Power Systems carries 8–12 items, the largest area after Circuit Analysis and Mathematics, tied with Digital Systems. It starts where AC circuit power leaves off and adds three-phase connections, lines, transformers and rotating machines.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Electrical and Computer
items · Power Systems8–12
problems on this page16
Checked
What the area covers
Four subtopics in the July 2020 specification:
Power theory: power factor, single and three phase, voltage regulation.
Transmission and distribution: real and reactive losses, efficiency, voltage drop, delta and wye connections.
Transformers: single-phase and three-phase connections, reflected impedance.
Motors and generators: synchronous, induction and dc machines.
16 power systems problems
0 / 16 answered · 0 correct
Question 1 of 16 · Power Systems · 8–12 items
A 4-pole, three-phase induction motor runs at 1750 rpm on a 60 Hz supply. What is its slip?
This divides the slip speed by the rotor speed (50/1750); slip is referred to synchronous speed.
This uses ns = 3600 rpm, the synchronous speed of a 2-pole machine.
This is n/ns, the per-unit speed, which equals 1 − s.
Question 2 of 16 · Power Systems · 8–12 items
A transformer has 500 turns on the primary and 50 turns on the secondary. A 4 Ω load is connected to the secondary. What impedance is seen at the primary?
Given
Primary turns Np = 500
Secondary turns Ns = 50
Secondary load Zs = 4 Ω
Find
Impedance referred to the primary (Ω)
Solution
Zp = (Np/Ns)2 · Zs
Zp = (500/50)2 × 4 Ω = 400 Ω
This squares the ratio upside down, (50/500)2 × 4; the high-turns side sees the larger impedance.
This uses the inverted ratio 50/500 without squaring it.
This scales by the turns ratio once; impedance scales with its square.
The no-load voltage at the receiving end of a transmission line is 13.2 kV. Under full load, the voltage drops to 12.6 kV. What is the voltage regulation?
Given
No-load voltage Vnl = 13.2 kV
Full-load voltage Vfl = 12.6 kV
Find
Voltage regulation (%)
Solution
VR = (Vnl − Vfl)/Vfl × 100 %
VR = (13.2 − 12.6)/12.6 × 100 % = 4.76 %
This divides by the no-load voltage; regulation is referred to the full-load voltage.
This is Vfl/Vnl, a voltage ratio, not the regulation.
This is Vnl/Vfl; regulation subtracts 1 (100 %) from that ratio.
Question 4 of 16 · Power Systems · 8–12 items
A balanced three-phase delta-connected load draws 24 kW at 480 V line-to-line with a power factor of 0.8. What is the phase current in each delta branch?
Given
Real power P = 24 kW
Line-to-line voltage VLL = 480 V
Power factor pf = 0.8
Find
Phase current in each delta branch (A)
Solution
Pph = 24 kW/3 = 8 kW, Vph = VLL = 480 V (delta)
Iph = Pph/(Vph · pf) = 8000/(480 × 0.8) = 20.8 A
This puts the whole 24 kW through one branch: 24,000/(480 × 0.8).
This is the line current, P/(√3VLLpf); in a delta each branch carries IL/√3.
Each branch takes 8 kW at 480 V: Iph = 8000/(480 × 0.8) = 20.8 A.
This divides the line current by 3; delta phase current is the line current divided by √3.
Question 5 of 16 · Power Systems · 8–12 items
A dc shunt motor with armature resistance 0.5 Ω draws 40 A from a 240 V supply. The field current is 2 A. What is the back EMF?
Given
Armature resistance Ra = 0.5 Ω
Line current I = 40 A
Supply voltage V = 240 V
Field current If = 2 A
Find
Back EMF Eb (V)
Solution
Ia = IL − If = 40 − 2 = 38 A
Eb = V − IaRa = 240 − 38 × 0.5 = 221 V
Ia = 40 − 2 = 38 A; Eb = V − IaRa = 240 − 38 × 0.5 = 221 V.
This is the terminal voltage; the armature resistance drop always lowers the back EMF.
This uses the generator relation V + IaRa; in a motor the back EMF is below the terminal voltage.
This uses the generator relation with the full 40 A line current; a motor subtracts the armature drop, and 2 A of the 40 A feeds the field.
Question 6 of 16 · Power Systems · 8–12 items
A three-phase, 6-pole induction motor operates at 1164 rpm on a 60 Hz supply. What is the frequency of the rotor currents?
This is the stator supply frequency, which the rotor sees only at standstill.
This is (1 − s)f, the electrical equivalent of the rotor speed; rotor currents are at the slip frequency sf.
This is the rotor speed in revolutions per second (1164/60), not the frequency of the rotor currents.
ns = 120 × 60/6 = 1200 rpm; s = (1200 − 1164)/1200 = 0.03; fr = sf = 0.03 × 60 = 1.8 Hz.
