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NICET EPT practice questions

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Question 1Electrical Safety, Codes, and Standards

Under NFPA 70E, the arc flash boundary is the distance at which a person without arc-rated protection could receive an incident energy of:

  1. 2.4 cal/cm²
  2. 8 cal/cm²
  3. 0.5 cal/cm²
  4. 1.2 cal/cm² Correct answer.

Explanation

NFPA 70E defines the arc flash boundary as the approach limit at which a person could receive a second-degree burn, which corresponds to an incident energy of 1.2 cal/cm². Enclosure doors are not a substitute for the boundary; anyone crossing it needs arc-rated protection.

Question 2Electrical Safety, Codes, and Standards

The OSHA standard that governs the control of hazardous energy (lockout/tagout) for general industry is:

  1. 29 CFR 1910.333
  2. 29 CFR 1910.147 Correct answer.
  3. 29 CFR 1910.269
  4. 29 CFR 1910.132

Explanation

29 CFR 1910.147, The Control of Hazardous Energy (Lockout/Tagout), sets the minimum requirements for isolating and securing energy sources during servicing and maintenance. Electrical work practices are addressed elsewhere, but 1910.147 is the governing lockout/tagout rule.

Question 3Electrical Safety, Codes, and Standards

Rubber insulating gloves must receive a dielectric retest at what maximum interval?

  1. Every 12 months
  2. Every 3 months
  3. Every 24 months
  4. Every 6 months Correct answer.

Explanation

ASTM D120 and NFPA 70E require rubber insulating gloves to be electrically retested at intervals not exceeding 6 months of use, in addition to the field air test performed before each use. Sleeves follow a 12-month retest interval, which is why the two are often confused.

Question 4Electrical Safety, Codes, and Standards

Under NFPA 70E, a qualified person is one who has:

  1. completed a 40-hour arc flash awareness course
  2. has five years of documented field experience
  3. demonstrated the skills and knowledge required Correct answer.
  4. holds a journeyman license issued by the state

Explanation

NFPA 70E defines a qualified person as one who has demonstrated skills and knowledge related to the construction and operation of electrical equipment and installations and has received safety training to recognize hazards and reduce risk. Qualification is task-specific and must be demonstrated, not merely assumed from tenure or credentials.

Question 5Electrical Safety, Codes, and Standards

A Class A ground-fault circuit interrupter intended for personnel protection trips at a ground-fault current of:

  1. 10 mA to 30 mA
  2. 1 mA to 3 mA
  3. 4 mA to 6 mA Correct answer.
  4. 15 mA to 20 mA

Explanation

Class A GFCIs are designed to trip when the ground-fault current exceeds 4 mA to 6 mA, a level chosen to protect people from ventricular fibrillation. Higher trip levels are used for equipment protection, not personnel protection.

Question 6Electrical Safety, Codes, and Standards

A test instrument marked CAT IV is intended for measurements at:

  1. de-energized equipment only
  2. branch circuits and fixed receptacles
  3. the service entrance or utility connection Correct answer.
  4. control circuits derived from a transformer

Explanation

IEC 61010-1 measurement categories describe the transient overvoltage environment at the point of measurement. CAT IV covers the origin of the installation, including the service entrance, meter socket, and utility connections, where available fault energy and transients are highest.

Question 7Electrical Safety, Codes, and Standards

The document that sets electrical safety requirements for employee workplaces, including shock and arc flash risk assessment, is:

  1. NFPA 70B
  2. NFPA 780
  3. NFPA 70E Correct answer.
  4. NFPA 70

Explanation

NFPA 70E, Standard for Electrical Safety in the Workplace, establishes the safety-related work practices, risk assessment procedures, and PPE requirements for employees exposed to electrical hazards. It is a workplace safety standard and is separate from the installation code.

Question 8Electrical Safety, Codes, and Standards

Arc-rated clothing worn under the NFPA 70E PPE Category 1 table must have a minimum arc rating of:

  1. 1.2 cal/cm²
  2. 8 cal/cm²
  3. 12 cal/cm²
  4. 4 cal/cm² Correct answer.

Explanation

The NFPA 70E PPE category tables set PPE Category 1 at a minimum arc rating of 4 cal/cm². Categories 2, 3, and 4 step up to 8, 25, and 40 cal/cm² respectively.

