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EWRB Electrical practice questions

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Question 1Electrical Theory and Practice

A 12 V DC supply is connected across a 4 Ω resistor. What current flows in the circuit?

  1. 48 A
  2. 0.33 A
  3. 16 A
  4. 3 A Correct answer.

Explanation

Ohm's law states I = V / R. Dividing 12 V by 4 Ω gives 3 A.

Question 2Electrical Theory and Practice

A 230 V single-phase heater draws 10 A. What is its power rating?

  1. 230 W
  2. 23 000 W
  3. 2300 W Correct answer.
  4. 23 W

Explanation

For a resistive single-phase load, P = V × I. Multiplying 230 V by 10 A gives 2300 W.

Question 3Electrical Theory and Practice

Two 6 Ω resistors are connected in series across a 12 V supply. What current flows?

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

Explanation

Series resistances add, so total resistance is 12 Ω. Using I = V / R gives 12 V ÷ 12 Ω = 1 A.

Question 4Electrical Theory and Practice

Two 10 Ω resistors are connected in parallel. What is the equivalent resistance?

  1. 20 Ω
  2. 10 Ω
  3. 5 Ω Correct answer.
  4. 2.5 Ω

Explanation

For two equal resistors in parallel, the equivalent is half of one value. Here 10 Ω ÷ 2 = 5 Ω.

Question 5Electrical Theory and Practice

What is the standard frequency of the low-voltage AC supply in New Zealand?

  1. 50 Hz Correct answer.
  2. 60 Hz
  3. 25 Hz
  4. 100 Hz

Explanation

New Zealand's electricity supply operates at 50 Hz, in common with most of the world outside North America.

Question 6Electrical Theory and Practice

A resistor is marked with the bands brown, black, red and gold. What is its value and tolerance?

  1. 1 kΩ ±5% Correct answer.
  2. 10 Ω ±5%
  3. 10 kΩ ±5%
  4. 100 Ω ±5%

Explanation

Brown = 1, black = 0 and red = ×100, giving 1000 Ω or 1 kΩ. Gold indicates a tolerance of ±5%.

Question 7Electrical Theory and Practice

At a circuit node, currents of 3 A and 5 A flow in, and 6 A flows out. What is the remaining current and its direction?

  1. 14 A out of the node
  2. 2 A into the node
  3. 8 A out of the node
  4. 2 A out of the node Correct answer.

Explanation

Kirchhoff's current law requires total current in to equal total current out. With 8 A in and 6 A out, a further 2 A must leave the node.

Question 8Electrical Theory and Practice

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

  1. 31.8 Ω Correct answer.
  2. 318 Ω
  3. 3.18 Ω
  4. 0.318 Ω

Explanation

Capacitive reactance is Xc = 1 / (2πfC). Substituting 50 Hz and 100 µF gives approximately 31.8 Ω.

Question 9Electrical Theory and Practice

A 0.1 H inductor is connected to a 50 Hz supply. What is its inductive reactance?

  1. 0.314 Ω
  2. 3.14 Ω
  3. 314 Ω
  4. 31.4 Ω Correct answer.

Explanation

Inductive reactance is XL = 2πfL. With f = 50 Hz and L = 0.1 H, XL is approximately 31.4 Ω.

Question 10Electrical Theory and Practice

A single-phase load takes 2 kW at a power factor of 0.8. What is the apparent power?

  1. 1.6 kVA
  2. 2.0 kVA
  3. 2.5 kVA Correct answer.
  4. 3.2 kVA

Explanation

Apparent power S = P / power factor. Dividing 2 kW by 0.8 gives 2.5 kVA.

Question 11Electrical Theory and Practice

A balanced three-phase load is connected to a 400 V supply and draws 10 A per line at unity power factor. What total power does it draw?

  1. 4.0 kW
  2. 2.31 kW
  3. 8.0 kW
  4. 6.93 kW Correct answer.

Explanation

For a balanced three-phase load, P = √3 × VL × IL × power factor. This gives 1.732 × 400 × 10 = 6928 W, about 6.93 kW.

Question 12Electrical Theory and Practice

A transformer with 1000 primary turns is supplied at 230 V. How many secondary turns are needed to produce 46 V?

