o e a corrente em um elemento do circuito sa˜o mostradas na
figura a seguir. Determine a natureza da carga e o valor do elemento.
Problems 277
i (mA)
v (V)
10
8
6
4
2
0
2
4
6
8
10
(b)
0 4 8 12 16 20
t
(ms)
Figure P5.38 (Cont.)
P5.39. Use MATLAB or manually produce plots of
the magnitudes of the impedances of a 10-
mH inductance, a 10-,F capacitance, and a
50-+ resistance to scale versus frequency for
the range from zero to 1000Hz.
P5.40. a.A certain element has a phasor voltage of
V = 50 45* V and current of I = 10 45* A.
The angular frequency is 1000 rad/s. Deter-
mine the nature and value of the element.
b. Repeat for V = 20 45* V and cur-
rent of I = 5 135* A. c. Repeat for V =
100 30* V and current of I = 5 120* A.
Section 5.4: Circuit Analysis with Phasors
and Complex Impedances
P5.41. Describe the step-by-step procedure for
steady-state analysis of circuits with sinu-
soidal sources. What condition must be true
of the sources?
*P5.42. Find the phasors for the current and for the
voltages of the circuit shown in Figure P5.42.
Construct a phasor diagram showing Vs, I,
VR, and VL. What is the phase relationship
between Vs and I?
100 *
0.2 H
V
vs(t) =
10 cos(500t) i
vL
vR +
+
+
Figure P5.42
P5.43. Change the inductance to 0.3H, and repeat
Problem P5.42.
*P5.44. Find the phasors for the current and the volt-
ages for the circuit shown in Figure P5.44.
Construct a phasor diagram showing Vs, I,
VR, and VC . What is the phase relationship
between Vs and I?
1000 *
1 mF
V
vs(t) =
10 cos(500t)
i
vC
vR ++
+
Figure P5.44
P5.45. Repeat Problem P5.44, changing the capaci-
tance value to 4 ,F.
*P5.46. Find the complex impedance in polar form
of the network shown in Figure P5.46 for
* = 500. Repeat for * = 1000 and * = 2000.
Z
10 mF
50 *
100 mH
Figure P5.46
Figura 13: Formas de onda de tensa˜o e corrente no elemento.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 32 / 36
Impedaˆncia complexa
Resumo
SECTION 10.4 THE PHASOR 387
The Capacitor
The final element to consider is the capacitor. The time-domain current-
voltage relationship is
i(t) = C dv(t)
dt
The equivalent expression in the frequency domain is obtained once more
by letting v(t) and i(t) be the complex quantities of Eqs. [16] and [17], tak-
ing the indicated derivative, suppressing e jωt , and recognizing the phasors V
and I. Doing this, we find
I = jωCV [21]
Thus, I leads V by 90° in a capacitor. This, of course, does not mean that a
current response is present one-quarter of a period earlier than the voltage
that caused it! We are studying steady-state response, and we find that the
current maximum is caused by the increasing voltage that occurs 90° earlier
than the voltage maximum.
The time-domain and frequency-domain representations are compared
in Fig. 10.15a and b. We have now obtained the V-I relationships for the
three passive elements. These results are summarized in Table 10.1, where
the time-domain v-i expressions and the frequency-domain V-I relation-
ships are shown in adjacent columns for the three circuit elements. All the
phasor equations are algebraic. Each is also linear, and the equations relat-
ing to inductance and capacitance bear a great similarity to Ohm’s law. In
fact, we will indeed use them as we use Ohm’s law.
■ FIGURE 10.15 (a) The time-domain and (b) the
frequency-domain relationships between capacitor
current and voltage.
