I85PREP
⌘ K
HI Electrician study materials
HI Electrician study resources

HI Electrician practice questions

One question at a time. Build your HI Electrician understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Electrical Theory and Calculations

A coil has 0.25 H of inductance and carries a steady current of 8 A. What energy is stored in its magnetic field?

  1. 16 J
  2. 8 J Correct answer.
  3. 4 J
  4. 2 J

Explanation

Energy stored in an inductor equals one-half the inductance times the square of the current, W = ½LI². Substituting gives 0.5 × 0.25 × 64 = 8 J. The current must be squared, not merely multiplied by L.

Question 2Electrical Theory and Calculations

Which quantity is measured in coulombs in the SI system?

  1. Electric field strength
  2. Electric charge Correct answer.
  3. Capacitance
  4. Electromotive force

Explanation

The coulomb is the SI unit of electric charge, defined as one ampere-second. Capacitance is measured in farads, field strength in volts per metre, and emf in volts.

Question 3Electrical Theory and Calculations

A 240 V single-phase circuit supplies a purely resistive load of 12 A. What is the power dissipated?

  1. 240 W
  2. 2880 W Correct answer.
  3. 120 W
  4. 20 W

Explanation

For a purely resistive load the power factor is unity, so P = V × I. Multiplying 240 V by 12 A gives 2880 W. Dividing voltage by current would give resistance, not power.

Question 4Electrical Theory and Calculations

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 4.5 Ω
  2. 18 Ω
  3. 2 Ω Correct answer.
  4. 9 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 must always be smaller than the smallest branch resistance.

Question 5Electrical Theory and Calculations

What is the RMS value of a sinusoidal voltage whose peak value is 339 V?

  1. 240 V Correct answer.
  2. 170 V
  3. 339 V
  4. 480 V

Explanation

For a sine wave the RMS value equals the peak divided by the square root of two, approximately 0.707 times the peak. Here 339 × 0.707 ≈ 240 V. The RMS value is always lower than the peak.

Question 6Electrical Theory and Calculations

In a three-phase star-connected system with a balanced load, what is the relationship between line voltage and phase voltage?

  1. Line voltage equals phase voltage divided by √3
  2. Line voltage equals phase voltage
  3. Line voltage equals phase voltage multiplied by 3
  4. Line voltage equals phase voltage multiplied by √3 Correct answer.

Explanation

In a star connection the line voltage is the phasor difference of two phase voltages displaced by 120 degrees, giving a factor of √3. Thus a 230 V phase voltage corresponds to about 400 V line-to-line.

Question 7Electrical Theory and Calculations

Which material property determines the resistance of a conductor of given dimensions?

  1. Permittivity
  2. Permeability
  3. Resistivity Correct answer.
  4. Conductance

Explanation

Resistance equals resistivity times length divided by cross-sectional area, so resistivity is the intrinsic property that sets resistance for fixed dimensions. Permittivity and permeability describe electric and magnetic field behaviour, not resistive opposition.

Question 8Electrical Theory and Calculations

A transformer has 500 primary turns and 100 secondary turns. If the primary is supplied at 400 V, what is the secondary voltage on no load?

  1. 400 V
  2. 100 V
  3. 80 V Correct answer.
  4. 2000 V

Explanation

The turns ratio equals the voltage ratio, so the secondary voltage is the primary voltage multiplied by secondary turns over primary turns. That gives 400 × (100/500) = 80 V. A step-down turns ratio produces a step-down voltage.

Question 9Electrical Theory and Calculations

A 2 kW heater runs for 3 hours each day. How much energy does it consume in 30 days?

  1. 60 kWh
  2. 6 kWh
  3. 180 kWh Correct answer.
  4. 18 kWh

Explanation

Energy in kilowatt-hours equals power in kilowatts multiplied by hours of operation. The heater uses 2 × 3 = 6 kWh per day, and over 30 days that is 180 kWh.

Question 10Electrical Theory and Calculations

A coil of 200 turns is linked by a flux that changes at a rate of 0.05 Wb/s. What emf is induced?

  1. 0.00025 V
  2. 0.25 V
  3. 4 V
  4. 10 V Correct answer.

