The new flashcard feature is useful to memorize questions.
?Change your study mode
At any time, you can change the study mode, and alternate between the practice mode and the exam mode. In practice mode, you can configure for example the number of questions or tests, and other parameters to help you study.
Randomized | 10 Questions per Test | 20 Minutes | 70% to pass|
To re-configure your study mode again and change - for example - the number of tests, whether you have random questions and all other configuration parameters.
?Simulator Configuration
Auto-scroll: You can use the automatic scrolling of the questionnaire that occurs as soon as you answer one or all of the answers to a question correctly. Auto scrolling is activated if you answer a single answer, or as soon as you answer all the mandatory answers. Learning Mode: During learning mode you can get a real time result for your answer.
Free Test
Question: / 10
20:00Min. left
?Restart the current test
To restart the current test by clearing all your answers and the time used up to now. Warning: all answers will be lost.
Question: / 10
0.0(0 Votes)
Quiz
Question 1/101/10
Electrical Theory and Calculations
Electrical Theory and Calculations
Electrical Theory and Calculations
A resistive heating element with a resistance of 16 ohms is connected to a 240-volt single-phase supply. How much power does the element consume?
Select the answer:Select the answer
1 correct answer
A.
3,600 W
B.
15 W
C.
3,840 W
D.
960 W
Explanation: Power in a purely resistive circuit can be found with any of the Ohm's law power formulas. The most direct approach when voltage and resistance are known is P = E squared / R. Here E = 240 V and R = 16 ohms, so P = (240 x 240) / 16 = 57,600 / 16 = 3,600 W. You can check the result another way: first find the current with I = E / R = 240 / 16 = 15 A, then multiply by the voltage, P = E x I = 240 x 15 = 3,600 W, or use P = I squared x R = 15 x 15 x 16 = 3,600 W. All three methods must agree, which is a good habit on the exam to catch keying errors on the calculator. Common mistakes include reporting the current (15 A) as the power, multiplying the voltage by the resistance, or dividing the voltage by the current a second time. Because a heating element is resistive, the power factor is 1.0 and the true power in watts equals the apparent power in volt-amperes, so no power factor correction is needed in this calculation.
Right Answer: A
Quiz
Question 2/102/10
Lightning Protection
Lightning Protection
Lightning Protection
A lightning protection system is being designed for an office building whose main roof is 82 feet above grade. According to NFPA 780 (2017), which class of materials is required for the system?
Select the answer:Select the answer
1 correct answer
A.
Class I materials, because the building is less than 100 feet tall
B.
Class I materials, provided the number of down conductors is doubled
C.
Either class, at the discretion of the lightning protection designer
D.
Class II materials
Explanation: NFPA 780 divides lightning protection materials into two classes based on structure height. Class I materials are permitted for ordinary structures not exceeding 75 feet in height. Class II materials are required for ordinary structures that exceed 75 feet in height. Class II main conductors, air terminals, and related components are larger and heavier than their Class I counterparts because taller structures are more likely to be struck and to experience higher-energy strikes, and the conductors must carry the lightning current over longer paths without damage. Since this building's roof is 82 feet above grade, it exceeds the 75-foot threshold and Class II materials are required. There is no 100-foot threshold in the standard, and adding extra down conductors does not permit undersized materials to be used. The material class is not a matter of designer preference; it is a requirement of the standard based on height. On the exam, remember the key number 75 feet: at or below it, Class I is acceptable, and above it, Class II is required. Material class requirements also govern items such as minimum strand size, conductor weight per 1,000 feet, and the dimensions of air terminals.
Right Answer: D
Quiz
Question 3/103/10
Burglar Alarm and Security Systems
Burglar Alarm and Security Systems
Burglar Alarm and Security Systems
A passive infrared (PIR) motion detector is being installed in an office. What does the PIR detector actually respond to?
Select the answer:Select the answer
1 correct answer
A.
A rapid change in infrared energy as a warm object moves across its detection zones
B.
A Doppler frequency shift in microwave energy reflected from a moving object
C.
A change in air pressure when a door or window is opened
D.
The high-frequency sound produced by breaking glass
Explanation: A passive infrared detector does not transmit any energy; it is passive because it only receives. Its pyroelectric sensor, combined with a segmented lens or mirror, divides the protected area into a pattern of detection zones. When an intruder whose body temperature differs from the background moves from one zone into another, the sensor sees a rapid change in received infrared energy and signals an alarm. Because the detector responds best to movement across its zones, PIRs are most sensitive to an intruder walking across the field of view and least sensitive to someone walking directly toward the unit, which affects mounting location. Proper installation avoids aiming the detector at heat sources such as forced-air registers, space heaters, sunlit windows, and machinery that cycles on and off, all of which can cause false alarms. The Doppler shift description applies to microwave motion detectors, which are active devices. Air pressure detection describes a different volumetric technology, and listening for breaking glass describes acoustic glass-break detectors. Understanding the operating principle helps the installer select the correct detector for each environment and place it to minimize false alarms.