Question 9Electrical Safety, Codes, and Standards

For exposed movable conductors energized at 480 V, the limited approach boundary in NFPA 70E is approximately:

  1. 0.3 m (1 ft)
  2. 3.0 m (10 ft) Correct answer.
  3. 1.1 m (3 ft 6 in)
  4. 1.5 m (5 ft)

Explanation

In the 0 V to 750 V row of the approach boundary table, exposed movable conductors carry a limited approach boundary of 3.0 m (10 ft). Exposed fixed circuit parts in the same voltage row use a much shorter distance of 1.1 m (3 ft 6 in).

Question 10Electrical Safety, Codes, and Standards

NFPA 70E requires an energized electrical work permit when:

  1. PPE Category 2 or higher is required
  2. work is performed within the restricted approach boundary Correct answer.
  3. the equipment is more than 25 years old
  4. any work is performed on a circuit above 50 volts

Explanation

When energized work is justified, NFPA 70E requires an energized electrical work permit for work on conductors or circuit parts that are not in an electrically safe work condition, including work inside the restricted approach boundary. The permit documents the justification, hazard analysis, shock and arc flash controls, and authorization.

Question 11Electrical Safety, Codes, and Standards

When installing temporary protective grounds on a de-energized circuit, the correct sequence is to:

  1. connect the conductor clamp first, then the ground clamp
  2. connect the conductor clamp only after testing for voltage at the ground end
  3. connect both clamps at the same time with a hot stick
  4. connect the ground clamp first, then the conductor clamp Correct answer.

Explanation

Temporary protective grounds are installed by connecting to the system ground or structural ground first and then to the de-energized conductor, so the worker is never handling an ungrounded clamp that could become energized. Removal reverses the sequence: the conductor clamp is removed first and the ground clamp last.

Question 12Electrical Safety, Codes, and Standards

NFPA 70E requires the arc flash risk assessment to be reviewed for accuracy at intervals not exceeding:

  1. 3 years
  2. 10 years
  3. 1 year
  4. 5 years Correct answer.

Explanation

The arc flash risk assessment must be reviewed periodically, at intervals not exceeding 5 years, to remain valid. It must also be reviewed whenever a major modification or renovation takes place that could affect the hazard.

Question 13Electrical Safety, Codes, and Standards

Insulated hand tools used for work on energized parts must be rated and marked for use at:

  1. 500 V
  2. 10,000 V
  3. 1,000 V Correct answer.
  4. 250 V

Explanation

Insulated hand tools are manufactured and marked to ASTM F1505 or IEC 60900, which rate them for use up to 1,000 V AC and mark them with the voltage rating and the standard. Tools without that marking are not acceptable for energized work.

Question 14Electrical Safety, Codes, and Standards

On a construction site, cord- and plug-connected tools that are not covered by an assured equipment grounding conductor program must be:

  1. protected by a ground-fault circuit interrupter Correct answer.
  2. bonded to a driven ground rod at the point of use
  3. tested with a megohmmeter before each shift
  4. double insulated and marked with a UL label

Explanation

OSHA's construction electrical rules require either GFCI protection for receptacles and cord sets or a documented assured equipment grounding conductor program of inspection and testing. If the assured grounding program is not in place, GFCI protection is the required alternative.

Question 15Electrical Safety, Codes, and Standards

Before a voltage detector is used to verify the absence of voltage, it must be:

  1. calibrated within the previous twelve months
  2. proven on a known source before and after Correct answer.
  3. rated for at least twice the system voltage
  4. equipped with an audible and visual indicator

Explanation

Verification of an electrically safe work condition requires proving the tester on a known live source both before and after the test, so a failed instrument cannot be mistaken for a de-energized circuit. Calibration and features such as audible indication support this but do not replace the before-and-after proof.

Question 16Electrical Safety, Codes, and Standards

In a fall-of-potential ground resistance test, the potential electrode is placed at what approximate distance to the current electrode?

  1. 75%
  2. 62% Correct answer.
  3. 25%
  4. 50%

Explanation

The fall-of-potential method places the potential probe at approximately 62% of the distance from the electrode under test to the remote current electrode. This location minimizes the effect of the resistance areas of the two outer electrodes overlapping.

Question 17Electrical Safety, Codes, and Standards

If the working distance from an arc source is doubled, the incident energy at the worker's position drops to approximately:

  1. 25% of the original value Correct answer.
  2. 50% of the original value
  3. 12% of the original value
  4. 75% of the original value

Explanation

Incident energy from a point source falls off approximately with the inverse square of distance, so doubling the distance reduces the energy to about one quarter. This is why increasing working distance is one of the most effective ways to reduce arc flash exposure.

