Quiz 1
ECE 476
1. Two balanced three-phase loads are in parallel.
a. Load 1 draws 10 kW at 0.8 PF lagging
b. Load 2 draws 20 kVA at 0.6 PF leading
The loads are supplied by a balanced three-phase 480 VLL source.
(a) Draw the power triangle for the combined load.SL = 23.59
kVA
QL = Q1 + Q2
= -8.5 kVAR
PL = P1 + P2 = 22 kW
21.12°
(b) Determine PF of the combined load.
cos21.12° = 0.933 leading
(c) Determine the magnitude of the line current from the source.
𝐼𝐿 = 𝑆𝐿
√3𝑉𝐿𝐿
= 23.59×103
√3×480 = 28.37 𝐴
(d) Y-connected inductors are now installed in parallel with the combined load. What value of
inductive reactance is needed in each leg of the Y to make the source power factor unity?
𝑄𝐼𝑛𝑑 = |𝑄𝐿| = 8.5 × 103 𝑉𝐴𝑅 = 3(𝑉𝐿𝑁)2
𝑋Y
𝑋Y = 3(480/√3)2
8.5×103 = 27.1 Ω

Quiz 2
ECE 476
Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗
1. A 60-Hz single –phase, two-wire overhead line has solid cylindrical copper conductors with 2.4
cm diameter. The conductors are arranged in a horizontal configuration with 3 m spacing.
Calculate the inductance of each conductor due to both internal and external flux linkages in
mH/km (40 points).
𝐿𝑥 = 𝐿𝑦 = 2 × 10−7𝐿𝑛 (𝐷
𝑟′) 𝐻/𝑚
𝐷 = 3𝑚
𝑟′ = 0.7788 × 𝑟 = 0.7788 × (0.024
2 ) = 9.346 × 10−3
𝐿𝑥 = 𝐿𝑦 = 2 × 10−7𝐿𝑛 ( 3
9.346 × 10−3) 𝐻
𝑚 (1000𝑚
𝑘𝑚 ) (1000𝑚𝐻
𝐻 ) = 1.154 𝑚𝐻/𝑘𝑚
2. The figure below is a completely transposed three-phase overhead transmission line with
bundled phase conductors. All conductors have a radius of 2 cm.0.4m
0.4m
0.4m
0.4m
12m 12m
a. Determine the inductance per phase in mH/km (40 points).
𝑟′ = 0.7788 ∗ 0.02 = 0.0156
𝑅𝑏 = √(𝑟′)(0.4)(0.4)(√2 × 0.4)
4
= 0.1938𝑚
𝐷𝑒𝑞 = √𝐷𝐴𝐵𝐷𝐵𝐶 𝐷𝐶𝐴
3 = √(12)(12)(24)
3 = 15.12𝑚
𝐿 = 0.2𝐿𝑛 (𝐷𝑒𝑞
𝑅𝑏
) = 0.2𝐿𝑛 ( 15.12
0.1938) = 0.8714 𝑚𝐻/𝑘𝑚
b. Find the inductive line reactance per phase in Ω/km at 60 Hz (20 points).
𝑋 = 𝜔𝐿 = 2𝜋60 × 0.8714 × 10−3 = 0.3285 Ω/𝑘𝑚

