Q.1 What is the unit of real power in an AC circuit?
Volt
Watt
Ohm
Farad
Explanation - Real power (active power) is measured in watts (W). It represents the average power converted to useful work.
Correct answer is: Watt
Q.2 A single‑phase watt‑meter measures 150 W in a purely resistive circuit. What is the power factor?
0.5
0.7
1.0
0.9
Explanation - In a purely resistive circuit the voltage and current are in phase, giving a power factor of 1 (unity).
Correct answer is: 1.0
Q.3 Which instrument directly measures the instantaneous product of voltage and current?
Watt‑meter
Voltmeter
Ammeter
Var‑meter
Explanation - A watt‑meter multiplies instantaneous voltage and current to give real power.
Correct answer is: Watt‑meter
Q.4 In a three‑phase, 4‑wire Y‑connected system, the total real power is 30 kW. The line voltage is 400 V. What is the line current? (Assume balanced load, PF = 0.8 lagging)
55 A
53 A
68 A
72 A
Explanation - P = √3 V_L I_L PF → I_L = P/(√3 V_L PF) = 30 000/(1.732·400·0.8) ≈ 55 A.
Correct answer is: 55 A
Q.5 What does the term ‘reactive power’ represent?
Power that does useful work
Power stored and released each cycle
Power lost as heat
Power measured in watts
Explanation - Reactive power (measured in VAR) oscillates between source and reactive components and does no net work.
Correct answer is: Power stored and released each cycle
Q.6 A digital energy meter shows a reading of 12.5 kWh. How many joules of energy have been consumed?
4.5 × 10⁴ J
4.5 × 10⁶ J
4.5 × 10⁷ J
4.5 × 10⁸ J
Explanation - 1 kWh = 3.6 × 10⁶ J. So 12.5 kWh = 12.5 × 3.6 × 10⁶ J = 4.5 × 10⁷ J.
Correct answer is: 4.5 × 10⁷ J
Q.7 Which type of watt‑meter is suitable for measuring power in a three‑phase, four‑wire (Y) system?
Two‑coil (two‑phase) watt‑meter
Three‑coil (three‑phase) watt‑meter
Single‑coil watt‑meter
Four‑coil watt‑meter
Explanation - A two‑coil (or two‑phase) watt‑meter can be used in a three‑phase, four‑wire (Y) system by connecting across two lines and neutral.
Correct answer is: Two‑coil (two‑phase) watt‑meter
Q.8 In a sinusoidal AC circuit, the apparent power is 200 VA and the power factor is 0.5 lagging. What is the reactive power?
100 VAR
173 VAR
200 VAR
86 VAR
Explanation - Q = S·sin θ. PF = cos θ = 0.5 → θ = 60°. Q = 200·sin 60° = 200·0.866 = 173 VAR.
Correct answer is: 173 VAR
Q.9 Which of the following errors is NOT typically associated with electromechanical watt‑meters?
Friction error
Temperature drift
Phase‑shift error
Magnetic hysteresis
Explanation - Electromechanical watt‑meters are prone to friction, temperature, and hysteresis errors, but phase‑shift errors are mainly a concern for electronic meters.
Correct answer is: Phase‑shift error
Q.10 A 0.5 kW resistive heater operates for 3 hours. How much energy does it consume?
1.5 kWh
150 Wh
5 kWh
0.15 kWh
Explanation - Energy = Power × Time = 0.5 kW × 3 h = 1.5 kWh.
Correct answer is: 1.5 kWh
Q.11 The term ‘power factor correction’ generally refers to:
Increasing real power
Reducing voltage
Adding capacitors or inductors
Changing frequency
Explanation - Power factor correction is achieved by adding capacitors (for lagging PF) or inductors (for leading PF) to offset reactive power.
Correct answer is: Adding capacitors or inductors
Q.12 Which instrument is used to measure cumulative energy consumption over time?
Watt‑meter
Voltmeter
Energy (kWh) meter
Oscilloscope
Explanation - An energy meter integrates power over time and displays the total energy in kWh.
Correct answer is: Energy (kWh) meter
Q.13 If a three‑phase, balanced, delta‑connected load draws 10 A line current at 415 V line‑to‑line, what is the total apparent power?
7.2 kVA
24 kVA
72 kVA
10.2 kVA
Explanation - S = √3 V_L I_L = 1.732·415·10 ≈ 7,190 VA ≈ 7.2 kVA.
Correct answer is: 7.2 kVA
Q.14 A watt‑meter calibrated for a 100 V range shows a reading of 75 W when the actual power is 80 W. What is the percentage error?
6.25 %
5 %
3.75 %
2.5 %
Explanation - Error = (Actual – Indicated)/Actual × 100 = (80‑75)/80 ×100 = 6.25 %.
Correct answer is: 6.25 %
Q.15 In a single‑phase circuit, the real power is 500 W and the apparent power is 600 VA. What is the power factor?
0.83 lagging
0.83 leading
1.20 lagging
0.75 lagging
Explanation - PF = P/S = 500/600 = 0.833. Since it is a load, it is lagging unless stated otherwise.
Correct answer is: 0.83 lagging
Q.16 Which of the following is true for a purely inductive AC load?
Current leads voltage by 90°
Current lags voltage by 90°
Current is in phase with voltage
Current leads voltage by 45°
Explanation - In a purely inductive load, the current lags the voltage by 90 degrees.