Question 7 of 16 · Power Systems · 8–12 items
A balanced wye-connected three-phase generator has a phase voltage of 7.2 kV. If it delivers 15 MW at 0.85 power factor lagging, what is the line current?
Given
Phase voltage Vph = 7.2 kV
Real power P = 15 MW
Power factor pf = 0.85 lagging
Find
Line current IL (A)
Solution
P = 3VphIph·pf, IL = Iph (wye)
IL = 15 × 106/(3 × 7200 × 0.85) = 817 A
This divides the correct current by √3 once more (817/√3 = 472 A), as if converting line to phase current in a delta.
This leaves out the 0.85 power factor: P/(3Vph).
I = P/(3Vph·pf) = 15 × 106/(3 × 7200 × 0.85) = 817 A; in a wye the line current equals the phase current.
This puts the phase voltage into the √3 formula, P/(√3Vphpf); √3 goes with the line-to-line voltage.
Question 8 of 16 · Power Systems · 8–12 items
A three-phase transformer bank consists of three single-phase transformers, each rated 7200:240 V. The primaries are connected in delta and the secondaries are connected in wye. What is the line-to-line voltage on the secondary side?
Given
Transformer rating = 7200:240 V each
Connection: delta primary, wye secondary
Find
Secondary line-to-line voltage (V)
Solution
Vph,sec = 240 V (winding voltage)
VLL,sec = √3 × 240 V = 415.7 V
This adds three windings in series (3 × 240); a wye puts two phases between lines, displaced by 120°.
Each secondary winding gives 240 V phase-to-neutral; in wye, VLL = √3 × 240 = 415.7 V.
This is the voltage across one secondary winding, the phase voltage of the wye.
This divides by √3, the delta-to-wye direction applied backwards.
Question 9 of 16 · Power Systems · 8–12 items
Two 230 kV buses are joined by a lossless line with a series reactance of 120.0 Ω per phase, and both voltages are held at 230 kV line-to-line. What are the maximum three-phase real power the line can carry and the power at a 30° power angle?
Given
Bus voltages VS = VR = 230 kV line-to-line
Series reactance X = 120.0 Ω per phase (lossless)
Power angle δ = 30°
Find
Three-phase Pmax and P at δ = 30° (MW)
Solution
P = (VSVR/X) sin δ (VLL values give the three-phase total)
This multiplies VLL2/X by 3; the factor 3 belongs with the phase voltage, not the line voltage.
This uses the phase voltage 132.8 kV in V2/X without multiplying by 3; with line-to-line voltages the formula gives the three-phase total directly.
This uses cos 30° instead of sin 30°; real power follows sin δ.
Pmax = VSVR/X = (230 kV)2/120 Ω = 441 MW; P = Pmax sin δ = 441 × sin 30° = 220 MW.
Question 10 of 16 · Power Systems · 8–12 items
A three-phase transformer rated 5 MVA, 13.8 kV/480 V, has a 5.0 % impedance. Assuming an infinite source and a bolted three-phase fault at the 480 V terminals, what is the secondary short-circuit current?
A single-phase load draws 7.50 kW at 0.70 power factor lagging from a 240 V, 60 Hz supply. What capacitance connected in parallel raises the power factor to 0.95 lagging?
Given
Load real power P = 7.50 kW
Initial power factor pf1 = 0.70 lagging
Target power factor pf2 = 0.95 lagging
Supply voltage V = 240 V
Frequency f = 60 Hz
Find
Parallel capacitance C (µF)
Solution
QC = P(tan θ1 − tan θ2) = 7500 × (1.0202 − 0.3287) = 5186 var
C = QC/(ωV2) = 5186/(2π × 60 × 2402) = 238.8 µF
This sizes the capacitor from the drop in apparent power (kVA), not in reactive power.
QC = 7500(tan θ1 − tan θ2) = 5186 var; C = QC/(ωV2) = 5186/(377 × 2402) = 238.8 µF.
This cancels all of the original 7651 var, correcting to unity instead of 0.95.
This uses f = 60 Hz in place of ω = 377 rad/s in QC = ωCV2.
Question 12 of 16 · Power Systems · 8–12 items
A single-phase AC motor delivers 20 hp from a 240 V rms supply with 80 % efficiency and a lagging power factor of 0.85. Taking 1 hp = 745.7 W, what rms current does it draw?
Given
Output power Pout = 20 hp
Supply voltage V = 240 V rms
Efficiency η = 80%
Power factor pf = 0.85 lagging
Conversion 1 hp = 745.7 W
Find
Supply current I (A)
Solution
Pin = Pout/η = (20 × 745.7 W)/0.80 = 18,643 W
I = Pin/(V · pf) = 18,643/(240 × 0.85) = 91.4 A
Pin = 20 × 745.7/0.80 = 18,643 W; I = Pin/(V·pf) = 18,643/(240 × 0.85) = 91.4 A.
