CSLB General Building (B) — All Questions

A heating element of 30 ohms is connected across 240 volts. What current flows?

• a.8 amperes✓
• b.0.125 ampere
• c.7,200 amperes
• d.12 amperes

Ohm's law gives I = E / R = 240 / 30 = 8 amperes. The 0.125-ampere answer inverts the division, 7,200 multiplies voltage by resistance instead of dividing, and 12 amperes would require a 20-ohm element.

What power is consumed by a 120-volt load drawing 12 amperes at unity power factor?

• a.1,440 watts✓
• b.10 watts
• c.132 watts
• d.1,200 watts

Power in a resistive single-phase circuit is P = E x I = 120 x 12 = 1,440 watts. The 10-watt answer divides instead of multiplying, 132 watts adds the two values, and 1,200 watts does not follow from the given numbers.

How much power is dissipated when 5 amperes flows through a 20-ohm resistor?

• a.100 watts
• b.500 watts✓
• c.4 watts
• d.2,000 watts

Using P = I squared x R, the calculation is 5 x 5 x 20 = 500 watts. The 100-watt answer multiplies current by resistance without squaring the current, 4 watts divides the wrong way, and 2,000 watts squares the resistance instead of the current.

Three resistors of 10, 15 and 25 ohms are connected in series across 120 volts. What is the circuit current?

• a.24 amperes
• b.2.4 amperes✓
• c.4.8 amperes
• d.0.42 ampere

Series resistances add: 10 + 15 + 25 = 50 ohms, so I = 120 / 50 = 2.4 amperes. The 24-ampere answer misplaces the decimal, 4.8 amperes would result from a 25-ohm total, and 0.42 ampere inverts the division.

A 20-ohm resistor and a 30-ohm resistor are connected in parallel. What is the total resistance?

• a.50 ohms
• b.12 ohms✓
• c.25 ohms
• d.10 ohms

For two resistors in parallel, R total = (R1 x R2) / (R1 + R2) = (20 x 30) / 50 = 600 / 50 = 12 ohms. The 50-ohm answer adds them as if they were in series, and 25 ohms averages them, but total parallel resistance is always less than the smallest branch resistance.

A balanced three-phase load draws 40 amperes at 480 volts with a power factor of 0.85. What is the true power?

• a.Approximately 33.3 kilowatts
• b.Approximately 28.3 kilowatts✓
• c.Approximately 19.2 kilowatts
• d.Approximately 16.3 kilowatts

True power for a three-phase load is P = 1.732 x E x I x PF = 1.732 x 480 x 40 x 0.85, or about 28,270 watts. The 33.3-kilowatt answer omits the power factor and gives apparent power in kVA instead, 19.2 kilowatts leaves out the 1.732 factor, and 16.3 kilowatts omits both.

On a 208Y/120-volt wye system, what is the relationship between line voltage and phase voltage?

• a.Line voltage equals phase voltage divided by 1.732
• b.Line voltage equals phase voltage
• c.Line voltage equals phase voltage multiplied by 3
• d.Line voltage equals phase voltage multiplied by 1.732✓

In a wye configuration the line voltage is the square root of three times the phase voltage: 120 x 1.732 = 208 volts, while line current equals phase current. In a delta configuration the relationship reverses, with line voltage equal to phase voltage and line current equal to 1.732 times phase current.

A balanced delta-connected load has a phase current of 20 amperes. What is the line current?

• a.20 amperes
• b.Approximately 34.6 amperes✓
• c.Approximately 11.5 amperes
• d.60 amperes

In a delta connection each line conductor feeds two phase windings, so line current equals 1.732 times phase current: 20 x 1.732 = 34.6 amperes. The 20-ampere answer describes a wye connection, 11.5 amperes divides instead of multiplying, and 60 amperes triples the value rather than applying the square root of three.

A single-phase transformer has a 480-volt primary and a 120-volt secondary. If the secondary carries 100 amperes, what is the approximate primary current?

