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Question 1Electrical theory and fundamentals
A 12 V DC supply is connected across a 4 Ω resistor. What current flows through the resistor?
- 48 A
- 16 A
- 0.33 A
- 3 A Correct answer.
Explanation
Ohm's law states that current equals voltage divided by resistance. Dividing 12 V by 4 Ω gives 3 A.
Question 2Electrical theory and fundamentals
A 240 V single-phase resistive load draws a current of 10 A. What power does the load consume?
- 4800 W
- 24 W
- 240 W
- 2400 W Correct answer.
Explanation
For a resistive load the power is the product of voltage and current. Multiplying 240 V by 10 A gives 2400 W.
Question 3Electrical theory and fundamentals
Resistors of 6 Ω and 4 Ω are connected in series across a 20 V DC supply. What current flows in the circuit?
- 5 A
- 2 A Correct answer.
- 0.5 A
- 10 A
Explanation
In series the resistances add, giving a total of 10 Ω. Ohm's law then gives 20 V divided by 10 Ω, which is 2 A.
Question 4Electrical theory and fundamentals
What is the period of a 50 Hz alternating waveform?
- 2 ms
- 10 ms
- 20 ms Correct answer.
- 50 ms
Explanation
Period is the reciprocal of frequency, so 1 divided by 50 Hz gives 0.02 s, which is 20 ms.
Question 5Electrical theory and fundamentals
A sinusoidal AC supply has an RMS value of 230 V. What is its peak value?
- 460 V
- 230 V
- 325 V Correct answer.
- 163 V
Explanation
For a sine wave the peak value equals the RMS value multiplied by √2. Multiplying 230 V by 1.414 gives approximately 325 V.
Question 6Electrical theory and fundamentals
Two 12 Ω resistors are connected in parallel across a 24 V supply. What current flows from the supply?
- 1 A
- 2 A
- 4 A Correct answer.
- 8 A
Explanation
Two equal resistors in parallel give half the value of one, so the equivalent resistance is 6 Ω. Dividing 24 V by 6 Ω gives 4 A.
Question 7Electrical theory and fundamentals
A single-phase load draws 10 kVA at a power factor of 0.8. What is the real power drawn by the load?
- 12.5 kW
- 6 kW
- 8 kW Correct answer.
- 10 kW
Explanation
Real power equals apparent power multiplied by power factor. Multiplying 10 kVA by 0.8 gives 8 kW.
Question 8Electrical theory and fundamentals
A balanced three-phase load draws 10 A per line from a 400 V three-phase supply at unity power factor. What total power is consumed?
- 2.3 kW
- 6.9 kW Correct answer.
- 4.0 kW
- 12.0 kW
Explanation
Three-phase power equals √3 multiplied by line voltage, line current and power factor. So √3 × 400 × 10 × 1 gives about 6.9 kW.
Question 9Electrical theory and fundamentals
An ideal 240 V / 24 V single-phase transformer has a primary current of 1 A. What is the secondary current?
- 10 A Correct answer.
- 2.4 A
- 1 A
- 0.24 A
Explanation
In an ideal transformer the voltage ratio equals the inverse of the current ratio. With a 10:1 step-down ratio the secondary current is ten times the primary current, so 10 A.
Question 10Electrical theory and fundamentals
A 50 m length of a conductor has a resistance of 0.5 Ω. What is the resistance of 100 m of the same conductor at the same temperature?
- 1.0 Ω Correct answer.
- 2.0 Ω
- 0.5 Ω
- 0.25 Ω
Explanation
Resistance is directly proportional to length when material and cross-sectional area are unchanged. Doubling the length to 100 m doubles the resistance to 1.0 Ω.
Question 11Electrical theory and fundamentals
What is the inductive reactance of a 0.1 H coil connected to a 50 Hz supply?
- 31.4 Ω Correct answer.
- 15.9 Ω
- 5 Ω
- 314 Ω
Explanation
Inductive reactance is 2π times frequency times inductance. So 2π × 50 × 0.1 gives approximately 31.4 Ω.
