CSLB General Building (B) — All Questions

You need to check whether the insulation of a 480 V motor winding has degraded to ground. Which test instrument is designed for this measurement?

• a.A standard digital multimeter set to the 200 ohm range
• b.A clamp-on ammeter
• c.A megohmmeter (insulation-resistance tester)✓
• d.A plug-in receptacle tester

A megohmmeter applies a high DC test voltage (e.g., 500-1000 V) and reads insulation resistance in the megohm range, revealing breakdown a standard DMM cannot. A DMM's ohm range uses only a few volts and tops out far too low. A clamp ammeter reads current, and a receptacle tester only checks 120 V outlet wiring.

A heating element rated at 24 ohms is connected across a 120 V supply. Using Ohm's law, what current flows through it?

• a.5 A✓
• b.0.2 A
• c.2,880 A
• d.144 A

Ohm's law gives I = V / R = 120 V / 24 ohms = 5 A. Option b inverts the formula (R / V). Option c multiplies V x R and d adds them, neither is Ohm's law.

You will take voltage measurements at the line terminals of a 240 V main service panel. What minimum measurement category (CAT) rating should your meter and leads carry for this location?

• a.CAT I
• b.CAT II
• c.No CAT rating is needed for 240 V
• d.CAT III (or CAT IV)✓

Service-entrance and main-panel/feeder work sits at CAT III-IV because available fault energy and transient spikes are highest there. CAT I is for protected electronic circuits and CAT II for cord-and-plug loads at receptacles. Using an under-rated meter risks an arc-blast if a transient occurs.

A 120 V branch circuit supplies a load drawing 1,500 W. What current does the load draw?

• a.18 A
• b.12.5 A✓
• c.1,380 A
• d.0.08 A

From P = V x I, I = P / V = 1,500 W / 120 V = 12.5 A. Option a multiplies incorrectly, c multiplies P x V, and d inverts to V / P.

Before performing a continuity test on a cable with an ohmmeter, what condition must exist?

• a.The circuit must be de-energized (no voltage present)✓
• b.The circuit must be energized at full load
• c.The conductor must be under 120 V
• d.The neutral must be bonded to a driven ground rod

An ohmmeter/continuity function supplies its own small test voltage; applying it to a live circuit gives false readings and can damage the meter or injure you. The conductor must be de-energized and isolated. Energizing (b, c) defeats the test, and grounding a rod (d) is unrelated.

A plug-in receptacle tester shows its 'hot/neutral reversed' indication at a 120 V outlet. What does this mean and what is the fix?

• a.The equipment ground is open; drive a new ground rod
• b.The receptacle is fine; the tester is faulty
• c.The hot and neutral conductors are swapped at the receptacle terminals; reconnect the ungrounded (hot) conductor to the brass screw and the neutral to the silver screw✓
• d.The voltage is too high; install a buck transformer

Reversed polarity means the ungrounded (hot) conductor is landed on the neutral (silver) terminal and vice-versa, leaving the device energized through the neutral side. Correct it by moving the hot to the brass screw and neutral to the silver screw. It is a wiring error, not an open ground (a) or over-voltage (d).

On a 120/240 V multiwire branch circuit, the shared neutral becomes open. What symptom is most characteristic?

• a.Both 120 V loads simply stop working with 0 V at each
• b.The two 120 V loads see unequal, shifting voltages, one above 120 V and one below, as their series combination divides the 240 V✓
• c.The breaker trips immediately every time
• d.Voltage to ground doubles equally on both lines to 240 V each

With the neutral open, the two line-to-neutral loads are placed in series across 240 V; the voltage divides inversely to their resistances, so the lighter-loaded side rises (often damaging equipment) and the heavier side drops. They do not both read 0 V (a). An open neutral does not by itself trip the breaker (c) or raise each line to 240 V to ground (d).

You test a 480 V motor with a 500 V megohmmeter and read 0.4 megohms. Using the common one-megohm-per-kilovolt-plus-one-megohm minimum guideline, is the insulation acceptable?

• a.Yes, 0.4 megohms easily exceeds the minimum
• b.Yes, any reading above 0 megohms passes
• c.Cannot tell without measuring current
• d.No, the reading is well below the ~1 megohm minimum, indicating degraded insulation✓

A widely used field minimum is about 1 megohm (or 1 megohm per kV of rating plus 1). At 0.4 megohms the motor is far below that floor, so the insulation is suspect and the motor should not be returned to service without further evaluation. A near-zero reading trends toward a winding-to-ground fault; 'any value above 0' (b) is not a valid criterion.

