Section 8-2 and 8-3: Average and Complex Power

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Section 8-2 and 8-3: Average and Complex Power
Problem 8.9 Determine the complex power, apparent power, average power
absorbed, reactive power, and power factor (including whether it is leading or
lagging) for a load circuit whose voltage and current at its input terminals are given
by:
(a) v(t) = 100 cos(377t − 30◦ ) V,
i(t) = 2.5 cos(377t − 60◦ ) A.
(b) v(t) = 25 cos(2π × 103 t + 40◦ ) V,
i(t) = 0.2 cos(2π × 103 t − 10◦ ) A.
(c) Vrms = 110∠60◦ V, Irms = 3∠45◦ A.
(d) Vrms = 440∠0◦ V, Irms = 0.5∠75◦ A.
(e) Vrms = 12∠60◦ V, Irms = 2∠−30◦ A.
Solution:
(a)
100
◦
Vrms = √ e− j30 V,
2
2.5 − j60◦
Irms = √ e
V.
2
100
◦
2.5
◦
◦
S = Vrms I∗rms = √ e− j30 × √ e j60 = 125e j30
2
2
S = |S| = 125 VA
(VA)
Pav = Re[S] = 125 cos 30◦ = 108.25 W
Q = Im[S] = 125 sin 30◦ = 62.5 VAR
φs = φv − φi = −30◦ + 60◦ = 30◦
(hence pf is lagging)
◦
pf = cos 30 = 0.866.
(b)
25
◦
Vrms = √ e j40 V,
2
0.2
◦
Irms = √ e− j10 A,
2
◦
S = Vrms I∗rms = 2.5e j50
(VA),
φs = 50◦ (lagging)
S = 2.5 VA,
Pav = 2.5 cos 50◦ = 1.61 W,
Q = 2.5 sin 50◦ = 1.92 VAR,
pf = cos 50◦ = 0.64 lagging.
(c)
◦
◦
◦
S = Vrms I∗rms = 110e j60 × 3e− j45 = 330e j15 VA.
S = 330 VA,
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Pav = 330 cos 15◦ = 318.76 W,
Q = 330 sin 15◦ = 85.41 VAR,
pf = cos 15◦ = 0.97 (lagging).
(d)
◦
◦
S = Vrms I∗rms = 440 × 0.5e− j75 = 220e− j75 VA.
S = 220 VA,
Pav = 220 cos(−75◦ ) = 56.94 W,
Q = 220 sin(−75◦ ) = −212.50 VAR,
pf = cos(−75◦ ) = 0.26 (leading).
(e)
◦
◦
◦
S = Vrms I∗rms = 12e j60 × 2e j30 = 24e j90 VA.
S = 24 VA,
Pav = 24 cos 90◦ = 0,
Q = 24 sin 90◦ = 24 VAR,
pf = cos 90◦ = 0
(purely inductive with I lagging V by 90◦ )
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Problem 8.11 In the circuit of Fig. P8.11, is (t) = 0.2 sin 105 t A, R = 20 Ω,
L = 0.1 mH, and C = 2 µ F. Show that the sum of the complex powers for the three
passive elements is equal to the complex power of the source.
+
is(t) _
R
L
C
Figure P8.11: Circuit for Problem 8.11.
Solution:
V
Is
IR
IL
20 Ω
j10 Ω
IC
+
_
−j5 Ω
Is = 0.2∠0◦ A
ZL = jω L = j105 × 10−4 = j10 Ω
−j
−j
= 5
= − j5 Ω.
ωC 10 × 2 × 10−6
ZC =
V
V
V
+
+
= Is = 0.2
20 j10 − j5
V = 1.79e− j63.4◦ V.
◦
V
1.79e− j63.4
IR =
=
A.
20
20
1
(1.79)2
= 0.08 VA.
SR = VI∗R =
2
2 × 20
V
1.79 − j153.4◦
IL =
=
e
j10
10
1
1.79 − j63.4◦ 1.79 j153.4◦
◦
SL = VI∗L =
e
×
e
= 0.16e j90 = 0 + j0.16 VA.
2
2
10
V
1.79 j26.6◦
IC =
=
e
A.
− j5
5
1
1
◦
1.79 − j26.6◦
◦
SC = VI∗C = 1.79e− j63.4 ×
e
= 0.32e− j90 = 0 − j0.32 VA.
2
2
5
ST = SR + SL + SC = 0.08 + j0.16 − j0.32 = 0.08 − j0.16 VA.
For the source,
◦
◦
1
1
Ss = VI∗s = 1.79e− j63.4 × 0.2 = 0.179e− j63.4
2
2
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= 0.179 cos 63.4◦ − j0.179 sin 63.4◦
= 0.08 − j0.16 VA.
Hence
ST = Ss .
