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Ten real questions from our PE Electrical and Computer: Power bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures9 knowledge areasFull worked solutions

1
medium
A. Three-phase circuits
A three-phase induction motor delivers 50 hp at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The motor is supplied at 480 V (line-to-line). The full-load line current is most nearly:
1. 57.3 A
2. 99.3 A
3. 52.2 A
4. 49.3 A
Show the answer & worked solution
Answer: A.57.3 A
$P_{out}=50\times746=37.3$ kW; $P_{in}=P_{out}/\eta=37.3/0.91=40.99$ kW; $S=P_{in}/pf=47.66$ kVA; $I=S/(\sqrt3\,V_{LL})=47.66\times10^3/(\sqrt3\times480)=57.3$ A. Distractors forget the efficiency (use shaft output as electrical input), forget the power factor (treat $P_{in}$ as kVA), and drop the $\sqrt3$ .
2
medium
E. Testing
A transformer turns-ratio (TTR) test set is connected to a single-phase transformer rated 13800–480 V on the nominal tap, so the nameplate voltage ratio equals the turns ratio. The set measures a ratio of 28.95. The ratio error relative to nameplate is most nearly — and does the transformer pass the ±0.5% acceptance criterion?
1. -0.7% — fail (outside ±0.5%)
2. +0.7% — fail (outside ±0.5%)
3. +0.2% — pass
4. +0.7% — pass
Show the answer & worked solution
Answer: B.+0.7% — fail (outside ±0.5%)
Nameplate ratio $=13800/480=28.75$ . Error $=\dfrac{28.95-28.75}{28.75}\times100=+0.7\%$ , which is outside the ±0.5% band, so the unit fails. Distractors keep the right number but flip the pass-band verdict, reverse the sign (calling a high ratio low), and report the raw ratio difference +0.2 as if it were already a percent.
3
medium
C. Special occupancies and systems
A hospital's Type 1 essential electrical system is divided per the governing code (rules given): the life safety branch carries only the functions essential for life safety (egress lighting, exit signs, alarm and emergency communication systems); the critical branch carries task illumination, fixed equipment, and selected receptacles serving patient care — both branches must be restored automatically within 10 seconds. The equipment branch carries major mechanical loads (medical air/vacuum, HVAC, pumps) on delayed-automatic or manual connection. To which portion of the system should the corridor egress illumination and exit signs along the means of egress be connected?
1. Nonessential (normal system) loads
2. Critical branch
3. Equipment branch
4. Life safety branch
Show the answer & worked solution
Answer: D.Life safety branch
The described load falls squarely in the life safety branch scope: functions essential for life safety — egress illumination, exit signs, alarm and emergency communication — restored automatically within 10 seconds. The distractors assign it to the wrong portion of the system: the critical branch serves task illumination, fixed equipment, and selected receptacles serving patient care, restored automatically within 10 seconds; the equipment branch serves major mechanical/three-phase loads (medical air and vacuum, HVAC, pumps) on delayed-automatic or manual connection; the nonessential (normal system) loads serves ordinary building loads that are permitted to remain de-energized on loss of normal power — none of which matches this load's function.
4
medium
B. Surge protection
A metal-oxide surge arrester is being selected for a 12.47 kV (line-to-line), effectively grounded distribution system that may operate continuously at 105% of nominal voltage. Standard arrester MCOV values are 2.55, 5.1, 7.65, 8.4, 10.2, 12.7, 15.3, 19.5, 22, and 24.4 kV. The minimum standard MCOV suitable for this application is:
1. 8.4 kV
2. 7.65 kV
3. 15.3 kV
4. 5.1 kV
Show the answer & worked solution
Answer: B.7.65 kV
Required $MCOV \ge 1.05\times12.47/\sqrt3 = 7.56$ kV; the smallest standard value at or above this is 7.65 kV. Distractors round DOWN to 5.1 kV (the arrester would conduct continuously and thermally run away), over-select 8.4 kV (raising the discharge voltage and eroding the protective margin for no reason), and size from the full line-to-line voltage instead of line-to-ground, forcing 15.3 kV.
5
medium
B. Insulation testing
An insulation-resistance test on a motor winding is recorded at three points: 160 megohm at 30 seconds, 200 megohm at 1 minute and 500 megohm at 10 minutes. Compute the Polarization Index (PI) and classify the insulation.
1. PI = 1.25 — questionable
2. PI = 2.5 — questionable
3. PI = 2.5 — good
4. PI = 0.4 — investigate
Show the answer & worked solution
Answer: C.PI = 2.5 — good
$PI=R_{10}/R_{1}=500/200=2.5$ , which is good per IEEE 43. Distractors invert the ratio; compute the dielectric absorption ratio $DAR=R_{1\,min}/R_{30\,s}=200/160=1.25$ instead — DAR is a different index taken off the early capacitive part of the curve, not the 1-to-10-minute polarization tail the PI measures; and misread the acceptance band.