  1. 50 turns
  2. 100 turns
  3. 200 turns Correct answer.
  4. 500 turns

Explanation

Turns ratio equals voltage ratio, so Ns = Np × Vs / Vp. This gives 1000 × 46 / 230 = 200 turns.

Question 13Electrical Theory and Practice

A star-connected three-phase supply has a phase voltage of 230 V. What is the line voltage?

  1. 400 V Correct answer.
  2. 230 V
  3. 132 V
  4. 460 V

Explanation

In a star connection, line voltage equals √3 times phase voltage. Multiplying 230 V by 1.732 gives about 400 V.

Question 14Electrical Theory and Practice

What is the main purpose of the MEN link in a New Zealand installation?

  1. To isolate the neutral during a fault
  2. To bond the main earth conductor to a water pipe
  3. To reduce voltage drop on final subcircuits
  4. To link the neutral bar to the earth bar Correct answer.

Explanation

The MEN link connects the neutral bar to the earth bar at the main switchboard. This creates a low-impedance earth-fault path so protective devices operate quickly.

Question 15Electrical Theory and Practice

A 30 mA residual current device provides additional protection against which of the following?

  1. Short-circuit current between phases
  2. Overload of the final subcircuit
  3. Electric shock from live parts Correct answer.
  4. Voltage surges from the supply network

Explanation

An RCD detects a difference between the current in the active and neutral conductors, which indicates earth leakage. A 30 mA device provides additional protection against electric shock.

Question 16Electrical Theory and Practice

Why is the current-carrying capacity of cables reduced when they are bunched together in a conduit?

  1. Mutual heating reduces each cable's heat dissipation Correct answer.
  2. The conduit adds inductive reactance to the circuit
  3. The cables' insulation resistance falls when grouped
  4. Grouping lowers the supply voltage to the cables

Explanation

Grouped cables cannot dissipate heat as effectively because each cable is warmed by its neighbours. The rating must therefore be derated to keep conductor temperatures within insulation limits.

Question 17Electrical Theory and Practice

To achieve discrimination between a 20 A upstream MCB and a 16 A downstream MCB, what characteristic must the upstream device have?

  1. A higher rating and a slower time-current curve Correct answer.
  2. An identical time-current curve to the downstream device
  3. A lower breaking capacity than the downstream device
  4. A faster tripping time at all current levels

Explanation

Discrimination means only the device nearest the fault should operate. The upstream device must have a higher current rating and a time-current characteristic that lies above the downstream device's curve.

Question 18Electrical Theory and Practice

A squirrel-cage induction motor started direct-on-line typically draws what multiple of full-load current at starting?

  1. 1 to 2 times
  2. 6 to 8 times Correct answer.
  3. 10 to 15 times
  4. 3 to 4 times

Explanation

Direct-on-line starting applies full voltage to the stator, and the starting current is typically six to eight times the motor's full-load current.

Question 19Electrical Theory and Practice

What is the minimum acceptable insulation resistance for a low-voltage installation when tested with a 500 V DC insulation tester?

  1. 10 MΩ
  2. 1 MΩ Correct answer.
  3. 5 MΩ
  4. 0.5 MΩ

Explanation

AS/NZS 3000 requires a minimum insulation resistance of 1 MΩ for a low-voltage installation. Readings above this indicate insulation in acceptable condition.

Question 20Electrical Theory and Practice

Which test confirms that a protective earthing conductor is electrically continuous from end to end?

  1. Insulation resistance test at 500 V DC
  2. Earth loop impedance test
  3. Polarity test using a voltmeter
  4. Continuity test with an ohmmeter Correct answer.

Explanation

A continuity test passes a small current through the conductor and measures resistance with a low-reading ohmmeter. A near-zero reading confirms the conductor is continuous.

Question 21Electrical Theory and Practice

A 100/5 A current transformer is used with an ammeter. If the primary current is 80 A, what current flows in the secondary?

  1. 4 A Correct answer.
  2. 5 A
  3. 16 A
  4. 0.4 A

Explanation

The current transformer ratio is 100:5, which simplifies to 20:1. Dividing 80 A by 20 gives a secondary current of 4 A.

Question 22Electrical Theory and Practice

In a balanced three-phase four-wire system supplying non-linear loads, what effect do triplen harmonics have on the neutral conductor?

