Análise de circuitos em estado permanente - Eletricidade e Energia UFJF

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Ana´lise de circuitos em estado
permanente
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 / 18
Ana´lise de circuitos em estado permanente
Procedimento passo a passo para a ana´lise de circuitos em regime per-
manente com fontes senoidais
1. Substitua as descric¸o˜es temporais da tensa˜o e da corrente pelos fasores
correspondentes. (Todas as fontes devem ter a mesma frequeˆncia)
2. Substitua as indutaˆncias e capacitaˆncias por suas impedaˆncias comple-
xas
ZL = jωL = ω 90
◦.
ZC =
1
jωC
=
1 −90
ωC
.
Resisteˆncias teˆm impedaˆncias iguais a`s suas resisteˆncias.
3. Analise o circuito utilizando as Leis de Kirchhoff e fac¸a os ca´lculos com
aritme´tica complexa.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 2 / 18
Exemplo 1
Encontre a corrente de regime permanente para o circuito mostrado na
figura abaixo. Encontre tambe´m o fasor de tensa˜o sobre cada elemento
do circuito e construa o diagrama fasorial.
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 233
Similarly, KCL can be applied to currents in phasor form. The sum of the phasor
currents entering a node must equal the sum of the phasor currents leaving.
Circuit Analysis Using Phasors and Impedances
We have seen that phasor currents and voltages are related by complex impedances,
and Kirchhoff s laws apply in phasor form. Except for the fact that the voltages,
currents, and impedances can be complex, the equations are exactly like those of
resistive circuits.
A step-by-step procedure for steady-state analysis of circuits with sinusoidal
sources is
1. Replace the time descriptions of the voltage and current sources with the
corresponding phasors. (All of the sources must have the same frequency.)
2. Replace inductances by their complex impedances ZL = j L = L 90* .
Replace capacitances by their complex impedances ZC = 1/(j C) = (1/ C)
90* . Resistances have impedances equal to their resistances.
3. Analyze the circuit by using any of the techniques studied in Chapter 2, and
perform the calculations with complex arithmetic.
Example 5.4 Steady-State AC Analysis of a Series Circuit
Find the steady-state current for the circuit shown in Figure 5.12(a). Also, nd the
phasor voltage across each element and construct a phasor diagram.
Solution From the expression given for the source voltage vs(t), we see that the Step 1: Replace the time
description of the voltage
source with the
corresponding phasor.
peak voltage is 100 V, the angular frequency is = 500, and the phase angle is 30 .
The phasor for the voltage source is
Vs = 100 30*
The complex impedances of the inductance and capacitance are Step 2: Replace inductances
and capacitances with their
complex impedances.ZL = j L = j500× 0.3 = j150
and
ZC = j
1
C
= j
1
500× 40× 10 6
= j50
R = 100 
I
Vs =
100 30°
VR
V
L
VC
j50
+j150
L =
0.3 H
100
i(t)
vs(t) =
100 cos(500t + 30°)
++
C = 40 mF
+
+
+
(a) (b)
Figure 5.12 Circuit for Example 5.4.
Figura 1: Circuito exemplo 1.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 3 / 18
Exemplo 1
1. Substituindo a fonte de tensa˜o pelo seu fasor correspondente
Vs = 100 30◦
2. Substituindo a indutaˆncia e a capacitaˆncia pelas suas respectivas indu-
taˆncias complexas
ZL = jωL = j500× 0,3 = j150 Ω
ZC = −j 1
ωC
= −j 1
500× 40 × 10−6 = −j50 Ω
3. Redesenhe o circuito no dom´ınio da frequeˆncia.
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 233
Similarly, KCL can be applied to currents in phasor form. The sum of the phasor
currents entering a node must equal the sum of the phasor currents leaving.
Circuit Analysis Using Phasors and Impedances
We have seen that phasor currents and voltages are related by complex impedances,
and Kirchhoff s laws apply in phasor form. Except for the fact that the voltages,
currents, and impedances can be complex, the equations are exactly like those of
resistive circuits.
A step-by-step procedure for steady-state analysis of circuits with sinusoidal
sources is
1. Replace the time descriptions of the voltage and current sources with the
corresponding phasors. (All of the sources must have the same frequency.)