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Problem 8.14 Determine S for the RL load in the circuit of Fig. P8.14, given that
Is = 4∠0◦ A, R1 = 10 Ω, R2 = 5 Ω, ZC = − j20 Ω, R = 10 Ω, and ZL = j20 Ω.
R2
a
R
R1
ZC
Is
ZL
b
Load
Figure P8.14: Circuit for Problem 8.14.
Solution:
V I
R2
a
R
R1
ZC
Is
ZL
b
1
1
1
V
+
+
= Is
ZC R1 R2 + R + ZL
1
1
1
+ +
=4
V
− j20 10 5 + 10 + j20
Solution gives
◦
V = 31.6 − j4.6 = 31.93e− j8.28 V.
For RL load:
1
Vab I∗
2
∗
1 (10 + j20)
V
=
V×
2 15 + j20
15 + j20
S=
= (5 + j10)
|V|2
(31.93)2
◦
j63.4◦
=
11.18e
×
= 18.24e j63.4 VA.
2
|15 + j20|
625
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Problem 8.20 The apparent power entering a certain load Z is 250 VA at a power
factor of 0.8 leading. If the rms phasor voltage of the source is 125 V at 1 MHz:
(a) Determine Irms going into the load
(b) Determine S into the load
(c) Determine Z
(d) The equivalent impedance of the load circuit should be of the form
Z = R + jω L or Z = R − j/ωC. Determine the value of L or C, whichever
is applicable.
Solution:
(a) From
S = Vrms Irms ,
Irms =
S
Vrms
=
250
= 2 A.
125
(b) pf = 0.8 leading means φz is negative. Hence,
φz = − cos−1 0.8 = −36.87◦ .
S = S cos φz + jS sin φz
= 250[cos(−36.87◦ ) + j sin(−36.87◦ )] = (200 − j150) VA.
(c)
2
Pav = Irms
R
R=
Pav
200
=
= 50 Ω.
2
Irms
4
Also,
2
Q = Irms
X
X=
Hence,
(d)
−j
= − j37.5, or
ωC
C=
Q
−150
=
= −37.5 Ω.
2
Irms
4
Z = R + jX = (50 − j37.5) Ω.
1
1
=
= 4.24 nF.
37.5ω
37.5 × 2π × 106
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Section 8-4: Power Factor
Problem 8.22 The RL load in Fig. P8.22 is compensated by adding the shunt
capacitance C so that the power factor of the combined (compensated) circuit is
exactly unity. How is C related to R, L, and ω in that case?
R
C
L
Figure P8.22: Circuit for Problem 8.22.
Solution: For the combined load, the impedance is
1
Z = (R + jω L) k
jω C
(R + jω L) ω−Cj
=
R + j ω L − ω1C
=
=
=
ω L − jR
ω RC − j(1 − ω 2 LC)
(ω L − jR)[ω RC + j(1 − ω 2 LC)]
[ω RC − j(1 − ω 2 LC)][ω RC + j(1 − ω 2 LC)]
[ω 2 RLC + R(1 − ω 2LC)] j[ω L(1 − ω 2 LC) − ω R2C]
+
ω 2 R2C2 + (1 − ω 2 LC)2
ω 2 R2C2 + (1 − ω 2 LC)2
For the pf to be unity, the imaginary component of Z has to be zero, which is realized
if
ω L(1 − ω 2 LC) − ω R2C = 0.
or
C=
L
.
R2 + ω 2 L2
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Problem 8.23 The generator circuit shown in Fig. P8.23 is connected to a distant
load via a long coaxial transmission line. The overall circuit can be modeled as
in Fig. P8.23(b), in which the transmission line is represented by an equivalent
impedance Zline = (5 + j2) Ω.
a
10 Ω
Vs
c
+
_
ZL = (50 + j40) Ω
Transmission line
b
d
(a) Transmission-line circuit
a
10 Ω
5Ω
j2 Ω
c
Transmission line
Vs
+
_
ZL = (50 + j40) Ω
C
(b) Equivalent circuit
b
d
Figure P8.23: Circuit for Problem 8.23.
(a) Determine the power factor of voltage source Vs .
(b) Specify the capacitance of a shunt capacitor C that would raise the power factor
of the source to unity when connected between terminals (a, b). The source
frequency is 1.5 kHz.
Solution:
(a) The power factor of the source is the same as the phase of the impedance
representing the entire circuit connected to Vs . Thus,
ZT = 10 + (5 + j2) + (50 + j40) = (65 + j42) Ω,
42
= 32.87◦ ,
65
pf = cos φZ = 0.84, lagging.
φZ = tan−1
(b) For the circuit to the right of terminals (a, b), the impedance—with a shunt
capacitance C—is:
−j
Zab =
k (55 + j42)
ωC
=
− jXC (55 + j42)
55 + j(42 − XC )
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where XC = ω1C .