• a.50 amperes
• b.100 amperes
• c.400 amperes
• d.25 amperes✓

Voltage and current transform inversely, so the 4 to 1 voltage ratio produces a 1 to 4 current ratio: 100 / 4 = 25 amperes on the primary. Neglecting losses, the volt-amperes match on both sides, 480 x 25 = 120 x 100 = 12,000 VA. The 400-ampere answer inverts the ratio and would represent an impossible power gain.

What is the secondary full-load current of a 25 kVA single-phase transformer with a 240-volt secondary?

• a.Approximately 60 amperes
• b.Approximately 104 amperes✓
• c.Approximately 208 amperes
• d.Approximately 52 amperes

For a single-phase transformer, I = VA / E = 25,000 / 240, or about 104 amperes. The 60-ampere answer divides by 480 volts, 208 amperes divides by 120 volts, and 52 amperes is half the correct value.

A 480-volt panelboard is mounted on a wall with a grounded concrete wall directly behind the working space. What is the minimum depth of working space?

• a.2 and 1/2 feet
• b.3 feet
• c.4 feet
• d.3 and 1/2 feet✓

For nominal voltages of 151 to 600 volts to ground, Condition 2 exists when exposed live parts face a grounded surface such as a concrete or masonry wall, and the required depth is 3 and 1/2 feet. Condition 1, with no live or grounded parts opposite, requires 3 feet, and Condition 3, with exposed live parts on both sides, requires 4 feet.2023 NEC §110.26(A)(1)

What is the minimum width of working space in front of electrical equipment?

• a.30 inches in all cases regardless of equipment width
• b.30 inches, or the width of the equipment, whichever is greater✓
• c.36 inches, or the width of the equipment, whichever is greater
• d.24 inches, measured from the centerline of the equipment

The working space must be at least 30 inches wide or as wide as the equipment, whichever is greater, and it must allow all doors and hinged panels to open at least 90 degrees. The space need not be directly centered on the equipment, but it must be continuous across the full width required.2023 NEC §110.26(A)(2)

Entrances at each end of the working space are generally required for equipment rated at least what value and over 6 feet wide?

• a.1,000 amperes
• b.800 amperes
• c.1,200 amperes✓
• d.600 amperes

Equipment rated 1,200 amperes or more and over 6 feet wide containing overcurrent, switching or control devices needs an entrance at each end of the working space so a worker is never trapped behind an arcing fault. The 1,000-ampere threshold belongs to service ground-fault protection, and 800 amperes is the ceiling for the next-standard-size overcurrent rule.2023 NEC §110.26(C)(2)

The dedicated equipment space above a panelboard extends from the top of the equipment to what height?

• a.To the structural ceiling in all cases
• b.3 feet above the equipment in all cases
• c.6 feet above the equipment or to the structural ceiling, whichever is lower✓
• d.6 and 1/2 feet above the finished floor

The dedicated space is the footprint of the equipment extended from the floor to a height of 6 feet above it or to the structural ceiling, whichever is lower, and no foreign piping, ducts or equipment may be installed in it. Suspended ceilings with removable panels are not considered structural ceilings, and the 6 and 1/2-foot dimension is the minimum headroom of working space, a separate requirement.2023 NEC §110.26(E)(1)

Which sequence correctly describes verifying that equipment is de-energized before beginning work?

• a.Test the meter on a known live source, test the de-energized conductors, then retest the meter on the known live source✓
• b.Test the de-energized conductors, then test the meter on a known live source
• c.Rely on the open position of the disconnect and the applied lock
• d.Test the de-energized conductors only, since the meter was checked at the start of the shift

The live-dead-live method proves the instrument is working both before and after the absence-of-voltage test, so a meter that failed during the check cannot lead a worker to treat energized conductors as safe. Locking and tagging the disconnect is required, but it does not by itself confirm that the conductors ahead of the work are actually de-energized, since backfeeds and mislabeled circuits are common.