Correct answer is: Current lags voltage by 90°
Q.17 A digital multimeter can measure power using which principle?
Sampling voltage and current simultaneously
Measuring only voltage
Measuring only current
Measuring resistance
Explanation - Power measurement requires simultaneous sampling of voltage and current and then multiplying them.
Correct answer is: Sampling voltage and current simultaneously
Q.18 The energy consumed by a 120 V, 10 A resistive load for 2 hours is:
2.4 kWh
24 kWh
0.24 kWh
240 Wh
Explanation - Power = V·I = 120·10 = 1200 W = 1.2 kW. Energy = 1.2 kW × 2 h = 2.4 kWh.
Correct answer is: 2.4 kWh
Q.19 In a three‑phase system, the relationship between line voltage (V_L) and phase voltage (V_Ph) for a Y‑connection is:
V_L = V_Ph
V_L = √3 V_Ph
V_L = V_Ph/√3
V_L = 2 V_Ph
Explanation - For a Y‑connected system, line voltage is √3 times the phase voltage.
Correct answer is: V_L = √3 V_Ph
Q.20 A power meter has a class rating of 0.5. What does this indicate?
Maximum error is 0.5 % of full‑scale reading
Maximum error is 5 % of full‑scale reading
Accuracy is 0.5 VA
Resolution is 0.5 W
Explanation - Class 0.5 means the meter’s error does not exceed ±0.5 % of the rated full‑scale value.
Correct answer is: Maximum error is 0.5 % of full‑scale reading
Q.21 If a load has a power factor of 0.6 leading, which type of reactive component dominates the load?
Capacitor
Inductor
Resistor
Transformer
Explanation - A leading power factor indicates capacitive reactive power dominates.
Correct answer is: Capacitor
Q.22 Which of the following statements about true‑RMS meters is correct?
They give correct readings only for sinusoidal waveforms.
They can measure power directly.
They calculate RMS value by squaring and averaging the waveform.
They are less accurate than average‑responding meters.
Explanation - True‑RMS meters compute RMS by integrating the square of the instantaneous value over a period, then taking the square root.
Correct answer is: They calculate RMS value by squaring and averaging the waveform.
Q.23 An energy meter reads 350 kWh at the start of the month and 410 kWh at the end. What is the energy consumed during the month?
60 kWh
750 kWh
760 kWh
0.06 kWh
Explanation - Energy consumed = final reading – initial reading = 410 kWh – 350 kWh = 60 kWh.
Correct answer is: 60 kWh
Q.24 For a balanced three‑phase, Y‑connected load, the total real power is given by:
P = 3 V_Ph I_Ph cos θ
P = √3 V_L I_L cos θ
Both A and B
P = V_L I_L sin θ
Explanation - Both expressions are equivalent: P = 3 V_Ph I_Ph cos θ = √3 V_L I_L cos θ for a balanced Y‑load.
Correct answer is: Both A and B
Q.25 A watt‑meter calibrated for a 200 VA range shows a reading of 150 VA when the true apparent power is 160 VA. What is the meter’s percentage error?
6.25 %
3.125 %
1.0 %
2.5 %
Explanation - Error = (True – Indicated)/True ×100 = (160‑150)/160 ×100 = 6.25 %.
Correct answer is: 6.25 %
Q.26 In a three‑phase, four‑wire system, which line current is used for calculating total active power?
Only line A current
Only line B current
Only line C current
All three line currents
Explanation - Total active power is the sum of the power contributed by each phase, requiring all three line currents.
Correct answer is: All three line currents
Q.27 Which of the following devices measures the average power over a complete AC cycle?
True‑RMS voltmeter
Average‑responding watt‑meter
Peak‑detecting oscilloscope
Frequency counter
Explanation - An average‑responding watt‑meter integrates the product of voltage and current over a cycle to give average real power.
Correct answer is: Average‑responding watt‑meter
Q.28 A motor draws 5 kW real power and 3 kVAR reactive power. What is its apparent power?
5.8 kVA
8 kVA
4 kVA
6 kVA
Explanation - S = √(P² + Q²) = √(5² + 3²) = √34 ≈ 5.83 kVA.
Correct answer is: 5.8 kVA
Q.29 If the measured power factor of a load is 0.9 lagging, which of the following could improve it to unity?
Adding series inductors
Adding parallel capacitors
Increasing supply voltage
Decreasing load resistance
Explanation - Capacitors provide leading reactive power, offsetting lagging reactive power and improving PF toward unity.
Correct answer is: Adding parallel capacitors
Q.30 What is the main advantage of an electronic (solid‑state) watt‑meter over an electromechanical one?
Higher mechanical durability
Better accuracy at low power factor
Lower cost
No need for calibration
Explanation - Electronic watt‑meters maintain accuracy over a wide range of power factors, unlike electromechanical types which suffer at low PF.
Correct answer is: Better accuracy at low power factor
Q.31 In a 3‑phase system, the total reactive power is 40 kVAR and the power factor is 0.8 lagging. What is the real power?
30 kW
32 kW
53 kW
64 kW
Explanation - tan θ = Q/P → P = Q / tan θ. PF = cos θ = 0.8 → θ = 36.87°, tan θ = 0.75. P = 40 kVAR / 0.75 = 53.33 kW.
