Free PE Electrical and Computer: Power Practice Questions

Sample Practice Questions

Try these 10 sample questions from the PE Electrical and Computer: Power practice bank. Each includes the answer and explanation. Continue with the interactive practice above to review the full question set.

1. 1An engineer adds a second high-impedance meter in parallel with an existing meter on a potential-transformer secondary. With the secondary voltage unchanged, how does this addition affect the total secondary burden?

   1. A. Current demand increases and equivalent load impedance decreases.
   2. B. Current demand decreases and equivalent load impedance increases.
   3. C. Current demand increases and equivalent load impedance increases.
   4. D. Current demand decreases and equivalent load impedance decreases.

   Answer: A

   Explanation: FIST states, "As devices are added in parallel to the secondary circuit, the burden (current requirement) increases (the load impedance decreases)". The added meter creates another current path at the same voltage.

   Reference: FIST 3-8: Operation, Maintenance, and Field Test Procedures for Protective Relays and Associated Circuits — Section 5.1, p. 15

2. 2An engineer reviews a building lightning-protection system. Which component carries intercepted lightning current from the roof network toward the earth-electrode system?

   1. A. Air terminal
   2. B. Down conductor
   3. C. Surge arrester
   4. D. Ground plate

   Answer: B

   Explanation: The FAA definition states, "The down conductor serves as the path to the EES from the roof system of air terminals and roof conductors or from an overhead ground wire." The other components provide interception, surge limitation, or earth connection rather than this roof-to-earth path.

   Reference: FAA-STD-019f: Lightning and Surge Protection, Grounding, Bonding, and Shielding Requirements for Facilities and Electronic Equipment — Section 3, Conductor, Down definition, p. 10

3. 3A workplace enclosure is isolated from its metal supply raceway by listed nonmetallic fittings to reduce grounding-circuit interference. Under 29 CFR 1910.305, what additional grounding provision is required?

   1. A. An internal insulated equipment grounding conductor to the enclosure.
   2. B. A neutral-conductor jumper from the enclosure to the local neutral bar.
   3. C. An independent grounding rod connected directly to the enclosure.
   4. D. A bonding conductor bypassing the isolating fitting along the raceway.

   Answer: A

   Explanation: The rule requires "an internal insulated equipment grounding conductor installed to ground the equipment enclosure." This preserves equipment grounding while maintaining the intended raceway isolation.

   Reference: 29 CFR 1910.305 — Wiring Methods, Components, and Equipment for General Use — 1910.305(a)(1)(ii)

4. 4An engineer measures 480 V line-to-line and 100 A line current at a balanced three-phase load with power factor 0.80. What is the total real input power?

   1. A. 22.2 kW
   2. B. 38.4 kW
   3. C. 66.5 kW
   4. D. 83.1 kW

   Answer: C

   Explanation: DOE states, "Total power (PT) is equal to three times the single-phase power." Using the accompanying balanced three-phase relation gives √3 × 480 × 100 × 0.80 = 66.5 kW. The alternatives divide by √3, omit √3, or omit power factor.

   Reference: DOE Fundamentals Handbook: Electrical Science, Volume 3 of 4 — Module 9, Power in Balanced 3φ Loads, p. 20

5. 5A power-electronics stage receives sinusoidal AC and supplies unidirectional power to a DC link. Which conversion function is it performing?

   1. A. Inversion
   2. B. Rectification
   3. C. DC chopping
   4. D. AC frequency conversion

   Answer: B

   Explanation: The handbook describes a rectifier as "a diode specifically designed or applied to convert power to dc." AC-to-DC conversion is rectification; the other functions describe different input-output conversions.

   Reference: Bureau of Mines Information Circular 9258: Mine Power Systems — Chapter 5, Rectifier Circuits

6. 6A four-pole induction motor operates from a 60-Hz supply and runs at 1,746 rpm under load. What is its slip?