Right Answer: A
Quiz
Question 4/104/10
NEC Wiring and Protection
NEC Wiring and Protection
NEC Wiring and Protection
According to the 2023 NEC, unless specifically permitted elsewhere in the Code (such as for motor circuits), what is the maximum rating of the overcurrent protective device permitted for a 12 AWG copper branch-circuit conductor?
Select the answer:Select the answer
1 correct answer
A.
20 amperes
B.
25 amperes
C.
30 amperes
D.
15 amperes
Explanation: The small conductor rule in 240.4(D) limits the overcurrent protection of the smallest and most commonly used conductors regardless of the ampacity shown in Table 310.16. Even though 12 AWG copper with 75 degrees C or 90 degrees C insulation has a table ampacity of 25 or 30 amperes, the overcurrent device protecting it may not exceed 20 amperes once any correction and adjustment factors have been applied. The companion limits are 15 amperes for 14 AWG copper and 30 amperes for 10 AWG copper; for aluminum and copper-clad aluminum, 12 AWG is limited to 15 amperes and 10 AWG to 25 amperes. These limits exist because termination temperature ratings, voltage drop and the realities of residential and commercial branch circuits make the higher table values unsafe as a protection basis for small conductors. The higher table ampacity is still useful, however: it is the starting point when derating for ambient temperature or more than three current-carrying conductors, so a 90 degrees C 12 AWG conductor can often be derated and still remain at or above 20 amperes. Exceptions to the small conductor rule are listed in 240.4(G), which covers specific equipment such as motors, air-conditioning equipment and fire alarm circuits, where other articles set the protection.
Right Answer: A
Quiz
Question 5/105/10
NEC Equipment for General Use
NEC Equipment for General Use
NEC Equipment for General Use
A 25-hp, 460-volt, three-phase squirrel-cage induction motor is used for continuous duty. Using the full-load current from Table 430.250 of the 2023 NEC, what is the minimum ampacity of the branch-circuit conductors supplying this single motor?
Select the answer:Select the answer
1 correct answer
A.
34.0 amperes
B.
39.1 amperes
C.
42.5 amperes
D.
47.6 amperes
Explanation: Section 430.22 requires conductors that supply a single motor used in a continuous-duty application to have an ampacity of not less than 125 percent of the motor full-load current rating. Per 430.6(A)(1), that full-load current must be taken from the tables in Article 430, not from the motor nameplate, whenever the motor is a general-purpose type covered by the tables. Table 430.250 lists 34 amperes for a 25-hp, 460-volt, three-phase induction motor. The calculation is 34 A x 1.25 = 42.5 amperes, so the conductors must have an ampacity of at least 42.5 amperes after any correction or adjustment factors. Using the table value without the 125 percent multiplier gives 34 amperes and undersizes the conductors. Using 115 percent (39.1 A) confuses the conductor rule with the overload rule for some motors or the disconnect rule, and 140 percent (47.6 A) is the maximum overload relay setting where the initial setting will not carry the starting current. The extra 25 percent margin allows the conductors to carry the continuous running current and any normal overload up to the overload relay setting without exceeding their temperature rating. Overcurrent protection for short circuits and ground faults is sized separately under 430.52.
Right Answer: C
Quiz
Question 6/106/10
NEC Wiring Methods and Materials
NEC Wiring Methods and Materials
NEC Wiring Methods and Materials
According to the 2023 NEC, direct-buried conductors and cables that emerge from grade must be protected by enclosures or raceways extending from the minimum cover depth required by Table 300.5 up to at least what height above finished grade?
Select the answer:Select the answer
1 correct answer
A.
6 ft
B.
8 ft
C.
10 ft
D.
