Circuitos em Corrente Alternada - Eletricidade e Energia UFJF

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Circuitos em Corrente Alternada
(CA)
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 de 2015
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 1 / 19
Suma´rio
1 Contextualizac¸a˜o
2 Ana´lise senoidal em estado permanente
3 Valor Eficaz
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 2 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustrate this procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)+Ri(t)+ 1
C
∫ t
0
i(t) dt +vc(0) = 0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustrate this procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)
+Ri(t)+
1
C
∫ t
0
i(t) dt +vc(0) = 0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustrate this procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)+Ri(t)+
1
C
∫ t
0
i(t) dt +vc(0) = 0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustratethis procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)+Ri(t)+ 1
C
∫ t
0
i(t) dt +vc(0) =
0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustrate this procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)+Ri(t)+ 1
C
∫ t
0
i(t) dt +vc(0) = 0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
Determine a corrente no circuito da Figura 1. Sabendo que o
capacitor possui uma carga inicial vc(0) = 1 V .
180 Chapter 4 Transients
3. Obtain the complete solution by adding the particular solution to the com-
plementary solution given by Equation 4.44, which contains the arbitrary
constant K.
4. Use initial conditions to nd the value of K.
We illustrate this procedure with an example.
Example 4.4 Transient Analysis of an RC Circuit with a Sinusoidal Source
Solve for the current in the circuit shown in Figure 4.14. The capacitor is initially
charged so that vC(0+) = 1 V.
Solution First, we write a voltage equation for t > 0. Traveling clockwise and
summing voltages, we obtainStep 1: Write the circuit
equation and reduce it to a
rst-order differential
equation. Ri(t)+
1
C
* t
0
i(t) dt + vC(0) 2 sin(200t) = 0
We convert this to a differential equation by taking the derivative of each term.
Of course, the derivative of the integral is simply the integrand. Because vC(0) is a
constant, its derivative is zero. Thus, we have
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (4.45)
Multiplying by C, we get
RC
di(t)
dt
+ i(t) = 400C cos(200t) (4.46)
Substituting values for R and C, we obtain
5× 10 3
di(t)
dt
+ i(t) = 400× 10 6 cos(200t) (4.47)
The second step is to nd a particular solution ip(t). Often,we start by guessing atStep 2: Find a particular
solution. the formof ip(t), possibly including some unknown constants.Then,we substitute our
guess into the differential equation and solve for the constants. In the present case,
since the derivatives of sin(200t) and cos(200t) are 200 cos(200t) and 200 sin(200t),
respectively, we try a particular solution of the formThe particular solution for a
sinusoidal forcing function
always has the form given by
Equation 4.48. ip(t) = A cos(200t)+ B sin(200t) (4.48)
Figure 4.14 A rst-order RC circuit
with a sinusoidal source. See
Example 4.4.
2 sin(200t)
R = 5 k
t = 0
vC(t)
vC(0) = 1 V
+
+
i(t)
C
1 mF
Figura 1: Circuito RC.
−2 sen(200t)+Ri(t)+ 1
C
∫ t
0
i(t) dt +vc(0) = 0
ou
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 3 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
A soluc¸a˜o da equac¸a˜o diferencial (1) e´ composta por duas parcelas.
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (1)
A soluc¸a˜o particular: devido a func¸a˜o forc¸ante
ip(t) = 200 cos(200t) + 200 sen(200t) µA (2)
A soluc¸a˜o complementar: resposta natural
ic(t) = −400e−t/RC µA (3)
A soluc¸a˜o completa: i(t) = ip(t) + ic(t)
i(t) = 200 cos(200t) + 200 sen(200t)− 400e−t/RC µA (4)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 4 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
A soluc¸a˜o da equac¸a˜o diferencial (1) e´ composta por duas parcelas.