This leaves out the power factor: I = Pin/V.
This uses the 14.9 kW output as the input power and ignores the 80 % efficiency.
This multiplies the output by the efficiency instead of dividing; the input power is larger than the output.
Question 13 of 16 · Power Systems · 8–12 items
An industrial facility is billed based on peak monthly kVA demand and total kWh consumption. The plant currently operates at a peak load of 500 kW with a power factor of 0.75. The facility manager proposes installing capacitor banks to improve the power factor to 0.95. Assuming the real power demand (kW) and hours of operation remain identical, which of the following describes the most likely impact on the utility bill?
Solution
S1 = 500/0.75 = 667 kVA → S2 = 500/0.95 = 526 kVA
kW × hours unchanged → kWh charge unchanged.
Real power stays at 500 kW; the demand charge falls because apparent power drops.
kWh depends on real power and hours, which stay the same; only the kVA-based charge falls.
This reverses the effect: the capacitors cut kVA, not kWh.
A 21 % cut in peak kVA is a significant change to a kVA-billed demand charge.
Peak kVA falls from 500/0.75 = 667 kVA to 500/0.95 = 526 kVA, while the kW and hours, and so the kWh, are unchanged.
Question 14 of 16 · Power Systems · 8–12 items
A single-phase transformer rated at 25 kVA has an efficiency of 96% at full load with unity power factor. What are the total losses at full load?
Given
Rating S = 25 kVA
Full-load efficiency η = 96 %
Power factor = 1.0
Find
Total full-load losses (kW)
Solution
Pout = S · pf = 25 kW, Pin = Pout/η = 26.04 kW
Ploss = Pin − Pout = 1.04 kW
This is the input power; the losses are the input minus the output.
This is 0.96 × 25, as if the rating were the input; it is an output, not a loss.
Pout = 25 kW at unity pf; Pin = 25/0.96 = 26.04 kW, so the losses are 26.04 − 25 = 1.04 kW.
This takes 4 % of the output; losses are 4 % of the input, which is larger than the output.
Question 15 of 16 · Power Systems · 8–12 items
A balanced three-phase motor on a 480 V line-to-line source draws 25 A line current at a power factor of 0.85. Its overall efficiency is 90 %, and it runs at 1750 rpm. What is the shaft output torque?
This uses the electrical input power and ignores the 90 % efficiency.
This divides by the efficiency instead of multiplying, so the output exceeds the input.
Question 16 of 16 · Power Systems · 8–12 items
A single-phase transformer has a turns ratio of 10:1 (primary to secondary). If the primary voltage is 2400 V, what is the secondary voltage under no-load conditions?
Given
Turns ratio = 10:1
Primary voltage Vp = 2400 V
Find
Secondary voltage Vs (V)
Solution
Vs = Vp(Ns/Np) = 2400 V/10 = 240 V
This divides by 100, applying the 10:1 ratio twice.
Vs = Vp × Ns/Np = 2400/10 = 240 V.
This is the primary voltage; a 10:1 transformer steps it down.
This multiplies by the ratio, as if the transformer stepped up.
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Three-phase in a few lines
For a balanced load, total power uses line quantities:
P = √3·VL·IL·cos θ S = √3·VL·IL
Wye: VL = √3·Vφ, IL = Iφ.
Delta: VL = Vφ, IL = √3·Iφ.
Balanced problems are easiest on a per-phase basis: convert to an equivalent wye, solve one phase, multiply power by three. Machines add a few more relations, starting with synchronous speed ns = 120f/p in rpm and slip s = (ns − n)/ns; the Handbook keeps them in the AC Machines section of its Electrical and Computer Engineering chapter (NCEES errata notice).
Common traps in power systems
Line or phase. The √3 sits on voltage in a wye and on current in a delta. Write down which quantity the stem gives before using it.
kW, kVAR, kVA. Real, reactive and apparent power form a right triangle; a power factor turns one into another. Transformer and generator ratings are in kVA.
Correction in kVAR first. A power-factor correction problem is solved as reactive power to cancel; capacitance comes after, from the voltage and frequency.
Poles, not pole pairs. In ns = 120f/p, p counts poles.
Which side of the transformer. An impedance moved across a transformer scales with the square of the turns ratio, and the direction matters.
Voltage regulation. It is measured against the full-load voltage: (VNL − VFL)/VFL.
Per-unit bases. Change the base before adding per-unit impedances from different nameplates.
Sources: Chapman, Electric Machinery Fundamentals; Glover, Overbye and Sarma, Power System Analysis and Design.
How it connects to other areas
Single-phase complex power is set up in Circuit Analysis; rectifiers, inverters and drives are Electronics. In this hub, three-phase and transformer questions are owned here. Electrical safety and grounding are part of Ethics and Professional Practice, on the FE practice test.