  1. They cancel in the neutral as in the phase currents
  2. They cause the neutral current to be zero
  3. They reduce neutral current to one third of phase current
  4. They add in the neutral, which can exceed phase current Correct answer.

Explanation

Third-harmonic and other triplen components are in phase in all three phases. They therefore add arithmetically in the neutral, which can carry more current than a phase conductor.

Question 23Electrical Theory and Practice

What is the purpose of a fault loop impedance test on a final subcircuit?

  1. To verify the polarity of the active and neutral conductors
  2. To confirm the protective device will disconnect in time Correct answer.
  3. To check the earth electrode resistance only
  4. To measure the insulation resistance of the cable

Explanation

Fault loop impedance determines the current that will flow during an earth fault. The measured value is compared with the maximum permitted so the protective device will operate within the required disconnection time.

Question 24Electrical Theory and Practice

Why is the neutral of a MEN distribution system earthed at multiple points?

  1. To allow the neutral to be disconnected at each installation
  2. To lower earth-return impedance and neutral voltage rise Correct answer.
  3. To increase the resistance of the earth return path
  4. To reduce the load current in the neutral conductor

Explanation

Multiple earthing of the neutral keeps the earth-return path impedance low and limits voltage rise on the neutral under fault conditions. This helps protective devices operate quickly.

Question 25Electrical Theory and Practice

A 2 kW appliance is used continuously for 3 hours. How much energy does it consume?

  1. 0.67 kWh
  2. 1.5 kWh
  3. 6 kWh Correct answer.
  4. 60 kWh

Explanation

Energy in kilowatt-hours equals power in kilowatts multiplied by time in hours. Multiplying 2 kW by 3 h gives 6 kWh.

Question 26Electrical Theory and Practice

A current of 0.5 A flows through a 12 Ω resistor. What is the voltage across the resistor?

  1. 12.5 V
  2. 6 V Correct answer.
  3. 24 V
  4. 0.04 V

Explanation

Ohm's law states that V = I × R, so the voltage is 0.5 A × 12 Ω = 6 V. Voltage is proportional to both current and resistance.

Question 27Electrical Theory and Practice

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

  1. 6 Ω
  2. 5 Ω
  3. 150 Ω
  4. 25 Ω Correct answer.

Explanation

In a series circuit the same current passes through every component and the total resistance is the sum of the individual resistances, so 10 Ω + 15 Ω = 25 Ω. Adding resistance in series always increases the total.

Question 28Electrical Theory and Practice

Two identical 10 Ω resistors are connected in parallel. What is the equivalent resistance of the pair?

  1. 20 Ω
  2. 10 Ω
  3. 5 Ω Correct answer.
  4. 0.2 Ω

Explanation

For two equal resistors in parallel the equivalent resistance is half of one value, so 10 Ω ÷ 2 = 5 Ω. Using product over sum gives the same result: (10 × 10) ÷ (10 + 10) = 5 Ω.

Question 29Electrical Theory and Practice

A resistive appliance draws 10 A from a 230 V single-phase supply. What power does it consume?

  1. 230 W
  2. 23 W
  3. 24 W
  4. 2.3 kW Correct answer.

Explanation

For a purely resistive load the power is P = V × I, so 230 V × 10 A = 2300 W, which is 2.3 kW. The power factor is unity so no correction is needed.

Question 30Electrical Theory and Practice

What is the period of an alternating supply with a frequency of 50 Hz?

  1. 0.2 s
  2. 20 ms Correct answer.
  3. 50 ms
  4. 2 ms

Explanation

Period and frequency are reciprocals, so T = 1/f = 1/50 = 0.02 s, which is 20 ms. One complete cycle therefore takes 20 milliseconds.

Question 31Electrical Theory and Practice

A sinusoidal supply has an RMS value of 230 V. What is its peak value?

  1. 230 V
  2. 163 V
  3. 325 V Correct answer.
  4. 460 V

Explanation

For a sine wave the peak value is the RMS value multiplied by √2, so 230 × 1.414 ≈ 325 V. The RMS value is the heating equivalent and is always lower than the peak.

Question 32Electrical Theory and Practice

In a New Zealand installation using standard fixed-wiring colours, which colour identifies the neutral conductor?