Question 12Electrical theory and fundamentals
What is the capacitive reactance of a 100 µF capacitor connected to a 50 Hz supply?
- 31.8 Ω Correct answer.
- 63.7 Ω
- 15.9 Ω
- 3.18 Ω
Explanation
Capacitive reactance is 1 divided by 2π times frequency times capacitance. Using 2π × 50 × 100 × 10⁻⁶ gives approximately 31.8 Ω.
Question 13Electrical theory and fundamentals
At a circuit junction, currents of 5 A and 3 A flow into the junction and one current of 6 A flows out. What is the magnitude of the remaining current flowing out of the junction?
- 2 A Correct answer.
- 4 A
- 14 A
- 8 A
Explanation
Kirchhoff's current law requires the total current entering a junction to equal the total current leaving it. The current entering is 8 A and 6 A already leaves, so a further 2 A must flow out.
Question 14Electrical theory and fundamentals
A coil with a resistance of 3 Ω and an inductive reactance of 4 Ω is connected in series to an AC supply. What is the impedance of the circuit?
- 5 Ω Correct answer.
- 25 Ω
- 1 Ω
- 7 Ω
Explanation
Resistance and reactance are 90° out of phase, so impedance is the square root of the sum of their squares. The square root of 9 plus 16 is 5 Ω.
Question 15Electrical theory and fundamentals
A 2 kW heater runs continuously for 3 hours. How much energy does it consume?
- 6 kWh Correct answer.
- 60 kWh
- 1.5 kWh
- 0.6 kWh
Explanation
Energy in kilowatt hours equals power in kilowatts multiplied by time in hours. Multiplying 2 kW by 3 hours gives 6 kWh.
Question 16Electrical theory and fundamentals
A single-phase circuit carries 10 A and the total resistance of the active and neutral conductors forming the loop is 0.05 Ω. What is the voltage drop in the conductors?
- 5 V
- 2 V
- 0.05 V
- 0.5 V Correct answer.
Explanation
Voltage drop equals current multiplied by the total loop resistance of the conductors. Multiplying 10 A by 0.05 Ω gives 0.5 V.
Question 17Electrical theory and fundamentals
In a delta-connected load the current in each phase winding is 10 A. What is the line current?
- 30 A
- 17.3 A Correct answer.
- 5.8 A
- 10 A
Explanation
For a delta connection the line current is √3 times the phase current. Multiplying 10 A by 1.732 gives approximately 17.3 A.
Question 18Electrical theory and fundamentals
In a series RLC circuit, at which condition does resonance occur?
- The capacitive reactance falls to zero
- The resistance equals the inductive reactance
- The inductive reactance is twice the capacitive reactance
- The inductive reactance equals the capacitive reactance Correct answer.
Explanation
Series resonance occurs at the frequency where the inductive and capacitive reactances are equal in magnitude. They then cancel, leaving only resistance to limit the current.
Question 19Electrical theory and fundamentals
A copper conductor has a resistance of 1.00 Ω at 20 °C. Using a temperature coefficient of 0.00393 per °C at 20 °C, what is its resistance at 70 °C?
- 0.80 Ω
- 1.20 Ω Correct answer.
- 1.00 Ω
- 1.50 Ω
Explanation
Resistance rises with temperature according to R = R₂₀[1 + α(T − 20)]. The 50 °C rise contributes 0.00393 × 50, giving a factor of 1.1965 and a resistance of about 1.20 Ω.
Question 20Electrical theory and fundamentals
A load of 100 kW operates at a lagging power factor of 0.8. How many kVAR of capacitance are required to raise the power factor to unity?
- 20 kVAR
- 80 kVAR
- 75 kVAR Correct answer.
- 60 kVAR
Explanation
The reactive power to be cancelled equals the load's reactive power, which is real power multiplied by tan θ for the original power factor. With tan θ equal to 0.75, the requirement is 75 kVAR.
Question 21Electrical theory and fundamentals
A balanced star-connected load is supplied from a 400 V three-phase supply. What is the voltage across each phase winding?
