Question 1

In DC potentiometry for electrical measurements, which of the following tasks does not require prior standardization (calibration) of the potentiometer scale?

Accepted Answer

Introduction / Context: DC potentiometers are precision instruments that compare unknown voltages to a reference, typically a standard cell. Before reading absolute voltages, a potentiometer is usually standardized so that the slide-wire scale reads in true volts. However, some measurement techniques use only ratios of balance lengths, which cancel the calibration constant. This question checks whether you know which applications do and do not require standardization. Given Data / Assumptions: Conventional DC potentiometer with slide wire and working battery. Standardization sets the current so that a known length corresponds to a known voltage (for example, 1.0186 V of a Weston cell). Resistance can be measured by comparing two drops produced by the same current through known and unknown resistors. Concept / Approach: When measuring resistance R_x by the drop across R_x and a standard R_s in the same circuit current I, the potentiometer balances two voltages that are proportional to IR_x and IR_s. The balance lengths l_x and l_s are proportional to those voltages with the same scale constant k (volts per unit length). Hence R_x / R_s = (k l_x) / (k l_s) = l_x / l_s, independent of k; therefore, standardization is unnecessary. By contrast, direct EMF measurements and instrument calibration require the absolute scale in volts, so standardization is essential. Measuring current from V = IR with a standard resistor also needs the absolute voltage value, hence requires standardization. Step-by-Step Solution: For resistance: measure l_x across R_x and l_s across R_s with the same current.Use R_x / R_s = l_x / l_s (the potentiometer constant cancels).Therefore no prior standardization is required for resistance measurement by ratio. Verification / Alternative check: Try a numerical example with any k; the computed resistance ratio is unchanged. Why Other Options Are Wrong: Unknown voltage and voltmeter calibration need the scale in true volts, so standardization is mandatory.Unknown current via V = IR also needs absolute volts, so standardization is required. Common Pitfalls: Assuming every potentiometer task must be standardized; not true for pure ratio methods. Final Answer: Measurement of resistance using the ratio of two balance lengths

Question 2

A moving-coil meter has 100 turns, coil length 40 mm, coil width 30 mm, and a uniform flux density B = 1 Wb/m^2. If the control (restoring) torque at full-scale is 240 × 10^-6 N·m, what is the full-scale current range of the meter?

Accepted Answer

Introduction / Context: Permanent-magnet moving-coil (PMMC) instruments balance electromagnetic deflecting torque against a spring control torque. At full-scale deflection, the two torques are equal, which allows the full-scale current to be determined from geometry and magnetic field data. This is a standard single-step substitution problem in electrical measurements. Given Data / Assumptions: Number of turns N = 100. Coil dimensions: length l = 0.04 m, width w = 0.03 m → area A = l * w = 0.0012 m^2. Flux density B = 1 Wb/m^2 (uniform in the air gap). Control torque at full scale T_c = 240 × 10^-6 N·m. At full-scale: deflecting torque T_d = control torque T_c. Concept / Approach: Deflecting torque for PMMC: T_d = N * B * I * A. Set T_d = T_c and solve for the full-scale current I_fs. This is simple algebra with consistent SI units. Step-by-Step Solution: Compute area A = 0.04 * 0.03 = 0.0012 m^2.Set N * B * I_fs * A = 240 × 10^-6.I_fs = (240 × 10^-6) / (100 * 1 * 0.0012) = 240 × 10^-6 / 0.12 = 2 × 10^-3 A.Therefore I_fs = 2 mA. Verification / Alternative check: Units: (Wb/m^2)(A)(m^2) = Wb·A, which in this torque relation reduces to N·m; dimensions are consistent. Why Other Options Are Wrong: 1 mA, 3 mA, and 4 mA do not satisfy the torque balance with the given parameters. Common Pitfalls: Forgetting to convert millimeters to meters or miscomputing the area. Final Answer: 2 mA

Question 3

Identify which instrument type is used both for system protection and for energy/metering applications in electrical power systems.