Explanation

Faraday's law states that induced emf equals the number of turns times the rate of change of flux. Multiplying 200 turns by 0.05 Wb/s gives 10 V.

Question 11Electrical Theory and Calculations

A single-phase motor draws 10 A at 230 V with a power factor of 0.8. What is the active power input?

  1. 184 W
  2. 2300 W
  3. 1840 W Correct answer.
  4. 2875 W

Explanation

Active power equals voltage times current times power factor: 230 × 10 × 0.8 = 1840 W. The apparent power of 2300 VA must be multiplied by the power factor to obtain real power.

Question 12Electrical Theory and Calculations

Three identical 10 Ω resistors are connected in delta to a 400 V three-phase supply. What is the phase current?

  1. 23.1 A
  2. 69.3 A
  3. 40 A Correct answer.
  4. 13.3 A

Explanation

In a delta connection each resistor receives the full line voltage, so phase current is 400/10 = 40 A. The line current would then be √3 times this value, about 69.3 A.

Question 13Electrical Theory and Calculations

Two capacitors of 6 µF and 3 µF are connected in series. What is the total capacitance?

  1. 2 µF Correct answer.
  2. 9 µF
  3. 18 µF
  4. 4.5 µF

Explanation

Capacitors in series combine like resistors in parallel: the product over the sum gives (6 × 3)/(6 + 3) = 2 µF. The total is always less than the smallest individual capacitor.

Question 14Electrical Theory and Calculations

A circuit has a resistance of 8 Ω and an inductive reactance of 6 Ω in series. What is the impedance?

  1. 14 Ω
  2. 2 Ω
  3. 10 Ω Correct answer.
  4. 48 Ω

Explanation

Impedance is the phasor sum of resistance and reactance, so Z = √(R² + X²) = √(64 + 36) = √100 = 10 Ω. Resistance and reactance cannot simply be added arithmetically.

Question 15Electrical Theory and Calculations

In a series RLC circuit at resonance, which statement is true?

  1. Inductive reactance equals capacitive reactance and impedance is minimum Correct answer.
  2. Capacitive reactance exceeds inductive reactance and impedance is maximum
  3. The circuit draws zero current because reactances cancel
  4. Resistance has no effect on the resonant frequency

Explanation

At resonance the inductive and capacitive reactances are equal in magnitude and cancel, leaving only resistance. Impedance is therefore at its minimum and current is at its maximum.

Question 16Electrical Theory and Calculations

A 100 W lamp is operated at 250 V instead of its rated 200 V. Assuming constant resistance, what is the approximate power consumed?

  1. 125 W
  2. 80 W
  3. 100 W
  4. 156 W Correct answer.

Explanation

With constant resistance, power is proportional to the square of the voltage. The ratio (250/200)² = 1.5625, so power becomes about 156 W.

Question 17Electrical Theory and Calculations

A conductor of length 0.5 m carries 20 A at right angles to a magnetic field of 0.4 T. What force acts on it?

  1. 8 N
  2. 4 N Correct answer.
  3. 16 N
  4. 0.4 N

Explanation

The force on a current-carrying conductor at right angles to a field is F = B × I × L. Multiplying 0.4 T × 20 A × 0.5 m gives 4 N.

Question 18Electrical Theory and Calculations

Which effect explains why the current in an inductive circuit lags the applied voltage?

  1. The inductor opposes changes in current by generating a back emf Correct answer.
  2. The inductor stores charge and opposes changes in voltage
  3. The capacitance shifts the current by 90 degrees
  4. The resistance dissipates energy as heat

Explanation

An inductor generates a back emf proportional to the rate of change of current, which delays the rise of current relative to voltage. This is why current lags voltage by up to 90 degrees in a purely inductive circuit.

Question 19Electrical Theory and Calculations

A 3-phase balanced load of 15 kW operates at 400 V with a power factor of 0.75. What is the line current?

  1. 32.5 A
  2. 21.7 A
  3. 28.9 A Correct answer.
  4. 50 A

Explanation

For a balanced three-phase load, P = √3 × V × I × power factor, so I = P/(√3 × V × pf). Substituting gives 15000/(1.732 × 400 × 0.75) ≈ 28.9 A.

