Q#2: The SI unit of electric charge, equal to the quantity of electricity transferred across a conductor by a current of one ampere in one second.
a. Amperage
b. Coulomb
c. Volt
d. Ohm
e. Watt
Ampere (A) (NOT Amperage) is the SI Unit for ELECTRIC CURRENT. It is a base unit.
Coulomb (C) is the SI Unit for ELECTRIC CHARGE.
1C = 1A x sec
Volt (V) is the SI Unit for ELECTRIC POTENTIAL DIFFERENCE.
1V = 1W / A = (kg)(m2) / (sec3)(A)
Ohm (Ω) is the SI Unit for ELECTRICAL RESISTANCE.
1Ω = 1V / A = (kg)(m2) / (sec3)(A2)
Watt (W) is the SI Unit for POWER.
1W = 1V x A = (kg)(m2) / (sec3)
Therefore, the correct answer is b. Coulomb.
Q#3: The overhead projection of service conductors extending from the nearest utility pole to a building.
a. Service entrance conductor
b. Service center
c. Service drop
d. Service lateral
The Service Entrance Conductors are the ACTUAL WIRING originating from the Service Point (where the service drop/lateral connects) to the Meter Socket/Disconnect Panel (to the house/structure).
Note that the Service drop (choice c.) and the Service lateral (choice d.) both SERVE THE SAME PURPOSE.
Their main difference is that the former is supplied OVERHEAD (hence, the need for a "drop") while the latter is supplied UNDERGROUND (hence, sideways or "lateral").
Q#4: A heavy conductor, usually in the form of a solid copper bar, used for collecting, carrying, and distributing large electric currents.
a. Cable
b. Cord
c. Bus
d. Loom
A cable (or sometimes also referred to as a cord) is an assembly of wires (one wire or bundled wires) used to carry electric current.
A loom is a weaved fabric-like sheathing used for managing cables and offer better aesthetics with some degree of protection.
A bus (or sometimes also referred to as a busbar) is a metallic uninsulated strip used for high current power distribution that has the ability to be tapped into at various points without splicing or jointing.
Q#5: The SI unit that measures the potential difference or electromotive force; analogous to pressure in water flow.
a. Voltage
b. Watt
c. Wattage
d. Volt
e. Ampere
Note that in my choices, b. Watt, d. Volt, and e. Ampere are all examples of SI Units (such as, meter, kilogram, etc.). On the other hand, both choices a. Voltage and c. Wattage are NOT units but only describes" the amount the unit represents (for example, length, weight, speed, etc.).
Ampere is an SI Unit for ELECTRIC CURRENT. It is analogous to flow rate in water (in L/s or gpm).
Watt is an SI Unit for POWER. It is Voltage (Volt) x Current (Ampere).
Volt is an SI Unit for POTENTIAL DIFFERENCE or FORCE. It is analogous to pressure in water flow (in kPa or psi).
Therefore, the correct answer is d. Volt.
Q#6: A No. 10 (AWG) sized wire is equivalent to a cross-sectional area of approx. _____.
a. 2.0 sq.mm.
b. 3.5 sq.mm.
c. 5.5 sq.mm.
d. 8.0 sq.mm.
AWG (American Wire Gauge) is the standard system of sizing wires in North America, usually denoted by the # symbol.
My tip to remember the corresponding sizes (cross-sectional area) in metric units of a wire specified in AWG is:
8 = 8
Meaning, a No. 8 AWG wire is approx. 8.0 sq.mm. in cross-sectional area.
Note that the rest of wire sizes in AWG are NOT equivalent to metric values of its cross-sectional area. The LOWER the AWG number, the THICKER the material; the HIGHER the AWG number, the THINNER the material. So,
Q#7: Which of the following can you use to find the electrical power (in Watts) of a certain load?
a. Current (in Amperes) x Current (in Amperes) x Resistance (in Ohms)
b. Potential (in Volts) x Current (in Amperes)
c. Potential (in Volts) x Potential (in Volts) / Resistance (in Ohms)
d. All of the above
Recall that:
Power (in Watts) = Potential (in Volts) x Current (in Amperes) <----Equation in Choice b.