Accepted Answer

Introduction / Context: Power systems require devices that simultaneously support accurate metering and reliable protection. Instrument transformers scale high currents and voltages to safe, measurable levels and provide galvanic isolation. They are indispensable at the interface of protection relays and revenue-grade meters. Given Data / Assumptions: Current transformers (CTs) and potential/voltage transformers (PTs or VTs) are passive, ratio-defined devices. Protective relays and meters connect to their secondary windings. Accuracy and burden ratings are specified to suit both protection and metering duties. Concept / Approach: Protection requires reliable reproduction of system quantities during faults (high primary values, transient behavior), while metering needs accuracy near nominal conditions. Special classes of instrument transformers (e.g., metering-class CTs, protection-class CTs) address each need; in practice, both protection relays and meters are fed by instrument transformers, sometimes dedicated, sometimes shared via separate cores. Step-by-Step Solution: Recognize that wattmeters, energy meters, and PF meters measure quantities but cannot directly connect to high primary levels safely.Instrument transformers step down quantities and isolate circuits, serving both metering inputs and protection relays.Therefore they are the common element used for both purposes. Verification / Alternative check: Typical substation schematic shows CTs feeding both relays and meters via separate secondary cores. Why Other Options Are Wrong: Wattmeters, energy meters, and PF meters are metering devices only; they do not provide isolation or scaling for protection. Common Pitfalls: Confusing the measuring instruments themselves with the auxiliary transformers that enable safe connections. Final Answer: Instrument transformers (current and potential transformers)

Question 4

Figure shows an RC potentiometer to measure ac voltage. It is desired that V0/Vi should be independent of frequency. The value of C should be

Accepted Answer

Since resistances are in the ratio of 10 : 1, capacitances should be in the ratio of 1 : 10 because XC ∝ 1/C.

Question 5

Two resistors R1 = 36 Ω ± 5% and R2 = 75 Ω ± 5% are connected in series. What is the combined resistance with its worst-case tolerance?

Accepted Answer

Introduction / Context: Tolerance analysis ensures that a design functions over component variations. For series resistors, the nominal value is the sum of resistances, while the worst-case absolute tolerance is the sum of the individual absolute tolerances. This problem reinforces translating percentage tolerances into absolute ohmic tolerances and combining them correctly. Given Data / Assumptions: R1 = 36 Ω with ±5% tolerance → ±1.8 Ω. R2 = 75 Ω with ±5% tolerance → ±3.75 Ω. Series connection: nominal R_total = R1 + R2. Concept / Approach: For worst-case analysis, add absolute tolerances algebraically for components in series: ΔR_total = ΔR1 + ΔR2. Percentage tolerances do not average; they accumulate as absolute terms in the same direction for worst-case bounds. Step-by-Step Solution: Compute nominal: R_total = 36 + 75 = 111 Ω.Compute absolute tolerances: ΔR1 = 0.05 * 36 = 1.8 Ω; ΔR2 = 0.05 * 75 = 3.75 Ω.Worst-case total tolerance: ΔR_total = 1.8 + 3.75 = 5.55 Ω.Therefore R_total = 111 ± 5.55 Ω. Verification / Alternative check: Lower bound: 111 − 5.55 = 105.45 Ω; Upper bound: 116.55 Ω. Both are consistent with extremal combinations. Why Other Options Are Wrong: ±0 Ω ignores tolerance; ±2.778 Ω and ±7.23 Ω mis-handle percentage-to-absolute conversion; ±10% is not the worst-case rule for unequal series parts. Common Pitfalls: Adding percentage tolerances directly instead of absolute values; forgetting to compute each part's absolute error first. Final Answer: 111 ± 5.55 Ω

Question 6

A speed transducer that uses a drag-cup rotor (e.g., drag-cup tachogenerator) has which principal advantage compared with conventional designs?