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (1)
A soluc¸a˜o particular: devido a func¸a˜o forc¸ante
ip(t) = 200 cos(200t) + 200 sen(200t) µA (2)
A soluc¸a˜o complementar: resposta natural
ic(t) = −400e−t/RC µA (3)
A soluc¸a˜o completa: i(t) = ip(t) + ic(t)
i(t) = 200 cos(200t) + 200 sen(200t)− 400e−t/RC µA (4)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 4 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
A soluc¸a˜o da equac¸a˜o diferencial (1) e´ composta por duas parcelas.
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (1)
A soluc¸a˜o particular: devido a func¸a˜o forc¸ante
ip(t) = 200 cos(200t) + 200 sen(200t)µA (2)
A soluc¸a˜o complementar: resposta natural
ic(t) = −400e−t/RC µA (3)
A soluc¸a˜o completa: i(t) = ip(t) + ic(t)
i(t) = 200 cos(200t) + 200 sen(200t)− 400e−t/RC µA (4)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 4 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Exemplo
A soluc¸a˜o da equac¸a˜o diferencial (1) e´ composta por duas parcelas.
R
di(t)
dt
+
1
C
i(t) = 400 cos(200t) (1)
A soluc¸a˜o particular: devido a func¸a˜o forc¸ante
ip(t) = 200 cos(200t) + 200 sen(200t) µA (2)
A soluc¸a˜o complementar: resposta natural
ic(t) = −400e−t/RC µA (3)
A soluc¸a˜o completa: i(t) = ip(t) + ic(t)
i(t) = 200 cos(200t) + 200 sen(200t)− 400e−t/RC µA (4)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 4 / 19
Contextualizac¸a˜o
Contextualizac¸a˜o
Soluc¸a˜o
Section 4.5 Second-Order Circuits 183
Figure 4.17 The complete solution
for Example 4.4.
i(t)
(mA)
t (ms)
300
400
200
100
0
100
200
300
400
0 20 40 60 80
Figure 4.18 The circuit for
Exercise 4.7.
2 cos(200t)
R = 5 k
t = 0
vC (t)
vC (0) = 0
+
+
i(t)
C
1 mF
Figure 4.19 The circuit for
Exercise 4.8.
10e t
R = 1 M
t = 0
vC (t)
vC(0) = 5 V
+
+
i(t) C
2 mF
Answer i(t) = 200 sin(200t)+ 200 cos(200t)+ 200e t/RC A, in which = RC =
5 ms. *
Exercise 4.8 Solve for the current in the circuit shown in Figure 4.19 after the switch
closes. [Hint:Try a particular solution of the form ip(t) = Ae
t .]
Answer i(t) = 20e t 15e t/2 A. *
4.5 SECOND-ORDER CIRCUITS
In this section, we consider circuits that contain two energy-storage elements. In
particular, we look at circuits that have one inductance and one capacitance, either
in series or in parallel.
Differential Equation
To derive the general form of the equations that we encounter in circuits with two
energy-storage elements, consider the series circuit shown in Figure 4.20(a).Writing
Figura 2: Soluc¸a˜o completa.
182 Chapter 4 Transients
Figure 4.15 The voltages and
currents for the circuit of Figure 4.14
immediately after the switch closes.
2 sin(0+) = 0
R = 5 k
vC (0+) = 1 V
vR(0+)
++
+
i(0+)
vC(0+) = 1. Consequently, the voltage across the resistor must be vR(0+) = 1 V.
Thus, we get
i(0+) =
vR(0+)
R
=
1
5000
= 200 A
Substituting t = 0 into Equation 4.54, we obtain
i(0+) = 200 = 200+K A (4.55)
Solving, we nd that K = 400 A. Substituting this into Equation 4.54, we have
the solution
i(t) = 200 cos(200t)+ 200 sin(200t) 400e t/RC A (4.56)
Plots of the particular solution and of the complementary solution are shown
in Figure 4.16. The time constant for this circuit is = RC = 5 ms. Notice that the
natural response decays to negligible values in about 25 ms.As expected, the natural
response has decayed in about ve time constants. Furthermore, notice that for a
sinusoidal forcing function, the forced response is also sinusoidal and persists after
the natural response has decayed.Notice that the forced
response is sinusoidal for a
sinusoidal forcing function.
A plot of the complete solution is shown in Figure 4.17.
