Power systems and transmission: board exam questions with solutions
On the REE exam under Electrical Engineering Professional Subjects. Try each question before you open its solution, then practise with fresh numbers below: 4 computed question types on this topic.
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Questions
1. A transformer has a reactance of 0.10 pu on its own 50 MVA base. What is its reactance on a 100 MVA system base at the same voltage base?
- 0.20 pu
- 0.05 pu
- 0.10 pu
- 0.40 pu
Show the answer and solution
Answer: A. 0.20 pu
Z_new = Z_old × (S_new/S_old) × (V_old/V_new)² = 0.10 × (100/50) × 1 = 0.20 pu.
2. What is the base impedance for a 13.8 kV, 100 MVA three-phase base?
- 0.138 Ω
- 1.904 Ω
- 0.525 Ω
- 5.713 Ω
Show the answer and solution
Answer: B. 1.904 Ω
Z_base = (kV_LL)²/MVA = 13.8²/100 = 1.904 Ω.
3. A 50 MVA, 13.8 kV generator with subtransient reactance 0.15 pu is unloaded at rated voltage when a bolted three-phase fault occurs at its terminals. What is the initial symmetrical fault current?
- 24.2 kA
- 2.09 kA
- 13.9 kA
- 8.05 kA
Show the answer and solution
Answer: C. 13.9 kA
I_base = 50 × 10⁶/(√3 × 13 800) = 2092 A. I_f = 1.0/0.15 = 6.67 pu, so I_f = 6.667 × 2092 = 13946 A ≈ 13.9 kA.
4. The Thevenin impedance at a bus is 0.08 pu on a 100 MVA base. What is the three-phase short-circuit MVA at the bus (prefault voltage 1.0 pu)?
- 8 MVA
- 800 MVA
- 125 MVA
- 1250 MVA
Show the answer and solution
Answer: D. 1250 MVA
Short-circuit MVA = MVA_base/Z_pu = 100/0.08 = 1250 MVA.
5. The receiving-end voltage of a distribution line is 13.8 kV at no load and 13.2 kV at full load, with sending-end voltage held constant. What is the voltage regulation?
- 4.55%
- 4.35%
- 0.60%
- 6.00%
Show the answer and solution
Answer: A. 4.55%
VR = (V_NL − V_FL)/V_FL × 100 = (13.8 − 13.2)/13.2 × 100 = 4.55%.
6. If the spacing between the conductors of a single-phase overhead line is increased, what happens to the line's inductance and capacitance per unit length?
- Both increase
- Inductance increases and capacitance decreases
- Both decrease
- Inductance decreases and capacitance increases
Show the answer and solution
Answer: B. Inductance increases and capacitance decreases
L ∝ ln(D/r′) rises with spacing D, while C ∝ 1/ln(D/r) falls with spacing.
7. In a power-flow (load-flow) study, which two quantities are specified at a generator (PV) bus?
- Voltage magnitude and angle
- Real and reactive power
- Real power and voltage magnitude
- Reactive power and voltage angle
Show the answer and solution
Answer: C. Real power and voltage magnitude
A PV bus fixes P and |V|; Q and δ are computed. The slack bus fixes |V| and δ, and a load (PQ) bus fixes P and Q.
8. A lossless line with series reactance 0.4 pu connects a sending bus at 1.05 pu to a receiving bus at 1.00 pu. What power angle δ is needed to transmit 1.2 pu of real power?
- 28.7°
- 62.8°
- 30.0°
- 27.2°
Show the answer and solution
Answer: D. 27.2°
P = Vs Vr sin δ / X, so sin δ = 1.2 × 0.4/(1.05 × 1.00) = 0.4571 and δ = 27.2°.
Worked examples with computed numbers
These numbers are fixed on this page; in practice they change every time.
9. A 200 kW load runs at 0.75 pf lagging. How much capacitive kVAR raises it to 0.9?
- 45.1 kVAR
- 79.5 kVAR
- 176 kVAR
- 30 kVAR
Show the answer and solution
Answer: B. 79.5 kVAR
Qc = P(tan θ₁ − tan θ₂) = 200(0.882 − 0.484) = 79.5 kVAR.
10. What is the base impedance on a 100 MVA, 138 kV base?
- 190.4 Ω
- 1380 Ω
- 1.38 Ω
- 110 Ω
Show the answer and solution
Answer: A. 190.4 Ω
Zbase = kV²/MVA = 138²/100 = 190.4 Ω.
11. A 500 kVA, 230 V three-phase transformer has 6% impedance. Estimate the three-phase fault current at its secondary terminals with an infinite source.
- 75.3 A
- 36.2 kA
- 20.9 kA
- 1.26 kA
Show the answer and solution
Answer: C. 20.9 kA
Rated I = S/(√3·V) = 1255 A; bolted three-phase fault ≈ I/Z(pu) = 1255/0.06 = 20.9 kA (infinite source).
12. A single-phase load 30 m away draws 20 A through 8 mm² copper (ρ = 0.0172 Ω·mm²/m). What is the voltage drop in the two wires (resistance only)?
- 1.29 V
- 20.6 V
- 2.23 V
- 2.58 V
Show the answer and solution
Answer: D. 2.58 V
Out and back: R = ρ(2L)/A = 0.0172 × 60/8 = 0.129 Ω; Vd = IR = 2.58 V.
Other topics in the same subjects
DC circuits · AC circuits · Three-phase systems · Transients (RC, RL, RLC) · Semiconductors and diodes · Power electronics · Instrumentation and measurement · Control systems · Transformers · DC machines · Induction motors · Synchronous machines · Protection: fuses, breakers, relays · Illumination · Power plant engineering · PEC: branch circuits, ampacity and protection
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