Accepted Answer

Introduction / Context: Drag-cup tachogenerators and similar speed sensors use a thin, lightweight conducting cup instead of a solid rotor. This design significantly reduces mass and inertia, improving dynamic response and lowering the torque required to drive the sensor—ideal for delicate shafts and low-speed applications. Given Data / Assumptions: Drag-cup rotor has very small moment of inertia compared with slotted or solid rotors. Sensor output is proportional to speed via electromagnetic induction. Application values low loading of the prime mover and fast response over extreme accuracy claims. Concept / Approach: The key performance driver is mechanical loading. The transducer should minimally disturb the system it measures. A drag-cup produces the needed EMF with negligible mechanical torque demand, hence low energy requirement. While accuracy can be good, it is not inherently 'very high' compared with all other sensor types, and cost depends on build quality, so 'all of the above' is not appropriate. Step-by-Step Solution: Identify property most directly tied to the drag-cup construction: low inertia.Relate low inertia to low energy/torque requirement and minimal shaft loading.Conclude option highlighting low energy requirement is correct. Verification / Alternative check: Compare dynamic response and starting torque specs of drag-cup vs. conventional tachos; drag-cups have markedly lower drive torque. Why Other Options Are Wrong: 'Very high accuracy' is not universally true; many factors affect accuracy (linearity, temperature, electronics).'Cheapest construction' is not guaranteed.'All of the above' bundles incorrect statements. Common Pitfalls: Assuming every advantage applies; always tie the benefit to the physical construction. Final Answer: Low energy requirement due to very low rotor inertia

Question 7

A digital voltmeter (0–999 counts) has a full-scale reading of 9.999 V. What is the resolution (smallest change it can display) on this range?

Accepted Answer

Introduction / Context: Digital meters display quantized values in counts. Knowing the range and the number of display counts lets you compute the resolution, which is the value represented by one count. This informs uncertainty and least significant digit (LSD) behavior in measurements. Given Data / Assumptions: Readout spans 0 to 999 counts. Full-scale reading is 9.999 V (typical 3½-digit DVM). Quantization is uniform. Concept / Approach: Resolution = Full-scale value / maximum count. For a 0–999 display on a 9.999 V range, one count equals approximately 9.999 / 999 V, which is essentially 0.01 V. Many instruments also quote resolution as 1 count (1 LSD). Step-by-Step Solution: Compute resolution: ΔV = 9.999 V / 999 ≈ 0.01001 V.Round to display granularity: ≈ 0.01 V. Verification / Alternative check: Noting that the instrument shows four digits with one decimal place (X.XXY), the smallest change is 0.01 V. Why Other Options Are Wrong: 1 V and 0.1 V are too coarse; 1 mV or 1 μV are far finer than the stated display capability. Common Pitfalls: Confusing resolution with accuracy; resolution is just the smallest step, not the measurement error. Final Answer: 0.01 V

Question 8

It is required to measure angular position. Considering these transducers—(1) circular potentiometer, (2) LVDT, (3) E pick-up, and (4) synchro—which devices are appropriate for angular position measurement?

Accepted Answer

Introduction / Context: Choosing the right transducer depends on the physical variable. Angular position is commonly measured using rotary devices that directly sense shaft angle and provide an electrical output proportional to that angle. This question asks you to identify suitable transducers from a mixed list of rotary and linear sensors. Given Data / Assumptions: (1) Circular (rotary) potentiometer: a resistive element with a wiper rotating with the shaft. (2) LVDT: Linear Variable Differential Transformer—intended for linear displacement. (3) E pick-up: an ambiguous term (often eddy-current or magnetic pick-up) typically suited to speed, proximity, or events, not absolute angle. (4) Synchro: a rotary transformer system for angular position/telemetry. Concept / Approach: Match sensor physics to the variable: rotary potentiometers and synchros directly produce signals proportional (or convertible) to shaft angle. LVDTs are for linear displacement, and generic E pick-ups usually do not deliver absolute angular position without auxiliary mechanics and signal processing. Step-by-Step Solution: Identify rotary-capable sensors: (1) circular potentiometer and (4) synchro.Exclude LVDT (linear-only) and generic E pick-up (not a standard absolute angle sensor).Therefore, the correct set is 1 and 4. Verification / Alternative check: Synchros are widely used in aerospace/industrial controls for shaft angle feedback; rotary pots are common in control knobs and servos. Why Other Options Are Wrong: Options including 2 or 3 incorrectly assume linear or event sensors measure angle directly. Common Pitfalls: Confusing linear with rotary transducers; assuming any magnetic pick-up provides absolute angle. Final Answer: 1 and 4

Question 9

A moving-coil instrument has internal resistance 0.6 Ω and gives full-scale deflection at 0.1 A. To convert it into an ammeter of 0–15 A range, what shunt resistance should be connected?