Exercise 4.7 Repeat Example 4.4 if the source voltage is changed to 2 cos(200t) and
the initial voltage on the capacitor is vC(0) = 0. The circuit with these changes is
shown in Figure 4.18.
Current
(mA)
t (ms)
300
400
200
100
0
100
200
300
400
0 20 40 60 80
Particular solution or forced response
Complementary solution or natural response
Figure 4.16 The complementary solution and the par-
ticular solution for Example 4.4.Figura 3: Soluc¸a˜o particular e soluc¸a˜o
complementar.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 5 / 19
Ana´lise senoidal em estado permanente
Ana´lise senoidal em estado permanente
Resposta natural decai a zero ao longo do tempo
Resposta forc¸ada para fontes senoidais persiste indefinidamente
Resposta em estado permanente
Maior interesse na analise de circuitos CA
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 6 / 19
Ana´lise senoidal em estado permanente
Tensa˜o e correntes senoidais
216 Chapter 5 Steady-State Sinusoidal Analysis
In Chapter 4, we saw that the response of a network has two parts: the forced
response and the natural response. In most circuits, the natural response decays
rapidly to zero. The forced response for sinusoidal sources persists inde nitely and,
therefore, is called the steady-state response. Because the natural response quickly
decays, the steady-state response is often of highest interest. In this chapter, we learn
ef cient methods for nding the steady-state responses for sinusoidal sources.
We also study three-phase circuits, which are used in electric power-distribution
systems. Most engineers who work in industrial settings need to understand three-
phase power distribution.
5.1 SINUSOIDAL CURRENTS AND VOLTAGES
A sinusoidal voltage is shown in Figure 5.1 and is given by
v(t) = Vm cos( t + ) (5.1)
where Vm is the peak value of the voltage, is the angular frequency in radians per
second, and is the phase angle.
Sinusoidal signals are periodic, repeating the same pattern of values in each
period T . Because the cosine (or sine) function completes one cycle when the angle
increases by 2 radians, we get
T = 2 (5.2)
The frequency of a periodic signal is the number of cycles completed in one
second. Thus, we obtain
f =
1
T
(5.3)
The units of frequency are hertz (Hz). (Actually, the physical units of hertz areWe refer to as angular
frequency with units of
radians per second and f
simply as frequency with
units of hertz (Hz).
equivalent to inverse seconds.) Solving Equation 5.2 for the angular frequency, we
have
=
2
T
(5.4)
Using Equation 5.3 to substitute for T , we nd that
= 2 f (5.5)
Figure 5.1 A sinusoidal voltage
waveform given by v(t) = Vm cos
( t + ). Note: Assuming that is in
degrees, we have tmax = 360 × T . For
the waveform shown, is 45 .
v(t)
t
tmax
Vm
Vm
T
Figura 4: Tensa˜o senoidal.
A tensa˜o mostrada na figura ao lado e´
dada por:
v(t) = Vm cos(ωt + θ) (5)
em que
Vm e´ o valor de pico
ω e´ a frequeˆncia angular em rad/s
θ e´ o aˆngulo de fase
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 7 / 19
Ana´lise senoidal em estado permanente
Tensa˜o e correntes senoidais
Sinais senoidais sa˜o perio´dicos, repetindo o mesmo padra˜o de valores
a cada per´ıodo T .
Frequeˆncia de um sinal perio´dico e´ o nu´mero de ciclos por segundo.