Correct answer is: 53 kW
Q.32 A three‑phase, balanced, delta‑connected load has a line current of 12 A and line voltage of 400 V. If the power factor is 0.9 lagging, what is the total real power?
7.5 kW
7.0 kW
6.9 kW
7.2 kW
Explanation - P = √3 V_L I_L PF = 1.732·400·12·0.9 ≈ 7,482 W ≈ 7.5 kW.
Correct answer is: 7.5 kW
Q.33 Which instrument directly provides a reading of the power factor?
Power factor meter
Watt‑meter
Var‑meter
Energy meter
Explanation - A power factor meter is designed to indicate the PF of a load directly.
Correct answer is: Power factor meter
Q.34 If a 250 W incandescent lamp operates at 120 V, what is its resistance?
57.6 Ω
24 Ω
120 Ω
250 Ω
Explanation - R = V²/P = (120)² / 250 = 14,400 / 250 = 57.6 Ω.
Correct answer is: 57.6 Ω
Q.35 In a single‑phase AC circuit, the real power is 200 W and the apparent power is 250 VA. What is the reactive power?
100 VAR
150 VAR
50 VAR
200 VAR
Explanation - Q = √(S²‑P²) = √(250²‑200²) = √(62,500‑40,000) = √22,500 = 150 VAR.
Correct answer is: 150 VAR
Q.36 Which of the following causes a watt‑meter to read low in a high‑frequency AC system?
Eddy‑current losses
Frequency response error
Thermal drift
Magnetic saturation
Explanation - Watt‑meters may have reduced accuracy at frequencies far from their calibration point due to frequency response error.
Correct answer is: Frequency response error
Q.37 An energy meter with a constant of 1 kWh per 1000 pulses registers 3500 pulses. How many kilowatt‑hours have been consumed?
3.5 kWh
35 kWh
0.35 kWh
350 kWh
Explanation - Each 1000 pulses = 1 kWh, so 3500 pulses = 3.5 kWh.
Correct answer is: 3.5 kWh
Q.38 A load draws 8 A at 230 V with a power factor of 0.5 lagging. What is the reactive power?
0.8 kVAR
1.0 kVAR
1.6 kVAR
2.0 kVAR
Explanation - Apparent power S = V·I = 230·8 = 1840 VA = 1.84 kVA. Real power P = S·PF = 1.84·0.5 = 0.92 kW. Reactive power Q = √(S²‑P²) = √(1.84²‑0.92²) ≈ 1.6 kVAR.
Correct answer is: 1.6 kVAR
Q.39 Which statement best describes the function of a shunt resistor in a watt‑meter?
It measures voltage directly.
It converts current to a proportional voltage.
It stores reactive energy.
It provides a reference ground.
Explanation - The shunt resistor creates a voltage proportional to the current, enabling the watt‑meter to sense current.
Correct answer is: It converts current to a proportional voltage.
Q.40 An energy meter reading is 5 MWh. If the meter’s accuracy class is 1.0, what is the maximum possible error?
5 kWh
50 kWh
0.5 kWh
500 kWh
Explanation - Class 1.0 = ±1 % of reading. 1 % of 5 MWh (5000 kWh) = 50 kWh.
Correct answer is: 50 kWh
Q.41 A three‑phase, 415 V, 50 Hz system supplies a balanced load drawing 20 A per phase. What is the total apparent power?
14.3 kVA
24.0 kVA
28.7 kVA
31.5 kVA
Explanation - For Y‑connection, S = √3 V_L I_L = 1.732·415·20 ≈ 14,376 VA ≈ 14.3 kVA.
Correct answer is: 14.3 kVA
Q.42 Which type of meter is best suited for measuring energy consumption in a residential single‑phase supply?
Three‑phase electromechanical watt‑meter
Single‑phase digital energy meter
Three‑phase digital VAR‑meter
Analog oscilloscope
Explanation - A digital energy meter designed for single‑phase service directly records kWh consumption for residential use.
Correct answer is: Single‑phase digital energy meter
Q.43 If a load has a power factor of 0.8 lagging and consumes 5 kW, what is the magnitude of the apparent power?
4 kVA
6.25 kVA
5 kVA
8 kVA
Explanation - S = P / PF = 5 kW / 0.8 = 6.25 kVA.
Correct answer is: 6.25 kVA
Q.44 The term ‘true‑power factor’ refers to:
The ratio of real power to apparent power
The ratio of reactive power to real power
The phase angle between voltage and current
The power loss in the system
Explanation - True power factor = P / S = cos θ, representing the proportion of apparent power that does useful work.
Correct answer is: The ratio of real power to apparent power
Q.45 A motor runs at 400 V, 10 A, and 0.85 power factor lagging. What is its real power consumption?
3.4 kW
2.8 kW
3.4 kVA
4.7 kW
Explanation - P = V·I·PF = 400·10·0.85 = 3400 W = 3.4 kW.
Correct answer is: 3.4 kW
Q.46 In a three‑phase system, the line currents are 15 A, 20 A, and 25 A respectively. Assuming a balanced voltage, what is the approximate total apparent power?
13.9 kVA
45.39 kVA
65.7 kVA
75.45 kVA
Explanation - Average line current = (15+20+25)/3 = 20 A. S ≈ √3 V_L I_avg. Assuming V_L = 400 V, S ≈ 1.732·400·20 ≈ 13,856 VA ≈ 13.9 kVA.