Which instrument lets you measure the current in a single energized conductor without disconnecting or breaking into the circuit?

• a.A clamp-on ammeter✓
• b.An in-line series ammeter on the DMM's 10 A jack
• c.A megohmmeter
• d.A non-contact voltage detector

A clamp meter senses the magnetic field around a conductor, so you simply clamp its jaws around one wire while the circuit stays energized and intact. A series ammeter (b) requires opening the circuit and inserting the meter. A megohmmeter reads insulation resistance and a non-contact detector only indicates presence of voltage, not current.

A DMM reads 12 V across a resistor and a clamp meter reads 3 A through it. What is the resistor's value?

• a.36 ohms
• b.4 ohms✓
• c.0.25 ohms
• d.15 ohms

R = V / I = 12 V / 3 A = 4 ohms. Option a multiplies V x I (that yields watts, 36 W), c inverts to I / V, and d adds the values.

A conductor run carries 10 A and the total round-trip resistance of the wire is 0.5 ohms. What is the voltage drop along the conductors?

• a.20 V
• b.0.05 V
• c.10.5 V
• d.5 V✓

Voltage drop Vd = I x R = 10 A x 0.5 ohms = 5 V. Option a divides I by R inverted, b divides R by I, and c adds the numbers, none is Ohm's law applied to the wire resistance.

Three resistors of 10 ohms, 20 ohms, and 30 ohms are wired in series. What is the total resistance the ohmmeter should read across the string?

• a.5.45 ohms
• b.10 ohms
• c.60 ohms✓
• d.0.183 ohms

Series resistances add: 10 + 20 + 30 = 60 ohms. Option a is the parallel result, and b/d ignore two of the resistors or invert the math.

Two 20 ohm resistors are connected in parallel. What total resistance will an ohmmeter read?

• a.10 ohms✓
• b.40 ohms
• c.20 ohms
• d.400 ohms

For two equal parallel resistors, the total is half of one: 20 ohms / 2 = 10 ohms. Option b is the series total, c is a single resistor, and d multiplies them.

The 2023 NEC added a requirement on how terminations must be tightened when the manufacturer provides a torque value. What does it require?

• a.Tighten firmly by hand until the conductor cannot be pulled out
• b.Use any screwdriver and tighten an extra quarter-turn past snug
• c.Torque is only required on conductors 4 AWG and larger
• d.Tighten to the manufacturer's specified torque using a calibrated torque tool✓

Where a manufacturer provides a torque value, the connection must be made with a calibrated torque tool (torque screwdriver/wrench) to that value, since both under- and over-torquing cause failures and heating. Hand-tight or 'quarter-turn' methods (a, b) are not the code rule, and the requirement is not limited to large conductors (c).2023 NEC §110.14(D)

A terminal on a device is marked 'CU' only. What does this mean for terminations?

• a.It may be used with copper or aluminum interchangeably
• b.It is listed for copper conductors only; aluminum must not be landed on it✓
• c.It requires antioxidant paste with any conductor
• d.It is rated for control wiring only

A 'CU'-only marking means the terminal is tested and listed for copper conductors exclusively; landing aluminum can cause overheating and failure from oxidation and dissimilar-metal effects. Only terminals marked 'CU-AL' or 'AL-CU' accept aluminum. The marking is about conductor material, not paste (c) or a voltage/control limit (d).2023 NEC §110.14

A newly connected three-phase motor runs backward. What is the standard corrective action?

• a.Reverse all three line leads at the motor
• b.Add a phase-shifting capacitor to one leg
• c.Interchange any two of the three line leads✓
• d.Increase the supply voltage by 10%

Swapping any two of the three phase conductors reverses the rotation of the motor's rotating magnetic field, so the motor turns the correct way. Reversing all three (a) leaves rotation unchanged. A capacitor (b) and higher voltage (d) do not change phase sequence.

A receptacle reads 120 V hot-to-neutral but 0 V hot-to-ground. Neutral-to-ground also reads 0 V and the ground pin has no continuity to the panel. What is the fault?

• a.Open equipment grounding conductor (open ground)✓
• b.Reversed polarity
• c.Open neutral
• d.A bolted line-to-line short

With hot-to-neutral normal at 120 V but hot-to-ground reading 0 V and no continuity from the ground pin back to the panel, the equipment grounding conductor is open. Reversed polarity (b) would still show voltage relationships swapped, an open neutral (c) would kill the hot-to-neutral reading, and a line-to-line short (d) would trip the breaker.

Testing a suspected conductor with an ohmmeter, you read 'OL' (over-limit / infinite). What does this indicate about the conductor?