f =
1
T
Hz (6)
Frequeˆncia angular
ω =
2pi
T
rad/s (7)
Portanto
ω = 2pif (8)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 8 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Esta´ diretamente ligado a` poteˆncia transferida por grandezas CA;
E´ atrave´s do valor eficaz que se pode comparar a poteˆncia associada a
grandezas CA com poteˆncias associadas a grandezas CC;
O valor eficaz de uma corrente alternada e´ o valor da
intensidade de uma corrente cont´ınua que produziria, numa
resisteˆncia, o mesmo efeito calor´ıfico que a corrente alternada
em questa˜o.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 9 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Esta´ diretamente ligado a` poteˆncia transferida por grandezas CA;
E´ atrave´s do valor eficaz que se pode comparar a poteˆncia associada a
grandezas CA com poteˆncias associadas a grandezas CC;
O valor eficaz de uma corrente alternada e´ o valor da
intensidade de uma corrente cont´ınua que produziria, numa
resisteˆncia, o mesmo efeito calor´ıfico que a corrente alternada
em questa˜o.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 9 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)Esta´ diretamente ligado a` poteˆncia transferida por grandezas CA;
E´ atrave´s do valor eficaz que se pode comparar a poteˆncia associada a
grandezas CA com poteˆncias associadas a grandezas CC;
O valor eficaz de uma corrente alternada e´ o valor da
intensidade de uma corrente cont´ınua que produziria, numa
resisteˆncia, o mesmo efeito calor´ıfico que a corrente alternada
em questa˜o.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 9 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Considere uma tensa˜o perio´dica v(t) com per´ıodo T a uma
resisteˆncia R;
A poteˆncia instantaˆnea dissipada na resisteˆncia e´:
p(t) =
v(t)2
R
(9)
Ja´ a energia consumida e´ dada por:
ET =
∫ T
0
p(t) dt
A poteˆncia me´dia dissipada em um per´ıodo T e´
Pm =
ET
T
=
1
T
∫ T
0
p(t) dt (10)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 10 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Considere uma tensa˜o perio´dica v(t) com per´ıodo T a uma
resisteˆncia R;
A poteˆncia instantaˆnea dissipada na resisteˆncia e´:
p(t) =
v(t)2
R
(9)
Ja´ a energia consumida e´ dada por:
ET =
∫ T
0
p(t) dt
A poteˆncia me´dia dissipada em um per´ıodo T e´
Pm =
ET
T
=
1
T
∫ T
0
p(t) dt (10)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 10 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Considere uma tensa˜o perio´dica v(t) com per´ıodo T a uma
resisteˆncia R;
A poteˆncia instantaˆnea dissipada na resisteˆncia e´:
p(t) =
v(t)2
R
(9)
Ja´ a energia consumida e´ dada por:
ET =
∫ T
0
p(t) dt
A poteˆncia me´dia dissipada em um per´ıodo T e´
Pm =
ET
T
=
1
T
∫ T
0
p(t) dt (10)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 10 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Considere uma tensa˜o perio´dica v(t) com per´ıodo T a uma
resisteˆncia R;
A poteˆncia instantaˆnea dissipada na resisteˆncia e´:
p(t) =
v(t)2
R
(9)
Ja´ a energia consumida e´ dada por:
ET =
∫ T
0
p(t) dt
A poteˆncia me´dia dissipada em um per´ıodo T e´
Pm =
ET
T
=
1
T
∫ T
0
p(t) dt (10)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 10 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Portanto
Pm =
1
T
∫ T
0
v(t)2
R
dt
Que pode ser rearranjada como
Pm =
[√
1
T
∫ T
0
v(t)2 dt
]2
R
Pode–se definir enta˜o o valor eficaz (RMS) de um sinal perio´dico
como:
VRMS =
√
1
T
∫ T
0
v(t)2 dt (11)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 11 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Portanto
Pm =
1
T
∫ T
0
v(t)2
R
dt
Que pode ser rearranjada como
Pm =
[√
1
T
∫ T
0
v(t)2 dt
]2
R
Pode–se definir enta˜o o valor eficaz (RMS) de um sinal perio´dico
como:
VRMS =
√
1
T
∫ T
0
v(t)2 dt (11)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 11 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Portanto
Pm =
1
T
∫ T
0
v(t)2
R
dt
Que pode ser rearranjada como
Pm =
[√
1
T
∫ T
0
v(t)2 dt
]2
R