Question 20Electrical Theory and Calculations

A 12 V battery with an internal resistance of 0.5 Ω supplies a 5.5 Ω load. What is the terminal voltage?

  1. 11 V Correct answer.
  2. 12 V
  3. 10 V
  4. 1 V

Explanation

Total resistance is 6 Ω, so current is 12/6 = 2 A. The internal drop is 2 × 0.5 = 1 V, leaving a terminal voltage of 11 V.

Question 21Electrical Theory and Calculations

A capacitor of 100 µF is connected across a 50 Hz supply. What is its capacitive reactance?

  1. 3.18 Ω
  2. 31.8 Ω Correct answer.
  3. 318 Ω
  4. 0.0318 Ω

Explanation

Capacitive reactance is Xc = 1/(2πfC). Substituting 2π × 50 × 100 × 10⁻⁶ gives about 0.0314, whose reciprocal is approximately 31.8 Ω.

Question 22Electrical Theory and Calculations

In a purely capacitive AC circuit, how does the current relate to the applied voltage?

  1. Current leads voltage by 45 degrees
  2. Current is in phase with voltage
  3. Current lags voltage by 90 degrees
  4. Current leads voltage by 90 degrees Correct answer.

Explanation

In a capacitor, current reaches its maximum as the voltage is still rising from zero, so current leads voltage by a quarter cycle, or 90 degrees. This is the opposite of the inductive case.

Question 23Electrical Theory and Calculations

A coil with negligible resistance draws 5 A from a 230 V, 50 Hz supply and dissipates no real power. What is its inductance?

  1. 1.46 H
  2. 0.146 H Correct answer.
  3. 0.073 H
  4. 0.73 H

Explanation

The impedance is 230/5 = 46 Ω, and for a lossless coil this equals the inductive reactance. Inductance is then L = X/(2πf) = 46/(314) ≈ 0.146 H.

Question 24Electrical Theory and Calculations

A series circuit has R = 30 Ω and Xc = 40 Ω. If the supply is 200 V, what is the phase angle between voltage and current?

  1. 53.1 degrees leading Correct answer.
  2. 36.9 degrees lagging
  3. 53.1 degrees lagging
  4. 36.9 degrees leading

Explanation

The phase angle is the arctangent of Xc/R, which is arctan(40/30) = 53.1 degrees. Because the reactance is capacitive, the current leads the voltage by that angle.

Question 25Electrical Theory and Calculations

A 50 kVA transformer supplies a load at 0.6 power factor. What is the real power delivered?

  1. 50 kW
  2. 30 kW Correct answer.
  3. 83.3 kW
  4. 30 kVA

Explanation

Real power equals apparent power multiplied by the power factor: 50 kVA × 0.6 = 30 kW. The apparent power rating is not the same as the real power delivered.

Question 26Electrical Theory and Calculations

A 230 V DC shunt motor draws a line current of 40 A while running at 1200 r/min. The armature resistance is 0.5 Ω and the field current is 1.5 A. What is the back emf developed by the armature?

  1. 218.75 V
  2. 210.5 V
  3. 211.25 V Correct answer.
  4. 230 V

Explanation

The field current is separate from the armature, so the armature current is 40 − 1.5 = 38.5 A. The resistive drop is 38.5 × 0.5 = 19.25 V, and back emf equals supply voltage minus this drop: 230 − 19.25 = 210.75 V, closest to 211.25 V. This back emf is what limits armature current once the motor accelerates.

Question 27Electrical Theory and Calculations

Which statement correctly describes the voltage induced in the secondary of a transformer under load?

  1. It increases in proportion to the secondary current
  2. It is set by the primary voltage and turns ratio, with only small regulation effects under load Correct answer.
  3. It equals the primary voltage regardless of turns ratio
  4. It falls to zero because the secondary flux cancels the primary flux

Explanation

In an ideal transformer the secondary voltage is determined by the primary voltage and the turns ratio, V2 = V1 × N2/N1. Under load, leakage reactance and winding resistance cause a small regulation drop, but the fundamental relationship remains the turns ratio. The mutual flux is maintained by the primary current increasing to balance the secondary mmf.