Simplifying,
Zab =
=
42XC − j55XC
· [55 − j(42 − XC)]
552 + (42 − XC)2
[42 × 55XC − 55XC(42 − XC )] − j[552 XC + 42XC(42 − XC )]
3025 + (42 − XC )2
For a pf of 1, the imaginary part of Zab should be zero,
552 XC + 42XC (42 − XC ) = 0.
Solution gives
XC = 114.02,
or
C=
1
1
=
= 930.5 nF.
ω XC 2π × 1.5 × 103 × 114.02
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Problem 8.25 Use the power information given for the circuit in Fig. P8.25 to
determine:
(a) Z1 and Z2
(b) the rms value of Vs .
0.6 Ω
j0.4 Ω
1.2 Ω
+
+
_ Vs
Z2
Z1
Vrms = 440
0o V
_
Load Z1 : 24 kW @ pf = 0.66 leading
Load Z2 : 18 kW @ pf = 0.82 lagging
Figure P8.25: Circuit for Problem 8.25.
Solution:
0.6 Ω
I1
Is
Vs
j0.4 Ω
1.2 Ω
V1
+
_
I2
Z1
Z2
+
Vrms = 440
_
(a) For load Z2 :
φZ2 = cos−1 0.82 = 34.9◦ .
Since φZ2 = φv2 − φi2 , and φv2 = 0,
=⇒
φi2 = −34.9◦ .
Pav2 = Vrms I2rms cos φZ2
18 × 103 = 440I2rms cos 34.9◦ ,
or
I2rms = 49.88 A,
and
S2 = Vrms I2rms = 440 × 49.88 = 21.95 kVA.
Also, I2rms = 49.88∠−34.9◦ A.
If Z2 = R2 + jX2 ,
Pav2 = I22rms R2
=⇒
R2 =
18 × 103
= 7.23 Ω
(49.88)2
Q2 = S2 sin φZ2 = 21.95 × 103 sin 34.9◦ = 12.56 kVAR
But
Q2 = I22rms X2
=⇒
X2 =
12.56 × 103
= 5.05 Ω.
(49.88)2
0o V
Hence,
Z2 = (7.23 + j5.05) Ω.
To determine Z1 , we first determine the voltage across it, V1rms :
V1rms = (1.2 + j0.4 + Z2 )I2rms
= (1.2 + j0.4 + 7.23 + j5.05)49.88e− j34.9
◦
= 500.8∠−2◦ V.
For load Z1 :
φZ1 = − cos−1 0.66 = −48.7◦
S1 =
I1rms =
Pav1
24
=
= 36.36 kVA
cos φZ1
0.66
36.36 × 103
S1
= 72.6 A
=
V1rms
500.8
φZ1 = φv1 − φi1
−48.7◦ = −2 − φi1
=⇒
φi1 = 46.7◦ .
I1rms = 72.6∠46.7◦ A.
Pav1 = I21rms R1
=⇒
R1 =
24 × 103
= 4.55 Ω
(72.6)2
Q1 = S1 sin φZ1 = 36.36 × 103 sin(−48.7◦ ) = −27.32 kVAR
X1 =
−27.32 × 103
Q1
=
= −5.18 Ω
(72.6)2
I12rms
Hence,
Z1 = (4.55 − j5.18) Ω.
(b) Given I1 and I2 , we can now determine Is :
Isrms = I1rms + I2rms
◦
= 72.6e j46.7 + 49.88e− j34.9
◦
= (90.7 + j24.3) A
Vsrms = V1rms + 0.6Isrms
◦
= 500.8e− j2 + 0.6(90.7 + j24.3) = (554.9 − j2.9) V.
Problem 8.27 For the circuit in Fig. P8.27, choose the load impedance ZL so that
the power dissipated in it is a maximum. How much power will that be?
−j2 Ω
V1
1Ω
6
o
0
j2 Ω
V2
a
+
V _
ZL
2Ω
b
Figure P8.27: Circuit for Problem 8.27.
Solution: To determine Vs of the equivalent source circuit, we remove ZL and
calculate Voc at terminals (a, b).
−j2 Ω
1Ω
6
0o
I 1 V1
j2 Ω
V2
+
+
V _
Voc
_
2Ω
At node V2 :
V2 − 6 V2 V2 − 6
+
+
=0
=⇒
V2 = 0.6(3 + j) V.
− j2
2
1 + j2
V2 − 6 V1 − 6
I1 =
=
1 + j2
1
Hence,
Vs = V1 =
V2 − 6
0.6(3 + j) − 6
+6 =
+ 6 = 5.7∠18.4◦ V.
1 + j2
1 + j2
To determine ZTh at terminals (a, b), we suppress the 6-V source and simplify the
circuit. The process leads to:
Zs = ZTh = (0.6 + j0.2) Ω
For maximum power transfer:
ZL = Z∗s = (0.6 − j0.2) Ω,
and
Pav (max) =
1 |Vs |2 1 (4.9)2
= ×
= 6.78 W.
8 RL
8
0.6
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