i = C
v
+
–
(a)
C
dv
dt I = j�CV
V
+
–
(b)
C
Ri
v+ –
Li
v+ –
Ci
v+ –
RI
V+ –
V+ –
V+ –
I
I
j�L
1/j�C
TABLE 10.1 Comparison of Time-Domain and Frequency-Domain 
● Voltage-Current Expressions
Time Domain Frequency Domain
v = Ri V = RI
v = L di
dt
V = jωLI
v = 1
C
∫
i dt V = 1jωC I
Kirchhoff’s Laws Using Phasors
Kirchhoff’s voltage law in the time domain is
v1(t) + v2(t) + · · · + vN (t) = 0
We now use Euler’s identity to replace each real voltage vi by a complex
voltage having the same real part, suppress e jωt throughout, and obtain
V1 + V2 + · · · + VN = 0
Thus, we see that Kirchhoff’s voltage law applies to phasor voltages just as
it did in the time domain. Kirchhoff’s current law can be shown to hold for
phasor currents by a similar argument.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 33 / 36
Impedaˆncia equivalente
Associac¸a˜o se´rie
CHAPTER 9 Sinusoids and Phasors 373
By following a similar procedure, we can show that Kirchhoff’s
current law holds for phasors. If we let i1, i2, . . . , in be the current
leaving or entering a closed surface in a network at time t , then
i1 + i2 + · · · + in = 0 (9.56)
If I1, I2, . . . , In are the phasor forms of the sinusoids i1, i2, . . . , in, then
I1 + I2 + · · · + In = 0 (9.57)
which is Kirchhoff’s current law in the frequency domain.
Once we have shown that both KVL and KCL hold in the frequency
domain, it is easy to do many things, such asimpedance combination,
nodal and mesh analyses, superposition, and source transformation.
9.7 IMPEDANCE COMBINATIONS
Consider the N series-connected impedances shown in Fig. 9.18. The
same current I flows through the impedances. Applying KVL around the
loop gives
V = V1 + V2 + · · · + VN = I(Z1 + Z2 + · · · + ZN) (9.58)
The equivalent impedance at the input terminals is
Zeq = VI = Z1 + Z2 + · · · + ZN
or
Zeq = Z1 + Z2 + · · · + ZN (9.59)
showing that the total or equivalent impedance of series-connected imped-
ances is the sum of the individual impedances. This is similar to the series
connection of resistances.
+ − + − + −
+
−
I Z1
Zeq
Z2 ZN
V1
V
V2 VN
Figure 9.18 N impedances in series.
+
−
+ −
I
+
−
Z1
V1
Z2V2V
Figure 9.19 Voltage division.
IfN = 2, as shown in Fig. 9.19, the current through the impedances
is
I = V
Z1 + Z2 (9.60)
Since V1 = Z1I and V2 = Z2I, then
V1 = Z1Z1 + Z2 V, V2 =
Z2
Z1 + Z2 V (9.61)
which is the voltage-division relationship.
Figura 14: Associac¸a˜o se´rie de impedaˆncias.
Zeq = Z1 + Z2 + · · ·+ ZN (20)
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Impedaˆncia equivalente
Associac¸a˜o em paralelo
374 PART 2 AC Circuits
In the same manner, we can obtain the equivalent impedance or
admittance of the N parallel-connected impedances shown in Fig. 9.20.
The voltage across each impedance is the same. Applying KCL at the
top node,
I = I1 + I2 + · · · + IN = V
(
1
Z1
+ 1
Z2
+ · · · + 1
ZN
)
(9.62)
The equivalent impedance is
1
Zeq
= I
V
= 1
Z1
+ 1
Z2
+ · · · + 1
ZN
(9.63)
and the equivalent admittance is
Yeq = Y1 + Y2 + · · · + YN (9.64)
This indicates that the equivalent admittance of a parallel connection of
admittances is the sum of the individual admittances.
I
+
−
I1 I2 IN
VI Z1 Z2 ZN
Zeq
Figure 9.20 N impedances in parallel.
When N = 2, as shown in Fig. 9.21, the equivalent impedance
becomes
Zeq = 1Yeq =
1
Y1 + Y2 =
1
1/Z1 + 1/Z2 =
Z1Z2
Z1 + Z2 (9.65)
Also, since
V = IZeq = I1Z1 = I2Z2
the currents in the impedances are
I1 = Z2Z1 + Z2 I, I2 =
Z1
Z1 + Z2 I (9.66)
which is the current-division principle.
I1 I2+
−
I Z1 Z2V
Figure 9.21 Current division.
The delta-to-wye and wye-to-delta transformations that we applied
to resistive circuits are also valid for impedances. With reference to Fig.
9.22, the conversion formulas are as follows.
Figura 15: Associac¸a˜o paralelo de impedaˆncias.
Zeq =
1
1/Z1 + 1/Z2 + · · ·+ 1/ZN (21)
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Exemplo 7
Determine a impedaˆncia complexa equivalente na forma polar do
circuito abaixo para ω = 500 rad/s. Repita para ω = 1000 rad/s.
100mH
50Ω
10uF
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