Accepted Answer

Introduction / Context: Moving-coil meters measure small currents. To extend their range, a low-value shunt resistor bypasses most of the current, leaving only the full-scale meter current through the movement. Correct calculation ensures accurate division of current and prevents damage to the meter. Given Data / Assumptions: Meter resistance R_m = 0.6 Ω. Meter full-scale current I_m = 0.1 A. Desired total range I_total = 15 A → shunt current I_s = 15 − 0.1 = 14.9 A. Meter and shunt are in parallel at the terminals. Concept / Approach: At full-scale: the same voltage appears across the meter and the shunt. Hence V_m = I_m * R_m = I_s * R_s. Solve for the shunt resistance R_s = (I_m * R_m) / I_s. Step-by-Step Solution: Compute V_m at full-scale: V_m = 0.1 * 0.6 = 0.06 V.Compute shunt resistance: R_s = 0.06 / 14.9 ≈ 0.0040268 Ω.Round to a practical value: R_s ≈ 0.004 Ω. Verification / Alternative check: Check current split: I_s = V_m / R_s ≈ 0.06 / 0.004 = 15 A; with 0.1 A through the meter, total ≈ 15.1 A (rounding explains the small difference). Why Other Options Are Wrong: 0.6 Ω, 0.06 Ω, and 0.1 Ω are far too large; they would route excessive current through the meter. Common Pitfalls: Using series instead of parallel formulas; forgetting that the voltage across meter and shunt is equal at the design point. Final Answer: 0.004 Ω

Question 10

Assertion–Reason: De Sauty’s bridge is suitable only for measuring a pure (lossless) capacitor. Reason: Most practical capacitors are nearly perfect, so the loss can be ignored.

Accepted Answer

Introduction / Context: De Sauty’s bridge is a classical AC bridge used to compare capacitances by balancing the ratio of two capacitors against the ratio of two resistors. Its usefulness depends on the assumption that the capacitors are ideal (lossless). Real-world capacitors exhibit dielectric loss, modeled by a series resistance or a loss tangent, which affects balance conditions. Given Data / Assumptions: Assertion (A): The bridge is suitable only for pure capacitors. Reason (R): Practical capacitors are mostly perfect and loss can be ignored. Understanding of loss angle and dissipation factor is assumed. Concept / Approach: De Sauty’s bridge balance condition is C1/C2 = R4/R3. This relation holds strictly only if both capacitors are free from dielectric loss, because the bridge has no provision to balance out the resistive loss component. When dielectric loss is present, bridges like Schering bridge are used instead, as they include a loss arm to balance both capacitance and dissipation factor. Step-by-Step Solution: Recognize that De Sauty’s bridge neglects loss resistance in the capacitors.Therefore, A is true: it is suitable only for pure capacitors.Evaluate R: The statement that 'capacitors are mostly perfect' is incorrect; practical capacitors have measurable dissipation, especially electrolytics and paper types.Hence, R is false. Verification / Alternative check: Compare with Schering bridge which explicitly measures loss angle (tan δ) alongside capacitance. Why Other Options Are Wrong: Any option treating the reason as true or explanatory contradicts practical capacitor behavior. Common Pitfalls: Assuming ideal components in precision measurement without checking the bridge's capability to account for losses. Final Answer: A is true R is false

Question 11

Bridge balance condition: In the given AC bridge circuit, Z1 = 200∠60° Ω, Z2 = 400∠−90° Ω, and Z3 = 300∠0° Ω. Determine the impedance Z4 that will balance the bridge.

