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Ten real questions from our PE Electrical and Computer: Power bank, drawn from all 9 of its study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures9 topicsFull 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:
Figure — drag or pinch to move around
1. 44.9 A
2. 49.3 A
3. 52.2 A
4. 57.3 A
Show the answer & worked solution
Answer: D.57.3 A
Shaft output: $P_{out}=50\times746\text{ W}=37.3$ kW. Electrical input: $P_{in}=P_{out}/\eta=37.3/0.91=40.99$ kW. The handbook's induction-motor power flow (§4.2.5) gives $P_{in}=\sqrt3\,V_TI_L\cos\theta$ , so $I_L=\dfrac{40.99\times10^3}{\sqrt3(480)(0.86)}=57.3$ A. Distractors: 52.2 A leaves out the efficiency, taking the shaft output as the electrical input; 49.3 A leaves out the power factor, treating the input kW as kVA; and 44.9 A divides the shaft output by $\sqrt3\,V_T$ with neither the efficiency nor the power factor.
2
medium
E. Testing
A transformer turns-ratio (TTR) test set is connected to a single-phase transformer rated 13800–480 V. The HV winding has four 2.5% taps, two above and two below rated voltage, and the de-energized tap changer is on the second tap above rated voltage. The set measures a ratio of 30.11. The acceptance criterion (IEEE C57.152) is ±0.5% of the voltage ratio the nameplate lists for the tap in use. The ratio error and the test result are most nearly:
Figure — drag or pinch to move around
1. -0.26% — pass
2. -0.08% — pass
3. +4.73% — fail (outside ±0.5%)
4. +10.24% — fail (outside ±0.5%)
Show the answer & worked solution
Answer: A.-0.26% — pass
The turns ratio is the no-load voltage ratio, $a=N_1/N_2=E_1/E_2$ (§4.3.1.1), so the reference is the nameplate ratio for the tap in service. The tap changer is on the second tap above rated voltage, 2 × 2.5% = 5% above, which sets the HV winding for $1.05\times13800=14490$ V, so $a=14490/480=30.188$ . Error $=\dfrac{30.11-30.188}{30.188}\times100=-0.26\%$ , which is inside the ±0.5% band, so the unit passes; the sign says the measured ratio is low. Distractors: compared with the rated-tap ratio $13800/480=28.75$ , ignoring the tap in service (+4.73%); applied the tap the wrong way, $0.95\times13800$ on the HV side (+10.24%); and reported the raw ratio difference $30.11-30.188$ as if it were already a percent (-0.08%).
3
medium
C. Special occupancies and systems
A hospital's Type 1 essential electrical system has a life safety branch and a critical branch, both restored automatically within 10 seconds of a loss of normal power, and an equipment branch on delayed-automatic or manual connection; loads outside those branches stay on the normal system. Per the NEC, to which portion of the system should the corridor egress illumination and exit signs in an inpatient nursing unit be connected?
Figure — drag or pinch to move around
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
This load belongs on the life safety branch: the life safety branch (517.33) carries only life-safety functions (illumination of the means of egress, exit signs, alarm and alerting systems, emergency communications) and is restored automatically within 10 seconds. Being in a patient-care unit does not move them to the critical branch: the critical branch's lighting is task illumination serving patient care (517.34), while illumination of the means of egress and exit signs are life-safety functions (517.33) wherever the corridor runs. The other choices misplace it: the critical branch (517.34) carries task illumination, fixed equipment, selected receptacles and special power circuits serving patient care, and nurse call, and is restored automatically within 10 seconds; the equipment branch (517.35) carries the listed mechanical loads (medical suction and air, smoke control, kitchen hood systems, sump pumps, ventilation and heating for patient areas) on delayed-automatic or manual connection; the normal system carries every load that no branch list names, and those loads may stay de-energized on loss of normal power. None of those scopes covers this load.