• a.A dead short (0 ohms) between the leads
• b.A perfectly good, low-resistance conductor
• c.An open circuit, the path is broken (infinite resistance)✓
• d.Exactly 1 megohm of resistance

'OL' or an infinity display means resistance is beyond the meter's range, an open (broken) path with no continuity. A dead short reads near 0 ohms (a), a good conductor reads a very low value (b), and OL is not a specific finite value like 1 megohm (d).

In a properly bonded system, what is the primary purpose of the equipment grounding conductor during a line-to-enclosure (ground) fault?

• a.To carry normal load current back to the source
• b.To provide a low-impedance fault-current path that lets the overcurrent device trip quickly✓
• c.To improve power factor
• d.To limit voltage drop on long runs

The EGC's job on a fault is to give fault current a low-impedance path back to the source so enough current flows to open the breaker/fuse fast, clearing the fault and de-energizing the enclosure. It does not carry normal load current (that's the neutral, a), and it has nothing to do with power factor (c) or normal-load voltage drop (d).

Before touching conductors you believe are de-energized, what is the safest way to verify your voltage tester is trustworthy?

• a.Assume it works if the battery indicator is on
• b.Test only the conductors in question once
• c.Shake the meter and listen for the beep
• d.Prove the tester on a known-live source, test the de-energized conductors, then re-prove it on the known-live source (live-dead-live)✓

The live-dead-live method proves the tester reads voltage on a known source before and after checking the suspect conductors, confirming the meter didn't fail mid-test. A battery light (a) or a single test (b) does not prove the tester actually works, and shaking it (c) proves nothing.

For routine voltage checks at 120 V cord-and-plug receptacle outlets on a branch circuit, what CAT rating is generally the minimum appropriate category?

• a.CAT II✓
• b.CAT 0
• c.No rating required
• d.CAT IV only

Receptacle-level, cord-and-plug load measurements correspond to CAT II. CAT III/IV are for distribution panels, feeders, and service equipment; using a higher-rated meter is fine, but CAT II is the minimum appropriate for the outlet. 'CAT 0' (b) is not a standard category and a rating is always required for live work.

You want to measure current with a DMM's internal ammeter function. Which practice is essential to avoid a dangerous fault?

• a.Connect the meter in parallel across the load like a voltmeter
• b.Connect the meter in series with the load, on the correct (fused) current jack, and never across a voltage source✓
• c.Leave the leads in the current jacks while also measuring voltage
• d.Use the 10 A jack for all measurements regardless of range

An ammeter is a near-short internally, so it must be placed in series with the load using the proper fused current jack; connecting it in parallel across voltage (a) creates a bolted fault. Leaving leads in the current jacks while probing voltage (c) is a classic cause of arc-blast, and blindly using the 10 A jack (d) can bypass the fuse or misread small currents.

A conductor with 0.2 ohms of resistance carries 15 A. How much power is dissipated as heat in that conductor?

• a.3 W
• b.75 W
• c.0.9 W
• d.45 W✓

Power in a resistance is P = I^2 x R = (15 A)^2 x 0.2 ohms = 225 x 0.2 = 45 W. Option a is I x R (a voltage, 3 V), b divides incorrectly, and c uses I x R x 0.3. Excess I^2R heating is why loose or undersized terminations overheat.

A homeowner reports a 'weird' outlet. Which instrument most quickly indicates common miswires (open ground, open neutral, reversed polarity) at a 120 V receptacle?

• a.A megohmmeter
• b.A clamp-on ammeter
• c.A plug-in receptacle (three-light) tester✓
• d.A phase-rotation meter

A plug-in three-light receptacle tester lights a coded pattern that flags open ground, open neutral, and reversed polarity instantly. A megohmmeter tests insulation, a clamp meter reads current, and a phase-rotation meter is for three-phase sequence, none of which quickly maps outlet miswires.

Two identical 60 W, 120 V lamps are accidentally wired in series across 120 V. Approximately what voltage will a DMM read across each lamp, and how will they glow?

• a.120 V each; both at full brightness
• b.About 60 V each; both glow dim✓
• c.120 V on one, 0 V on the other
• d.240 V each; both very bright

Identical lamps in series split the 120 V roughly equally, so each sees about 60 V and both glow dimly at reduced power. They cannot each get the full 120 V in series (a, d), and equal lamps do not divide 120/0 (c). Unequal loads would divide the voltage unequally.

You must measure current on a circuit feeding electronic (nonlinear) loads such as LED drivers and VFDs. Which meter feature gives an accurate reading?