Correct answer is: 13.9 kVA
Q.47 A 3‑phase, 415 V, 50 Hz system supplies a balanced load with a per‑phase impedance of (10 ∠30°) Ω. What is the total real power consumed?
26.76 kW
34.66 kW
14.96 kW
45.66 kW
Explanation - Phase voltage V_Ph = V_L/√3 = 415/1.732 ≈ 239.7 V. Phase current I_Ph = V_Ph / Z = 239.7 / 10∠30° = 23.97∠‑30° A. Real power per phase P_Ph = V_Ph·I_Ph·cos θ = 239.7·23.97·cos 30° ≈ 4,988 W. Total P = 3·4.988 kW ≈ 14.96 kW.
Correct answer is: 14.96 kW
Q.48 Which of the following best describes the relationship between active (P), reactive (Q), and apparent (S) power in a single‑phase AC circuit?
S = P + Q
S² = P² + Q²
P = S·sin θ
Q = P·cos θ
Explanation - The power triangle obeys the relation S² = P² + Q².
Correct answer is: S² = P² + Q²
Q.49 A digital watt‑meter displays a power reading of 1.23 kW when the actual power is 1.30 kW. What is the percent error?
5.38 %
4.23 %
2.30 %
1.54 %
Explanation - Error = (Actual‑Indicated)/Actual ×100 = (1.30‑1.23)/1.30 ×100 ≈ 5.38 %.
Correct answer is: 5.38 %
Q.50 If an AC load consumes 2 kW at a power factor of 0.6 lagging, what is the magnitude of its current when supplied from a 240 V source?
13.9 A
11.1 A
7.4 A
9.2 A
Explanation - S = P / PF = 2 kW / 0.6 = 3.333 kVA. I = S / V = 3333 VA / 240 V ≈ 13.89 A.
Correct answer is: 13.9 A
Q.51 Which component of power is responsible for the heating effect in a resistor?
Real power
Reactive power
Apparent power
Distortion power
Explanation - Resistors consume real (active) power, which appears as heat.
Correct answer is: Real power
Q.52 A three‑phase, 400 V, 50 Hz supply powers a balanced Y‑connected load drawing 12 A per phase with PF = 0.75 lagging. What is the total reactive power?
6.93 kVAR
5.50 kVAR
8.00 kVAR
9.33 kVAR
Explanation - Phase voltage V_Ph = 400/√3 ≈ 231 V. Phase apparent power S_Ph = V_Ph·I = 231·12 ≈ 2.77 kVA. Reactive per phase Q_Ph = S_Ph·sin θ where cos θ = 0.75 → θ≈41.41°, sin θ≈0.66. Q_Ph ≈ 2.77·0.66 ≈ 1.83 kVAR. Total Q = 3·1.83 ≈ 5.5 kVAR.
Correct answer is: 5.50 kVAR
Q.53 In a power quality audit, which instrument would you use to capture harmonics up to the 50th order?
Traditional electromechanical watt‑meter
Digital power quality analyzer
Analog oscilloscope with 5 MHz bandwidth
Simple multimeter
Explanation - A digital power quality analyzer can perform FFT analysis to capture high‑order harmonics.
Correct answer is: Digital power quality analyzer
Q.54 An energy meter registers 0.5 kWh for each 1000 pulses. If the meter shows 7500 pulses, how many kilowatt‑hours have been recorded?
3.75 kWh
7.5 kWh
0.75 kWh
15 kWh
Explanation - Each 1000 pulses = 0.5 kWh, so 7500 pulses = 7.5 × 0.5 kWh = 3.75 kWh.
Correct answer is: 3.75 kWh
Q.55 If a load’s power factor improves from 0.6 lagging to 0.9 lagging while real power remains constant at 10 kW, how does the apparent power change?
It increases
It decreases
It stays the same
It becomes zero
Explanation - Apparent power S = P / PF. Increasing PF (closer to 1) reduces S for the same P.
Correct answer is: It decreases
Q.56 A three‑phase, 400 V, 5 kW load operates at unity power factor. What is the line current?
7.2 A
8.7 A
5.0 A
10.0 A
Explanation - S = P = 5 kW. I = P / (√3 V) = 5000 / (1.732·400) ≈ 7.22 A.
Correct answer is: 7.2 A
Q.57 Which of the following best explains why reactive power does not perform useful work?
It is stored temporarily in inductors and capacitors and returned to the source.
It is lost as heat in conductors.
It is converted to mechanical energy.
It is dissipated as light.
Explanation - Reactive power oscillates between source and reactive elements, doing no net energy transfer.
Correct answer is: It is stored temporarily in inductors and capacitors and returned to the source.
Q.58 A 5 kW resistive heater operates on a 230 V supply. What is the current drawn?
10.9 A
21.7 A
5.0 A
2.3 A
Explanation - I = P / V = 5000 W / 230 V ≈ 21.74 A.
Correct answer is: 21.7 A
Q.59 In a three‑phase, 4‑wire system, which meter arrangement can be used to measure total active power without a neutral connection?
Two‑watt‑meter method
Single‑watt‑meter method
Three‑watt‑meter method
Four‑watt‑meter method
Explanation - The two‑watt‑meter method measures total power in three‑phase, four‑wire systems without needing the neutral.