2. Replace inductances by their complex impedances ZL = j L = L 90* .
Replace capacitances by their complex impedances ZC = 1/(j C) = (1/ C)
90* . Resistances have impedances equal to their resistances.
3. Analyze the circuit by using any of the techniques studied in Chapter 2, and
perform the calculations with complex arithmetic.
Example 5.4 Steady-State AC Analysis of a Series Circuit
Find the steady-state current for the circuit shown in Figure 5.12(a). Also, nd the
phasor voltage across each element and construct a phasor diagram.
Solution From the expression given for the source voltage vs(t), we see that the Step 1: Replace the time
description of the voltage
source with the
corresponding phasor.
peak voltage is 100 V, the angular frequency is = 500, and the phase angle is 30 .
The phasor for the voltage source is
Vs = 100 30*
The complex impedances of the inductance and capacitance are Step 2: Replace inductances
and capacitances with their
complex impedances.ZL = j L = j500× 0.3 = j150
and
ZC = j
1
C
= j
1
500× 40× 10 6
= j50
R = 100 
I
Vs =
100 30°
VR
V
L
VC
j50
+j150
L =
0.3 H
100
i(t)
vs(t) =
100 cos(500t + 30°)
++
C = 40 mF
+
+
+
(a) (b)
Figure 5.12 Circuit for Example 5.4.
Figura 2: Circuito exemplo 1 no dom´ınio da frequeˆncia.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 4 / 18
Exemplo 1
Como os treˆs elementos esta˜o em se´rie, pode–se agrupar as impedaˆncias
em uma impedaˆncia equivalente
Zeq = R + ZL + ZC
Zeq = 100 + j150− j50 = 100 + j100 Ω
Convertendo para polar
Zeq = 141,4 45◦ Ω
O fasor de corrente pode ser obtido dividindo o fasor de tensa˜o pela
impedaˆncia equivalente
I =
Vs
Zeq
=
100 30◦
141,4 45◦
= 0,707 −15◦ A
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 5 / 18
Exemplo 1
A corrente como func¸a˜o do tempo e´ dada por
i(t) = 0,707 cos(500t − 15◦) A
O fasor de tensa˜o sobre cada elemento e´ obtida multiplicando a corrente
pela respectiva impedaˆncia
VR = R × I = 100× 0,707 −15◦ = 70,7 −15◦ V
VL = ZL × I = ωL 90◦ × I = 150 90◦ × 0,707 −15◦ = 106,1 75◦ V
VC = ZC ×I = 1
ωC
−90◦×I = 50 −90◦×0,707 −15◦ = 35,4 −105◦ V
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 6 / 18
Exemplo 1
Diagrama fasorial
234 Chapter 5 Steady-State Sinusoidal Analysis
The transformed circuit is shown in Figure 5.12(b). All three elements are inStep 3: Use complex
arithmetic to analyze the
circuit.
series.Thus,we nd the equivalent impedance of the circuit by adding the impedances
of all three elements:
Zeq = R+ ZL + ZC
Substituting values, we have
Zeq = 100+ j150 j50 = 100+ j100
Converting to polar form, we obtain
Zeq = 141.4 45*
Now, we can nd the phasor current by dividing the phasor voltage by the
equivalent impedance, resulting in
I =
Vs
Z
=
100 30*
141.4 45*
= 0.707 15*
As a function of time, the current is
i(t) = 0.707 cos(500t 15 )
Next, we can nd the phasor voltage across each element by multiplying the
phasor current by the respective impedance:
VR = R× I = 100× 0.707 15* = 70.7 15*
VL = j L× I = L 90* × I = 150 90* × 0.707 15*
= 106.1 75*
VC = j
1
C
× I =
1
C
90* × I = 50 90* × 0.707 15*
= 35.4 105*
The phasor diagram for the current and voltages is shown in Figure 5.13. Notice
that the current I lags the source voltageVsby 45 .As expected, the voltageVR and
current I are in phase for the resistance. For the inductance, the voltageVL leads the
current I by 90 . For the capacitance, the voltage VC lags the current by 90 .
Figure 5.13 Phasor diagram
for Example 5.4.
Vs
VL
VR
I
VC
35.4
15*
30*
100
106.1
75*
Figura 3: Diagrama fasorial.