- 693 V
- 231 V Correct answer.
- 133 V
- 400 V
Explanation
In a star connection the phase voltage is the line voltage divided by √3. Dividing 400 V by 1.732 gives approximately 231 V.
Question 22Electrical theory and fundamentals
A 100 µF capacitor is charged to 400 V DC. How much energy is stored in the capacitor?
- 8 J Correct answer.
- 0.8 J
- 16 J
- 4 J
Explanation
Energy stored in a capacitor is one half of the capacitance multiplied by the square of the voltage. Half of 100 × 10⁻⁶ F times 160 000 V² gives 8 J.
Question 23Electrical theory and fundamentals
A balanced star-connected load of 10 Ω per phase is connected to a 400 V three-phase supply. What total power is consumed?
- 48 kW
- 4.0 kW
- 8.0 kW
- 16 kW Correct answer.
Explanation
The phase voltage is 400/√3, about 231 V, giving a phase current of 23.1 A. Total power is three times the phase current squared times the phase resistance, which is about 16 kW.
Question 24Electrical theory and fundamentals
According to Faraday's law, the EMF induced in a coil is proportional to which quantity?
- The rate of change of flux linkage with the coil Correct answer.
- The resistance of the coil
- The inductance of the coil divided by the current
- The steady magnetic flux linking the coil
Explanation
Faraday's law states that the induced EMF depends on how quickly the flux linking the coil changes, not on the flux itself. A steady flux therefore produces no EMF once it is established.
Question 25Electrical theory and fundamentals
A coil with a resistance of 8 Ω and an inductive reactance of 6 Ω is connected to a 100 V AC supply. What true power is dissipated by the coil?
- 600 W
- 800 W Correct answer.
- 1000 W
- 1600 W
Explanation
The impedance is the square root of 8² plus 6², which is 10 Ω, so the current is 10 A. True power is dissipated only in the resistance and equals current squared times resistance, giving 800 W.
Question 26Electrical theory and fundamentals
A resistor of 6 Ω carries a current of 2 A. What is the voltage across the resistor?
- 8 V
- 24 V
- 3 V
- 12 V Correct answer.
Explanation
Ohm's law states that voltage equals current multiplied by resistance, V = I × R. Substituting the values gives 2 A × 6 Ω = 12 V.
Question 27Electrical theory and fundamentals
A current of 2 A flows for 30 s. How much charge passes a point in the circuit?
- 15 C
- 32 C
- 120 C
- 60 C Correct answer.
Explanation
Charge is the product of current and time, Q = I × t. Therefore Q = 2 A × 30 s = 60 C.
Question 28Electrical theory and fundamentals
Three resistors of 10 Ω, 20 Ω and 30 Ω are connected in series. What is the total resistance?
- 6 Ω
- 20 Ω
- 60 Ω Correct answer.
- 600 Ω
Explanation
In a series circuit the total resistance is the sum of the individual resistances. Adding 10 Ω, 20 Ω and 30 Ω gives 60 Ω.
Question 29Electrical theory and fundamentals
A sinusoidal supply has a frequency of 50 Hz. What is its period?
- 2 ms
- 20 ms Correct answer.
- 50 ms
- 25 ms
Explanation
The period is the reciprocal of the frequency, T = 1/f. For 50 Hz this is 1/50 = 0.02 s, which equals 20 ms.
Question 30Electrical theory and fundamentals
An AC supply has an RMS voltage of 230 V. What is its peak voltage, assuming a sinusoidal waveform?
- 460 V
- 325 V Correct answer.
- 163 V
- 230 V
Explanation
For a sine wave the peak value equals the RMS value multiplied by √2. Thus 230 V × 1.414 is approximately 325 V.
Question 31Electrical theory and fundamentals
Two capacitors of 6 µF and 3 µF are connected in series. What is the total capacitance?
- 18 µF
- 4.5 µF
- 2 µF Correct answer.
- 9 µF
Explanation
For capacitors in series the reciprocals add: 1/C = 1/6 + 1/3 = 1/2. Inverting gives a total capacitance of 2 µF.