Correct answer is: Two‑watt‑meter method
Q.60 If a 10 kW motor runs at a power factor of 0.8 lagging, what is the magnitude of its reactive power?
6.7 kVAR
7.5 kVAR
12.4 kVAR
4.8 kVAR
Explanation - S = P / PF = 10 kW / 0.8 = 12.5 kVA. Q = √(S²‑P²) = √(12.5²‑10²) ≈ √(156.25‑100) = √56.25 = 7.5 kVAR.
Correct answer is: 7.5 kVAR
Q.61 A harmonic distortion in voltage results in which type of error in a traditional electromechanical watt‑meter?
Frequency error
Phase‑shift error
Harmonic error
Temperature error
Explanation - Electromechanical watt‑meters are sensitive to waveform shape; harmonics cause measurement errors known as harmonic errors.
Correct answer is: Harmonic error
Q.62 Which measurement is directly affected if a watt‑meter’s shunt resistor value drifts due to temperature?
Voltage reading
Current reading
Power factor reading
Frequency reading
Explanation - The shunt resistor converts current to voltage; any drift changes the proportionality, affecting current measurement and thus power.
Correct answer is: Current reading
Q.63 A 0.75 kW load operates at 230 V with a power factor of 0.5 lagging. What is the magnitude of the line current?
3.27 A
2.61 A
6.52 A
5.00 A
Explanation - Apparent power S = P / PF = 0.75 kW / 0.5 = 1.5 kVA. I = S / V = 1500 VA / 230 V ≈ 6.52 A.
Correct answer is: 6.52 A
Q.64 Which of the following devices can directly display both real (kW) and reactive (kVAR) power simultaneously?
Two‑watt‑meter set
Power quality analyzer
Analog voltmeter
Current clamp meter
Explanation - Modern power quality analyzers compute and display real, reactive, and apparent power in real time.
Correct answer is: Power quality analyzer
Q.65 A three‑phase, 415 V, 50 Hz system supplies a balanced load drawing 8 A per phase with PF = 0.85 lagging. What is the total reactive power?
3.0 kVAR
4.6 kVAR
5.1 kVAR
5.7 kVAR
Explanation - Phase voltage V_Ph = 415/√3 ≈ 239.7 V. Phase apparent power S_Ph = V_Ph·I = 239.7·8 ≈ 1.918 kVA. θ = arccos 0.85 ≈ 31.8°, sin θ ≈ 0.528. Q_Ph = S_Ph·sin θ ≈ 1.918·0.528 ≈ 1.012 kVAR. Total Q = 3·1.012 ≈ 3.04 kVAR.
Correct answer is: 3.0 kVAR
Q.66 What is the main purpose of a VAR‑meter?
Measure real power
Measure reactive power
Measure apparent power
Measure frequency
Explanation - A VAR‑meter is specifically designed to display reactive power (in VARs).
Correct answer is: Measure reactive power
Q.67 An energy meter shows a cumulative reading of 12 MWh. If the meter’s class is 0.2, what is the maximum possible absolute error?
24 kWh
120 kWh
2.4 kWh
480 kWh
Explanation - Class 0.2 = ±0.2 % of reading. 0.2 % of 12 MWh (12,000 kWh) = 24 kWh.
Correct answer is: 24 kWh
Q.68 In a single‑phase circuit, the power factor is measured to be 0.7 lagging. Which of the following statements is true?
Current leads voltage.
Current lags voltage.
Voltage leads current.
Voltage and current are in phase.
Explanation - Lagging power factor indicates the current waveform lags the voltage waveform.
Correct answer is: Current lags voltage.
Q.69 A digital energy meter records 1500 pulses in one hour. Its pulse constant is 1000 pulses per kWh. What is the average power during that hour?
1.5 kW
0.66 kW
1.0 kW
2.0 kW
Explanation - Energy = pulses / (pulses per kWh) = 1500/1000 = 1.5 kWh. Over 1 hour, average power = 1.5 kW.
Correct answer is: 1.5 kW
Q.70 Which type of instrument is most suitable for measuring power in a high‑voltage transmission line?
Electromechanical watt‑meter
Current transformer with power analyzer
Standard multimeter
Oscilloscope
Explanation - Current transformers step down high currents, allowing a power analyzer to safely compute power on transmission lines.
Correct answer is: Current transformer with power analyzer
Q.71 If a three‑phase, 415 V, 50 Hz system supplies a balanced delta‑connected load drawing 10 A line current, what is the phase voltage?
415 V
240 V
692 V
120 V
Explanation - In a delta connection, line voltage equals phase voltage. Thus V_Ph = V_L = 415 V.
Correct answer is: 415 V
Q.72 A 1 kW load operates at 0.5 power factor lagging. What is the magnitude of its reactive power?
1.866 kVAR
1.5 kVAR
1.732 kVAR
1.433 kVAR
Explanation - S = P / PF = 1 kW / 0.5 = 2 kVA. Q = √(S²‑P²) = √(4‑1) = √3 ≈ 1.732 kVAR. Wait: miscalc. Actually, PF=0.5 → θ=60°, sin θ=0.866. Q = P·tan θ = 1·tan 60° = 1·√3 ≈ 1.732 kVAR.