• a.A true-RMS meter✓
• b.An averaging (non-true-RMS) meter
• c.An analog swinging-needle meter only
• d.Any meter reads distorted waveforms accurately

Nonlinear loads draw distorted, non-sinusoidal current, and only a true-RMS meter computes the real heating value accurately. Averaging meters (b) assume a clean sine wave and under- or over-read on distorted waveforms. An analog meter (c) is also averaging-type, and not all meters handle distortion (d).

A 6 ohm and a 3 ohm resistor are connected in parallel. What total resistance should an ohmmeter read?

• a.9 ohms
• b.4.5 ohms
• c.2 ohms✓
• d.18 ohms

Parallel: R = (R1 x R2) / (R1 + R2) = (6 x 3) / (6 + 3) = 18 / 9 = 2 ohms. The total is always less than the smallest resistor. Option a is the series sum, b the simple average, and d the product only.

During commissioning you want to confirm that a metal equipment enclosure is effectively bonded. Which check best verifies a low-impedance bond to ground?

• a.Measure 120 V from enclosure to a hot conductor
• b.Confirm the enclosure is painted for corrosion resistance
• c.Verify the enclosure feels cool to the touch
• d.Measure a very low resistance/continuity between the enclosure and the grounding-electrode/EGC point✓

A low-resistance continuity reading between the enclosure and the grounding system confirms an effective bond that can carry fault current. Reading 120 V to a hot (a) is a shock hazard test, not a bond check; paint (b) actually impairs bonding at contact surfaces; and temperature (c) tells nothing about the ground path.

In a series circuit, 24 V is applied across a 4 ohm and an 8 ohm resistor in series. What voltage will a DMM read across the 8 ohm resistor?

• a.16 V✓
• b.8 V
• c.24 V
• d.2 V

Total R = 4 + 8 = 12 ohms, so I = 24 V / 12 ohms = 2 A. Across the 8 ohm resistor, V = I x R = 2 A x 8 ohms = 16 V (the 4 ohm drops the remaining 8 V). The larger series resistor drops the larger share of voltage.

A branch-circuit breaker trips instantly every time it is reset, with nothing plugged in. Continuity from hot to ground reads near 0 ohms. What is the most likely fault?

• a.Simple overload from too many appliances
• b.An open neutral
• c.A short circuit (hot-to-ground or hot-to-neutral) in the wiring✓
• d.Undersized service transformer

Instant tripping with no load, plus a near-0 ohm hot-to-ground reading, points to a bolted short circuit providing a very low-resistance fault path. An overload (a) needs connected load and trips on a time delay, an open neutral (b) would not trip the breaker, and transformer sizing (d) does not cause instantaneous branch trips.

Three parallel loads on a 120 V circuit draw 4 A, 6 A, and 5 A respectively. What total current should a clamp meter read on the feeder conductor?

• a.5 A
• b.15 A✓
• c.120 A
• d.1.6 A

Currents in parallel branches add: 4 + 6 + 5 = 15 A on the common feeder. Option a is only one branch, c confuses voltage with current, and d inverts the sum. This is why the feeder must be sized for the combined load.

A non-contact voltage tester (NCVT) does not light up near a conductor you must work on. Before treating the conductor as dead, what is the correct action?

• a.Trust the NCVT completely and begin work
• b.Assume the conductor is a neutral and cut it
• c.NCVTs are infallible, so no further check is needed
• d.Confirm de-energization with a contact voltage tester using live-dead-live, since NCVTs can miss voltage (shielding, ganged wires, low sensitivity)✓

An NCVT is only a first indicator; it can fail to detect voltage due to shielding, cable geometry, or dead batteries, so it must never be the sole basis for calling a conductor dead. Verify with a contact tester using the live-dead-live method before touching conductors. Trusting it blindly (a, c) or guessing it's a neutral (b) can be fatal.

A single-phase load operates at 240 V and draws 20 A at unity power factor. What is its power in kilowatts?

• a.12 kW
• b.0.083 kW
• c.4.8 kW✓
• d.48 kW

P = V x I = 240 V x 20 A = 4,800 W = 4.8 kW at unity power factor. Option a inverts, b divides V by I, and d misplaces the decimal. Confirming measured watts against the nameplate is part of functional commissioning.

Before connecting a three-phase motor to a new feeder, why would you use a phase-rotation (phase-sequence) meter on the supply?

• a.To determine the phase sequence (A-B-C vs A-C-B) so the motor will rotate in the intended direction✓
• b.To measure insulation resistance of the feeder
• c.To read the exact current the motor will draw
• d.To verify the neutral is bonded

A phase-rotation meter identifies the sequence of the three phases so you can connect the motor for correct rotation before energizing, avoiding damage from running backward (e.g., pumps, compressors). It does not measure insulation (b), load current (c), or neutral bonding (d).