Note que
A corrente I esta´ atrasada da
tensa˜o Vs de 45
◦.
A tensa˜oVR e a corrente I esta˜o
em fase.
A tensa˜o VL esta´ adiantada da
corrente I em 90◦.
A tensa˜o VC esta´ atrasada da
corrente I em 90◦.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 7 / 18
Exemplo 2
Encontre a tensa˜o vC (t) em estado permanente. Determine o fasor de
corrente atrave´s de cada elemento, e construa o diagrama fasorial das
correntes e da tensa˜o da fonte.
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 235
Example 5.5 Series and Parallel Combinations of Complex Impedances
Consider the circuit shown in Figure 5.14(a). Find the voltage vC(t) in steady state.
Find the phasor current through each element, and construct a phasor diagram
showing the currents and the source voltage.
Solution The phasor for the voltage source is Vs = 10 90* . [Notice that vs(t) is Step 1: Replace the time
description of the voltage
source with the
corresponding phasor.
a sine function rather than a cosine function, and it is necessary to subtract 90 from
the phase.] The angular frequency of the source is = 1000. The impedances of the
inductance and capacitance are
ZL = j L = j1000× 0.1 = j100
and Step 2: Replace inductances
and capacitances with their
complex impedances.ZC = j
1
C
= j
1
1000× 10× 10 6
= j100
The transformed network is shown in Figure 5.14(b).
To ndVC ,wewill rst combine the resistanceand the impedanceof thecapacitor Step 3: Use complex
arithmetic to analyze the
circuit.
in parallel. Then, we will use the voltage-division principle to compute the voltage
across the RC combination. The impedance of the parallel RC circuit is
ZRC =
1
1/R+ 1/ZC
=
1
1/100+ 1/( j100)
=
1
0.01+ j0.01
=
1 0*
0.01414 45*
= 70.71 45*
Converting to rectangular form, we have
ZRC = 50 j50
The equivalent network is shown in Figure 5.14(c).
Vs =
10 90°
VC
+
+j100
j100100
vs(t) =
10 sin(1000t)
vC
+
L = 0.1 H
C =
10 mF
R =
100
+ +
(a) (b)
IC
IRI
ZRCVC
Vs =
10 90°
+j100
I
+
+
(c)
Figure 5.14 Circuit for Example 5.5.
Figura 4: Circuito exemplo 2.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 8 / 18
Exemplo 2
1. Substituindo a fonte de tensa˜o pelo seu fasor correspondente
Vs = 10 −90◦
2. Substituindo a indutaˆncia e a capacitaˆncia pelas suas respectivas indu-
taˆncias complexas
ZL = jωL = j1000× 0,1 = j100 Ω
ZC = −j 1
ωC
= −j 1
1000× 10 × 10−6 = −j100 Ω
3. Redesenhe o circuito no dom´ınio da frequeˆncia.
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 235
Example 5.5 Series and Parallel Combinations of Complex Impedances
Consider the circuit shown in Figure 5.14(a). Find the voltage vC(t) in steady state.
Find the phasor current through each element, and construct a phasor diagram
showing the currents and the source voltage.
Solution The phasor for the voltage source is Vs = 10 90* . [Notice that vs(t) is Step 1: Replace the time
description of the voltage
source with the
corresponding phasor.
a sine function rather than a cosine function, and it is necessary to subtract 90 from
the phase.] The angular frequency of the source is = 1000. The impedances of the
inductance and capacitance are
ZL = j L = j1000× 0.1 = j100
and Step 2: Replace inductances
and capacitances with their
complex impedances.ZC = j
1
C
= j
1
1000× 10× 10 6
= j100
The transformed network is shown in Figure 5.14(b).
To ndVC ,wewill rst combine the resistanceand the impedanceof thecapacitor Step 3: Use complex
arithmetic to analyze the
circuit.
in parallel. Then, we will use the voltage-division principle to compute the voltage
across the RC combination. The impedance of the parallel RC circuit is
ZRC =
1
1/R+ 1/ZC
=
1
1/100+ 1/( j100)
=
1
0.01+ j0.01
=
1 0*
0.01414 45*
= 70.71 45*
Converting to rectangular form, we have
ZRC = 50 j50
The equivalent network is shown in Figure 5.14(c).