Pode–se definir enta˜o o valor eficaz (RMS) de um sinal perio´dico
como:
VRMS =
√
1
T
∫ T
0
v(t)2 dt (11)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 11 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
A poteˆncia me´dia e´ dada por
Pm =
V 2RMS
R
(12)
De maneira ana´loga
IRMS =
√
1
T
∫ T
0
i(t)2 dt (13)
Portanto
Pm = RI
2
RMS (14)
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 12 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Considere a seguinte tensa˜o senoidal
v(t) = Vm cos(ωt + θ)
O valor RMS e´ dado por
VRMS =
√
1
T
∫ T
0
Vm
2 cos2(ωt + θ) dt
Utilizando a identidade trigonome´trica
cos(z )2 =
1
2
+
1
2
cos(2z )
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 13 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Considere a seguinte tensa˜o senoidal
v(t) = Vm cos(ωt + θ)
O valor RMS e´ dado por
VRMS =
√
1
T
∫ T
0
Vm
2 cos2(ωt + θ) dt
Utilizando a identidade trigonome´trica
cos(z )2 =
1
2
+
1
2
cos(2z )
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 13 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Considere a seguinte tensa˜o senoidal
v(t) = Vm cos(ωt + θ)
O valor RMS e´ dado por
VRMS =
√
1
T
∫ T
0
Vm
2 cos2(ωt + θ) dt
Utilizando a identidade trigonome´trica
cos(z )2 =
1
2
+
1
2
cos(2z )
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 13 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Substituindo (13) em (13)
VRMS =
√
V 2m
2T
∫ T
0
[1 + cos(2ωt + 2θ)] dt
Integrando, tem–se
VRMS =
√
V 2m
2T
[
t +
1
2ω
sen(2ωt + 2θ)
]T
0
Substituindo os limites
VRMS =
√
V 2m
2T
[
T +
1
2ω
sen(2ωT + 2θ)− 1
2ω
sen(2θ)
]
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 14 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Que se reduz a
VRMS =
Vm√
2
(15)
O valor RMS de uma senoide e´ o valor de pico dividido pela raiz
quadrada de dois.
Isto na˜o e´ verdade para outras formas de onda perio´dicas, tais como
ondas triangulares e quadradas.
Usualmente, apenas o valor RMS e´ fornecido.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 15 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
Que se reduz a
VRMS =
Vm√
2
(15)
O valor RMS de uma senoide e´ o valor de pico dividido pela raiz
quadrada de dois.
Isto na˜o e´ verdade para outras formas de onda perio´dicas, tais como
ondas triangulares e quadradas.
Usualmente, apenas o valor RMS e´ fornecido.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 15 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Valor eficaz de uma senoide
v(t)
t
Vm
VRMS
Figura 5: Valor eficaz de uma senoide graficamente
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 16 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 1
Tensa˜o de alimentac¸a˜o residencial em Juiz de Fora:
Fase-neutro: VRMS = 127 V
Valor de pico: Vm = 127×
√
2 = 179,6 V .
Fase-fase: VRMS = 220 V
Valor de pico: Vm = 220×
√
2 = 311,12 V .
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 17 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 1
Tensa˜o de alimentac¸a˜o residencial em Juiz de Fora:
Fase-neutro: VRMS = 127 V
Valor de pico: Vm = 127×
√
2 = 179,6 V .
Fase-fase: VRMS = 220 V
Valor de pico: Vm = 220×
√
2 = 311,12 V .
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 17 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 1
Tensa˜o de alimentac¸a˜o residencial em Juiz de Fora:
Fase-neutro: VRMS = 127 V
Valor de pico: Vm = 127×
√
2 = 179,6 V .
Fase-fase: VRMS = 220 V
Valor de pico: Vm = 220×
√
2 = 311,12 V .
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 17 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 1
Tensa˜o de alimentac¸a˜o residencial em Juiz de Fora:
Fase-neutro: VRMS = 127 V
Valor de pico: Vm = 127×
√
2 = 179,6 V .
Fase-fase: VRMS = 220 V
Valor de pico: Vm = 220×
√
2 = 311,12 V .
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 17 / 19
Valor Eficaz
ValorEficaz ou RMS (root mean square)
Exemplo 2
Calcule a poteˆncia me´dia dissipada em uma resisteˆncia de 50 Ω,
considerando que a tensa˜o v(t) = 100 cos(2× 50× pit) V e´ aplicada
sobre a resisteˆncia.