Question 18Electrical Safety, Codes, and Standards

A circuit breaker installed in switchgear must have an interrupting rating that is:

  1. determined by the serving utility in all cases
  2. based on the transformer kVA rating alone
  3. at least equal to the available fault current Correct answer.
  4. no greater than the available fault current

Explanation

NEC 110.9 requires equipment intended to interrupt current at fault levels to have an interrupting rating at nominal circuit voltage not less than the available fault current. A breaker applied above its interrupting rating can fail violently while attempting to clear a fault.

Question 19Electrical Safety, Codes, and Standards

After a DC high-potential test on a cable is completed, the conductor must be:

  1. wiped with a solvent to remove surface leakage
  2. left isolated until the next scheduled outage
  3. discharged to ground and bonded for a period Correct answer.
  4. re-energized immediately to confirm that the test passed

Explanation

A DC test charges the cable capacitance, and the stored energy can remain after the source is removed. The conductor must be discharged to ground and then held at ground for a period long enough for absorbed charge to dissipate before it is handled or reconnected.

Question 20Electrical Safety, Codes, and Standards

When PPE is selected from an incident energy analysis, the arc rating of the selected clothing must be:

  1. at least half of the calculated incident energy
  2. no more than the calculated incident energy
  3. within 10 percent of the calculated incident energy
  4. at least equal to the calculated incident energy Correct answer.

Explanation

Arc-rated clothing must have a system arc rating equal to or greater than the incident energy calculated for the task and working distance. Clothing with a lower arc rating offers no assurance that the wearer will avoid a second-degree burn.

Question 21Electrical Safety, Codes, and Standards

Two multimeters share a 600 V rating, but one is marked CAT III and the other CAT IV. The CAT IV instrument is built to:

  1. show a higher basic DC accuracy
  2. withstand higher transient overvoltages Correct answer.
  3. include a non-contact voltage sensor
  4. carry a higher continuous current

Explanation

Measurement categories are defined by the transient overvoltage the instrument can withstand at a given working voltage, and CAT IV covers the highest transient environment. Accuracy, current capacity, and features are unrelated to the category marking.

Question 22Electrical Safety, Codes, and Standards

Which two quantities are required to determine incident energy by the incident energy analysis method?

  1. fault current and device clearing time Correct answer.
  2. system voltage and conductor ampacity
  3. conductor size and electrode resistance
  4. transformer impedance and ambient temperature

Explanation

The incident energy analysis method uses the bolted fault current at the equipment and the time the upstream protective device takes to clear the arc, along with gap, working distance, and electrode configuration. Clearing time is often the dominant variable because incident energy is proportional to arc duration.

Question 23Electrical Safety, Codes, and Standards

Barricades used to keep unqualified persons out of an arc flash area should be placed at:

  1. the restricted approach boundary
  2. the equipment enclosure door
  3. the limited approach boundary
  4. the arc flash boundary Correct answer.

Explanation

NFPA 70E requires barricades to be placed so that unqualified persons do not cross the arc flash boundary, in addition to barricading approach boundaries for shock protection. The arc flash boundary is the limit at which ordinary clothing no longer provides adequate protection.

Question 24Electrical Safety, Codes, and Standards

Which requirement applies to underlayers worn beneath arc-rated outer garments?

  1. Underlayers must be made of non-melting materials Correct answer.
  2. Underlayers must be 100 percent cotton in every case
  3. Underlayers must match the arc rating of the outer layer
  4. Underlayers must be tested as a system with the outer layer

Explanation

NFPA 70E allows non-arc-rated underlayers only if they will not melt and do not contribute fuel to the arc; meltable synthetics must not be worn next to the skin. Natural-fiber clothing is common but is not the only acceptable choice.

Question 25Electrical Safety, Codes, and Standards

The arc flash PPE category tables in NFPA 70E may be used only when:

  1. the task and equipment match the table parameters Correct answer.
  2. the equipment is 600 V or less with a main breaker
  3. an incident energy analysis has already been completed
  4. available fault current is below 100 kA in all cases

Explanation

The tables are valid only within their stated limits, which include equipment type, voltage, task, and a range of available fault current and clearing time. Where the equipment falls outside those parameters, an incident energy analysis must be performed instead.