Vs =
10 90°
VC
+
+j100
j100100
vs(t) =
10 sin(1000t)
vC
+
L = 0.1 H
C =
10 mF
R =
100
+ +
(a) (b)
IC
IRI
ZRCVC
Vs =
10 90°
+j100
I
+
+
(c)
Figure 5.14 Circuit for Example 5.5.
Figura 5: Circuito exemplo 2 no dom´ınio da frequeˆncia.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 9 / 18
Exemplo 2
Para determinar VC, primeiramente deve-se associar a resisteˆncia e o
capacitor que esta˜o em paralelo
ZRC =
1
1/R + 1/ZC
=
1
1/100 + 1/(−j100)
ZRC =
1
0,01 + j0,01
=
1 0◦
0,01414 45◦
= 70,71 −45◦ Ω
O circuito equivalente e´ mostrado na figura a seguir
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 235
Example 5.5 Series and Parallel Combinations of Complex Impedances
Consider the circuit shown in Figure 5.14(a). Find the voltage vC(t) in steady state.
Find the phasor current through each element, and construct a phasor diagram
showing the currents and the source voltage.
Solution The phasor for the voltage source is Vs = 10 90* . [Notice that vs(t) is Step 1: Replace the time
description of the voltage
source with the
corresponding phasor.
a sine function rather than a cosine function, and it is necessary to subtract 90 from
the phase.] The angular frequency of the source is = 1000. The impedances of the
inductance and capacitance are
ZL = j L = j1000× 0.1 = j100
and Step 2: Replace inductances
and capacitances with their
complex impedances.ZC = j
1
C
= j
1
1000× 10× 10 6
= j100
The transformed network is shown in Figure 5.14(b).
To ndVC ,wewill rst combine the resistanceand the impedanceof thecapacitor Step 3: Use complex
arithmetic to analyze the
circuit.
in parallel. Then, we will use the voltage-division principle to compute the voltage
across the RC combination. The impedance of the parallel RC circuit is
ZRC =
1
1/R+ 1/ZC
=
1
1/100+ 1/( j100)
=
1
0.01+ j0.01
=
1 0*
0.01414 45*
= 70.71 45*
Converting to rectangular form, we have
ZRC = 50 j50
The equivalent network is shown in Figure 5.14(c).
Vs =
10 90°
VC
+
+j100
j100100
vs(t) =
10 sin(1000t)
vC
+
L = 0.1 H
C =
10 mF
R =
100
+ +
(a) (b)
IC
IRI
ZRCVC
Vs =
10 90°
+j100
I
+
+
(c)
Figure 5.14 Circuit for Example 5.5.
Figura 6: Circuito equivalente exemplo 2.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 10 / 18
Exemplo 2
A corrente I que flui pelo circuito equivalente e´ dada por
I =
Vs
ZL + ZRC
=
10 −90◦
j100 + 50 − j50 =
10 −90◦
70,71 45◦
= 0,1414 −135◦ A
A tensa˜o VC pode ser escrita como
VC = I× ZRC = 0,1414 −135◦ × 70,71 −45◦ = 10 −180◦ V
Reescrevendo a tensa˜o no capacitor no dom´ınio do tempo
vC (t) = 10 cos(1000t − 180◦) = −10 cos(1000t) V
A corrente no resistor e no capacitor sa˜o respectivamente
IR =
VC
R
=
10 −180◦
100
= 0,1 −180◦ A
IC =
VC
ZC
=
10 −180◦
−j100 =
10 −180◦
100 −90◦ = 0,1 −90
◦ A
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 11 / 18
Exemplo 2
Diagrama fasorial
236 Chapter 5 Steady-State Sinusoidal Analysis
Now, we use the voltage-division principle to obtain
VC = Vs
ZRC
ZL + ZRC
= 10 90*
70.71 45*
j100+ 50 j50
= 10 90*
70.71 45*
50+ j50
= 10 90*
70.71 45*
70.71 45*
= 10 180*
Converting the phasor to a time function, we have
vC(t) = 10 cos(1000t 180 ) = 10 cos(1000t)Next, we compute the current in each element yielding
IR
IC
Vs
I
Figure 5.15 Phasor diagram
for Example 5.5.
