CSLB General Building (B) — All Questions

A single-family dwelling has 2,000 sq ft of habitable floor area. Using the standard calculation method at 3 VA per sq ft, what is the general lighting and general-use receptacle load before any demand factors?

• a.6,000 VA✓
• b.4,000 VA
• c.8,000 VA
• d.3,000 VA

General lighting is figured at 3 VA per sq ft: 2,000 sq ft x 3 VA = 6,000 VA. This value is calculated before applying the general-lighting demand factors.

In the standard dwelling calculation, what is the minimum load to include for two required small-appliance branch circuits plus one laundry branch circuit, before any demand factor?

• a.3,000 VA
• b.4,500 VA✓
• c.6,000 VA
• d.1,500 VA

Each small-appliance and laundry circuit is figured at 1,500 VA. Two small-appliance circuits (2 x 1,500 = 3,000) plus one laundry circuit (1,500) equals 4,500 VA.

A dwelling has 7,200 VA general lighting, 3,000 VA small-appliance, and 1,500 VA laundry. Applying the standard demand factor (first 3,000 VA at 100%, remainder at 35%), what is the net general lighting load?

• a.11,700 VA
• b.7,200 VA
• c.6,045 VA✓
• d.4,095 VA

Sum = 7,200 + 3,000 + 1,500 = 11,700 VA. First 3,000 at 100% = 3,000; remaining 8,700 at 35% = 3,045. Net = 3,000 + 3,045 = 6,045 VA.

Using the demand table for household electric ranges, what is the demand load for one 12-kW electric range in a dwelling (Column C, one appliance)?

• a.12,000 VA
• b.9,600 VA
• c.6,000 VA
• d.8,000 VA✓

For a single household range not over 12 kW, the Column C demand is 8 kW = 8,000 VA. The 12-kW nameplate is reduced to an 8-kW calculated demand for one range.

What is the demand load for two 12-kW household electric ranges served from one feeder (range demand table, Column C, two appliances)?

• a.11,000 VA✓
• b.16,000 VA
• c.12,000 VA
• d.19,200 VA

Two ranges not over 12 kW have a Column C demand of 11 kW = 11,000 VA. This is less than the 24,000 VA connected total because of diversity.

A single household range is rated 16 kW. Using the 5% increase per kW over 12 kW rule on the 8-kW base, what is the demand load?

• a.8,000 VA
• b.9,600 VA✓
• c.16,000 VA
• d.10,400 VA

16 kW is 4 kW over 12 kW, so add 5% x 4 = 20% to the 8-kW base: 8,000 x 1.20 = 9,600 VA.

A dwelling has one 4-kW electric clothes dryer. What value must be used as the dryer demand load in the service calculation?

• a.4,000 VA
• b.4,500 VA
• c.5,000 VA✓
• d.5,500 VA

The dryer load is the nameplate rating or 5,000 VA, whichever is larger. Since 4,000 VA is below 5,000 VA, 5,000 VA must be used.2023 NEC 220.54

A dwelling has four fastened-in-place appliances (dishwasher, disposal, water heater, compactor) totaling 6,000 VA. What is the demand load after applying the fastened-appliance demand factor?

• a.6,000 VA
• b.3,000 VA
• c.7,500 VA
• d.4,500 VA✓

Where four or more fastened-in-place appliances are on the same feeder, a 75% demand factor may be applied: 6,000 x 0.75 = 4,500 VA.2023 NEC 220.53

A dwelling's total calculated load is 24,000 VA on a 240-V, single-phase service. What is the calculated service current?

• a.100 A✓
• b.120 A
• c.83 A
• d.167 A

Current equals volt-amperes divided by voltage: 24,000 VA / 240 V = 100 A.

Using the 75 C column, what is the smallest copper conductor rated to carry a 100-A service load?

• a.#4 AWG
• b.#3 AWG✓
• c.#2 AWG
• d.#1 AWG

At 75 C, #4 Cu is rated 85 A and #3 Cu is rated 100 A. The #3 AWG copper is the smallest conductor that meets the 100-A requirement.2023 NEC Table 310.16

For a 200-A dwelling service using the 83% conductor allowance, what is the smallest copper ungrounded conductor (75 C) permitted for the main power feeder?

• a.#3/0 AWG
• b.#1/0 AWG
• c.#2/0 AWG✓
• d.#4/0 AWG

200 A x 0.83 = 166 A required ampacity. At 75 C, #2/0 Cu is rated 175 A, which satisfies 166 A, so #2/0 is the smallest permitted.2023 NEC 310.12

An electric range has a Column C demand of 8,000 VA. Using the 70% neutral demand for ranges, what is the range contribution to the feeder neutral load?

• a.8,000 VA
• b.4,000 VA
• c.7,000 VA
• d.5,600 VA✓

The feeder neutral for a household range may be figured at 70% of the range demand: 8,000 x 0.70 = 5,600 VA.

From the motor full-load current table, what is the full-load current of a 25-HP, 480-V, three-phase induction motor?

• a.34 A✓
• b.40 A
• c.28 A
• d.42 A

The table value for a 25-HP, 460-480-V three-phase motor is 34 A. Table full-load current (FLC), not nameplate, is used for conductor and protection sizing.2023 NEC Table 430.250

A single continuous-duty motor has a table full-load current of 34 A. What minimum conductor ampacity is required for its branch circuit?

• a.34 A
• b.42.5 A✓
• c.40 A
• d.50 A

Motor branch-circuit conductors must be at least 125% of the FLC: 34 A x 1.25 = 42.5 A minimum ampacity.2023 NEC 430.22

A motor with a service factor of 1.15 has a nameplate full-load current of 34 A. What is the maximum overload protection setting at 125%?