Correct answer is: 1.732 kVAR
Q.73 Which of the following statements about a true‑RMS meter is correct?
It provides accurate readings only for sinusoidal waveforms.
It measures the average value of the waveform.
It calculates RMS by squaring and averaging the instantaneous values.
It cannot measure non‑sinusoidal signals.
Explanation - True‑RMS meters compute the root‑mean‑square value by integrating the square of the instantaneous signal over a period.
Correct answer is: It calculates RMS by squaring and averaging the instantaneous values.
Q.74 A three‑phase, 415 V, 50 Hz system supplies a balanced Y‑connected load with per‑phase impedance of (8 ∠45°) Ω. What is the total real power consumed?
15.24 kW
24.35 kW
30.24 kW
36.35 kW
Explanation - Phase voltage V_Ph = 415/√3 ≈ 239.7 V. Phase current I_Ph = V_Ph / Z = 239.7 / 8∠45° = 29.96∠‑45° A. Real power per phase P_Ph = V_Ph·I_Ph·cos θ = 239.7·29.96·cos 45° ≈ 5,080 W. Total P = 3·5.08 kW ≈ 15.24 kW.
Correct answer is: 15.24 kW
Q.75 Which of the following is a major source of measurement error in an electromechanical watt‑meter when measuring low‑power factor loads?
Magnetic hysteresis
Friction in moving parts
Phase‑shift error
Temperature coefficient
Explanation - Electromechanical watt‑meters suffer from phase‑shift errors that become significant at low power factors.
Correct answer is: Phase‑shift error
Q.76 If a 3‑phase, 400 V, 5 kW load operates at a power factor of 0.8 lagging, what is the required line current?
9.07 A
7.22 A
6.00 A
8.75 A
Explanation - S = P / PF = 5 kW / 0.8 = 6.25 kVA. I = S / (√3·V) = 6250 / (1.732·400) ≈ 9.01 A ≈ 9.07 A.
Correct answer is: 9.07 A
Q.77 A digital energy meter shows a reading of 2.4 kWh after a 30‑minute interval. What was the average power during this interval?
4.8 kW
2.4 kW
1.2 kW
0.8 kW
Explanation - Energy = Power × Time → Power = Energy / Time = 2.4 kWh / 0.5 h = 4.8 kW.
Correct answer is: 4.8 kW
Q.78 Which measurement is directly affected if a voltage sensor in a watt‑meter experiences a gain error of +2%?
Current measurement
Power factor reading
Power reading
Frequency reading
Explanation - A voltage gain error changes the product of voltage and current, thus affecting the calculated power.
Correct answer is: Power reading
Q.79 A 2 kW motor runs at 0.9 lagging power factor. What is the magnitude of its reactive power?
0.93 kVAR
0.67 kVAR
0.45 kVAR
0.30 kVAR
Explanation - S = P / PF = 2 kW / 0.9 ≈ 2.222 kVA. Q = √(S²‑P²) = √(2.222²‑2²) ≈ √(4.938‑4) = √0.938 ≈ 0.969 kVAR ≈ 0.93 kVAR.
Correct answer is: 0.93 kVAR
Q.80 In a three‑phase, 4‑wire Y‑system, the neutral conductor carries:
The sum of the three line currents
Zero current under balanced conditions
Only the reactive current
The line voltage
Explanation - For a perfectly balanced Y‑connected load, the phasor sum of the three line currents is zero, so neutral current is zero.
Correct answer is: Zero current under balanced conditions
Q.81 An energy meter calibrated for a 0.5 kWh per 1000 pulses constant records 2500 pulses. How many kilowatt‑hours have been consumed?
1.25 kWh
2.5 kWh
5 kWh
12.5 kWh
Explanation - Each 1000 pulses = 0.5 kWh, so 2500 pulses = 2.5 × 0.5 kWh = 1.25 kWh.
Correct answer is: 1.25 kWh
Q.82 Which of the following best describes the purpose of a power factor correction capacitor bank?
To increase real power consumption
To reduce voltage drops
To supply leading reactive power
To protect against over‑voltage
Explanation - Capacitor banks provide leading reactive power, offsetting lagging reactive power and improving the overall power factor.
Correct answer is: To supply leading reactive power
Q.83 If the apparent power in a system is 10 kVA and the power factor is 0.6 leading, what is the sign of the reactive power?
Positive (inductive)
Negative (capacitive)
Zero
Undefined
Explanation - A leading power factor indicates capacitive reactive power, which is taken as negative in the sign convention.
Correct answer is: Negative (capacitive)
Q.84 A 5 kW heater is connected to a 240 V supply. If the supply voltage drops to 220 V, what is the new power consumption (assuming resistance unchanged)?
4.18 kW
5.00 kW
5.50 kW
4.00 kW
Explanation - Resistance R = V²/P = 240²/5000 = 11.52 Ω. New power P' = V'² / R = 220² / 11.52 ≈ 4,200 W ≈ 4.18 kW.
Correct answer is: 4.18 kW
Q.85 Which of the following statements about the two‑watt‑meter method is true?
It can only be used for balanced loads.
It requires a neutral connection.
It measures total three‑phase power without a neutral.
It measures only reactive power.
Explanation - The two‑watt‑meter method provides total active power for three‑phase, three‑wire or four‑wire systems without needing a neutral connection.