Accepted Answer

Introduction / Context: AC bridges are widely used for measuring impedances with accuracy. The condition of balance for a bridge is that the ratio of impedances in one arm must equal the ratio in the opposite arm. This problem tests understanding of phasor impedances and complex ratio arithmetic. Given Data / Assumptions: Z1 = 200∠60° Ω Z2 = 400∠−90° Ω Z3 = 300∠0° Ω Z4 = ? (to be found for bridge balance) Standard balance condition: Z1 / Z2 = Z3 / Z4 Concept / Approach: For a balanced bridge, the product of opposite arms must be equal: Z1 * Z4 = Z2 * Z3. Therefore, Z4 = (Z2 * Z3) / Z1. This requires multiplication and division of complex numbers in polar form, which involves multiplying/dividing magnitudes and adding/subtracting angles. Step-by-Step Solution: Z2 * Z3 = (400∠−90°) * (300∠0°) = (400 * 300) ∠(−90° + 0°) = 120000∠−90°.Divide by Z1: Z4 = (120000∠−90°) / (200∠60°).Magnitudes: 120000 / 200 = 600.Angles: (−90° − 60°) = −150°.So Z4 = 600∠−150° Ω. Verification / Alternative check: Check ratio: Z1 / Z2 = 200∠60° / 400∠−90° = 0.5∠150°. Similarly, Z3 / Z4 = 300∠0° / 600∠−150° = 0.5∠150°. Both sides equal, confirming the balance condition is satisfied. Why Other Options Are Wrong: 150∠30° Ω: incorrect magnitude and angle, does not satisfy balance condition. 400∠−90° Ω: simply repeats Z2, not derived from balance equation. 300∠90° Ω: wrong phase shift; fails ratio check. 600∠−150° Ω: correct value derived from balance condition (this is the answer). Common Pitfalls: Mixing up balance condition (Z1/Z2 = Z3/Z4) with (Z1Z4 = Z2Z3). Both are equivalent but must be applied carefully. Forgetting to subtract angles properly when dividing in polar form. Neglecting to check with ratio verification after solving. Final Answer: 600∠−150° Ω

Question 12

Moving-coil conversion to ammeter: A 0.5 Ω, 0.1 A full-scale movement is to be converted into a 0–10 A ammeter. Calculate the required shunt resistance that must be connected in parallel with the coil.

Accepted Answer

Introduction / Context: Converting a sensitive moving-coil instrument into a high-range ammeter is a classic instrumentation task. The idea is to bypass most of the load current through a low-value shunt so that the delicate movement always carries only its rated full-scale current. This problem checks shunt sizing using basic parallel-circuit relationships and voltage equality across parallel branches. Given Data / Assumptions: Coil (movement) resistance, Rm = 0.5 Ω. Full-scale current of movement, Im = 0.1 A. Desired ammeter range: 0–10 A, so maximum total current I = 10 A. Shunt is ideal (negligible inductance) and connected in parallel with the movement. Concept / Approach: At full-scale, the coil must carry Im and the shunt carries the remaining current Is = I − Im. Since the coil and shunt are in parallel, the voltage drop across each is equal. Therefore, Rs is determined by equating the coil's full-scale voltage to the shunt voltage at Is. This ensures protection of the movement and correct overall range. Step-by-Step Solution: Compute coil full-scale voltage: Vm = Im * Rm = 0.1 * 0.5 = 0.05 V.Compute shunt full-scale current: Is = 10 − 0.1 = 9.9 A.Use equality of parallel voltages: Vm = Is * Rs.Solve for shunt: Rs = Vm / Is = 0.05 / 9.9 ≈ 0.00505 Ω.Round to standard: Rs ≈ 0.005 Ω. Verification / Alternative check: Check power in the movement at FS: P = Im^2 * Rm = 0.01 * 0.5 = 0.005 W = 5 mW, a reasonable level. The shunt voltage equals 0.05 V; multiplying by Is gives 0.495 W dissipated in the shunt—consistent with a small, low-ohm, higher-wattage resistor selection. Why Other Options Are Wrong: 0.004 Ω / 0.008 Ω: would not maintain the correct coil current at 10 A total; coil would be under/over-driven. 0.05 Ω / 0.1 Ω: far too large; the coil would be forced to carry more than 0.1 A before reaching 10 A total. Common Pitfalls: Forgetting that the coil's voltage at full scale sets the shunt voltage. Using series-resistor formulas instead of parallel-voltage equality. Final Answer: 0.005 Ω

Question 13

Standards and accuracy classes: As per Bureau of Indian Standards (BIS), there are exactly six accuracy classes of electrical measuring instruments. Evaluate the statement.