• a.34 A
• b.39.1 A
• c.42.5 A✓
• d.45.9 A

For a motor with service factor 1.15 or greater, overload is set at up to 125% of nameplate FLA: 34 x 1.25 = 42.5 A.2023 NEC 430.32

For a motor with a 34-A full-load current, what is the maximum inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection (250%, then next higher standard size)?

• a.70 A
• b.80 A
• c.100 A
• d.90 A✓

Inverse-time breaker: 250% of FLC = 34 x 2.50 = 85 A. Where this does not correspond to a standard size, the next higher standard rating (90 A) is permitted.2023 NEC 430.52

A feeder supplies three motors with full-load currents of 34 A, 28 A, and 22 A. What minimum conductor ampacity is required for the feeder?

• a.92.5 A✓
• b.105 A
• c.84 A
• d.110 A

Feeder ampacity = 125% of the largest motor FLC plus the sum of the others: (34 x 1.25) + 28 + 22 = 42.5 + 50 = 92.5 A.2023 NEC 430.24

What is the full-load primary current of a 75-kVA, 480-V, three-phase transformer?

• a.156 A
• b.90 A✓
• c.104 A
• d.72 A

Three-phase current = VA / (1.732 x V): 75,000 / (1.732 x 480) = 90.2 A, about 90 A.

What is the full-load secondary current of a 75-kVA, three-phase transformer with a 208-V secondary?

• a.156 A
• b.180 A
• c.208 A✓
• d.90 A

Secondary current = 75,000 / (1.732 x 208) = 208.2 A, about 208 A.

A 75-kVA transformer has a 90-A primary current (over 9 A). With primary-only protection at 125% max, what is the primary overcurrent device rating (next higher standard permitted)?

• a.110 A
• b.100 A
• c.150 A
• d.125 A✓

125% of 90 A = 112.5 A. Where this does not match a standard rating, the next higher standard device (125 A) is permitted for primary-only protection.2023 NEC 450.3(B)

A 120-V branch circuit carries 16 A over a one-way run of 80 ft using #12 Cu (circular mils = 6,530, K = 12.9). What is the approximate voltage drop?

• a.5.1 V✓
• b.3.0 V
• c.2.5 V
• d.7.4 V

Single-phase VD = (2 x K x I x L) / CM = (2 x 12.9 x 16 x 80) / 6,530 = 33,024 / 6,530 = 5.06 V.

Using the previous circuit (5.06-V drop on a 120-V circuit), what is the approximate percent voltage drop?

• a.2.1%
• b.4.2%✓
• c.6.0%
• d.3.0%

Percent VD = 5.06 / 120 = 0.042 = 4.2%, which exceeds the 3% recommended for branch circuits.

A 208-V three-phase feeder carries 40 A over 150 ft using #6 Cu (CM = 26,240, K = 12.9). What is the approximate voltage drop?

• a.3.0 V
• b.7.2 V
• c.5.1 V✓
• d.9.0 V

Three-phase VD = (1.732 x K x I x L) / CM = (1.732 x 12.9 x 40 x 150) / 26,240 = 134,050 / 26,240 = 5.11 V.

For a 112.5-kVA, 208-V, three-phase transformer with 2% impedance, what is the approximate available fault current at the secondary terminals (infinite primary)?

• a.6,250 A
• b.20,000 A
• c.10,000 A
• d.15,617 A✓

Secondary FLA = 112,500 / (1.732 x 208) = 312.3 A. Isc = FLA / %Z = 312.3 / 0.02 = 15,617 A.

A 45-kVA, 208-V, three-phase transformer has 2% impedance. What is the approximate available fault current at its secondary (infinite primary)?

• a.6,245 A✓
• b.3,120 A
• c.12,500 A
• d.9,000 A

Secondary FLA = 45,000 / (1.732 x 208) = 124.9 A. Isc = 124.9 / 0.02 = 6,245 A.

Six current-carrying THHN conductors share one raceway. If the 90 C ampacity of #6 Cu is 75 A, what is the adjusted ampacity after the fill adjustment factor for 4-6 conductors?

• a.75 A
• b.60 A✓
• c.55 A
• d.52.5 A

For 4 to 6 current-carrying conductors, apply an 80% adjustment factor: 75 A x 0.80 = 60 A.2023 NEC Table 310.15(C)(1)

A #3 Cu THHN conductor has a 90 C ampacity of 110 A. In a 50 C ambient (correction factor 0.82), what is the corrected ampacity?

• a.110 A
• b.100 A
• c.90 A✓
• d.82 A

Corrected ampacity = 110 A x 0.82 = 90.2 A, about 90 A, before comparing to the termination temperature limit.2023 NEC Table 310.15(B)(1)

A #2 Cu THHN (90 C ampacity 130 A) runs with 6 current-carrying conductors (0.80 factor) in a 40 C ambient (0.91 factor). What is the adjusted-and-corrected ampacity?

• a.130 A
• b.115 A
• c.104 A
• d.95 A✓

Apply both factors to the 90 C ampacity: 130 x 0.80 x 0.91 = 94.6 A, about 95 A.

An office has 45 general-use receptacle outlets. At 180 VA per outlet, what is the receptacle load before demand factors?

• a.8,100 VA✓
• b.4,500 VA
• c.9,000 VA
• d.5,400 VA

Each general-use receptacle outlet is figured at 180 VA: 45 x 180 = 8,100 VA.2023 NEC 220.14(I)

A commercial building has a 30,000-VA receptacle load. Applying the demand (first 10 kVA at 100%, remainder at 50%), what is the demand load?

• a.30,000 VA
• b.20,000 VA✓
• c.15,000 VA
• d.25,000 VA

First 10,000 VA at 100% = 10,000; remaining 20,000 at 50% = 10,000. Total demand = 20,000 VA.