Question 26Electrical Safety, Codes, and Standards

Under OSHA lockout/tagout rules, a lock and tag applied to an energy-isolating device to protect a worker may be removed by:

  1. the shift supervisor after a visual check of the equipment
  2. any qualified person on the same crew
  3. the employee who applied it Correct answer.
  4. the site safety supervisor at shift change

Explanation

OSHA 29 CFR 1910.147 requires that each lock and tag be removed by the authorized employee who applied it, because that person is the only one who knows the work is complete and the equipment is safe to re-energize. Removal by anyone else is permitted only under a documented, employer-specific release procedure that includes verifying the owner is not at the equipment.

Question 27Electrical Safety, Codes, and Standards

A 60 A branch circuit is wired with copper conductors. Per NEC Table 250.122, the minimum size copper equipment grounding conductor for this circuit is:

  1. 12 AWG
  2. 6 AWG
  3. 8 AWG
  4. 10 AWG Correct answer.

Explanation

NEC Table 250.122 sizes the equipment grounding conductor by the rating of the overcurrent device protecting the circuit, not by the phase conductor size. For a 60 A overcurrent device the table requires 10 AWG copper; 8 AWG applies at 100 A and 6 AWG at 200 A.

Question 28Electrical Safety, Codes, and Standards

At a subpanel fed from a service, the installer bonds the equipment grounding terminal bar to the neutral bar. What is the principal effect of this connection?

  1. The grounded conductor loses its required identification at the panel.
  2. Objectionable current flows on grounding paths and metal enclosures during normal operation. Correct answer.
  3. The feeder overcurrent device will respond only to ground faults.
  4. The grounding electrode conductor must be increased in size to compensate.

Explanation

NEC 250.24(A)(5) permits the grounded conductor to be bonded to the enclosure and grounding terminal only at the service disconnecting means (or at a separately derived system). Bonding again at a downstream panel creates a parallel return path, so part of the normal neutral current travels on equipment grounding conductors and metal enclosures, which is objectionable current under NEC 250.6 and a shock and heating hazard.

Question 29Power System Fundamentals and Circuit Analysis

A 120 V DC source is connected across a 24 Ω resistor. What current flows through the resistor?

  1. 2,880 A
  2. 5 A Correct answer.
  3. 0.2 A
  4. 20 A

Explanation

Ohm's law states that current equals voltage divided by resistance (I = V/R). Dividing 120 V by 24 Ω gives 5 A. The other answers come from inverting the ratio or multiplying the two quantities.

Question 30Power System Fundamentals and Circuit Analysis

Three resistors of 10 Ω, 15 Ω, and 25 Ω are connected in series. What is the total resistance of the combination?

  1. 16.7 Ω
  2. 50 Ω Correct answer.
  3. 40 Ω
  4. 2.0 Ω

Explanation

In a series circuit the total resistance is the sum of the individual resistances. Adding 10 Ω, 15 Ω, and 25 Ω gives 50 Ω. There is only one path, so every resistor carries the same current and the resistances simply add.

Question 31Power System Fundamentals and Circuit Analysis

Two resistors of different values are connected in parallel across an ideal 12 V source. What is the voltage across the larger resistor?

  1. More than 12 V
  2. Dependent on the resistance ratio
  3. 12 V Correct answer.
  4. Less than 12 V

Explanation

In a parallel circuit every branch is connected directly across the same two nodes, so all branches see the full source voltage. The resistance values change the branch currents, not the branch voltages. Each resistor therefore has 12 V across it.

Question 32Power System Fundamentals and Circuit Analysis

A sinusoidal voltage has a peak value of 170 V. What is its RMS value?

  1. 170 V
  2. 120 V Correct answer.
  3. 85 V
  4. 340 V

Explanation

For a sine wave, the RMS value equals the peak value divided by the square root of 2, or about 0.707 times the peak. Multiplying 170 V by 0.707 gives approximately 120 V. RMS is the value that produces the same heating effect as an equivalent DC voltage.

Question 33Power System Fundamentals and Circuit Analysis

A 24 V DC source is applied across an 8 Ω resistor. How much power is dissipated in the resistor?

  1. 0.33 W
  2. 3 W
  3. 192 W
  4. 72 W Correct answer.

Explanation

Power can be found from P = V²/R, which for 24 V and 8 Ω gives 576/8 = 72 W. The same result follows from first finding I = 3 A and then using P = I²R. Power in a resistor always appears as heat.