4
medium
B. Surge protection
A metal-oxide surge arrester is being selected for a 13.2 kV (line-to-line), solidly grounded, four-wire multigrounded-neutral 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:
Figure — drag or pinch to move around
1. 7.65 kV
2. 8.4 kV
3. 10.2 kV
4. 15.3 kV
Show the answer & worked solution
Answer: B.8.4 kV
NEC 2020 242.42(A) requires the arrester rating to cover the maximum continuous operating voltage at its point of application, taken phase-to-ground on a solidly grounded system: the arrester is connected line-to-ground, and a ground fault does not hold the healthy phases at line-to-line voltage. That voltage is the line-to-neutral value raised by the continuous allowance, $MCOV \ge 1.05\times13.2/\sqrt3 = 1.05\times7.62 = 8.00$ kV, and the smallest listed MCOV at or above it is 8.4 kV; a lower rating would conduct continuously and overheat. Distractors: dropped the 105% allowance and sized to the nominal $V_{LN} = 7.62$ kV, giving 7.65 kV; used the phase-to-phase basis that applies only to impedance-grounded or ungrounded systems, $1.05\times13.2 = 13.86$ kV, giving 15.3 kV; and applied the 125% factor that 242.42(B) sets for silicon-carbide arresters to this metal-oxide unit, $1.25\times8.00 = 10.00$ kV, giving 10.2 kV.
5
medium
A. Instrument transformers & metering
A balanced three-phase, three-wire load is measured by the two-wattmeter method. The meters read $W_1=12$ kW and $W_2=4$ kW. The load power factor is most nearly:
Figure — drag or pinch to move around
1. 0.536
2. 0.655
3. 0.756
4. 0.894
Show the answer & worked solution
Answer: C.0.756
In the two-wattmeter method on a balanced three-wire load, the sum of the readings is the total real power and $\sqrt3$ times their difference is the total reactive power, so the readings alone fix the power-factor angle. $P=W_1+W_2=16$ kW and $\tan\theta=\sqrt3(W_1-W_2)/(W_1+W_2)=\sqrt3(12-4)/16=0.866\Rightarrow\theta=40.9^\circ$ , so $PF=\cos\theta=0.756$ . The distractors dropped the $\sqrt3$ , $\tan\theta=(W_1-W_2)/(W_1+W_2)$ (0.894); reported $\sin\theta$ , the reactive factor, as the power factor (0.655); and added $P$ and $Q$ arithmetically for the apparent power, $P/(P+Q)$ (0.536).
6
medium
A. Power electronics (converters, inverter-based resources, VFDs)
A three-phase six-pulse diode bridge rectifier is supplied at 480 V line-to-line. With ideal devices and negligible commutation overlap, the average DC output voltage is most nearly:
Figure — drag or pinch to move around
1. 648 V
2. 679 V
3. 917 V
4. 1120 V
Show the answer & worked solution
Answer: A.648 V
The handbook's uncontrolled three-phase full-wave bridge (§3.2.2.1) gives $V_0=\dfrac{3V_{m,L-L}}{\pi}=0.955\,V_{m,L-L}$ , where $V_{m,L-L}$ is the PEAK line-to-line voltage. The 480 V rating is rms, so $V_{m,L-L}=\sqrt2\times480=678.8$ V and $V_0=0.955\times678.8=648$ V (about $1.35\,V_{LL}$ ). Distractors: the peak line-to-line voltage $\sqrt2\,V_{LL}$ instead of the average (679 V); the 1.35 rms factor applied to the peak $\sqrt2\,V_{LL}$ , converting twice (917 V); and the line-to-neutral factor 2.34 applied to the line-to-line voltage (1120 V).
7
medium
A. Overcurrent protection
A 13.8 kV feeder carries a maximum continuous load of 420 A. Its 51 phase time-overcurrent relay is fed from wye-connected 600:5 CTs, and company practice sets the phase pickup at $125\%$ of maximum load. The required relay tap (pickup) setting, in secondary amperes, is most nearly:
Figure — drag or pinch to move around
1. 0.875 A
2. 2.53 A
3. 3.5 A
4. 4.38 A
Show the answer & worked solution
Answer: D.4.38 A
Primary pickup $=420\times1.25=525$ A, set above the maximum load so the relay never operates on load current; reflected through the 600:5 CTs, tap $=525\times\dfrac{5}{600}=4.38$ A secondary. With wye-connected CTs the relay sees the line current one-for-one through the ratio, so no $\sqrt3$ enters. The other options set the tap right at full load, leaving out the margin (3.5 A); divide by 600 instead of 600/5, dropping the 5 A secondary (0.875 A); and divide by $\sqrt3$ as though converting a line current to a phase current (2.53 A).
