Question 1

A single-phase energy meter has a rating of 1200 revolutions per kilowatt-hour (kWh). If an electric appliance rated at 500 W operates continuously for 4 hours, how many revolutions will the meter complete?

Accepted Answer

Introduction / Context: Energy meters (also called watt-hour meters) measure the energy consumed by electrical loads. They are calibrated such that a fixed number of revolutions of the meter disc corresponds to one kilowatt-hour of electrical energy. This question involves computing the number of revolutions for a given power, time, and meter constant. Given Data / Assumptions: Meter constant: 1200 revolutions per 1 kWh. Load power: 500 W = 0.5 kW. Time of operation: 4 hours. Concept / Approach: Energy consumed (kWh) = Power (kW) × Time (hours). Number of revolutions = Energy consumed × meter constant. By substituting the given data, we can directly calculate the revolutions. Step-by-Step Solution: Step 1: Power = 500 W = 0.5 kW.Step 2: Time = 4 h.Step 3: Energy consumed = 0.5 × 4 = 2 kWh.Step 4: Revolutions = 2 × 1200 = 2400 revolutions. Verification / Alternative check: If load was 1 kW for 1 hour, energy = 1 kWh → 1200 revolutions. Our case is 2 kWh, so 2 × 1200 = 2400. This matches. Why Other Options Are Wrong: 1200 corresponds to 1 kWh, but here energy is 2 kWh.1800 and 2100 are arbitrary mismatches.1500 is not consistent with the meter constant. Common Pitfalls: Forgetting to convert watts to kilowatts or ignoring operating hours. Final Answer: 2400 revolutions

Question 2

If the reference sound pressure P0 is 2 × 10^-5 N/m², what actual sound pressure corresponds to a sound level of 90 dB?

Accepted Answer

Introduction / Context: Sound levels in decibels (dB) are logarithmic measures of sound pressure relative to a reference level, usually 2 × 10^-5 N/m² in air. This question requires calculation of the actual sound pressure corresponding to 90 dB. Given Data / Assumptions: Reference sound pressure P0 = 2 × 10^-5 N/m². Level Lp = 90 dB. Formula: Lp = 20 log10(P / P0). Concept / Approach: Rearranging the formula: P = P0 × 10^(Lp/20). Substituting values gives the actual sound pressure. This is a standard conversion in acoustics. Step-by-Step Solution: Step 1: Lp = 90 dB.Step 2: P = 2 × 10^-5 × 10^(90/20).Step 3: 90/20 = 4.5.Step 4: 10^4.5 ≈ 31622.8.Step 5: P ≈ 2 × 10^-5 × 31622.8 ≈ 0.632 N/m². Verification / Alternative check: Checking unit consistency: Pressure remains in N/m². Calculation matches published sound pressure levels. Why Other Options Are Wrong: 0.707, 0.835, and 0.925 are close but incorrect approximations.0.500 is significantly below the correct value. Common Pitfalls: Forgetting to divide by 20 (instead of 10) in the formula for pressure levels. Final Answer: 0.632 N/m²

Question 3

Assertion (A): A hot-wire ammeter has a cramped (non-linear) scale. Reason (R): The heat produced in the wire is proportional to the square of the current passing through it.

Accepted Answer

Introduction / Context: Hot-wire ammeters are based on the heating effect of current. Their scale is not uniform but cramped at lower values due to the square-law relationship between current and heating. Understanding the assertion and reasoning requires connecting the thermal principle with scale graduation. Given Data / Assumptions: Heating effect: H ∝ I^2 × R × t. Deflection of pointer is proportional to heat produced. Scale markings depend on deflection response. Concept / Approach: If heat ∝ I^2, then deflection θ ∝ I^2. Thus, small increases in current cause disproportionately small increases in deflection at the lower end, resulting in cramped spacing of scale divisions. Step-by-Step Solution: Step 1: Heat H = I^2 × R × t.Step 2: Deflection θ ∝ H.Step 3: Hence θ ∝ I^2.Step 4: The quadratic nature produces cramped (non-linear) scale at the start. Verification / Alternative check: Experimental hot-wire ammeter curves show scale compressions at low readings. Why Other Options Are Wrong: Since both A and R are true and linked directly, other combinations are invalid. Common Pitfalls: Confusing linear instruments (PMMC) with thermal instruments; forgetting quadratic law. Final Answer: Both A and R are true and R is the correct explanation of A

Question 4

In a differential amplifier using JFETs, the differential output offset may arise due to which of the following causes?