Question 34Power System Fundamentals and Circuit Analysis

A conductor is replaced with one of the same material and cross-sectional area but twice the length. Its resistance becomes:

  1. Unchanged
  2. One-half as large
  3. Twice as large Correct answer.
  4. Four times as large

Explanation

Resistance is directly proportional to length and inversely proportional to cross-sectional area (R = ρL/A). Doubling the length while holding material and area constant doubles the resistance. Only a change in area or material would alter the result differently.

Question 35Power System Fundamentals and Circuit Analysis

A transformer has 800 turns on its primary winding and 200 turns on its secondary winding. If the primary is energized at 480 V, what is the secondary voltage?

  1. 60 V
  2. 120 V Correct answer.
  3. 1,920 V
  4. 240 V

Explanation

The voltage ratio of a transformer equals the turns ratio: Vp/Vs = Np/Ns. The turns ratio here is 800/200 = 4, so the secondary voltage is 480 V divided by 4, or 120 V. Because the turns decrease, this is a step-down transformer.

Question 36Power System Fundamentals and Circuit Analysis

What is the period of a 60 Hz sinusoidal waveform?

  1. 0.6 s
  2. 0.00167 s
  3. 0.0167 s Correct answer.
  4. 6 s

Explanation

Period is the reciprocal of frequency: T = 1/f. Taking the reciprocal of 60 Hz gives 0.01667 s, or about 16.7 ms. This is the time required for one complete cycle.

Question 37Power System Fundamentals and Circuit Analysis

A 12 V source supplies a series combination of a 4 Ω resistor and an 8 Ω resistor. What current flows in the circuit?

  1. 3 A
  2. 0.5 A
  3. 1.33 A
  4. 1 A Correct answer.

Explanation

The series resistance is 4 Ω + 8 Ω = 12 Ω, and the same current flows through every element. Applying Ohm's law gives I = 12 V / 12 Ω = 1 A. That 1 A produces a 4 V drop and an 8 V drop, which sum to the source voltage.

Question 38Power System Fundamentals and Circuit Analysis

A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is the equivalent resistance of the pair?

  1. 2 Ω Correct answer.
  2. 4.5 Ω
  3. 9 Ω
  4. 0.5 Ω

Explanation

For two resistors in parallel, the equivalent resistance is the product divided by the sum: (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance. The same answer follows from 1/R = 1/6 + 1/3.

Question 39Power System Fundamentals and Circuit Analysis

At a junction, 5 A and 3 A enter the node while 6 A leaves through one branch. What current must leave through the remaining branch?

  1. 14 A
  2. 8 A
  3. 6 A
  4. 2 A Correct answer.

Explanation

Kirchhoff's current law requires that the total current entering a node equal the total current leaving it. The total entering is 5 A + 3 A = 8 A, and 6 A of that already leaves through one branch. The remaining branch must therefore carry 2 A.

Question 40Power System Fundamentals and Circuit Analysis

A 24 V source is connected across three series resistors. The measured drops across two of them are 8 V and 6 V. What is the voltage drop across the third resistor?

  1. 38 V
  2. 10 V Correct answer.
  3. 14 V
  4. 4 V

Explanation

Kirchhoff's voltage law requires the sum of the voltage drops around a closed loop to equal the applied voltage. The two known drops total 14 V, so the remaining drop is 24 V − 14 V = 10 V. All drops together must return to the source value.

Question 41Power System Fundamentals and Circuit Analysis

A 100 V source is applied to a series combination of a 20 kΩ resistor and an 80 kΩ resistor. What voltage appears across the 80 kΩ resistor?

  1. 20 V
  2. 64 V
  3. 80 V Correct answer.
  4. 50 V

Explanation

In a series voltage divider, each resistor's voltage is proportional to its share of the total resistance. The 80 kΩ element is 80 percent of the 100 kΩ total, so it drops 80 percent of 100 V, or 80 V. The remaining 20 V appears across the 20 kΩ resistor.

Question 42Power System Fundamentals and Circuit Analysis

A load draws 10 kVA at a power factor of 0.8 lagging. What is the real power consumed by the load?

  1. 10 kW
  2. 6 kW
  3. 12.5 kW
  4. 8 kW Correct answer.

Explanation

Real power equals apparent power multiplied by power factor: P = S × pf. Multiplying 10 kVA by 0.8 gives 8 kW. The remaining 6 kvar is reactive power that does no useful work.