Current (in Amperes) = Potential (in Volts) / Resistance (in Ohms) <----The Ohm's Law
Substituting Current in the Power equation with Current based on Ohm's Law, we would have:
Power (in Watts) = Potential (in Volts) x [ Potential (in Volts) / Resistance (in Ohms) ], or
Power (in Watts) = Volts x Volts / Ohms <----Equation in Choice c.
Substituting Potential in the Power equation with Potential based on Ohm's Law , we would also have:
Power (in Watts) = [ Current (in Amperes) x Resistance (in Ohms) ] x Current (in Amperes), or
Power (in Watts) = Amps x Amps x Ohms <----Equation in Choice a.
So, it turns out that all the given choices all equate to Power (in Watts). Let's check with some hypothetical values as an example,
Given a current of 5A, a resistance of 46Ω, and a potential of 230V, find the power (disregarding any power factors):
Using equation in Choice a.) Power = (Current)2 x Resistance = (5A)2 x 46Ω = 1,150W
Using equation in Choice b.) Power = Potential x Current = 230V x 5A = 1,150W
Using equation in Choice c.) Power = (Potential)2 / Resistance = (230V)2 / 46Ω = 1,150W
Therefore, correct answer is d. All of the above.
Q#8: What is the minimum circuit wire size (copper conductors) that can be used for a circuit protected by a 30-ampere breaker?
a. 8.0 sq.mm.
b. 2.0 sq.mm.
c. 5.5 sq.mm.
d. 3.5 sq.mm.
Note that temperature rating is primarily dependent on the type of wire insulation (NOT the wire size). The size AND the temperature rating (as well as the type of conductor material) then determines its maximum ampacity.
For example, Thermoplastic High Heat-Resistant Nylon-Jacketed (THHN) wires have a 90°C (194°F) temp. rating, Thermoplastic Heat-Resistant Water-Resistant Insulated (THW) wires have a 75°C (167°F) temp. rating, while Thermoplastic Water-Resistant Insulated (TW) wires only have a 60°C (140°F) temp. rating.
The higher the temperature rating (based on insulation) of a given wire size, the higher the ampacity of its same wire size.
The rated ampacities for wire sizes in my question above is based on the 60°C (140°F) minimum temp. rating for copper wire materials.
General objective is: Breaker must trip BEFORE the Wire burns up (i.e., wire ampacity rating > breaker rating). So,
#8 AWG (approx. 8.0mm^2) min. copper wire are used in 40A circuits
#10 AWG (approx. 5.5mm^2) min. copper wire are used in 30A circuits
#12 AWG (approx. 3.5mm^2) min. copper wire are used in 20A circuits
#14 AWG (approx. 2.0mm^2) min. copper wire are used in 15A circuits
Therefore, the correct answer is c. 5.5 sq.mm. (or #10 AWG).
Q#9: You are to design the vanity lighting using 2"-dia. halogen fixture frames. Which of the following bulbs/lamps can you specify for this purpose?
a. 2U CFL E27 Bulbs
b. MR16 GU5.3 Bulbs
c. Spiral ESL E14 Bulbs
d. PAR38 Halogen Lamps
This is a 2"-diameter halogen fixture frame as described in my question above:
Compact Fluorescent Lamp (CFL), or sometimes also known as Energy-Saving Lamp (ESL), is a type of lamp commonly identified with the number of U-shaped fluorescent tubes such as "2U", "3U", and "4U". Other than U-shaped tubes, there are also specially-designed tubes such as Spiral, Butterfly, Candle, Globe, etc.
These CFLs and ESLs commonly have screw-type connector bases. Screw-type connectors are commonly identified with an "E" prefix, which stands for Edison Screw (developed by Thomas Edison). The "27" in E27 is one of the standard-sized screw base which measures 27mm in base diameter. E14 is a smaller-sized screw base with a 14mm base diameter. Other common sizes are E40 (Mogul) and E11 (Candelabra)
This is an example of a 2U CFL Bulb with E27 Socket (choice a. in my question):
This is an example of a Spiral ESL Bulb with E14 Socket (choice c. in my question):
As you can see, the above bulbs may have its 27mm and 14mm base diameter fit inside the 2"-diameter halogen frame, but the actual tube will protrude outside the frame and the bulbs' body will obstruct the installation into the frame due to their size differences.