I =
Vs
ZL + ZRC
=
10 90*
j100+ 50 j50
=
10 90*
50+ j50
=
10 90*
70.71 45*
= 0.1414 135*
IR =
VC
R
=
10 180*
100
= 0.1 180*
IC =
VC
ZC
=
10 180*
j100
=
10 180*
100 90*
= 0.1 90*
The phasor diagram is shown in Figure 5.15.
Node-Voltage Analysis
We can perform node-voltage analysis by using phasors in a manner parallel to that
of Chapter 2.We illustrate with an example.
Example 5.6 Steady-State AC Node-Voltage Analysis
Use the node-voltage technique to nd v1(t) in steady state for the circuit shown in
Figure 5.16(a).
2 sin(100t) 1.5 cos(100t)
(b)(a)
10 *
2000 mF
0.1 H
v2(t)v1(t)
2 90+ 1.5 0+10 *
j5 *
+j10 *
V1 V2
Figure 5.16 Circuit for Example 5.6.
Figura 7: Diagrama fasorial.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 12 / 18
Exemplo 3
Considere o circuito mostrado a Figura abaixo. (a) Determine i(t); (b)
Desenhe o diagrama fasorial; (c) qual e´ a relac¸a˜o de fase entre vs(t) e
i(t).
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 237
Solution The transformed network is shown in Figure 5.16(b). We obtain two
equations by applying KCL at node 1 and at node 2. This yields
V1
10
+
V1 V2
j5
= 2 90*
V2
j10
+
V2 V1
j5
= 1.5 0*
These equations can be put into the standard form
(0.1+ j0.2)V1 j0.2V2 = j2
j0.2V1 + j0.1V2 = 1.5
Now, we solve for V1 yielding
V1 = 16.1 29.7*
Then, we convert the phasor to a time function and obtain
v1(t) = 16.1 cos(100t + 29.7 )
Mesh-Current Analysis
In a similar fashion, you can use phasors to carry out mesh-current analysis in ac
circuits. Exercise 5.11 gives you a chance to try this approach.
Exercise 5.9 Consider the circuit shown in Figure 5.17(a). a. Find i(t). b. Construct
a phasor diagram showing all three voltages and the current. c. What is the phase
relationship between vs(t) and i(t)?
Answer a. i(t) = 0.0283 cos(500t 135 ); b. the phasor diagram is shown in
Figure 5.17(b); c. i(t) lags vs(t) by 45 . *
Exercise 5.10 Find the phasor voltage and the phasor current through each element
in the circuit of Figure 5.18.
Answer V = 277 56.3* , IC = 5.55 33.7* , IL = 1.39 146.3* ,
IR = 2.77 56.3* . *
I
V
L
= 7.07 45°V
R
= 7.07 135°
V
s
= 10 90°
250
0.5 H
+vs(t) =
10 sin(500t)
v
L
v
R
i(t)
+
+
(a) (b)
Figure 5.17 Circuit and phasor diagram for Exercise 5.9.
Figura 8: Circuito exemplo 3.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 13 / 18
Exemplo 3
Resposta
(a) i(t) = 0.0283 cos(500t − 135◦) A;
(b) O diagrama fasorial e´ ilustrado abaixo;
(c) i(t) esta´ atrasada de vs(t) em 45
◦
Section 5.4 Circuit Analysis with Phasors and Complex Impedances 237
Solution The transformed network is shown in Figure 5.16(b). We obtain two
equations by applying KCL at node 1 and at node 2. This yields
V1
10
+
V1 V2
j5
= 2 90*
V2
j10
+
V2 V1
j5
= 1.5 0*
These equations can be put into the standard form
(0.1+ j0.2)V1 j0.2V2 = j2
j0.2V1 + j0.1V2 = 1.5
Now, we solve for V1 yielding
V1 = 16.1 29.7*
Then, we convert the phasor to a time function and obtain
v1(t) = 16.1 cos(100t + 29.7 )
Mesh-Current Analysis
In a similar fashion, you can use phasors to carry out mesh-current analysis in ac
circuits. Exercise 5.11 gives you a chance to try this approach.
Exercise 5.9 Consider the circuit shown in Figure 5.17(a). a. Find i(t). b. Construct
a phasor diagram showing all three voltages and the current. c. What is the phase
relationship between vs(t) and i(t)?
