Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 10(10).
For the circuit in Fig. 12.43, determine the current in the neutral line.
Figure 12.43
Chapter 12, Solution 10(10).
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia =
Van
220 ∠0°
= 6.55∠36.53°
=
Z A + 2 27 − j20
Ib =
Vbn
220 ∠ - 120°
=
= 10 ∠ - 120°
ZB + 2
22
Ic =
Vcn
220 ∠120°
=
= 16.92 ∠97.38°
ZC + 2
12 + j5
For phase b,
For phase c,
The current in the neutral line is
I n = -(I a + I b + I c )
or
- In = Ia + Ib + Ic
- I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78)
I n = 1.91 − j12.02 = 12.17 ∠ - 81° A
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 12(12).
Solve for the line currents in the Y-∆ circuit of Fig. 12.45. Take Z∆ = 60∠45°Ω.
Figure 12.45
Chapter 12, Solution 12(12).
Convert the delta-load to a Wye-load and apply per-phase analysis.
Ia
110∠0° V
ZY =
Ia =
+
-
ZY
Z∆
= 20 ∠45° Ω
3
110∠0°
= 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
I c = I a ∠120° = 5.5∠75° A
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 22(19).
Find the line currents Ia, Ib, and Ic in the three-phase network of Fig. 12.53 below.
Take Z∆ = 12 - j15Ω, ZY = 4 + j6 Ω, and Zl = 2 Ω.
Figure 12.50
Chapter 12, Solution 22(19).
Convert the D-connected source to a Y-connected source.
Vp
208
Van =
∠ - 30° =
∠ - 30° = 120 ∠ - 30°
3
3
Convert the D-connected load to a Y-connected load.
Z
(4 + j6)(4 − j5)
Z = Z Y || ∆ = (4 + j6) || (4 − j5) =
3
8+ j
Z = 5.723 − j0.2153
ZL
Van
Ia
+
-
Ia =
Z
Van
120 ∠30°
=
= 15.53∠ - 28.4° A
Z L + Z 7.723 − j0.2153
I b = I a ∠ - 120° = 15.53∠ - 148.4° A
I c = I a ∠120° = 15.53∠91.6° A
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 25(21).
In the circuit of Fig. 12.54, if Vab = 440 ∠10°, Vbc = 440∠250°,
Vca = 440 ∠130° V, find the line currents.
Figure 12.51
Chapter 12, Solution 25(21).
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
Ia =
where
440 ∠(10° − 30°)
3 ZY
Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78°
Ia =
440 ∠ - 20°
3 (14.32 ∠ - 24.78°)
= 17.74∠4.78° A
I b = I a ∠ - 120° = 17.74∠ - 115.22° A
I c = I a ∠120° = 17.74∠124.78° A
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 47(39).
The following three parallel-connected three-phase loads are fed by a balanced threephase source.
Load 1: 250 kVA, 0.8 pf lagging
Load 2: 300 kVA, 0.95 pf leading
Load 3: 450 kVA, unity pf
If the line voltage is 13.8 kV, calculate the line current and the power factor of the source.
Assume that the line impedance is zero.
Chapter 12, Solution 47(39).
pf = 0.8 (lagging) 
→ θ = cos -1 (0.8) = 36.87°
S1 = 250 ∠36.87° = 200 + j150 kVA
pf = 0.95 (leading) 
→ θ = cos -1 (0.95) = -18.19°
S 2 = 300 ∠ - 18.19° = 285 − j93.65 kVA
pf = 1.0 
→ θ = cos -1 (1) = 0°
S 3 = 450 kVA
S T = S1 + S 2 + S 3 = 935 + j56.35 = 936.7 ∠3.45° kVA
S T = 3 VL I L
IL =
936.7 × 10 3
3 (13.8 × 10 3 )
= 39.19 A rms
pf = cos θ = cos(3.45°) = 0.9982 (lagging)
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 12, Problem 81(67).
A professional center is supplied by a balanced three-phase source. The center
has four plants, each a balanced three-phase load as follows:
Load 1: 150 kVA at 0.8 pf leading
Load 2: 100 kW at unity pf
Load 3: 200 kVA at 0.6 pf lagging
Load 4: 80 kW and 95 kVAR (inductive)
If the line impedance is 0.02 + j0.05 Ω per phase and the line voltage at the loads
is 480 V, find the magnitude of the line voltage at the source.
Chapter 12, Solution 81(67).
pf = 0.8 (leading) 
→ θ1 = -36.87°
S 1 = 150 ∠ - 36.87° kVA
pf = 1.0 
→ θ 2 = 0°
S 2 = 100 ∠0° kVA
pf = 0.6 (lagging) 
→ θ3 = 53.13°
S 3 = 200∠53.13° kVA
S 4 = 80 + j95 kVA
S = S1 + S 2 + S 3 + S 4
S = 420 + j165 = 451.2∠21.45° kVA
S = 3 VL I L
S
451.2 × 10 3
= 542.7 A
IL =
=
3 VL
3 × 480
For the line,
S L = 3 I 2L Z L = (3)(542.7) 2 (0.02 + j0.05)
S L = 17.67 + j44.18 kVA
At the source,
S T = S + S L = 437.7 + j209.2
S T = 485.1∠25.55° kVA
S
485.1 × 10 3
VT = T =
= 516 V
3 IL
3 × 542.7
Copyright ©2004 The McGraw-Hill Companies Inc.
Chapter 12, Prob. 84. Find the magnitude and phase angle of currents Ia, Ib, Ic and
In if Van = V p ∠0°, VL = Vab = 440 Volts .
n Neutral
∆
Connected
Chapter 12, Solution 84.
We first find the magnitude of the various currents.
S
4000
For the motor: I L =
=
= 5.248 A
3 VL 440 3
Q c 1800
=
= 4.091 A
For the capacitor: I C =
VL
440
P
800
440
For the lighting: Vp =
= 254 V
I Li = Li =
= 3.15 A
Vp 254
3
Consider the figure below.
Ia
a
IC
+
Vab
b
c
n
If Van = Vp ∠0° ,
I1
Ib
-jXC
I2
I3
Ic
ILi
In
Vab = 3 Vp ∠30°
R
Vcn = Vp ∠120°
IC =
Vab
= 4.091∠120°
-j X C
Vab
= 5.249∠(θ + 30°) where θ = cos -1 (0.72) = 43.95°
Z∆
I 1 = 5.249 ∠73.95°
I 2 = 5.249 ∠ - 46.05° I 3 = 5.249∠193.95°
Vcn
= 3.15∠120°
Thus,
I Li =
R
I1 =
I a = I 1 + I C = 5.249∠73.95° + 4.091∠120°
I a = 8.608∠93.96° A
I b = I 2 − I C = 5.249∠ - 46.05° − 4.091∠120°
I b = 9.271∠ - 52.16° A
I c = I 3 + I Li = 5.249∠193.95° + 3.15∠120°
I c = 6.827 ∠167.6° A
I n = - I Li = 3.15∠ - 60° A