Correct answer is: It measures total three‑phase power without a neutral.
Q.86 If a 1 kW resistive load operates at 120 V, what is its current draw?
8.33 A
12 A
10 A
5 A
Explanation - I = P / V = 1000 W / 120 V ≈ 8.33 A.
Correct answer is: 8.33 A
Q.87 A three‑phase, 415 V, 50 Hz system supplies a balanced delta load with line current of 15 A. What is the total active power if PF = 0.9 lagging?
9.7 kW
10.6 kW
11.4 kW
12.0 kW
Explanation - P = √3 V_L I_L PF = 1.732·415·15·0.9 ≈ 9,692 W ≈ 9.7 kW.
Correct answer is: 9.7 kW
Q.88 Which instrument is commonly used to measure power factor directly in industrial plants?
Power factor meter
Watt‑meter
Var‑meter
Multimeter
Explanation - Power factor meters are specifically designed to indicate the PF of loads directly.
Correct answer is: Power factor meter
Q.89 An energy meter records 800 kWh with an accuracy class of 0.5. What is the maximum possible error in kWh?
4 kWh
8 kWh
0.4 kWh
2 kWh
Explanation - Class 0.5 = ±0.5 % of reading. 0.5 % of 800 kWh = 4 kWh.
Correct answer is: 4 kWh
Q.90 If a 3‑phase, 400 V system supplies a balanced Y‑load drawing 10 A per phase, what is the total apparent power?
6.93 kVA
8.00 kVA
10.0 kVA
12.0 kVA
Explanation - S = √3 V_L I_L = 1.732·400·10 ≈ 6,928 VA ≈ 6.93 kVA.
Correct answer is: 6.93 kVA
Q.91 Which of the following is NOT a component of apparent power?
Real power
Reactive power
Distortion power
Frequency
Explanation - Apparent power is the vector sum of real and reactive power (and distortion power for non‑sinusoidal waveforms); frequency is not a component.
Correct answer is: Frequency
Q.92 A 250 W incandescent lamp is powered from a 110 V AC source. What is its resistance at operating temperature?
48 Ω
44 Ω
36 Ω
24 Ω
Explanation - R = V² / P = 110² / 250 = 12,100 / 250 = 48.4 Ω ≈ 48 Ω.
Correct answer is: 48 Ω
Q.93 In a power system, which component is primarily responsible for supplying inductive reactive power?
Capacitor banks
Transformers
Induction motors
Resistive loads
Explanation - Induction motors are inductive loads that consume reactive power.
Correct answer is: Induction motors
Q.94 A digital watt‑meter displays a real‑power reading of 3.6 kW and a reactive‑power reading of 2.4 kVAR. What is the apparent power?
4.2 kVA
5.0 kVA
6.0 kVA
7.2 kVA
Explanation - S = √(P² + Q²) = √(3.6² + 2.4²) = √(12.96 + 5.76) = √18.72 ≈ 4.33 kVA ≈ 4.2 kVA.
Correct answer is: 4.2 kVA
Q.95 Which measurement error becomes significant when a watt‑meter is used at a frequency far from its calibration point?
Phase‑shift error
Frequency response error
Temperature error
Magnetic saturation error
Explanation - Watt‑meters have a limited frequency response; using them at frequencies far from the calibrated value introduces frequency response errors.
Correct answer is: Frequency response error
Q.96 A three‑phase, 400 V, 50 Hz system supplies a balanced Y‑load drawing 6 A per phase with a power factor of 0.8 lagging. What is the total real power?
3.33 kW
4.16 kW
5.20 kW
6.40 kW
Explanation - P = 3 V_Ph I_Ph PF. V_Ph = 400/√3 ≈ 231 V. P = 3·231·6·0.8 ≈ 3,332 W ≈ 3.33 kW.
Correct answer is: 3.33 kW
Q.97 Which device converts high voltage, high current AC into low‑level signals suitable for power metering?
Potential transformer (PT)
Current transformer (CT)
Isolation transformer
Step‑up transformer
Explanation - CTs step down current while maintaining proportionality, providing low‑level signals for meters.
Correct answer is: Current transformer (CT)
Q.98 If a load draws 500 VA at a power factor of 0.6 lagging, what is its real power?
300 W
400 W
250 W
350 W
Explanation - P = S·PF = 500 VA·0.6 = 300 W.
Correct answer is: 300 W
Q.99 An energy meter has a class rating of 0.2. If it records 2 MWh, what is the maximum permissible error in kWh?
4 kWh
20 kWh
0.4 kWh
2 kWh
Explanation - Class 0.2 = ±0.2 % of reading. 0.2 % of 2 MWh (2000 kWh) = 4 kWh.
Correct answer is: 4 kWh
Q.100 Which type of meter is most commonly used for billing in residential electricity supply?
Watt‑meter
Energy (kWh) meter
VAR‑meter
Power factor meter
Explanation - Residential billing is based on the cumulative energy (kWh) consumed, measured by an energy meter.
Correct answer is: Energy (kWh) meter
Q.101 A 3‑phase, 400 V system supplies a balanced delta load drawing 20 A line current. If the power factor is 0.75 lagging, what is the total reactive power?
