Problem set #6, EE 223, 3/4/2003 – 3/11/2003
Chapter 12, Problem 3.
A six-phase power system is developed as part of a high-current dc magnet power supply. Write the phase
voltages for (a) positive phase sequence; (b) negative phase sequence.
(a) positive phase sequence
Van = |Vp| ∠ 0o
Vbn = |Vp| ∠ -60o
Vcn = |Vp| ∠ -120
Vdn = |Vp| ∠ -180o
o
Ven = |Vp| ∠ -240o
Vfn = |Vp| ∠ -300o
(b) negative phase sequence
Van = |Vp| ∠ 0o
o
Vbn = |Vp| ∠ 60
Vcn = |Vp| ∠ 120o
Vdn = |Vp| ∠ 180o
Ven = |Vp| ∠ 240o
Vfn = |Vp| ∠ 300o
Chapter 12, Problem 5.
For a particular circuit, it is known that V12 =100
V42 =80
(a) V25 = V24 + V45
,
, and V35 =-j 120, all in volts. Find (a) V25;(b) V13.
= -80 ∠ 120o + 60 ∠ 75o
= 40 – j69.28 + 15.53 + j57.96 = 55.53 – j11.32
= 56.67 ∠ -11.52o V
(a) V13 = V12 + V25 + V53
V45 =60
= 100 + 55.53 – j11.32 + j120
= 155.53 + j108.7
= 189.8 ∠ 34.95o V
A
Chapter 12, Problem 9.
In the balanced three-wire single-phase system of Fig.
12.30, let VAN=220 Vat 60 Hz.
(a) What size should C be to provide a unity-powerfactor load?
(b) How many kVA does C handle?
5+ j2Ω
C
N
5+ j2Ω
B
VAN = 220 Vrms, 60 Hz
(a)
PF = 1 ∴ IAN =
220∠0°
= 40.85+ ∠ − 21.80° A; IAB = j 377C × 440
5 + j2
∴ IaA = 40.85cos 21.80° + j (377C440 − 40.85sin 21.80°)
∴C =
(b)
40.85sin 21.80°
= 91.47 µ F
377 × 440
IAB = 377 × 91.47 ×10−6 × 440 = 15.172 A ∴ VA = 440 × 15.172 = 6.676 kVA
Chapter 12, Problem 12.
Let Van = 2300
V in the balanced system shown in Fig.
12.31, and set Rw =2 Ω Assume positive phase sequence
with the source supplying a total
complex power of S =100 +j30 kVA.
Find (a) IaA;(b) VAN ;(c) Zp ;(d) the transmission efficiency.
↑ Van = 2300∠0° Vrms , R w = 2 Ω, + seq., Stot = 100 + j 30 kVA
(a)
1
(100, 000 + j 30, 000) = 2300 I∗aA ∴ IaA = 15.131∠ − 16.699° A
3
(b)
VAN = 2300 − 2 × 15.131∠ − 16.699° = 2271∠0.2194° V
c)
Z p = VAN / IaA =
(d)
trans. eff. =
2271∠0.2194°
= 143.60 + j 43.67 Ω
15.131∠ − 16.699°
143.60
= 0.9863, or 98.63%
145.60