Multifaceted Reflectors (MR) and Parabolic Aluminized Reflectors (PAR) are typically designed for halogen/incandescent lighting (although, nowadays, there are bulbs that also feature both fluorescent and L.E.D. outputs).
The "16" in MR16 and the "38" in PAR38 are standard measurements of the sizes of their FRONT frame diameter (the actual bulb/lamp), measured in 1/8 of an inch. So, an MR16 bulb would have a 16 x 1/8" = 2" front frame; a PAR38 lamp would have a 38 x 1/8" = 4-3/4" front frame.
As such, MR16 bulbs are typically used for smaller lighting outlets (such as cabinet lighting, mirror lighting, etc.) while PAR38 bulbs are typically used for larger lighting outlets, usually outdoors (such as garden lighting, etc.)
Note that there are other common sizes of MR lamps other than MR16, such as MR20, MR11, and MR8. Other common sizes of PAR lamps other than PAR38, include PAR64, PAR56, PAR38, PAR30.
This is an example of a PAR38 Halogen Lamp with a screw-type/Edison base (choice d. in my question):
Notice the diameter of the front is 38 / 8" = 4-3/4"
MR bulbs are usually available in different types of base connectors. The GU5.3 is a type of bi-pin connector located at its base. The "5.3" is the distance between the two PINS in millimeters (more specifically, 5.33mm). Another common type of bi-pin connector is the GU10 (10mm distance between pins but with a TWIST-LOCK, similar to a starter). There are also MR16 bulbs that have screw-type (Edison) base connectors (which is similar to the ones used in CFL's/ESL's I have shown above).
This is an example of an MR16 Halogen Lamp with a GU5.3 base (choice b. in my question):
Notice the diameter is 16 / 8" = 2"
Another example of a MR16 Halogen Lamp, this time, with a GU10 twist-lock base (similar to a starter):
Notice the diameter is also 16 / 8" = 2"
So, in my question, the 2"-diameter lighting fixture frame for the vanity lighting can only fit an MR16 bulb type (regardless of the base connector). The rest of the bulbs in my other given choices simply will NOT fit in the 2"-diameter frame.
Therefore, the correct answer is b. MR16 GU5.3 Bulbs.
Q#10: If the conductors are specified as "Type Z", which of the following should you use?
a. Thermoplastic and fibrous outer braid wires
b. Perfluoro-alkoxy wires
c. Modified ethylene tetrafluoroethylene wires
d. Cross-linked polyethylene high heat-resistant water-resistant wires
Thermoplastic and fibrous outer braid wires = Type TBS
Perfluoro-alkoxy wire = Type PFA or Type PFAH
Cross-linked polyethylene high heat-resistant water-resistant wires = Type XHHW
Therefore, the correct answer is c. Modified ethylene tetrafluoroethylene wires.
Q#11: The portion of the calculated lighting load to which the demand factor of 100% applies in a dwelling unit.
a. First 3,000 volt-amperes or less
b. First 5,000 volt-amperes or less
c. First 12,500 volt-amperes or less
d. All or Total volt-amperes as calculated for lighting load
According to the Philippine Electrical Code (PEC) on Lighting Load Demand Factors for Dwelling Units:
"First 3,000 Volt-Amperes or less ===> 100% Demand Factor
From 3,001 Volt-Amperes to 120,000 Volt-Amperes ===> 35% Demand Factor
Remainder over 120,000 Volt-Amperes ===> 25% Demand Factor"
(Table 2.20.3.3 of the PEC)
Therefore, the correct answer is a. First 3,000 volt-amperes or less.
Q#12: Any of several conducting rods installed at the top of a structure and grounded to divert lightning away from the structure.
a. Spark Gap
b. Antenna
c. Air Terminal
d. Lightning Arrester
Note that:
Air Terminal = Finial = Strike Termination Device = Lightning Rod
while:
Lightning Arrester = Lightning Diverter (NOT a lightning rod)
Q#13: What is the maximum number of 3-pole circuit breakers (excluding the main) that can be fitted in one (1) panelboard?
a. 42
b. 21
c. 24
d. 14
According to the Philippine Electrical Code (PEC) on Number of Overcurrent Devices on One Panelboard,
"Not more than 42 overcurrent devices (other than those provided for in the mains) of a lighting and appliance branch-circuit panelboard shall be installed in any one cabinet or cutout box...For the purposes of this article, a 2-pole circuit breaker shall be considered two overcurrent devices; a 3-POLE CIRCUIT BREAKER shall be considered THREE OVERCURRENT DEVICES."