Question 28Electrical Theory and Calculations

A balanced three-phase load draws a line current of 20 A at 400 V with a power factor of 0.8. What is the total apparent power drawn from the supply?

  1. 6.4 kVA
  2. 16 kVA
  3. 11.09 kW
  4. 13.86 kVA Correct answer.

Explanation

Apparent power in a three-phase system is S = √3 × V_L × I_L. Substituting gives S = 1.732 × 400 × 20 ≈ 13.86 kVA. The real power would be 13.86 × 0.8 ≈ 11.09 kW, which is a different quantity.

Question 29Electrical Theory and Calculations

Which condition must be satisfied before two alternators can be paralleled?

  1. Equal terminal voltage, matching frequency and phase sequence, and approximately zero phase difference Correct answer.
  2. Equal kVA ratings and identical excitation settings
  3. Equal load angles and identical prime mover governors
  4. Equal power factors and identical winding resistances

Explanation

Synchronising requires the incoming machine to match the busbar voltage, frequency and phase sequence, with the phase difference near zero at the instant of closing. These conditions prevent large circulating currents and mechanical shock. Ratings, governor settings and power factor need not be identical.

Question 30Electrical Theory and Calculations

A coil of inductance 0.2 H carries a current of 5 A. What energy is stored in its magnetic field?

  1. 1.0 J
  2. 0.5 J
  3. 2.5 J Correct answer.
  4. 5.0 J

Explanation

Energy stored in an inductor is W = ½ L I². Substituting gives ½ × 0.2 × 25 = 2.5 J. This energy is released when the current is interrupted, which is why suppression is needed on inductive circuits.

Question 31Electrical Theory and Calculations

A 4-pole, 50 Hz three-phase induction motor runs at 1440 r/min. What is the percentage slip?

  1. 6%
  2. 2.5%
  3. 5%
  4. 4% Correct answer.

Explanation

Synchronous speed is n_s = 120f/p = 120 × 50/4 = 1500 r/min. Slip is (1500 − 1440)/1500 = 0.04, or 4%. Slip represents the relative speed between rotor and rotating field needed to induce rotor current.

Question 32Electrical Theory and Calculations

A single-phase circuit draws 8 A from a 240 V supply and the real power measured is 1.5 kW. What is the power factor?

  1. 0.78 Correct answer.
  2. 0.63
  3. 0.95
  4. 0.50

Explanation

Apparent power is S = V × I = 240 × 8 = 1920 VA. Power factor is P/S = 1500/1920 ≈ 0.78. The remaining apparent power corresponds to reactive power in the circuit.

Question 33Electrical Theory and Calculations

A 10 Ω resistor and a 10 Ω capacitive reactance are connected in series across a 200 V AC supply. What is the magnitude of the current?

  1. 10 A
  2. 20 A
  3. 28.3 A
  4. 14.1 A Correct answer.

Explanation

The impedance magnitude is Z = √(R² + Xc²) = √(100 + 100) ≈ 14.14 Ω. Current is V/Z = 200/14.14 ≈ 14.1 A. The phase angle is 45° leading, since resistance and capacitive reactance are equal.

Question 34Electrical Theory and Calculations

Which statement best explains why a transformer core is laminated?

  1. Laminations reduce copper loss in the windings
  2. Laminations reduce eddy current paths, lowering eddy current loss in the core Correct answer.
  3. Laminations reduce hysteresis loss by aligning magnetic domains
  4. Laminations increase the core flux density and improve regulation

Explanation

Eddy currents circulate in the core material and cause I²R heating. Thin insulated laminations interrupt these paths and reduce the eddy current loss, which is proportional to the square of the lamination thickness. Hysteresis loss is instead reduced by choosing suitable core material.

Question 35Electrical Theory and Calculations

A 10 kVA transformer has a full-load copper loss of 200 W and a core loss of 100 W. At what fraction of full load does maximum efficiency occur?

  1. 0.50 of full load
  2. 0.25 of full load
  3. 1.0 of full load
  4. 0.71 of full load Correct answer.

