Examples Chapter #12, EE 223, 2003: 20, 25, 28
Chapter 12, Problem 20.
Each load in the circuit of Fig. 12.31 is composed of a
1.5-H inductor in parallel with a 100-µF capacitor and
a 1-kΩ resistor. The resistance labeled Rw = 0 Ω.
V
Using positive phase sequence with Vab = 115
at f = 60 Hz, determine the rms line current and the
total power delivered to the load. Verify your answers
with an appropriate PSpice simulation.
Working from the single-phase equivalent,
1  115∠0 o 
 ∠-30o = 46.9 ∠-30o V rms

Van rms =
3
2 
1.5 H → j565 Ω, 100 µF → -j26.5 Ω and 1 kΩ → 1 kΩ.
These three impedances appear in parallel, with a combined value of 27.8 ∠ -88.4o Ω.
Thus, |Irms| = 46.9/ 27.8 = 1.69 A rms
Zload = 27.8 ∠–88.4o = 0.776 – j 27.8 Ω, so Pload = (3)(1.69)2 (0.776) = 3* 2.22 W = 6.66W.
Chapter 12, Problem 25.
The source in Fig. 12.33 is balanced and exhibits
(+)phase sequence. Find
(a) IaA;(b) IbB ;(c) IcC ;(d) the total complex power
supplied by the source.
(a)
120 3 ∠30°
= 20.78∠30° A
10
120 3 ∠ − 90°
120 3 ∠150°
=
= −41.57 A; ICA =
= 20.78∠ − 120° A
− j10
j5
Van = 120∠0° ∴ Vab = 120 3 ∠30°, etc., IAB =
IBC
IaA = IAB − ICA = 20.78(1∠30° − 1∠ − 120°) = 40.15∠45° A rms
(b)
IbB = −41.57 − 20.78∠30° = 60.47∠ − 170.10° A rms
(c)
I cC = 20.78∠ − 120° + 41.57 = 36.00∠ − 30° A rms
(d)
∗
∗
∗
Stot = VAB IAB
+ VBC IBC
+ VCA ICA
= 120 3 ∠30°× 20.78∠ − 30° + 120 3 ∠ − 90°(−41.57) +
120 3 ∠150°× 20.78∠120° = 4320 + j 0 + 0 + j8640 + 0 − j 4320 = 4320 + j 4320 VA
Chapter 12, Problem 28.
For the three-phase system of Fig. 12.34, assume a
balanced source with a positive phase sequence. If
the operating frequency is 60 Hz, find the
magnitude of (a) VAN ;(b) VBN ;(c) VCN . Verify
your answers with an appropriate PSpice
simulation.
15 mH → j5.65 Ω,
0.25 mF → -j10.6 Ω
VAB = 120 3 ∠30 o V
VBC = 120 3 ∠ − 90 o V
o
VCA = 120 3 ∠ − 210 V
The unbalanced load makes it necessary to first determine the actual phase voltages using nodal or mesh analysis.
Defining three clockwise mesh currents I1, I2 and I3 corresponding to sources VAB, VBC and VCA, respectively, we
may write:
VAB = (10 + j5.65) I1 – 10 I2 – j5.65 I3
VBC = -10 I1 + (10 – j10.6) I2 + j10.6 I3
VCA = - j5.65 I1 + j10.6 I2 + (j5.65 – j10.6) I3
[1]
[2]
[3]
Note: This linear system has an eigenvalue at zero resulting in a singular matrix A and the solution I = A-1V
depends on the algorithm used as well as your CPU. To ensure correct (numerical) results for the currents:
Replace one of the equations with 0 = I1 + I2 + I3. This avoids a singular A-matrix and allows the computation of
its inverse. Solving using MATLAB or a scientific calculator, we find that I1 = 24.11 ∠-20.20o A,
I2 = 13.59 ∠55.49o A, and I3 = 30.46 ∠-174.58o. The solution for the currents results in the following voltages:
(a) VAN = j5.65(I1 – I3) = 300.7 ∠ 84.13o V,
(b) VBN = 10(I2 – I1) = 245.7 ∠ 127.4o V,
(c) VCN = -j10.6 (I3-I2) = 429.8 ∠ 110.3o V,
so VAN = 300.7 V
so VBN = 245.7 V
so VCN = 429.8 V
Another possibility: Set one of the currents equal to zero and use two equations to find the remaining two currents.
PSpice Simulation Results
(agree with hand calculations)
FREQ
VM(A,N) VP(A,N)
6.000E+01 3.007E+02 8.410E+01
FREQ
VM(B,N) VP(B,N)
6.000E+01 2.456E+02 1.274E+02
FREQ
VM(C,N) VP(C,N)
6.000E+01 4.297E+02 1.103E+02