Question 32Electrical theory and fundamentals
Two inductors of 6 H and 3 H are connected in parallel, with no mutual coupling. What is the total inductance?
- 2 H Correct answer.
- 4.5 H
- 9 H
- 3 H
Explanation
For inductors in parallel the rule is the product divided by the sum, L = (L1 × L2)/(L1 + L2). This gives 18/9 = 2 H.
Question 33Electrical theory and fundamentals
In an AC circuit, power factor is defined as which ratio?
- True power divided by apparent power Correct answer.
- Apparent power divided by true power
- Reactive power divided by true power
- True power divided by reactive power
Explanation
Power factor is the ratio of true power in watts to apparent power in volt-amperes. It is also equal to the cosine of the phase angle between voltage and current.
Question 34Electrical theory and fundamentals
A series circuit has an inductance of 0.1 H and a capacitance of 10 µF. At what frequency does resonance occur?
- 159 Hz Correct answer.
- 1590 Hz
- 15.9 Hz
- 50 Hz
Explanation
Resonance occurs at f = 1/(2π√(LC)). Substituting L = 0.1 H and C = 10 µF gives approximately 159 Hz.
Question 35Electrical theory and fundamentals
A single-phase transformer has 600 turns on the primary and is supplied at 240 V. How many secondary turns are needed for a secondary voltage of 12 V?
- 300
- 30 Correct answer.
- 120
- 60
Explanation
The turns ratio equals the voltage ratio, Ns/Np = Vs/Vp. Here Ns = 600 × (12/240) = 30 turns.
Question 36Electrical theory and fundamentals
A 240 V DC shunt motor has a back EMF of 235 V and an armature resistance of 0.5 Ω. What is the armature current?
- 470 A
- 5 A
- 10 A Correct answer.
- 480 A
Explanation
The armature current is set by the net voltage across the armature resistance, Ia = (V − Eb)/Ra. This gives (240 − 235)/0.5 = 10 A.
Question 37Electrical theory and fundamentals
A semiconductor diode is forward biased when which condition is met?
- No voltage is applied across the junction
- The anode is positive with respect to the cathode Correct answer.
- The reverse voltage exceeds the peak inverse rating
- The cathode is positive with respect to the anode
Explanation
Forward bias occurs when the anode is made positive relative to the cathode. This reduces the depletion region and allows conventional current to flow.
Question 38Electrical theory and fundamentals
A 12 V DC source is connected to a 6 Ω resistor in series with a 3 Ω resistor. What are the Thevenin voltage and Thevenin resistance of the circuit seen by the 3 Ω resistor?
- 12 V and 9 Ω
- 4 V and 9 Ω
- 4 V and 2 Ω Correct answer.
- 8 V and 2 Ω
Explanation
Remove the 3 Ω load: the open-circuit voltage across its terminals is 12 × 3/(6 + 3) = 4 V. With the source replaced by a short, the resistance is 6 Ω in parallel with 3 Ω, giving 2 Ω.
Question 39Electrical theory and fundamentals
At higher frequencies, the skin effect in a conductor causes which change?
- Current to distribute evenly, reducing AC resistance
- The conductor's DC resistance to fall to zero
- Current to concentrate near the surface, increasing effective AC resistance Correct answer.
- Voltage to rise along the conductor without any current change
Explanation
The skin effect is the tendency of alternating current to crowd toward the outer surface of a conductor. This reduces the effective cross-sectional area carrying current, so the effective AC resistance increases with frequency.
Question 40Electrical theory and fundamentals
A series RLC circuit has R = 10 Ω, and at resonance XL = XC = 100 Ω. The supply is 10 V AC. What is the voltage across the capacitor at resonance?
- 1000 V
- 1 V
- 10 V
- 100 V Correct answer.
Explanation
At resonance the inductive and capacitive reactances cancel, so the impedance is equal to R = 10 Ω. The current is 10 V / 10 Ω = 1 A, and the capacitor voltage is 1 A × 100 Ω = 100 V.