7 ft
Explanation: Section 300.5(D)(1) addresses conductors and cables that are direct buried and then rise out of the ground, such as a UF cable feeding a post light or a USE service lateral climbing a pole. Where they emerge from grade, they are vulnerable to lawn equipment, vehicles, and general physical damage, so the Code requires them to be protected by enclosures or raceways. That protection must extend from the minimum cover depth required by Table 300.5 below grade up to a point at least 8 ft above finished grade. Under no circumstances is the protection required to extend more than 18 in. below grade. The raceway used for this protection must be suitable for the location; if it is subject to physical damage, rigid metal conduit, intermediate metal conduit, Schedule 80 PVC, or an equivalent should be selected. The 8 ft dimension is a frequently tested value because it differs from other common figures such as the 7 ft height used for some guarding rules or the 10 ft spacing used for raceway supports. Remember also that conductors entering a building must be protected to the point of entrance, so the 8 ft rule is a minimum, not a limit.
Right Answer: B
Quiz
Question 7/107/10
NEC Special Occupancies and Special Equipment
NEC Special Occupancies and Special Equipment
NEC Special Occupancies and Special Equipment
A grain elevator handles grain that produces combustible dust which may be suspended in the air in quantities sufficient to produce ignitible mixtures. Under the class system of Article 500 of the 2023 NEC, how is this location classified?
Select the answer:Select the answer
1 correct answer
A.
Class I
B.
Class II
C.
Class III
D.
Zone 0
Explanation: Article 500 classifies hazardous locations according to the properties of the materials present. Class I locations are those in which flammable gases, flammable liquid-produced vapors, or combustible liquid-produced vapors are or may be present in quantities sufficient to produce explosive or ignitible mixtures. Class II locations are those that are hazardous because of the presence of combustible dust, such as grain dust, flour, sugar, coal dust, and metal dusts. Class III locations are hazardous because of the presence of easily ignitible fibers or materials producing combustible flyings, such as cotton lint in textile mills or wood shavings in woodworking plants, but in which the fibers are not likely to be in suspension in the air in ignitible quantities. Each class is then divided into Division 1 (where the hazard is normally present) and Division 2 (where it is present only under abnormal conditions). Zone 0 belongs to the alternative zone classification system of Article 505 for flammable gases and vapors, where an ignitible concentration is present continuously or for long periods. Combustible dust zones under Article 506 are designated Zone 20, 21, and 22, so neither Zone 0 nor Class I applies to grain dust.
Right Answer: B
Quiz
Question 8/108/10
Fire Alarm Systems
Fire Alarm Systems
Fire Alarm Systems
A designer is laying out spot-type smoke detectors on a smooth, flat ceiling in an open office using the nominal 30 ft spacing recognized by NFPA 72-2019. In addition to the nominal spacing, every point on the ceiling must be within what maximum horizontal distance of a detector?
Select the answer:Select the answer
1 correct answer
A.
15 ft
B.
21 ft
C.
30 ft
D.
25 ft
Explanation: For spot-type smoke detectors on smooth ceilings, NFPA 72-2019 permits a nominal spacing of 30 ft between detectors where no other performance-based design is used. Because rooms are rarely perfect squares that match a 30 ft grid, the code adds a second test: all points on the ceiling must be within a distance equal to 0.7 times the selected spacing (0.7S) of the nearest detector. With S = 30 ft, 0.7 x 30 = 21 ft. The 0.7 factor comes from the geometry of a square: the corner of a 30 ft by 30 ft square is about 21.2 ft from its center (half the diagonal, 30 x 1.414 / 2), so each detector effectively protects a circle of about 21 ft radius. This rule lets the designer stretch spacing in one direction (for example, in a long narrow corridor) as long as no point on the ceiling is farther than 21 ft from a detector. Using 15 ft (one-half spacing) confuses the rule with the half-spacing distance to walls, and 30 ft or 25 ft would leave the corners of the protected area outside the detector's effective coverage circle.
Right Answer: B
Quiz
Question 9/109/10
Life Safety Code
Life Safety Code
Life Safety Code
According to NFPA 101-2018, emergency lighting facilities for means of egress must be arranged to provide the required illumination automatically for a minimum of how long after failure of normal lighting?
Select the answer:Select the answer
1 correct answer
A.
1 hour
B.
1.5 hours
C.
2 hours
D.
3 hours
Explanation: NFPA 101-2018 requires emergency illumination to be provided for a minimum of 1.5 hours (90 minutes) in the event of failure of normal lighting. The duration is intended to cover the time needed to evacuate occupants from the building and to support firefighters and other emergency personnel during the early stages of an incident. Emergency lighting must activate automatically, without manual intervention, upon loss of normal power, and the code permits a short delay of not more than 10 seconds for the transfer to emergency power. The source can be unit equipment with internal batteries, a central battery system, or a generator installed in accordance with NFPA 110 and NFPA 111, and the wiring must comply with NEC Article 700 where it is part of an emergency system. During the 1.5 hour period, illumination is permitted to decline from the initial levels, but not below 0.6 ft-candle average and 0.06 ft-candle at any one point at the end of the period. One hour is too short, while 2 or 3 hours exceed the minimum, although some specific occupancies or owners may choose longer durations.
