Electrical Engineering Multiple Choice and Calculation Questions

Part 1 – Multiple-Choice Questions
For the following questions, select which answer is TRUE.
Question 1 (# marks)
The per unit value of Z33kV line = (1.82 +j 2.61) Ω on a 10 MVA Base is:
a) (0.01 + j0.024) pu
b) (0.02 + j0.03) pu
c) (0.01 + j0.045) pu
d) (0.02 + j0.024) pu
Z pu= actual impedance/ base impedance
Actual impedance = 1.82 + j2.61 ohms
Base impedance = base voltage2 (kV)/ base MVA
= 332 / 10
= 108.9
Z pu= 1.82 + j2.61 / 108.9
= (0.02 + j0.024) pu
Answer d is true
Question 2 (# marks)
A 500 MVA transformer has R = 0.4% and X = 5.9%. Calculate in rectangular form the per unit
values on a 100MVA base:
a) (0.0001 + j0.012)llpu
b) (0.002 + j0.012) pu
c) (0.0008 + j0.012) pu
d) (0.001 - j0.012) pu
Zpu.new = Zpu.old* (base MVAnew )/ (base MVAold )
= (0.004 + j0.059) * 100/500
= (0.008 + j0.012) pu
Answer c is true
Question 3 (# marks)
A 132 kV bus has a balanced 3Φ system fault level of 500MVA at A, adopting a 1000MVA base
calculate the magnitude of IA 3Φ SC fault level (in kA):
a) 1.26 kA
b) 2.19 kA

c) 3.79 kA
d) 1.13 kA
Short circuit current, IASC = (fault MVA * 103)/√3 * base KV A
Substituting for the values in equation above
IASC = (500 * 103)/√3 * 132
= 2186 A
= 2.19 kA
Answer b is true
Question 4 (# marks)
The line voltage applied to the primary side of a 1MVA 3Φ 33/11kV transformer with a 3Φ short
circuit applied to the 11kV side equals 1625 V. The line current for this short circuit test equals 17.25
A. The impedance in per unit values ∣Z1MVA transformer∣ is:
a) 0.45 pu
b) 0.05 pu
c) 0.55 pu
d) 0.60 pu
Per unit impedance of the transformer, Z1MVA tx pu = VZ/Vn
VZ is the voltage that was applied at the LV side
Vn is rated line to line voltage of the transformer
Z1MVA txpu = 1625/33000
= 0.05p.u
Answer d is true
Question 5 (# marks)
An unbalanced 3Φ star-connected load comprises impedances ZA = 1.2/30°Ω, ZB = 1.5/20°Ω, ZC =
1.8/45°Ω. This is connected to a 3Φ supply VA= 230/0° V with a positive phase sequence. Calculate
ZTHEVENIN:
a) 2.02 /30.8 ° Ω
b) 1.01 /30.8° Ω
c) 0.98 /30.8° Ω
d) 0.49 /30.8° Ω
For the star connected loads ZA, ZB and ZC are all in parallel, thus the equivalent impedance of the
loads with the voltage sources shorted is
ZTHEVENIN = (1/1.2 < 30 + 1/1.5<20 + 1/1.8<45)-1

= 0.49<30.80
Ω
Answer d is true
Question 6 (# marks)
In Q5, calculate VON for ZN = ∞ (i.e. open circuit neutral)
a) 27.5 /85.1° V
b) 55 /85.1° V
c) 53.4 /-10.2° V
d) 106.8 /-10.2° V
ION = IA + IB + IC
= (240<0/1.2<30) + (240<240/1.5<20) + (240<120/1.8<45)
= 85.41<28.30 A
= 112.85<-410 A
VON= ION * ZTHEV
= 112.85<-41 * 0.49<30.8
= 53.4<-10.20 V
Answer c is true
Question 7 (# marks)
A two-wattmeter measurement is made of a balanced 3Φ load showing WA = 10.5 kW and WC = 15.9
kW. Calculate Q and pf:
a) Q = 17.41 kVAr, pf = 0.94
b) Q = 5.43 kVAr, pf = 0.89
c) Q = 9.35 kVAr, pf = 0.94
d) Q = 5.43 kVAr, pf = 0.94
For two wattmeter method the following formula is used to calculate the phase angle
Ø = Tan-1 ( √3∗wc−wa
wc +wa )
= Tan-1 ( √3∗15.9−10.5
15.9+10.5 )
= 19.500
Power factor = cos Ø = cos 19.5 = 0.94
Total power = 10.5 + 15.9 = 26.4 kW
From power angle, reactive power, Q = real power*tan Ø
= 26.4 * tan 19.5
= 9.35 kVAr.
Answer c is true

Part 2 – Calculation Questions
Question 8 (4 marks)
For the following unbalanced 3Φ voltages Va= 240/15° V, Vb= 340/-65° V, Vc = 400/105°V
calculate:
(a) the positive sequence component Va1
Figure 1. balanced systems
The theory of symmetrical components solves a problem that occurs when there is unbalanced
conditions. A set on unbalanced currents or voltages can be resolved into sets of symmetrical
balanced phasors components namely: positive, negative and zero sequence. For the positive
sequence, the following formula is used to compute symmetrical components
Va1 = 1/3 (Va + a * Vb + a2 * Vc)
= 1/3 { 240<15 + (1<120 * 340<-65) + (1<240 * 400<105)}
= 282.4 < 16.3 0 V
(b) the negative sequence component Va2
For negative sequence, the following formula is used to compute symmetrical components
Va2 = 1/3 (Va + a2 * Vb + a* Vc)
= 1/3 {240<15 + (1<240 * 340<-65) + (1<120* 400<105)}
= 144.7 < -153.88 0 V
(c) the zerosequence component Va0
For zero sequence, the following formula is used to compute symmetrical components
Va0 = 1/3 (Va + Vb+ Vc)
=1/3 {240<15 + 340<-65 + 400<105}
= 102.02 <27.30 V
(d) show Va1 + Va2 +Va0 = Va
Va1 + Va2 +Va0 = 282.4 < 16.3 + 144.7 < -153.88 + 102.02 < 27.3
= 240 < 150 V