Accepted Answer

Introduction / Context: Accuracy class specifies the permissible measurement error of an instrument, typically as a percentage of full-scale or reading under standard conditions. Various BIS/IEC standards define accuracy classes for different categories such as voltmeters, ammeters, energy meters, transducers, and protection relays. This question tests whether the number of accuracy classes is a fixed universal count across all BIS instrument standards. Given Data / Assumptions: BIS follows international harmonization with IEC for many instrumentation standards. Different instruments (indicating, integrating, protection, energy meters) are covered by distinct standards. Accuracy classes can include values like 0.05, 0.1, 0.2, 0.5, 1.0, 1.5, 2.5, 5 (and others), depending on the device and standard revision. Concept / Approach: It is incorrect to claim a single number (six) covers all instruments. The list and granularity of accuracy classes vary by instrument type and standard. For example, energy meters have classes such as 0.2, 0.5, 1.0, 2.0; indicating instruments often use 0.5, 1.0, 1.5, 2.5; protection relays and CTs/VTs have their own class schemes. Therefore, the given statement is too restrictive and false in general context. Step-by-Step Solution: Identify claim: “there are 6 accuracy classes” → universal statement.Check across categories: each category uses its own set—more or less than six are common.Conclude universality claim is false. Verification / Alternative check: Cross-referencing BIS/IEC documents for indicating instruments, energy meters, and instrument transformers shows differing class enumerations and tolerances, confirming that a fixed count does not apply universally. Why Other Options Are Wrong: “True” contradicts the variability across standards. “True only for …” options are misleading; even within a category, multiple classes exist and can exceed six across revisions. Common Pitfalls: Memorizing a single list and assuming it applies to all instruments. Confusing common classroom examples with official standard scopes. Final Answer: False

Question 14

Piezoelectric measurement chain: A force transducer has charge sensitivity 20 pC/N. The combined transducer + charge amplifier gain is 50 mV/N. Determine the amplifier gain in mV/pC.

Accepted Answer

Introduction / Context: Piezoelectric force sensors generate electric charge proportional to applied force. A charge amplifier converts this charge to a voltage with a known charge-to-voltage gain. Understanding how to relate sensor charge sensitivity and amplifier gain to the overall volts-per-newton factor is central to instrument calibration. Given Data / Assumptions: Charge sensitivity of transducer: 20 pC/N. Total chain gain: 50 mV/N at the output. Charge amplifier is linear, so V_out = (gain) * Q. Concept / Approach: The overall volts-per-newton response is the product of the sensor’s charge-per-newton and the amplifier’s volts-per-coulomb. Therefore, gain_amp (mV/pC) = (overall mV/N) / (pC/N). This simple ratio directly yields the charge amplifier's gain independent of force magnitude used for calibration. Step-by-Step Solution: Overall sensitivity S_total = 50 mV/N.Transducer sensitivity S_q = 20 pC/N.Amplifier gain G = S_total / S_q = (50 mV/N) / (20 pC/N) = 2.5 mV/pC. Verification / Alternative check: Check units: (mV/N) divided by (pC/N) → mV/pC, correct. Try a sample: 10 N produces Q = 200 pC; with G = 2.5 mV/pC, V_out = 200 * 2.5 = 500 mV = 0.5 V; also 10 N * 50 mV/N = 0.5 V—consistent. Why Other Options Are Wrong: 1, 1.5, 4, 0.25 mV/pC do not satisfy the product required to achieve 50 mV/N with 20 pC/N charge sensitivity. Common Pitfalls: Multiplying instead of dividing by the charge sensitivity. Missing unit consistency when moving between pC, mV, and N. Final Answer: 2.5 mV/pC

Question 15

Guaranteed accuracy and limiting error: A 0–25 A ammeter has ±1% of full-scale accuracy. If it reads 5 A, what is the limiting error expressed as a percentage of the measured value?

Accepted Answer

Introduction / Context: Guaranteed accuracy is often specified as a percentage of full-scale (FS). When the reading is far below full-scale, the same absolute error corresponds to a larger percentage of the reading. Understanding this distinction is crucial for correct error reporting and for deciding appropriate instrument ranges. Given Data / Assumptions: Ammeter range: 0–25 A. Accuracy: ±1% of full-scale. Measured current: 5 A. Concept / Approach: Absolute error at any reading = (accuracy% of FS) * FS. Limiting error as a percentage of the measured value = (absolute error / reading) * 100%. This converts a full-scale-based tolerance to a reading-based percentage, which is commonly requested in test reports. Step-by-Step Solution: Compute absolute error: ±1% of 25 A = ±0.25 A.Convert to percentage of reading: (0.25 / 5) * 100% = 5%.Hence the limiting error at 5 A is ±5% of the indicated value. Verification / Alternative check: If the meter read 25 A, the same ±0.25 A would be ±1% of reading, matching the specification. Lower readings inflate the percent-of-reading error, which is a key reason to select the smallest meter range that safely covers the expected value. Why Other Options Are Wrong: 2%, 2.5%, 4%: underestimate the impact of applying a full-scale accuracy to a sub-range reading. 1%: confuses accuracy-of-FS with accuracy-of-reading. Common Pitfalls: Reporting “±1%” at all readings without converting to percentage of the actual reading. Mixing up full-scale and of-reading specifications. Final Answer: 5%