Answer a. i(t) = 0.0283 cos(500t 135 ); b. the phasor diagram is shown in
Figure 5.17(b); c. i(t) lags vs(t) by 45 . *
Exercise 5.10 Find the phasor voltage and the phasor current through each element
in the circuit of Figure 5.18.
Answer V = 277 56.3* , IC = 5.55 33.7* , IL = 1.39 146.3* ,
IR = 2.77 56.3* . *
I
V
L
= 7.07 45°V
R
= 7.07 135°
V
s
= 10 90°
250
0.5 H
+vs(t) =
10 sin(500t)
v
L
v
R
i(t)
+
+
(a) (b)
Figure 5.17 Circuit and phasor diagram for Exercise 5.9.
Figura 9: Diagrama fasorial.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 14 / 18
Exemplo 4
Determine os fasor de tensa˜o e corrente em cada elemento do circuito
abaixo.238 Chapter 5 Steady-State Sinusoidal Analysis
5 cos(200t) 100100 mF 1 H
iC iL
iR
v
+
Figure 5.18 Circuit for Exercise 5.10.
100 cos(1000t) 100
i1(t) i2(t)
+
0.1 H
0.1 H 5 mF
Figure 5.19 Circuit for Exercise 5.11.
Exercise 5.11 Solve for the mesh currents shown in Figure 5.19.
Answer i1(t) = 1.414 cos(1000t 45 ), i2(t) = cos(1000t). *
5.5 POWER IN AC CIRCUITS
Consider the situation shown in Figure 5.20. A voltage v(t) = Vm cos( t) is applied
to a network composed of resistances, inductances, and capacitances (i.e., an RLC
network). The phasor for the voltage source is V = Vm 0* , and the equivalent
impedance of the network is Z = |Z| * = R+ jX . The phasor current is
I =
V
Z
=
Vm 0*
|Z| *
= Im * (5.52)
where we have de ned
Im =
Vm
|Z|
(5.53)
Before we consider the power delivered by the source to a general load, it
is instructive to consider a pure resistive load, a pure inductive load, and a pure
capacitive load.
Figure 5.20 A voltage source
delivering power to a load impedance
Z = R+ jX .
+
R
jX
Im u
Vm 0° Z u
Figura 10: Circuito exemplo 4.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 15 / 18
Exemplo 4
Resposta
V = 277 −56,31◦ V
IC = 5,54 33,69
◦ A
IL = 1,385 −146,3◦ A
IR = 2,77 −56,3◦ A
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 16 / 18
Exemplo 5
Determine as correntes i1(t) e i2(t).
238 Chapter 5 Steady-State Sinusoidal Analysis
5 cos(200t) 100100 mF 1 H
iC iL
iR
v
+
Figure 5.18 Circuit for Exercise 5.10.
100 cos(1000t) 100
i1(t) i2(t)
+
0.1 H
0.1 H 5 mF
Figure 5.19 Circuit for Exercise 5.11.
Exercise 5.11 Solve for the mesh currents shown in Figure 5.19.
Answer i1(t) = 1.414 cos(1000t 45 ), i2(t) = cos(1000t). *
5.5 POWER IN AC CIRCUITS
Consider the situation shown in Figure 5.20. A voltage v(t) = Vm cos( t) is applied
to a network composed of resistances, inductances, and capacitances (i.e., an RLC
network). The phasor for the voltage source is V = Vm 0* , and the equivalent
impedance of the network is Z = |Z| * = R+ jX . The phasor current is
I =
V
Z
=
Vm 0*
|Z| *
= Im * (5.52)
where we have de ned
Im =
Vm
|Z|
(5.53)
Before we consider the power delivered by the source to a general load, it
is instructive to consider a pure resistive load, a pure inductive load, and a pure
capacitive load.
Figure 5.20 A voltage source
delivering power to a load impedance
Z = R+ jX .
+
R
jX
Im u
Vm 0° Z u
Figura 11: Circuito exemplo 5.
Pedro Machado de Almeida (UFJF) ENE077 Novembro 2015 17 / 18
Exemplo 5
Resposta
i1(t) = 1,414 cos(1000t − 45◦) A
i2(t) = 1 cos(1000t) A
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