  1. Blue
  2. Red
  3. Green/yellow
  4. Black Correct answer.

Explanation

In New Zealand fixed wiring the neutral conductor is black and the protective earthing conductor is green/yellow. Phase conductors are identified by red, white or blue.

Question 33Electrical Theory and Practice

What is the normal function of a miniature circuit breaker in a final subcircuit?

  1. Reduce the supply voltage to a level that is safe for the user
  2. Provide isolation only when operated by hand
  3. Limit the current to a preset value during normal load
  4. Clear an overload or short-circuit current Correct answer.

Explanation

A miniature circuit breaker combines a thermal element that responds to sustained overload with a magnetic element that trips very quickly on short-circuit current. Both actions open the circuit automatically.

Question 34Electrical Theory and Practice

A 6 Ω resistor is connected in series with a parallel pair of 12 Ω resistors. What is the total resistance of the circuit?

  1. 18 Ω
  2. 12 Ω Correct answer.
  3. 6 Ω
  4. 9 Ω

Explanation

The parallel pair of equal 12 Ω resistors has an equivalent resistance of 6 Ω. Adding the series 6 Ω resistor gives 6 Ω + 6 Ω = 12 Ω.

Question 35Electrical Theory and Practice

A 12 V supply is connected across a 4 Ω resistor in series with an 8 Ω resistor. What is the voltage across the 8 Ω resistor?

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

Explanation

The total resistance is 12 Ω, so the current is 12 V ÷ 12 Ω = 1 A. The voltage across the 8 Ω resistor is therefore 1 A × 8 Ω = 8 V.

Question 36Electrical Theory and Practice

A series circuit contains a 0.1 H inductor and a 10 µF capacitor. At what frequency do they resonate?

  1. 50 Hz
  2. 1.59 kHz
  3. 159 Hz Correct answer.
  4. 100 Hz

Explanation

Resonance occurs where XL equals XC, at f = 1/(2π√(LC)). Substituting gives 1/(2π√(0.1 × 10 × 10⁻⁶)) ≈ 159 Hz.

Question 37Electrical Theory and Practice

A transformer with a 230 V primary supplies 24 V at 2 A to a load. Ignoring losses, what current flows in the primary winding?

  1. 2 A
  2. 4.6 A
  3. 1.04 A
  4. 0.21 A Correct answer.

Explanation

The secondary load is 24 V × 2 A = 48 VA, and in an ideal transformer the primary VA is the same. The primary current is therefore 48 VA ÷ 230 V ≈ 0.21 A.

Question 38Electrical Theory and Practice

A single-phase load of 5 kVA operates at a power factor of 0.6 lagging. What is the reactive power drawn?

  1. 3 kvar
  2. 2 kvar
  3. 4 kvar Correct answer.
  4. 5 kvar

Explanation

Real power is 5 kVA × 0.6 = 3 kW. Reactive power is found from the power triangle: √(5² − 3²) = 4 kvar.

Question 39Electrical Theory and Practice

A balanced delta-connected load draws 10 A in each phase winding from a three-phase supply. What is the line current?

  1. 5.8 A
  2. 10 A
  3. 30 A
  4. 17.3 A Correct answer.

Explanation

In a delta connection each line feeds two windings, so the line current is √3 times the phase current. Here 1.732 × 10 A ≈ 17.3 A.

Question 40Electrical Theory and Practice

A motor draws 5 kW from the supply and delivers 4 kW of mechanical output. What is its efficiency?

  1. 75%
  2. 125%
  3. 80% Correct answer.
  4. 95%

Explanation

Efficiency is output divided by input, expressed as a percentage: 4 kW ÷ 5 kW × 100 = 80%. The remaining 1 kW is lost as heat, windage and other losses.

Question 41Electrical Theory and Practice

A 230 V final subcircuit has a calculated voltage drop of 6.9 V at full load. What percentage drop is this?

  1. 30%
  2. 6.9%
  3. 0.3%
  4. 3% Correct answer.

Explanation

Percentage voltage drop is the drop divided by the supply voltage, multiplied by 100: 6.9 ÷ 230 × 100 = 3%. This is within the usual limit for a low-voltage final subcircuit.

Question 42Electrical Theory and Practice

A thermal overload relay in a motor starter is intended primarily to protect the motor against which condition?