Accepted Answer

Introduction / Context: Differential amplifiers are widely used for rejecting common-mode signals and amplifying only differential inputs. In practice, imperfections cause offset voltage even with zero input difference. Understanding these causes is crucial in analog design. Given Data / Assumptions: Amplifier uses JFET input transistors. Possible causes include device mismatches, resistance mismatches, and supply fluctuations. Concept / Approach: Differential offset occurs when the symmetry of the circuit is disturbed. Mismatch between the FET's threshold voltages or transconductance, unequal resistors in load or bias circuits, and variation in supply voltage all contribute to output offset. Step-by-Step Solution: Check device mismatch: Idss or Vgs(th) differences shift balance.Check resistance mismatch: unequal load or bias resistors distort symmetry.Supply variation: affects bias currents and can create imbalance. Verification / Alternative check: Simulation of a differential pair with matched vs mismatched parameters confirms output offset. Why Other Options Are Wrong: Options excluding any of the three causes are incomplete. Common Pitfalls: Assuming ideal symmetry; ignoring supply sensitivity. Final Answer: Mismatch between FET parameters, resistance mismatch, and variation in supply voltage

Question 5

A full wave rectifier type meter is used to measure the wave shape shown in figure. The voltmeter is calibrated to read rms value of sine wave. The reading of voltmeter will be

Accepted Answer

Peak value of sine wave = 100 x 1.11 = 111 V.

Question 6

A moving-iron instrument provides accurate readings under what frequency conditions?

Accepted Answer

Introduction / Context: Moving-iron (MI) instruments operate by the deflection caused due to the magnetic field created by current. Their response depends on inductance and coil impedance, making them frequency-dependent for AC measurement. Given Data / Assumptions: MI instruments can measure both AC and DC. Inductance of the coil creates reactance that increases with frequency. Concept / Approach: At low frequencies (including DC), the impedance is nearly resistive. As frequency increases, inductive reactance grows and the current for a given applied RMS voltage decreases, causing errors. Thus, MI instruments remain accurate only up to a certain limiting frequency. Step-by-Step Solution: Impedance = √(R^2 + (ωL)^2).At small ω, Z ≈ R, hence accurate.At higher ω, inductive effect increases → error grows. Verification / Alternative check: Practical MI voltmeters are specified up to ~100–200 Hz for accuracy. Why Other Options Are Wrong: Low or high frequencies alone are misleading; ”one frequency” is irrelevant; resonance does not apply here. Common Pitfalls: Assuming MI meters are broadband instruments; they are not. Final Answer: All frequencies up to a certain maximum value

Question 7

The loss-of-charge method is particularly suitable for measuring medium resistance values. Is this statement true or false?

Accepted Answer

Introduction / Context: The loss-of-charge method is a classical technique for measuring medium resistance values using the discharge of a capacitor through the unknown resistance. It is commonly taught in electrical measurement courses. Given Data / Assumptions: Principle: capacitor discharges through the resistance to be measured. Voltage across the capacitor decreases exponentially with time. By measuring time for a known voltage drop, resistance can be computed. Concept / Approach: The method uses the relation V = V0 e^(-t/RC). Taking logs, R = t / (C ln(V0/V)). Accuracy depends on time measurement and insulation. Suitable resistances are typically 10^5 to 10^7 ohms (medium range). Step-by-Step Solution: Charge a capacitor to V0.Connect resistance under test.Measure time for voltage to fall to V.Use relation R = t / (C ln(V0/V)). Verification / Alternative check: High resistances → leakage dominates; low resistances → discharge too fast; medium resistances ideal. Why Other Options Are Wrong: ”False” ignores well-established usage; voltage dependency is not the deciding factor; temperature only affects accuracy slightly. Common Pitfalls: Trying to use this method for too high resistances where leakage interferes. Final Answer: True

Question 8

Which statement correctly characterizes an impulse voltage used in high-voltage testing?