Quiz 3
ECE 476
Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗
A generator bus (slack bus) supplies a 100 MW, 50 Mvar load through a lossless transmission line with
parameters on 100 MVA base below. Also, there is a 10 Mvar shunt capacitor at the load bus.
X per unit: 0.2 & B per unit: 0.2
1. Determine the bus admittance matrix (Ybus) for the system.
10 Mvar = 0.1 p.u.
Assuming 1 pu bus voltage, 𝑌𝑐𝑎𝑝 = 𝑆
𝑉2 = 0.1
12 = 0.1 𝑝. 𝑢.
𝑌11 = 1
𝑗0.2 + 𝑗 0.2
2 , 𝑌12 = − 1
𝑗0.2, 𝑌21 = − 1
𝑗0.2, 𝑌22 = 1
𝑗0.2 + 𝑗 0.2
2 + 𝑗0.1
𝑌𝑏𝑢𝑠 = [−𝑗4.9 𝑗5
𝑗5 −𝑗4.8]
2. Calculate the voltage magnitude and angle at the load bus using the Newton-Rapson method. Use the
flat-start. Provide the values after 2 iterations. General power balance equations are given below.
𝑃𝑖 = ∑|𝑉𝑖||𝑉𝑘|(𝐺𝑖𝑘𝑐𝑜𝑠𝜃𝑖𝑘 + 𝐵𝑖𝑘𝑠𝑖𝑛𝜃𝑖𝑘) = 𝑃𝐺𝑖 − 𝑃𝐷𝑖
𝑛
𝑘=1
𝑄𝑖 = ∑|𝑉𝑖||𝑉𝑘|(𝐺𝑖𝑘𝑠𝑖𝑛𝜃𝑖𝑘 − 𝐵𝑖𝑘𝑐𝑜𝑠𝜃𝑖𝑘) = 𝑄𝐺𝑖 − 𝑄𝐷𝑖
𝑛
𝑘=1
𝑃2(𝑥) = |𝑉2|(5𝑠𝑖𝑛𝜃2) + 1 = 0
𝑄2(𝑥) = |𝑉2|(−5𝑐𝑜𝑠𝜃2) + |𝑉2|2(4.8) + 0.5 = 0
𝐽(𝑥) = [5|𝑉2|𝑐𝑜𝑠𝜃2 5𝑠𝑖𝑛𝜃2
5|𝑉2|𝑠𝑖𝑛𝜃2 −5𝑐𝑜𝑠𝜃2 + 9.6|𝑉2|]
𝑥(0) = [0
1], 𝑓(𝑥(0)) = [ 1
0.3], 𝐽(𝑥(0)) = [5 0
0 4.6]
𝑥(1) = [0
1] − [5 0
0 4.6]
−1
[ 1
0.3] = [ −0.2
0.9348]
𝑓(𝑥(1)) = [ 0.9348 × 5 sin(−0.2) + 1
0.9348 × −5 cos(−0.2) + 0.93482(4.8) + 0.5] = [0.0714
0.1137]
𝐽(𝑥(1)

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Quiz 4
ECE 476
Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗
1. A 1000-MVA 20-kV, 60-Hz three-phase generator is connected through a 1000-MVA 20-kV
Δ/138-kV Y transformer to a 138-kV circuit breaker and a 138-kV transmission line. The
generator reactances are 𝑋𝑑
′′=0.15, 𝑋𝑑
′ =0.45 and 𝑋𝑑=1.8 pu. The transformer series reactance
is 0.1 pu; transformer losses and exciting current are neglected. A 3-phase short-circuit occurs
on the line side of the circuit breaker when the generator is operated at rated terminal
voltage and at no-load. Determine the substransient current through the breaker in pu and
in kA rms ignoring any dc offset.
A. The base current on the high voltage side of the transformer is:
𝐼𝑏𝑎𝑠𝑒,𝐻𝑉 = 𝑆3∅
√3𝑉𝐿𝐿
= 1000
√3 × 138 = 4.184 𝑘𝑉𝐴
𝐼" = 𝐸𝑞
𝑋𝑑
′′ + 𝑋𝑇𝑅
= 1
0.15 + 0.1 = 4 𝑝. 𝑢. = 4 × 4.184 = 16.735 𝑘𝐴
2. Determine Ybus in per-unit for the circuit below using a 1000 MVA system base. Per unit values
for the generators and transformers are given on their own base so you need to first convert
some of them to the system MVA base.G1 G2
500 MVA, 13.8 kV
X”=0.2 pu
1000 MVA, 20 kV
X”=0.17 pu
500 MVA
13.8 Δ/500 Y kV
X=0.12 pu
1000 MVA
20 Δ/500 Y kV
X=0.10 pu
T1 T2
Δ Y Y Δ500 kV Line
X = 50 Ω
1 2
Hint: 𝑋𝐺1
′′ = 0.2 × 1000
500 = 0.4 𝑝𝑢, 𝑍𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒
2
𝑆𝑏𝑎𝑠𝑒
= 5002
1000 = 250 Ω, 𝑋𝑙𝑖𝑛𝑒 = 50
250 = 0.2 𝑝𝑢
A. 𝑋𝑇1

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Power System Analysis And Design, 6th Edition Test Bank

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