Soluc¸a˜o
sabendo que:
VRMS =
Vm√
2
=
100√
2
= 70,71 V
a poteˆncia me´dia e´ dada por
Pm =
V 2RMS
R
=
(70,71)2
50
= 100 W
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 18 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 2
Calcule a poteˆncia me´dia dissipada em uma resisteˆncia de 50 Ω,
considerando que a tensa˜o v(t) = 100 cos(2× 50× pit) V e´ aplicada
sobre a resisteˆncia.
Soluc¸a˜o
sabendo que:
VRMS =
Vm√
2
=
100√
2
= 70,71 V
a poteˆncia me´dia e´ dada por
Pm =
V 2RMS
R
=
(70,71)2
50
= 100 W
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 18 / 19
Valor Eficaz
Valor Eficaz ou RMS (root mean square)
Exemplo 2
Poteˆncia instantaˆnea.
p(t) =
v(t)2
R
=
1002 cos2(100pit)
50
= 200 cos2(100pit) W
220 Chapter 5 Steady-State Sinusoidal Analysis
v(t) (V)
p(t) (W)
t
(ms)
t
(ms)
100
10 20 30 40 50
100
(a)
(b)
Pavg = 100
0
200
0 10 20 30 40 50
Figure 5.2 Voltage and power versus time for Example 5.1.
The power as a function of time is given by
p(t) =
v2(t)
R
=
1002 cos2(100 t)
50
= 200 cos2(100 t) W
A plot of p(t) versus time is shown in Figure 5.2(b). Notice that the power uctuates
For a sinusoidal current
owing in a resistance, power
uctuates periodically from
zero to twice the average
value.
from 0 to 200W. However, the average power is 100W, as we found by using the rms
value.
RMS Values of Nonsinusoidal Voltages or Currents
Sometimes we need to determine the rms values of periodic currents or voltages
that are not sinusoidal. We can accomplish this by applying the de nition given by
Equation 5.12 or 5.14 directly.
Example 5.2 RMS Value of a Triangular Voltage
The voltage shown in Figure 5.3(a) is known as a triangular waveform. Determine
its rms value.
Solution First, we need to determine the equations describing the waveform
between t = 0 and t = T = 2 s. As illustrated in Figure 5.3(b), the equations for
the rst period of the triangular wave are
v(t) =
*
3t for 0 t 1
6 3t for 1 t 2
Figura 6: Tensa˜o.
220 Chapter 5 Steady-State Sinusoidal Analysis
v(t) (V)
p(t) (W)
t
(ms)
t
(ms)
100
10 20 30 40 50
100
(a)
(b)
Pavg = 100
0
200
0 10 20 30 40 50
Figure 5.2 Voltage and power versus time for Example 5.1.
The power as a function of time is given by
p(t) =
v2(t)
R
=
1002 cos2(100 t)
50
= 200 cos2(100 t) W
A plot of p(t) versus time is shown in Figure 5.2(b). Notice that the power uctuates
For a sinusoidal current
owing in a resistance, power
uctuates periodically from
zero to twice the average
value.
from 0 to 200W. However, the average power is 100W, as we found by using the rms
value.
RMS Values of Nonsinusoidal Voltages or Currents
Sometimes we need to determine the rms values of periodic currents or voltages
that are not sinusoidal. We can accomplish this by applying the de nition given by
Equation 5.12 or 5.14 directly.
Example 5.2 RMS Value of a Triangular Voltage
The voltage shown in Figure 5.3(a) is known as a triangular waveform. Determine
its rms value.
Solution First, we need to determine the equations describing the waveform
between t = 0 and t = T = 2 s. As illustrated in Figure 5.3(b), the equations for
the rst period of the triangular wave are
v(t) =
*
3t for 0 t 1
6 3t for 1 t 2
Figura 7: Poteˆncia instantaˆnea.
Pedro Machado de Almeida (UFJF) ENE077 Novembro de 2015 19 / 19
	Contextualização
	Análise senoidal em estado permanente
	Valor Eficaz