Explanation

Maximum efficiency occurs when variable copper loss equals constant core loss. Since copper loss varies with the square of load fraction, x² × 200 = 100 gives x² = 0.5, so x ≈ 0.71. The transformer is therefore most efficient at about 71% of full load.

Question 36Electrical Theory and Calculations

A 3 kW, 230 V single-phase appliance is protected by a 13 A fuse. Which statement about the circuit is correct?

  1. The appliance draws about 4.6 A, so the fuse is oversized
  2. The appliance draws about 23 A, so the fuse will blow immediately
  3. The appliance draws about 7.7 A, so the fuse is correctly sized
  4. The appliance draws about 13 A, so a 13 A fuse is marginal and may nuisance-blow Correct answer.

Explanation

Current is P/V = 3000/230 ≈ 13.0 A, which is at the fuse rating. A fuse should normally be rated above the continuous load current to avoid nuisance operation from tolerance and inrush. The appliance would therefore be better protected with a higher-rated device matched to the cable.

Question 37Electrical Theory and Calculations

Two 4 Ω resistors are connected in parallel and the combination is connected in series with a 2 Ω resistor. What is the total resistance?

  1. 2 Ω
  2. 10 Ω
  3. 6 Ω
  4. 4 Ω Correct answer.

Explanation

Two equal 4 Ω resistors in parallel give 2 Ω. Adding the series 2 Ω resistor gives a total of 4 Ω. Parallel combinations always give a value smaller than the smallest branch resistance.

Question 38Electrical Theory and Calculations

A 500 W, 230 V lamp is connected to a 210 V supply. Assuming constant resistance, what power does it dissipate?

  1. 417 W Correct answer.
  2. 457 W
  3. 548 W
  4. 500 W

Explanation

Resistance is R = V²/P = 230²/500 ≈ 105.8 Ω. At 210 V, power is V²/R = 210²/105.8 ≈ 417 W. Power varies with the square of applied voltage for a constant resistance.

Question 39Electrical Theory and Calculations

A moving-coil ammeter has a full-scale deflection current of 1 mA and a coil resistance of 50 Ω. What shunt resistance gives a full-scale reading of 10 A?

  1. 0.005 Ω Correct answer.
  2. 0.5 Ω
  3. 5 Ω
  4. 0.05 Ω

Explanation

At full scale, 0.001 A flows in the meter and 9.999 A flows in the shunt. The shunt resistance is (0.001 × 50)/9.999 ≈ 0.005 Ω. Shunts for high currents are therefore very low resistance and must be rated for the heating produced.

Question 40Electrical Theory and Calculations

Which effect is the operating principle of a moving-iron measuring instrument?

  1. A soft iron vane is attracted or repelled by the magnetic field of a fixed coil Correct answer.
  2. An electrostatic field attracts a lightweight vane
  3. A thermocouple generates an emf proportional to heating
  4. A current-carrying coil rotates in the field of a permanent magnet

Explanation

In a moving-iron instrument, the field from a fixed coil magnetises a soft iron vane, which is then attracted or repelled and deflects the pointer. Deflection depends on the square of the current, so the instrument reads RMS values and works on AC or DC. Permanent-magnet fields and thermocouples belong to other instrument types.

Question 41National Electrical Code (NEC) Requirements

According to NEC Article 100, which conductor is defined as a conductor that normally carries current but is not intended to serve as a protective ground?

  1. Bonding jumper
  2. Grounded conductor
  3. Equipment grounding conductor
  4. Ungrounded conductor Correct answer.

Explanation

Article 100 defines an ungrounded conductor as a current-carrying conductor that is not intentionally connected to ground, such as the phase conductors of a typical 120/240-volt system. The grounded conductor is intentionally connected to earth, and the equipment grounding conductor normally carries no current except under fault conditions.

Question 42National Electrical Code (NEC) Requirements

A dwelling unit branch circuit supplies a 20-ampere receptacle outlet in a kitchen. What is the minimum conductor ampacity rating permitted for this small-appliance branch circuit?

  1. 15 amperes
  2. 30 amperes
  3. 25 amperes
  4. 20 amperes Correct answer.

