QUIZ SHOW

Question 1

When no electrical equipment of the exact size or rating required is available, ________ may be used.

a. the next lower standard size
b. any available size
c. the next larger standard size
d. none of these

Answer 1

c. the next larger standard size

Question 2

The air conditioning load has a demand load of _____________.

a. 80%
b. 90%
c. 75%
d. 100%

Answer 2

d. 100%

Question 3

The workspace about electrical equipment shall be adequate to permit at least ______ degree opening of doors or hinged panels.

Answer 3

90 deg

Question 4

What is the maximum fuse rating allowed by the code to protect the single-phase motor that draws 20 A at full load against a short circuit but at the same time will not fall at the start?

Answer 4

60 A

Fuse Rating:
300% of FLC
Therefore,
20 A (3) = 60 A

Question 5

Motors with a marked service factor of less than 1.15, shall have an overload protection equal to _______ percent of the motor’s FLA.

a. 100
b. 125
c. 120
d. 115

Answer 5

d. 115

Question 6

Conductors supplying two or more motors shall have an ampacity equal to the sum of the FLA rating of all motors plus ________ of the highest motor FLA in the group.

Answer 6

25%

Question 7

A 15-A or 20-A branch circuit shall be permitted to supply lighting units and other utilization devices. The rating of any one cord and plug-connected appliance shall NOT exceed a certain percentage of the branch circuit rating of _____________.

Answer 7

80%

Question 8

The circuit conductor between the service entrance equipment or isolated generating plant and the branch circuit overload device or devices.

Answer 8

Feeder

Question 9

A short length of a conductor is used to make a connection between terminals or around a break in a circuit.

Answer 9

Jumper

Question 10

The branch circuit for driving equipment is the larger of either the VA rating of the nameplate or ________________.

a. 5000 W
b. 3500 W
c. 4000 W
d. 5500 W

Answer 10

a. 5000 W

Question 11

A dwelling unit having a floor area not more than 50 square meters shall be permitted to have a single 20-A, 2-wire branch circuit provided the total load shall not exceed ________.

Answer 11

3680 VA

Question 12

For barber shops and beauty parlors, the generated lighting load per square meter of area shall be _________.

Answer 12

24 VA/m^2

Question 13

For receptacle outlets, each single duplex, or triplex receptacle on one strap/yoke shall be considered not less than __________.

Answer 13

180 VA

Question 14

Service entrance using copper conductors shall have sufficient capacity and shall NOT be smaller than _________ copper conductor.

Answer 14

8.0 mm^2

Question 15

At least how many entrance(s) shall be provided to give access to the working space for electrical equipment?

Answer 15

One (1)

Question 16

The overhead conductors between the utility electric supply system and the service point.

Answer 16

Service Drop

Question 17

This PEC may require new products, constructions, or materials that may not yet available at the time this Code is adopted. In such event, the ________ may permit the use of the products, constructions, or materials that comply with the most recent previous edition of this Code adopted by the National Building Code.

Answer 17

OBO/EE

Question 18

Any current in excess of the rated current of the equipment or the ampacity of a conductor. It may result from overload, short circuit, or ground fault.

Answer 18

Overcurrent

Question 19

Motor A has a full load current of 8A and motor B, 10A. What is the ampacity of the feeder conductor supplying these two motors?

a. 9 A
b. 20.5 A
c. 18 A
d. 24.2 A

Answer 19

b. 20.5 A

125% of 10 A + 8 A = 20.5 A

Question 20

A device designed to open and close the circuit by nonautomatic means and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating.

Answer 20

Circuit Breaker

Question 21

The service conductors between terminals of the service equipment and a point usually outside the building, clear of building walls, where joined by a tap of splice to service drop or overhead service conductors.

Answer 21

Overhead Service Entrance Conductor

Question 22

Operation of equipment in excess of normal, full-load rating, or of a conductor in excess of rated ampacity, when it persists for a sufficient length of time, would cause damage or dangerous overheating.

Answer 22

Overload

Question 23

The prospective symmetrical fault current at a nominal voltage to which an apparatus or system can be connected without sustaining damage exceeding defined acceptance criteria.

Answer 23

Short-Circuit Current Rating

Question 24

An electrically conductive path from the point of a ground fault on a wiring system through normally non-current-carrying conductors, equipment, or the earth to an electrical supply source.

Answer 24

Ground-Fault Circuit Path

Question 25

It consists of making an allowance of watts per square meter of area to be illuminated according to the illumination desired on the assumption of an average figure of the overall efficiency of the system.

Answer 25

Watt per square meter method

Question 26

PROBLEM 1: In a group of four motors, one motor draws 10 amperes, one draws 45 amperes, and two draws 75 amperes. What size of the conductor must be used for the feeder circuit?

Answer 26

Total FLC + 25% FLC of the HIGHEST MOTOR RATING (10+45+75+75) A + 25%(75 A) = 223.75 A = 224 A

Question 27

PROBLEM 2: Three, 230 volts, 1-phase squirrel cage induction motors each drawing a current of 50 amperes are supplied by copper feeder conductors. The maximum rating of inverse time delay circuit breaker for feeder protection is ___________.

Answer 27

Total FLC + 250% FLC of the HIGHEST MOTOR RATING (50+50) A + 250% (50 A) = 225 A

Question 28

PROBLEM 3: Determine the lighting load requirement by PEC for a bungalow house measuring 10m X 15m. It has a total of 4500 VA loads for small appliances and laundry circuits. By applying the demand factor, what will be the total load of the dwelling unit?

Answer 28

Lighting Load: Floor Area Method: 24 VA/m^2 x Floor Area (in m^2) 24 VA/m^2 x 150m^2 = 3600 VA Sub-total = Lighting Load + Small Appliances Load: 3600 VA + 4500 VA = 8100 VA Application of Demand Factor: First 3000 VA @ 100% DF = 3000 VA Remainder @ 35% DF = 1785 VA Total Load: 3000 VA + 1785 VA = 4785 VA