Accepted Answer

Introduction / Context: Impulse voltages simulate lightning or switching surges to test insulation. They are defined by fast rise and relatively slower fall, and crucially, they do not reverse polarity during the impulse. Given Data / Assumptions: Standard impulse notation: e.g., 1.2/50 μs (rise/fall times). Used for flashover and dielectric strength tests. Concept / Approach: An impulse is a single-polarity, short-duration voltage. Rise time is much shorter than fall time. Chopped impulses can and do occur (e.g., due to flashover across a gap). RMS vs average is not a defining property and varies with waveform shape. Step-by-Step Solution: Identify unidirectional nature → correct key property.Reject statements implying symmetry of times or sinusoidal behavior.Acknowledge chopped impulse events are real in practice. Verification / Alternative check: Standards (e.g., IEC 60060) describe unidirectional standard lightning impulses. Why Other Options Are Wrong: (b) is false (chopped impulses occur); (c) false (rise ≪ fall); (d) not a general truth; (e) false (not sinusoidal). Common Pitfalls: Confusing AC test voltages with transient impulse tests. Final Answer: It is a unidirectional voltage

Question 9

A resistance strain gauge has gauge factor G = 2 mounted on a specimen under stress of 1050 kg/cm². If the gauge resistance is 1000 Ω, estimate the change in resistance ΔR (assume steel with E ≈ 2 × 10^6 kg/cm²).

Accepted Answer

Introduction / Context: Strain gauges convert mechanical strain into a resistance change. The core relation is ΔR/R = G * ε, where ε is axial strain and G is the gauge factor. To get ε from stress σ, Hooke's law ε = σ / E is used for linear elastic behavior. Given Data / Assumptions: Gauge factor G = 2. Stress σ = 1050 kg/cm². Gauge resistance R = 1000 Ω. Assume material modulus E ≈ 2 × 10^6 kg/cm² (typical steel order of magnitude). Concept / Approach: Compute strain ε from σ and E, then apply ΔR = R * G * ε. This is the standard small-strain, linear-elastic estimation widely used in experimental stress analysis. Step-by-Step Solution: ε = σ / E = 1050 / (2 × 10^6) = 5.25 × 10^-4ΔR/R = G * ε = 2 * 5.25 × 10^-4 = 1.05 × 10^-3ΔR = R * (ΔR/R) = 1000 × 1.05 × 10^-3 ≈ 1.05 Ω ≈ 1 Ω Verification / Alternative check: Using E = 2.1 × 10^6 kg/cm² gives ΔR ≈ 1.0 Ω; either way the closest option is 1 Ω. Why Other Options Are Wrong: 2–4 Ω would require roughly double to quadruple the computed strain; 0.5 Ω is about half the computed change. Common Pitfalls: Forgetting to convert stress to strain using E; misapplying gauge factor relation. Final Answer: 1 Ω

Question 10

In three-phase power measurement using the two-wattmeter method, one wattmeter reads zero. What is the load power factor?

Accepted Answer

Introduction / Context: The two-wattmeter method measures total power in a 3-wire, three-phase system. For a balanced load with displacement angle φ, the wattmeter readings depend on φ, providing a convenient way to infer power factor from observed readings. Given Data / Assumptions: Balanced three-phase system. Wattmeter readings: W1 = V_L I_L cos(30° + φ), W2 = V_L I_L cos(30° − φ). Total power P = W1 + W2. Concept / Approach: When one wattmeter reads zero, its cosine term is zero. Setting cos(30° ± φ) = 0 yields the critical φ for which a reading vanishes. That angle translates directly to a power factor value cos φ. Step-by-Step Solution: Let W1 = 0 ⇒ cos(30° + φ) = 0 ⇒ 30° + φ = 90° ⇒ φ = 60°Thus pf = cos φ = cos 60° = 0.5 Verification / Alternative check: At φ = 60°, the other wattmeter reads V_L I_L cos(30° − 60°) = V_L I_L cos(−30°) = V_L I_L (√3/2), nonzero, as expected. Why Other Options Are Wrong: pf = 1.0 gives equal positive readings; pf = 0 gives equal and opposite readings; pf = 0.8 does not null either meter. Common Pitfalls: Mixing up the ”equal and opposite” condition (pf = 0) with the ”one meter zero” condition (pf = 0.5). Final Answer: 0.5

Measurements and Instrumentation Questions