8
medium
A. Machine types and applications
The 4-pole, 60 Hz induction motor shown runs at a full-load slip of 4% while delivering 150 lb-ft of torque to its load. Its full-load shaft output is most nearly:
Figure — drag or pinch to move around
1. 36.8 hp
2. 49.4 hp
3. 51.4 hp
4. 98.7 hp
Show the answer & worked solution
Answer: B.49.4 hp
Synchronous speed (§4.2.1.2): $n_s=\dfrac{120f}{p}=\dfrac{120(60)}{4}=1800$ rpm. Shaft speed from the slip (§4.2.1.3): $n=n_s(1-s)=1800(1-0.04)=1728$ rpm. The handbook's power relation (§4.2.1.1) is in SI units, so convert the torque with the §1.2 table: $T=150\times1.356=203.4$ N·m, then $P=\dfrac{Tn}{9549}=\dfrac{203.4(1728)}{9549}=36.81$ kW, and $36.81\times1.341=49.4$ hp. (The US-customary shortcut $\text{hp}=T\,n/5252$ , with $T$ in lb-ft, gives the same result.) Distractors: 51.4 hp uses the 1800 rpm synchronous speed instead of the 1728 rpm shaft speed; 36.8 hp is the correct 36.8 kW reported as hp, without the kW-to-hp conversion; and 98.7 hp puts the 2 pole pairs in place of the 4 poles in $120f/p$ , doubling the speed.
9
medium
A. Voltage drop
A three-phase, 480 V feeder runs 320 ft one way to a balanced load drawing 290 A at 0.95 lagging power factor. Each phase conductor has $R=0.0258\ \Omega$ /1,000 ft and $X=0.039\ \Omega$ /1,000 ft. By the approximate method, the voltage drop as a percent of the nominal system voltage is most nearly:
Figure — drag or pinch to move around
1. 0.71%
2. 0.86%
3. 1.23%
4. 1.42%
Show the answer & worked solution
Answer: C.1.23%
Use the handbook's approximate line-to-neutral drop for a balanced three-phase load (§5.1.1), $VD_{ln}\approx I\left[R\cdot PF+X\sin(\cos^{-1}PF)\right]$ , with $R$ and $X$ the per-phase ohms of the 320 ft run: $R=0.0258(320/1{,}000)=0.008256\ \Omega$ and $X=0.039(320/1{,}000)=0.01248\ \Omega$ . With $\sin(\cos^{-1}0.95)=0.312$ , $VD_{ln}=290(0.007843+0.003897)=290(0.01174)=3.40$ V. The line-to-line drop is $\sqrt3\times3.40=5.90$ V, so $\%VD=5.90/480\times100=1.23\%$ of the nominal voltage. The other options come from the line-to-neutral drop divided by the line-to-line voltage; the unity-PF form $2LRI/(K\cdot1{,}000)$ with $K=2/\sqrt3$ , which ignores the power factor and the reactance; and the single-phase factor 2 in place of $\sqrt3$ .
10
medium
B. Symmetrical components
During a ground fault, the phase-to-neutral voltages measured at a bus are $V_a = 35\angle{0}^\circ$ V, $V_b = 105\angle{-128}^\circ$ V, and $V_c = 125\angle{120}^\circ$ V, as shown in the phasor diagram. The zero-sequence voltage $V_{a0}$ is most nearly:
Figure — drag or pinch to move around
1. $22.6\angle{-170.7}^\circ$ V
2. $31.9\angle{164.5}^\circ$ V
3. $88.1\angle{-3.2}^\circ$ V
4. $95.6\angle{164.5}^\circ$ V
Show the answer & worked solution
Answer: B. $31.9\angle{164.5}^\circ$ V
The zero-sequence voltage is the first row of the handbook's symmetrical-component transformation (§3.1.2): $V_{a0}=\tfrac13(V_a+V_b+V_c)$ , a plain phasor sum with no $a$ -operator rotations. In rectangular form $V_a=35+j0$ , $V_b=-64.6-j82.7$ and $V_c=-62.5+j108.3$ V, so $V_a+V_b+V_c=-92.1+j25.5$ V $=$ $95.6\angle{164.5}^\circ$ V, and $V_{a0}=$ $31.9\angle{164.5}^\circ$ V. Distractors: $95.6\angle{164.5}^\circ$ V is the phasor sum itself, $3V_{a0}$ (the residual voltage), with the factor $\tfrac13$ left off; $22.6\angle{-170.7}^\circ$ V is $V_{a2}$ : it uses the third (negative-sequence) row, $\tfrac13(V_a+a^2V_b+aV_c)$ ; and $88.1\angle{-3.2}^\circ$ V is $V_{a1}$ : it uses the second (positive-sequence) row, $\tfrac13(V_a+aV_b+a^2V_c)$ .

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