(Sec. 4.8.3.6 of the PEC)
In my question, if one panelboard has a maximum of 42 overcurrent devices, then,
Maximum # of 1-Pole Circuit Breakers = 42 / 1 = 42 sets of 1-Pole Circuit Breakers (excluding Main)
Maximum # of 2-Pole Circuit Breakers = 42 / 2 = 21 sets of 2-Pole Circuit Breakers (excluding Main)
Maximum # of 3-Pole Circuit Breakers = 42 / 3 = 14 sets of 3-Pole Circuit Breakers (excluding Main)
Therefore, the correct answer is d. 14.
Q#14: What is the minimum distance of a signboard from an electric or telephone post?
a. 1 meter
b. 3 meters
c. 4 meters
d. 900 millimeters
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD1096),
"SIGNBOARDS shall NOT obstruct any window or emergency exit and shall NOT be closer than 1.00 METER from ELECTRIC AND TELEPHONE POSTS and wires."
(Sec. 2004.d.iii.6 of the 2004 IRR of PD1096)
Therefore, the correct answer is a. 1 meter.
Q#15: A switch needed to control lights from three different locations.
a. Single Pole-Double Throw switch
b. Three-way switch
c. Three-gang switch
d. Four-way switch
Take note: The "2", "3", and "4", in 2-way, 3-way, and 4-way switch does NOT always mean 2, 3, and 4 LOCATIONS (depending on terminologies used in a specific country). The "2", "3", and "4", actually means there are 2, 3, and 4 NODES in the actual switch.
2-way switches = regular Single Pole-Single Throw (2-node) switches that control light in a single location. (May be referred to as "1-way" switch in some countries)
3-way switch = a Single Pole-Double Throw (3-node) switch, hence, when you connect 2 sets of 3-way switches together in different locations, toggling EITHER switch turns the light on or off.
4-way switch = a Double Pole-Double Throw switch (4-node), hence, if a 4-way switch is located between the 2 sets of 3-way switches, the 4-way switch becomes an "intermediate switch" that adds another on-off toggle in another location.
Below is an example of a wiring diagram of a lighting fixture controlled in 4 different locations - which needs 2 sets of 3-way switches + 2
sets of 4-way switches:
So, the more 4-way switches you add BETWEEN the 2 pcs of 3-way switches, the greater the number of different locations you can control the light (or outlet) from.
To give you an idea, here are the quantities and types of switches you will need in order to control a light (or outlet) from several different locations:
1 location = 1 set of 2-way switch (or "1-way" in some countries)
2 different locations = 2 sets of 3-way switches
3 different locations = 2 sets of 3-way swtiches + 1 set of 4-way switch
4 different locations = 2 sets of 3-way switches + 2 sets of 4-way switches
5 different locations = 2 sets of 3-way switches + 3 sets of 4-way switches
6 different locations = 2 sets of 3-way switches + 4 sets of 4-way switches
1,000 different locations = 2 sets of 3-way switches + 998 sets of 4-way switches
On the other hand, the term "gang" in 1-Gang, 2-Gang, 3-Gang, etc. only means the number of actual switch modules installed on a single plate. In short, it only pertains to the physical appearance and NOT directly related to switching operations of 1-Way/2-Way, 3-Way or 4-Way switches I have described above.
Therefore, the correct answer is d. Four-way switch.
Q#16: What is the maximum height of a receptacle outlet located above a countertop?
a. 19.685 inches
b. 60 cm
c. 1 ft. (approx.)
d. 1.35 meters
According the Philippine Electrical Code (PEC) on Wiring and Protection,
"Receptacle Outlet Location. Receptacle outlets shall be located above, but NOT MORE THAN 500 MM ABOVE, the COUNTERTOP."
(Sec. 2.10.3.3.c.5 of the PEC)
So, converting to English units, 500mm = 19.685" MAXIMUM height of a receptacle outlet above a countertop.
Therefore, the correct answer is a. 19.685 inches (or 500mm).
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