Right Answer: B
Quiz
Question 10/1010/10
OSHA Construction Safety
OSHA Construction Safety
OSHA Construction Safety
An electrical contractor is supplying portable power tools on a construction site from 120-volt, single-phase, 20-ampere receptacle outlets that are part of a temporary power pole and are not part of the permanent building wiring. Under OSHA 29 CFR 1926 Subpart K, what must the employer provide to protect workers from ground faults at these outlets?
Select the answer:Select the answer
1 correct answer
A.
A written hazard communication program covering electrical tools
B.
Ground-fault circuit interrupters, or alternatively an assured equipment grounding conductor program
C.
Double-insulated tools only, with no other protection required
D.
A surge protective device installed at the temporary service
Explanation: OSHA's construction electrical standard (29 CFR 1926.404(b)(1)) requires the employer to protect employees on construction sites from ground faults. For all 120-volt, single-phase, 15- and 20-ampere receptacle outlets that are not part of the permanent wiring of the building or structure and that are in use by employees, the employer must use ground-fault circuit interrupters. As an alternative, the employer may establish a written assured equipment grounding conductor program that is administered by one or more competent persons and that includes inspections and tests of cord sets, receptacles and cord-and-plug connected equipment. Using double-insulated tools is good practice, but it does not by itself satisfy the ground-fault protection requirement because cords and receptacles can still be damaged. A surge protective device limits transient overvoltages and has nothing to do with detecting current leaking to ground through a worker's body. A hazard communication program addresses chemical hazards, not electrical shock. On an exam, remember the pairing: GFCI protection or an assured equipment grounding conductor program for temporary receptacles used by workers.
36-FL-UN: Unlimited Practice test unlocks all online simulator questions
Thank you for choosing the free version of the 36-FL-UN: Unlimited practice test! Further deepen your knowledge on Florida DBPR Electrical Simulator; by unlocking the full version of our 36-FL-UN: Unlimited Simulator you will be able to take tests with over 312 constantly updated questions and easily pass your exam. 98% of people pass the exam in the first attempt after preparing with our 312 questions.
What to expect from our 36-FL-UN: Unlimited practice tests and how to prepare for any exam?
The 36-FL-UN: Unlimited Simulator Practice Tests are part of the Florida DBPR Electrical Database and are the best way to prepare for any 36-FL-UN: Unlimited exam. The 36-FL-UN: Unlimited practice tests consist of 312 questions and are written by experts to help you and prepare you to pass the exam on the first attempt. The 36-FL-UN: Unlimited database includes questions from previous and other exams, which means you will be able to practice simulating past and future questions. Preparation with 36-FL-UN: Unlimited Simulator will also give you an idea of the time it will take to complete each section of the 36-FL-UN: Unlimited practice test . It is important to note that the 36-FL-UN: Unlimited Simulator does not replace the classic 36-FL-UN: Unlimited study guides; however, the Simulator provides valuable insights into what to expect and how much work needs to be done to prepare for the 36-FL-UN: Unlimited exam.
36-FL-UN: Unlimited Practice test therefore represents an excellent tool to prepare for the actual exam together with our Florida DBPR Electrical practice test . Our 36-FL-UN: Unlimited Simulator will help you assess your level of preparation and understand your strengths and weaknesses. Below you can read all the quizzes you will find in our 36-FL-UN: Unlimited Simulator and how our unique 36-FL-UN: Unlimited Database made up of real questions:
Info quiz:
Quiz name:36-FL-UN: Unlimited
Total number of questions:312
Number of questions for the test:50
Pass score:80%
You can prepare for the 36-FL-UN: Unlimited exams with our mobile app. It is very easy to use and even works offline in case of network failure, with all the functions you need to study and practice with our 36-FL-UN: Unlimited Simulator.
Use our Mobile App, available for both Android and iOS devices, with our 36-FL-UN: Unlimited Simulator . You can use it anywhere and always remember that our mobile app is free and available on all stores.
Our Mobile App contains all 36-FL-UN: Unlimited practice tests which consist of 312 questions and also provide study material to pass the final 36-FL-UN: Unlimited exam with guaranteed success.
Our 36-FL-UN: Unlimited database contain hundreds of questions and Florida DBPR Electrical Tests related to 36-FL-UN: Unlimited Exam. This way you can practice anywhere you want, even offline without the internet.