Question 41Wiring Rules and installation standards (AS/NZS 3000)
In a MEN installation, the link between the neutral bar and the earth bar is installed at the:
- Each socket outlet
- Each distribution board
- The main switchboard Correct answer.
- The point of supply transformer
Explanation
The MEN link is made at the main switchboard so that the installation neutral and protective earth are bonded at one point only. This provides the low-impedance fault path required for protective devices while avoiding parallel neutral currents in earth conductors.
Question 42Wiring Rules and installation standards (AS/NZS 3000)
The rated residual operating current of an RCD used to provide additional protection for socket outlets is:
- 10 mA
- 300 mA
- 100 mA
- 30 mA Correct answer.
Explanation
Additional protection for socket outlets is provided by a 30 mA RCD. This sensitivity is intended to reduce the risk of electric shock from contact with live parts.
Question 43Wiring Rules and installation standards (AS/NZS 3000)
The active conductor of a single-phase flexible cord is coloured:
- Blue
- Green/yellow
- Black
- Brown Correct answer.
Explanation
In a single-phase flexible cord, brown identifies the active, blue identifies the neutral and green/yellow identifies the protective earth. These colours allow the conductors to be correctly terminated.
Question 44Wiring Rules and installation standards (AS/NZS 3000)
The protective earthing conductor in a fixed wiring installation is identified by:
- Green/yellow Correct answer.
- Brown
- Blue
- Green with a yellow stripe
Explanation
The protective earthing conductor is identified by green and yellow. The combination is reserved for earthing and must not be used for active or neutral conductors.
Question 45Wiring Rules and installation standards (AS/NZS 3000)
The maximum voltage drop permitted by AS/NZS 3000 from the point of supply to any point in an installation is:
- 10%
- 7.5%
- 5% Correct answer.
- 2.5%
Explanation
AS/NZS 3000 sets a maximum voltage drop of 5% between the point of supply and any point in the installation. This limit helps ensure connected equipment operates correctly.
Question 46Wiring Rules and installation standards (AS/NZS 3000)
The main purpose of the MEN system is to:
- Provide a low-impedance fault path Correct answer.
- Reduce voltage drop on long circuits
- Balance the load across the three phases
- Prevent nuisance tripping of RCDs
Explanation
The MEN system bonds the neutral and protective earth at the main switchboard, creating a low-impedance path for fault current. This allows protective devices to operate quickly under earth-fault conditions.
Question 47Wiring Rules and installation standards (AS/NZS 3000)
An RCD installed in a final subcircuit provides protection primarily against:
- Surge voltage
- Earth leakage current Correct answer.
- Overload current
- Short-circuit current
Explanation
An RCD detects a difference between active and neutral current, which indicates current flowing to earth. It is designed to protect against earth leakage and electric shock, not overload or short-circuit current.
Question 48Wiring Rules and installation standards (AS/NZS 3000)
A main switchboard in a domestic installation must be:
- Readily accessible Correct answer.
- Mounted on a wooden board
- Located in a bathroom
- Installed inside a ceiling space
Explanation
Switchboards must be accessible so that the main switch and protective devices can be operated and maintained safely. They must not be installed in locations where access is restricted.
Question 49Wiring Rules and installation standards (AS/NZS 3000)
When several cables are bunched together and enclosed in conduit, the current-carrying capacity must be:
- Determined only by the cable colour
- Left unchanged
- Increased to compensate for heat
- Reduced using a grouping factor Correct answer.
Explanation
Bunched cables cannot dissipate heat as readily as a single cable, so a grouping factor is applied to reduce the tabulated current-carrying capacity. This prevents excessive conductor temperature.
Question 50Wiring Rules and installation standards (AS/NZS 3000)
A cable rated for 40°C ambient but installed where the ambient temperature is 50°C must have its current-carrying capacity:
- Selected directly from the 40°C table
- Reduced by a derating factor Correct answer.
- Increased by 10%
- Kept the same
Explanation
Higher ambient temperature reduces the cable's ability to dissipate heat, so a derating factor must be applied. The cable's current-carrying capacity is reduced accordingly.