Question 16

LVDT with amplifier: An LVDT outputs 2 mV for a 0.5 mm displacement and feeds a 0–5 V voltmeter through a gain-250 amplifier. Determine the instrument sensitivity in V/mm.

Accepted Answer

Introduction / Context: A Linear Variable Differential Transformer (LVDT) is a highly sensitive displacement transducer. In practice, its small output is amplified before display or digitization. Sensitivity of the overall instrument chain (display volts per millimetre) is a key calibration figure for users. Given Data / Assumptions: LVDT raw output: 2 mV at 0.5 mm displacement. Amplifier gain: 250 (unitless voltage gain). Voltmeter range: 0–5 V (for context; not needed directly for sensitivity computation). Concept / Approach: Total sensitivity equals (amplifier gain) * (sensor output per mm). First compute the LVDT's millivolts per millimetre, then multiply by the gain to obtain the final volts per millimetre at the meter terminals. Step-by-Step Solution: LVDT raw sensitivity: 2 mV / 0.5 mm = 4 mV/mm.After amplification: 4 mV/mm * 250 = 1000 mV/mm = 1 V/mm.Therefore, instrument sensitivity = 1 V/mm. Verification / Alternative check: At 0.5 mm, the chain output is 1 V/mm * 0.5 mm = 0.5 V. Also, raw 2 mV * 250 = 0.5 V—consistent. This leaves ample headroom on a 0–5 V meter for larger displacements within linear range, improving resolution. Why Other Options Are Wrong: 0.1 V/mm and 0.5 V/mm: understate the amplified sensitivity by factors of 10 and 2. 5 V/mm: overstates the gain result by 5×. 0.02 V/mm: confuses raw LVDT output with post-gain value. Common Pitfalls: Forgetting to convert millivolts to volts after applying gain. Confusing raw sensor sensitivity with system sensitivity. Final Answer: 1 V/mm

Question 17

Two-wattmeter method in 3-phase systems: Under what condition will one wattmeter show a negative reading when measuring total power of a 3-phase load?

Accepted Answer

Introduction / Context: The two-wattmeter method measures total power in a 3-phase, 3-wire system irrespective of load balance, provided the supply is balanced. A well-known characteristic is that one wattmeter may read negative at low power factors. Recognizing the condition for a negative wattmeter helps diagnose load characteristics and power factor. Given Data / Assumptions: Balanced 3-phase source; load can be balanced or unbalanced. Two wattmeters connected in the standard manner. Power factor angle φ relates to PF by PF = cos φ. Concept / Approach: Two-wattmeter readings for a 3-phase system are W1 = V_L I_L cos(30° − φ) and W2 = V_L I_L cos(30° + φ). One reading becomes negative when its cosine term is negative. For W2, cos(30° + φ) becomes negative when 30° + φ > 90°, i.e., φ > 60°, which corresponds to PF < 0.5. This condition applies whether the power factor is lagging (inductive) or leading (capacitive); it depends only on the magnitude of φ exceeding 60°. Step-by-Step Solution: Set condition for negative: 30° + φ > 90° ⇒ φ > 60°.Translate to PF: PF = cos φ < cos 60° = 0.5.Therefore, one wattmeter will read negative when PF < 0.5 (leading or lagging). Verification / Alternative check: Vector diagrams confirm that the current phasor moves far enough relative to the wattmeter voltage reference to flip the sign of one cosine term beyond φ = 60°, producing a negative torque and reading. Why Other Options Are Wrong: “Less than 0.5 lagging”: too restrictive; leading also produces the effect. “Greater than 0.5 lagging” and “unity”: both lead to positive readings for both meters. “Only purely capacitive”: incorrect; inductive loads with PF < 0.5 also cause negativity. Common Pitfalls: Assuming the sign depends on lag vs lead; it depends on the magnitude of φ. Misapplying formulas for 4-wire systems in which the method is different. Final Answer: when power factor is less than 0.5

Question 18

Type of household energy meter: Identify whether a domestic electricity (kWh) meter is indicating, recording, or integrating in nature.