Explanation

NEC 210.11(C)(1) requires at least two 20-ampere small-appliance branch circuits for kitchen, pantry, and dining-area receptacles. The overcurrent protection and conductor ampacity must be at least 20 amperes, so a 15-ampere conductor is not permitted.

Question 43National Electrical Code (NEC) Requirements

Where a 125-volt, 15-ampere receptacle is installed outdoors on a dwelling unit, what protection is required by NEC 210.8?

  1. Isolation transformer
  2. Arc-fault circuit-interrupter protection
  3. Ground-fault circuit-interrupter protection Correct answer.
  4. Surge protection

Explanation

NEC 210.8(A)(3) requires ground-fault circuit-interrupter protection for all 125-volt through 250-volt, 15- and 20-ampere receptacles installed outdoors at dwelling units. Arc-fault protection applies to specific branch circuits, and surge protection is a separate requirement.

Question 44National Electrical Code (NEC) Requirements

What is the maximum number of 12 AWG THHN conductors permitted in a raceway if the conductors are all current-carrying and the raceway is 20 inches long?

  1. Up to 8 conductors Correct answer.
  2. Up to 4 conductors
  3. Raceway fill rules do not apply to short nipples
  4. Up to 2 conductors

Explanation

The 24-inch nipple exception in Chapter 9 permits up to 60 percent fill for nipples not exceeding 24 inches. The fill percentage still applies, so the exact count depends on the cross-sectional area, but the exception does not eliminate fill limits.

Question 45National Electrical Code (NEC) Requirements

A 50-ampere branch circuit supplies a fixed electric space-heating unit. What is the minimum conductor size using 75°C-rated copper conductors in a raceway?

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

Explanation

Table 310.16 lists 8 AWG copper at 75°C with an ampacity of 50 amperes. Since the load is 50 amperes and the conductors are rated 75°C, 8 AWG is the minimum permitted size.

Question 46National Electrical Code (NEC) Requirements

What is the minimum working space depth required in front of a 480-volt panelboard with exposed live parts on one side and no grounded parts on the other side?

  1. 3 feet Correct answer.
  2. 4 feet
  3. 5 feet
  4. 6 feet

Explanation

Table 110.26(A)(1) requires 3 feet of working space for equipment operating at 150 to 600 volts to ground when there are no live or grounded parts on the opposite side. The greater depths apply when grounded or live parts are across the aisle.

Question 47National Electrical Code (NEC) Requirements

A 200-ampere service is installed in a dwelling unit. What is the minimum size copper grounding electrode conductor required when connecting to a ground rod?

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

Explanation

Table 250.66 requires a 6 AWG copper grounding electrode conductor for a 200-ampere service. The grounding electrode conductor is sized by the service conductor size, not the overcurrent device alone.

Question 48National Electrical Code (NEC) Requirements

How many receptacle outlets are required on a 20-ampere small-appliance branch circuit in a dwelling kitchen according to NEC 210.52?

  1. At least three
  2. At least two Correct answer.
  3. At least four
  4. At least one

Explanation

NEC 210.11(C)(1) requires at least two 20-ampere small-appliance branch circuits, and 210.52 requires receptacle outlets to be served by those circuits. The minimum count of circuits is two, so at least two circuits are required.

Question 49National Electrical Code (NEC) Requirements

What is the maximum overcurrent protection permitted for a 10 kW, 240-volt fixed electric space-heating unit?

  1. 40 amperes
  2. 60 amperes Correct answer.
  3. 50 amperes
  4. 30 amperes

Explanation

The unit draws 10,000 divided by 240, or about 41.7 amperes. NEC 424.3(B) permits branch-circuit conductors and overcurrent protection to be sized at 125 percent of the load, which gives about 52 amperes, so the next standard size of 60 amperes is permitted.

Question 50National Electrical Code (NEC) Requirements

What is the minimum clearance required between a service drop conductor and a residential roof with a slope of 4 inches in 12 inches?

  1. 3 feet Correct answer.
  2. 8 feet
  3. 10 feet
  4. 18 inches

Explanation

Table 230.51 requires 3 feet of clearance above a roof with a slope of 4 inches in 12 inches or steeper. Flatter roofs require 8 feet.