6
medium
B. Relays, switches, Boolean and ladder logic
The figure shows a three-wire motor-control circuit: an NC Stop pushbutton and an NO Start pushbutton in series with the NC overload (OL) contact and the M contactor coil, with an M auxiliary contact connected in parallel with the Start pushbutton (seal-in). The motor is initially off and the Start pushbutton is pressed and released. What does the M contactor do?
1. M does not pick up until the Stop pushbutton is also pressed
2. M cannot pick up at all, because the M auxiliary contact is open until the coil energizes
3. M picks up when Start is pressed and remains energized after release — the M auxiliary contact seals in around the Start pushbutton
4. M is energized only while Start is held and drops out as soon as it is released
Show the answer & worked solution
Answer: C.M picks up when Start is pressed and remains energized after release — the M auxiliary contact seals in around the Start pushbutton
At rest the NC Stop and NC OL contacts are closed. Pressing Start completes the rung, M energizes, and the M auxiliary (NO) contact closes in parallel with Start; on release, coil current flows through that seal-in path, so M stays picked up. The first distractor ignores the seal-in branch (two-wire control behavior); the second wrongly places the M auxiliary contact in series with Start instead of in parallel; the third misreads the NC Stop as a contact that must be actuated to close.
7
medium
B. Protective relaying (differential, distance, undervoltage, pilot)
A line distance (21) relay protects a line whose positive-sequence impedance is $24\ \Omega$ primary. The relay is supplied by 1200:5 CTs and 14400:120 V VTs. With Zone 1 set to 80% of the line, the Zone 1 reach in secondary (relay) ohms is most nearly:
1. 38.4 ohm secondary
2. 9.6 ohm secondary
3. 19.2 ohm secondary
4. 48 ohm secondary
Show the answer & worked solution
Answer: A.38.4 ohm secondary
CTR $=240$ , VTR $=14400/120=120$ , so $Z_{sec}=Z_{pri}\times\dfrac{CTR}{VTR}=24\times2=48\ \Omega$ . Zone 1 $=0.8\times48=38.4\ \Omega$ secondary — deliberately short of the remote bus so measurement errors cannot cause an instantaneous overreach trip for the next line's faults. Distractors invert the CTR/VTR factor, forget the 80% margin (setting Zone 1 to the full line), and leave the reach in primary ohms.
8
medium
C. Power factor correction
A plant load draws 600 kW at 0.72 lagging power factor. A 300 kvar capacitor bank (standard-size units) is switched onto the plant bus. The resulting power factor is most nearly:
1. 0.949
2. 0.907
3. 0.894
4. 0.564
Show the answer & worked solution
Answer: B.0.907
$Q_1=P\tan(\cos^{-1}0.72)=600\times0.9639=578.3$ kvar. Remaining $Q=578.3-300=278.3$ kvar, so $PF_{new}=\cos\left(\tan^{-1}\dfrac{278.3}{600}\right)=0.907$ . Distractors add the capacitor kvar instead of subtracting (sign error), treat the bank rating itself as the remaining reactive power, and divide the remaining kvar by the kVA instead of the kW.
9
medium
A. Machine types and applications
The 4-pole, 60 Hz induction motor shown delivers 150 lb-ft of torque at its full-load speed of 1746 rpm. Its shaft output is most nearly:
1. 451 hp
2. 51.4 hp
3. 49.9 hp
4. 7.94 hp
Show the answer & worked solution
Answer: C.49.9 hp
$hp=\dfrac{TN}{5252}=\dfrac{150\times1746}{5252}=49.9$ hp. Distractors divide by 33000 without the $2\pi$ (the 5252 already folds in $2\pi/33000$ ), invert the relation as $5252T/N$ , and use the 1800 rpm synchronous speed instead of the actual shaft speed.
10
medium
B. Symmetrical components
The phase voltages on an unbalanced feeder are measured as $V_a = 120\angle 0^\circ$ V, $V_b = 90\angle -120^\circ$ V, and $V_c = 60\angle 120^\circ$ V. The zero-sequence component $V_0$ is most nearly:
1. $90\angle 0^\circ$ V
2. $17.3\angle -30^\circ$ V
3. $52\angle -30^\circ$ V
4. $17.3\angle 30^\circ$ V
Show the answer & worked solution
Answer: B. $17.3\angle -30^\circ$ V
$V_0=\tfrac13(V_a+V_b+V_c)$ . The phasor sum is $52\angle -30^\circ$ V, so $V_0=$ $17.3\angle -30^\circ$ V. Distractors: $52\angle -30^\circ$ V forgets the factor 1/3; $90\angle 0^\circ$ V averages the magnitudes and ignores the angles; $17.3\angle 30^\circ$ V applies $a$ -operator rotations (a negative-sequence calculation) that do not belong in $V_0$ .

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