   1. A. 1.5%
   2. B. 3.0%
   3. C. 6.0%
   4. D. 97.0%

   Answer: B

   Explanation: The handbook identifies slip as relative motion between rotor and rotating flux and gives Eq. 6.5. It states, "As slip is required to produce rotor induction, a squirrelcage motor may approach but never obtain synchronous speed." Here ns = 1,800 rpm and s = (1,800 − 1,746)/1,800 = 3%. The alternatives halve or double the slip, or report the rotor-speed fraction.

   Reference: Bureau of Mines Information Circular 9258: Mine Power Systems — Chapter 6, Elementary Three-Phase Motor, Eqs. 6.4–6.5

7. 7A 2,400/240-V ideal single-phase transformer supplies 20 A at its rated secondary voltage. What primary current does the load draw?

   1. A. 0.2 A
   2. B. 2.0 A
   3. C. 20 A
   4. D. 200 A

   Answer: B

   Explanation: The handbook states, "the currents are related inversely to the turns." The 10:1 primary-to-secondary voltage ratio therefore gives a primary current of 20/10 = 2.0 A. Multiplying by the ratio gives 200 A; ignoring it gives 20 A; applying it twice gives 0.2 A.

   Reference: Bureau of Mines Information Circular 9258: Mine Power Systems — Chapter 3, Ideal Transformer, equations 3.26–3.27

8. 8A 240-V DC feeder carries 20 A to a load. The design permits a maximum 4-V drop across the outgoing and return conductors together. What is the maximum resistance of each conductor if they have equal resistance?

   1. A. 0.05 ohm
   2. B. 0.10 ohm
   3. C. 0.20 ohm
   4. D. 0.40 ohm

   Answer: B

   Explanation: The handbook explains that series-circuit current adjusts until "the sum of voltage drops in the circuit is equal to the sum of voltage sources in the circuit." With equal outgoing and return resistances, ΔV = 2IR. Thus Rmax = 4/(2 × 20) = 0.10 ohm per conductor. Omitting the return gives 0.20; doubling rather than halving the loop result gives 0.40; counting four conductor drops gives 0.05.

   Reference: Bureau of Mines Information Circular 9258: Mine Power Systems — Chapter 2, Kirchhoff's voltage law and series circuits

9. 9An engineer reviews overcurrent protection on a 480-V industrial circuit subject to OSHA 1910.304(f)(1). The device would interrupt the grounded conductor, and the motor-running-overload exception does not apply. What condition must the design satisfy?

   1. A. Open every circuit conductor simultaneously.
   2. B. Open the grounded conductor before the other conductors.
   3. C. Open the grounded conductor after a fixed time delay.
   4. D. Open the grounded conductor through a separate fuse.

   Answer: A

   Explanation: OSHA states that overcurrent devices may not interrupt the grounded conductor "unless all conductors of the circuit are opened simultaneously." The stated exception is excluded, so simultaneous opening is required; independent fusing or sequential opening does not meet the condition.

   Reference: 29 CFR 1910.304 — Wiring Design and Protection — 29 CFR 1910.304(f)(1)(ii)

10. 10A CT burden measurement taken at a relay panel gives 4 V at 5 A. The round-trip resistance of the leads between that panel and the CT terminals is 0.20 ohm. What external burden impedance should the engineer use at the CT terminals? Treat the circuit as resistive.

    1. A. 0.60 ohm
    2. B. 0.80 ohm
    3. C. 1.00 ohm
    4. D. 4.20 ohms

    Answer: C

    Explanation: FIST states, "The external burden consists of the impedance of the CT secondary load plus the secondary circuit lead resistance." The panel measurement is 4/5 = 0.80 ohm; adding 0.20 ohm gives 1.00 ohm. Subtraction, omitting the leads, or treating volts as ohms gives the other results.

    Reference: FIST 3-8: Operation, Maintenance, and Field Test Procedures for Protective Relays and Associated Circuits — Appendix D, CT External Burden, p. 85