  1. Sustained overload Correct answer.
  2. A short-circuit fault
  3. An earth fault
  4. Reverse power flow

Explanation

A thermal overload relay uses bimetal elements that respond to sustained overcurrent, protecting the winding from gradual overheating. It is deliberately too slow to clear short-circuit currents, which are handled by the fuses or circuit breaker.

Question 43Electrical Theory and Practice

What is the purpose of equipotential bonding of metallic pipework and structural metal in an installation?

  1. Provide a low-impedance path for lightning currents to reach earth
  2. Reduce the operating current drawn by the circuit
  3. Keep all metalwork at a common potential Correct answer.
  4. Allow the residual current device to be left out

Explanation

Bonding connects extraneous and exposed conductive parts so that they sit at substantially the same potential. If a fault raises the potential of one part, the others rise with it and the touch voltage between them stays low.

Question 44Electrical Theory and Practice

A squirrel-cage motor started by a star-delta starter draws a starting current that is approximately what proportion of its direct-on-line starting current?

  1. One half of the DOL starting current
  2. The same value as with DOL starting
  3. Twice the DOL starting current
  4. One third of the DOL current Correct answer.

Explanation

In star-delta starting the motor begins in star, so each winding receives 1/√3 of the line voltage. The resulting line current is one third of the direct-on-line value, with torque reduced to one third as well.

Question 45Electrical Theory and Practice

In a purely capacitive AC circuit, what is the phase relationship between current and voltage?

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

Explanation

In a capacitor the current must flow to build up charge, so it reaches its maximum a quarter cycle before the voltage does. The current therefore leads the voltage by 90 electrical degrees.

Question 46Electrical Theory and Practice

A 100 kW load operates at a power factor of 0.7 lagging. What reactive power must a capacitor bank supply to raise the power factor to 0.95?

  1. 69 kvar Correct answer.
  2. 10 kvar
  3. 33 kvar
  4. 100 kvar

Explanation

The required reactive power is P(tan φ1 − tan φ2). For 0.7 the tangent is 1.020 and for 0.95 it is 0.329, so 100 × (1.020 − 0.329) ≈ 69 kvar.

Question 47Electrical Theory and Practice

A three-phase supply feeds a six-pulse rectifier, injecting a significant 5th harmonic current. What effect does this harmonic have on an induction motor on the same supply?

  1. It creates a stationary field that causes no extra heating
  2. It flows only in the neutral conductor
  3. It sets up a negative-sequence field that heats the rotor Correct answer.
  4. It creates a positive-sequence field that adds torque

Explanation

The 5th harmonic has a negative phase sequence, so it produces an air-gap field rotating against the rotor. This causes additional rotor losses, heating and reduced output torque.

Question 48Electrical Theory and Practice

A series RLC circuit is driven at its resonant frequency and has a high quality factor. Which statement about the circuit at resonance is true?

  1. The current is limited only by the inductive and capacitive reactances
  2. The voltages across L and C can exceed the supply voltage Correct answer.
  3. The impedance is at its maximum and the current is at its minimum
  4. The circuit draws only reactive current from the supply

Explanation

At resonance XL equals XC and the two reactances cancel, leaving only the resistance, so impedance is minimum and current is maximum. The voltages across L and C are equal and opposite but can each be Q times the supply voltage.

Question 49Electrical Theory and Practice

A 230 V final subcircuit has a measured earth fault loop impedance of 1.15 Ω. What magnitude of earth fault current would flow for a bolted fault at the far end?

  1. 230 A
  2. 100 A
  3. 150 A
  4. 200 A Correct answer.

Explanation

The prospective fault current is the nominal voltage divided by the loop impedance: 230 V ÷ 1.15 Ω = 200 A. This value is then compared with the tripping characteristics of the protective device.

Question 50Electrical Theory and Practice

A copper conductor has a resistance of 0.50 Ω at 20 °C. Using a temperature coefficient of 0.00393 per °C, what is its resistance at 70 °C?

  1. 0.50 Ω
  2. 0.52 Ω
  3. 0.60 Ω Correct answer.
  4. 0.45 Ω

Explanation

Resistance rises with temperature according to R2 = R1[1 + α(T2 − T1)]. Here 0.50 × [1 + 0.00393 × 50] ≈ 0.60 Ω.