Using the rule of thumb of 1 megohm minimum per 1,000 V of rating, what is the minimum acceptable insulation resistance for a 4,160 V motor?

• a.0.24 megohms
• b.1 megohm
• c.416 megohms
• d.About 4 to 5 megohms✓

The 1-megohm-per-kV guideline gives roughly 4,160 V / 1,000 = about 4.16 megohms, often rounded up (some add 1 megohm) to about 4-5 megohms minimum. Readings below this suggest degraded insulation. Option a inverts the ratio and c multiplies incorrectly.

Why does over-torquing a lug or terminal screw create a hazard, even though the connection feels 'extra tight'?

• a.Over-torque improves conductivity and is always safer
• b.Over-torque can crush strands, deform the lug, and damage threads, raising resistance and causing a hot connection or eventual failure✓
• c.Over-torque only matters on aluminum, never copper
• d.Torque has no effect on connection reliability

Both under- and over-torque are harmful: over-torquing can crack lugs, crush or shear conductor strands, and strip threads, increasing contact resistance and heat over time. That is why the NEC requires using the manufacturer's specified value with a calibrated tool. It applies to copper and aluminum alike (c), and torque very much affects reliability (d).2023 NEC §110.14(D)

A homeowner reports that throughout the house, some lights dim while others brighten, and the effect shifts as loads switch on and off. Which service-level fault best explains this?

• a.An open (loose or corroded) service neutral✓
• b.A single tripped GFCI receptacle
• c.One burned-out light bulb
• d.A reversed-polarity outlet in one bedroom

Whole-house dimming/brightening that shifts with load is the classic signature of an open service neutral: the two 120 V legs float in series across 240 V and divide unevenly. A tripped GFCI (b) or a bad bulb (c) affects only one location, and a single reversed outlet (d) does not cause house-wide voltage swings.

Two meters are both marked CAT III, but one is rated 600 V and the other 1000 V. What does the voltage number add to the CAT rating?

• a.Nothing; only the CAT number matters
• b.It is the meter's battery voltage
• c.It is the maximum resistance the meter can read
• d.It is the maximum working voltage to ground for that category; the meter must meet both the CAT level and the voltage of the circuit✓

The CAT level defines the transient/impulse withstand for a location, and the voltage rating states the maximum working voltage to ground for that category, both must suit the circuit. A CAT III 600 V meter is not appropriate where CAT III 1000 V is required. The number is not battery voltage (b) or a resistance range (c).

By mistake you clamp your ammeter around BOTH the hot and neutral conductors of a single-phase 120 V branch circuit that is drawing 8 A. What will the clamp meter most likely read?

• a.8 A
• b.16 A
• c.Approximately 0 A✓
• d.120 A

The hot and neutral carry equal and opposite currents, so their magnetic fields cancel when both are inside the jaw, giving a reading near 0 A. To read true load current you must clamp only one conductor. A non-zero reading around both would actually indicate leakage/ground-fault current.

A motor-control resistor of 15 ohms must carry 2 A during operation. What voltage drop will a DMM read across it?

• a.7.5 V
• b.30 V✓
• c.0.13 V
• d.17 V

V = I x R = 2 A x 15 ohms = 30 V. Option a divides R by I, c inverts to I / R, and d adds the two numbers, none is Ohm's law.

During commissioning of a new bathroom circuit, how should you verify a GFCI receptacle provides ground-fault protection?

• a.Press the built-in TEST button (and/or use a GFCI tester) and confirm it trips and removes power, then RESET✓
• b.Measure 120 V hot-to-neutral only and assume it works
• c.Check that the LED is lit and skip tripping it
• d.Short the hot to ground with a screwdriver to force a trip

The correct functional test is to operate the built-in TEST button (or an external GFCI tester that injects a small fault) and confirm the device trips and de-energizes the outlet, then reset it. A normal voltage reading (b) or a lit indicator (c) does not prove the ground-fault sensing works. Deliberately shorting to ground (d) is dangerous and unnecessary.

A 120 V heater is rated 600 W. What is the operating resistance of its element (assume purely resistive)?

• a.5 ohms
• b.0.2 ohms
• c.24 ohms✓
• d.72,000 ohms

Using R = V^2 / P = (120 V)^2 / 600 W = 14,400 / 600 = 24 ohms. Equivalently, I = P / V = 5 A, then R = V / I = 120 / 5 = 24 ohms. Option a is the current in amps, and d multiplies V^2 x P.