Solution 11.68
Let
cLR SSSS ++=
where
0jRI
2
1
jQP2
oRRR +=+=S
Solution 11.69
Refer to the circuit shown in Fig. 11.88.
(a) What is the power factor?
(b) What is the average power dissipated?
(c) What is the value of the capacitance that will give a unity power factor
when connected to the load?
Figure 11.88
For Prob. 11.69.
Solution
(a) Given that
1520 j+=Z
(b)
4.646,42.195,6
)440(
2
2
j+=
== Z
V
S
(c) For unity power factor,
°=θ 0
, which implies that the reactive power due to the
120 Vrms
440 Vrms
60 Hz
(20+j15) Ω
Solution 11.70
Design a problem to help other students to better understand power factor correction.
Problem
An 880-VA, 220-V, 50-Hz load has a power factor of 0.8 lagging. What value of parallel
capacitance will correct the load power factor to unity?
Solution
If the power factor is to be unity, the reactive power due to the capacitor is
Solution 11.71
(a) For load 1,
Q1 = 60 kVAR, pf = 0.85 or θ1 = 31.79˚
Hence,
Thus, Z = 0.05017∠2˚Ω or [50.14 + j1.7509] mΩ.
Solution 11.72
(a)
1
1
2.4
cos 3.0 kVA
cos 0.8
P
PS S
θθ
= → = = =
1
1
1.5 2.122 kVA
P
S
= = =
Solution 11.73
(a)
kVA7j1022j15j10 +=+−=S
(b)
240
000,7j000,10
** +
==→= V
S
IIVS
(c)
°=
=θ 35
10
7
tan
1–
1
,
°==θ 26.16)96.0(cos-1
2
(d)
222 jQP+=S
,
kW10PP
12
==
Solution 11.74
(a)
°==θ 87.36)8.0(cos-1
1
°==θ 19.18)95.0(cos-1
(b)
°=θ 95.25
2
,
°=θ 0
1
Solution 11.75
Consider the power system shown in Fig. 11.90. Calculate:
(a) the total complex power
(b) the power factor
(c) the parallel capacitance necessary to establish a unity power factor
Figure 11.90
For Prob. 11.75.
Solution
Step 1. S1 = (440)2/(40–j30)*, S2 = (440)2/(10+j10)*, and S3 = (440)2/10.
(a) S1 = 193,600/(50∠36.87°) = 3,872∠–36.87° = (3,097.6–j2,323.2) VA, S2 =
Solution 11.76
The wattmeter reads the real power supplied by the current source. Consider the circuit
below.
8j23j4
12
303 ooo VVV +=
−
−
+°∠
–j3 Ω
4 Ω
Vo
Solution 11.77
The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150
Ω.
°∠→ 0120)t2cos(120
,
2=ω
Consider the following circuit.
(15)(-j5)
The wattmeter reads
j8 Ω
6 Ω
I
Solution 11.78
Find the wattmeter reading of the circuit shown in Fig. 11.93 below.
Figure 11.93
For Prob. 11.78.
Solution
The wattmeter reads the power absorbed by the element to its right side.
Consider the following circuit.
)3j–)(4(
4j53j–||44j5 −
++=++=Z
10 Ω
I
170 sin(4t) V
Solution 11.79
The wattmeter reads the power supplied by the source and partly absorbed by the 40–
Ω
resistor.
The frequency-domain circuit is shown below.
20 Io
–
At node 1,
At node 2,
Solving (1) and (2) yields V1 = 1.5568 –j4.1405
Solution 11.80
The circuit of Fig. 11.95 portrays a wattmeter connected into an ac network.
(a) Find the load current.
(b) Calculate the wattmeter reading.
Figure 11.95
For Prob. 11.80.
Solution
240∠0° V
|Z
L
| = 20 Ω
pf = 0.8
Solution 11.81
Design a problem to help other students to better understand how to correct power factor
Problem
A 120-V rms, 60-Hz electric hair dryer consumes 600 W at a lagging pf of 0.92.
Calculate the rms–valued current drawn by the dryer.
How would you power factor correct this to a value of 0.95?
Solution
P = 600 W,
0.92 23.074o
pf
θ
= → =
To correct this to a pf = 0.95, I would add a capacitor in parallel with the hair dryer
(remember, series compensation will increase the power delivered to the load and
probably burn out the hair dryer.
Solution 11.82
(a)
0,000,5 11 == QP
(c )
865.0
220,34
600,29 === S
P
pf