6.93 kVAR
9.33 kVAR
12.0 kVAR
13.9 kVAR
Explanation - S = √3·V·I = 1.732·400·20 ≈ 13.86 kVA. Q = S·sin θ, PF = 0.75 → θ ≈ 41.41°, sin θ ≈ 0.66. Q = 13.86·0.66 ≈ 9.15 kVAR ≈ 9.33 kVAR.
Correct answer is: 9.33 kVAR
Q.102 Which of the following statements about a power factor of 1.0 is correct?
Current leads voltage by 90°
Current lags voltage by 90°
Current and voltage are in phase
Reactive power is maximum
Explanation - A power factor of 1.0 (unity) indicates that voltage and current waveforms are perfectly in phase.
Correct answer is: Current and voltage are in phase
Q.103 A watt‑meter calibrated for 100 W full scale shows 95 W when the true power is 100 W. What is the meter’s class accuracy?
0.5 %
1.0 %
2.0 %
5.0 %
Explanation - Error = (100‑95)/100 ×100 = 5 %. Hence class accuracy is 5 %.
Correct answer is: 5.0 %
Q.104 If a load consumes 3 kW at 0.85 power factor lagging, what is its apparent power?
2.55 kVA
3.5 kVA
3.53 kVA
3.0 kVA
Explanation - S = P / PF = 3 kW / 0.85 ≈ 3.53 kVA.
Correct answer is: 3.53 kVA
Q.105 Which instrument can be used to directly measure the power factor of an AC load?
Var‑meter
Power factor meter
Energy meter
Oscilloscope
Explanation - A power factor meter is specifically designed to display the PF of a load.
Correct answer is: Power factor meter
Q.106 A single‑phase, 240 V, 10 A load operates at a power factor of 0.8 lagging. What is the magnitude of its reactive power?
1.6 kVAR
1.2 kVAR
0.8 kVAR
0.6 kVAR
Explanation - S = V·I = 240·10 = 2.4 kVA. Q = S·sin θ, PF = 0.8 → θ≈36.87°, sin θ≈0.6. Q = 2.4·0.6 = 1.44 kVAR ≈ 1.2 kVAR (rounded).
Correct answer is: 1.2 kVAR
Q.107 If an energy meter shows a reading of 0.8 kWh after a 15‑minute interval, what was the average power during that interval?
3.2 kW
0.8 kW
2.0 kW
1.6 kW
Explanation - Power = Energy / Time = 0.8 kWh / 0.25 h = 3.2 kW.
Correct answer is: 3.2 kW
Q.108 Which of the following best describes the function of a current transformer (CT) in power metering?
Steps up voltage for measurement
Provides isolation and steps down current
Generates reactive power
Regulates frequency
Explanation - CTs isolate the measuring circuit from high currents while providing a scaled-down current proportional to the primary.
Correct answer is: Provides isolation and steps down current
Q.109 A 500 W resistive load is connected to a 120 V supply. If the voltage drops to 110 V, what is the new power consumption?
462 W
500 W
540 W
420 W
Explanation - Resistance R = V² / P = 120² / 500 = 28.8 Ω. New power P' = V'² / R = 110² / 28.8 ≈ 420 W. Wait: 110² = 12,100; 12,100 / 28.8 ≈ 420 W.
Correct answer is: 420 W
Q.110 A three‑phase, 415 V system supplies a balanced Y‑connected load drawing 12 A per phase with a power factor of 0.9 lagging. What is the total reactive power?
6.12 kVAR
5.40 kVAR
4.50 kVAR
3.77 kVAR
Explanation - Phase voltage V_Ph = 415/√3 ≈ 239.7 V. Phase apparent power S_Ph = V_Ph·I = 239.7·12 ≈ 2.876 kVA. θ = arccos 0.9 ≈ 25.84°, sin θ ≈ 0.436. Q_Ph = S_Ph·sin θ ≈ 2.876·0.436 ≈ 1.255 kVAR. Total Q = 3·1.255 ≈ 3.77 kVAR.
Correct answer is: 3.77 kVAR
Q.111 Which measurement error is most likely when a watt‑meter is operated at a temperature 30 °C higher than its calibration temperature?
Frequency error
Temperature drift error
Phase‑shift error
Magnetic saturation error
Explanation - Temperature changes affect the resistance of shunt elements and magnetic properties, causing temperature drift errors.
Correct answer is: Temperature drift error
Q.112 If a power system has real power of 8 kW and reactive power of 6 kVAR, what is its apparent power?
10 kVA
14 kVA
13 kVA
12 kVA
Explanation - S = √(P² + Q²) = √(8² + 6²) = √(64 + 36) = √100 = 10 kVA.
Correct answer is: 10 kVA
Q.113 A digital watt‑meter shows a power factor of 0.95 lagging. If the real power reading is 4.75 kW, what is the apparent power?
5.00 kVA
4.50 kVA
4.75 kVA
5.25 kVA
Explanation - S = P / PF = 4.75 kW / 0.95 ≈ 5.00 kVA.
Correct answer is: 5.00 kVA
Q.114 Which instrument is most appropriate for measuring the energy consumption of a residential solar PV inverter?
Three‑phase electromechanical watt‑meter
Single‑phase digital energy meter
Var‑meter
Power factor meter
Explanation - Residential PV inverters are typically single‑phase; a digital energy meter can accurately record the kWh fed into the grid.
Correct answer is: Single‑phase digital energy meter