Question 43Power System Fundamentals and Circuit Analysis

A balanced three-phase load draws 100 A per line at 480 V line-to-line with a power factor of 0.9. What is the total real power?

  1. 74.8 kW Correct answer.
  2. 43.2 kW
  3. 129.6 kW
  4. 48.0 kW

Explanation

Three-phase real power is P = √3 × V(line) × I(line) × power factor. Substituting gives 1.732 × 480 V × 100 A × 0.9, which equals about 74.8 kW. Failing to include the √3 factor is the most common error.

Question 44Power System Fundamentals and Circuit Analysis

What is the capacitive reactance of a 10 µF capacitor at 60 Hz?

  1. 3.77 Ω
  2. 265 Ω Correct answer.
  3. 26.5 Ω
  4. 60 Ω

Explanation

Capacitive reactance is Xc = 1/(2πfC). Using f = 60 Hz and C = 10 × 10⁻⁶ F gives 1/(2π × 60 × 10⁻⁶) ≈ 265 Ω. Reactance falls as capacitance or frequency increases.

Question 45Power System Fundamentals and Circuit Analysis

An inductor has a reactance of 40 Ω at 60 Hz. What is its reactance at 120 Hz?

  1. 40 Ω
  2. 160 Ω
  3. 80 Ω Correct answer.
  4. 20 Ω

Explanation

Inductive reactance is directly proportional to frequency (XL = 2πfL). Doubling the frequency from 60 Hz to 120 Hz therefore doubles the reactance to 80 Ω. Inductance itself does not change with frequency.

Question 46Power System Fundamentals and Circuit Analysis

A transformer with a 4:1 voltage step-down ratio delivers 100 A to its secondary load. What current flows in the primary winding?

  1. 100 A
  2. 50 A
  3. 25 A Correct answer.
  4. 400 A

Explanation

In an ideal transformer, primary volt-amperes equal secondary volt-amperes, so the current ratio is the inverse of the voltage ratio. With a 4:1 step-down, the primary current is one-fourth of the secondary current, or 25 A. This is why step-down transformers allow smaller primary conductors.

Question 47Power System Fundamentals and Circuit Analysis

In a balanced delta-connected load, the line-to-line voltage is 480 V. What is the voltage across each phase winding of the load?

  1. 480 V Correct answer.
  2. 240 V
  3. 277 V
  4. 831 V

Explanation

In a delta connection each phase winding is connected directly between two line conductors. Therefore the phase voltage equals the line voltage, which is 480 V. The √3 relationship applies to current in a delta, and to voltage in a wye.

Question 48Power System Fundamentals and Circuit Analysis

A balanced three-phase, four-wire wye system supplies a balanced linear load. What is the current in the neutral conductor?

  1. 1.73 times the phase current
  2. Three times the phase current
  3. Zero Correct answer.
  4. Equal to the line current

Explanation

In a balanced three-phase system the three phase currents are equal in magnitude and displaced by 120 degrees. Their phasor sum is zero, so no neutral current flows even though the neutral is connected. Neutral current appears only when the load becomes unbalanced or carries harmonic content.

Question 49Power System Fundamentals and Circuit Analysis

A 12 V source is in series with a 6 Ω resistor, and a 3 Ω resistor shunts the output terminals. What is the Thevenin voltage looking into those terminals?

  1. 6 V
  2. 4 V Correct answer.
  3. 8 V
  4. 12 V

Explanation

The Thevenin voltage is the open-circuit voltage across the output terminals, which here is set by the 6 Ω and 3 Ω divider. The 3 Ω resistor receives 3/(6+3) of the source, giving 12 V × 1/3 = 4 V. The Thevenin resistance would then be 6 Ω in parallel with 3 Ω, or 2 Ω.

Question 50Power System Fundamentals and Circuit Analysis

A series RLC circuit has a resistance of 10 Ω, an inductive reactance of 40 Ω, and a capacitive reactance of 40 Ω. What is the magnitude of the total impedance?

  1. 40 Ω
  2. 0 Ω
  3. 80 Ω
  4. 10 Ω Correct answer.

Explanation

Impedance magnitude is the phasor sum of resistance and net reactance: |Z| = √(R² + (XL − XC)²). Here XL and XC are equal, so the net reactance is zero and the impedance is simply the 10 Ω resistance. This is the resonant condition of a series RLC circuit.