Accepted Answer

Introduction / Context: Domestic energy meters installed by utilities display cumulative electrical energy consumption, typically in kilowatt-hours (kWh). Distinguishing among indicating, recording, and integrating instruments is part of basic measurement science and is often featured in introductory electrical engineering curricula. Given Data / Assumptions: Household meter accumulates energy use over time. Display may be electromechanical dials or electronic LCDs, but the functional principle is the same. Energy is the time integral of instantaneous power. Concept / Approach: An indicating instrument shows instantaneous value (e.g., voltmeter, ammeter). A recording instrument logs a time history (chart recorders). An integrating instrument accumulates a quantity over time; a kWh meter integrates power (kW) with respect to time to yield energy (kWh). Therefore, a household energy meter is an integrating instrument. Step-by-Step Solution: Define energy: E = ∫ P(t) dt.Meters implement E accumulation mechanically (Ferraris disk) or digitally (sampling and computation).The register displays cumulative kWh, not instantaneous kW or a plotted trace. Verification / Alternative check: Modern smart meters still report totalized kWh between billing cycles; instantaneous power is optional telemetry but the billing core remains integrated energy, confirming the classification. Why Other Options Are Wrong: Indicating: would show instantaneous power only. Recording: implies a continuous time record; traditional meters do not plot a chart of P(t). None/null-type: not applicable to kWh metering function. Common Pitfalls: Confusing real-time demand (kW) displays in smart meters with the legal billing quantity (kWh). Final Answer: integrating instrument

Question 19

Operating domain of induction wattmeter: For which type of circuit(s) is an induction (AC) wattmeter suitable?

Accepted Answer

Introduction / Context: Induction-type measuring instruments (e.g., induction wattmeters and induction energy meters) exploit the interaction of alternating magnetic fields to produce torque proportional to electrical power. Recognizing their operating limitation is necessary when choosing instruments for AC versus DC measurements. Given Data / Assumptions: Induction wattmeter uses two alternating fluxes with phase displacement to generate torque proportional to VI cos φ. DC signals provide constant (non-alternating) fields. Proper operation requires time-varying flux to induce eddy currents in the moving element (e.g., aluminum disk). Concept / Approach: In DC, with steady current and voltage, there is no alternating flux to sustain eddy currents and hence little or no torque in induction mechanisms. Therefore induction wattmeters are inherently AC-only devices. They are widely used on both single-phase and three-phase AC, but the principle does not work on pure DC without auxiliary choppers/inverters (which are not part of the instrument itself). Step-by-Step Solution: Identify principle: torque ∝ product of alternating fluxes and induced currents.DC lacks alternating flux → torque collapses.Conclusion: suitable only for AC circuits (single-phase or polyphase). Verification / Alternative check: Ferraris-disk energy meters and induction wattmeters populate AC metering extensively; DC metering uses dynamometer or electronic wattmeters, not induction mechanisms. Why Other Options Are Wrong: DC only / both AC and DC: contradict the induction principle. AC 3-phase only: induction wattmeters also work on single-phase AC. “DC with ripple”: not a defined operating mode; any torque would be erratic and not calibrated. Common Pitfalls: Assuming all wattmeters are universal; only dynamometer/electronic types handle DC correctly. Final Answer: ac circuit only

Question 20

Slide-wire potentiometer calibration: A 1.0185 V standard cell balances at 60 cm. With an unknown EMF connected, balance is at 82 cm. Compute the unknown EMF.

Accepted Answer

Introduction / Context: A DC slide-wire potentiometer measures an unknown EMF by balancing it against a known potential drop along a uniform resistance wire. Once calibrated using a standard cell, the instrument provides highly accurate voltage measurements without drawing current from the unknown source at balance. This question requires applying the uniform scale proportionality. Given Data / Assumptions: Standard cell Es = 1.0185 V balances at length Ls = 60 cm. Unknown EMF Ex balances at length Lx = 82 cm. Wire is uniform: voltage drop is proportional to length. Measurements are taken under null (no current drawn) conditions. Concept / Approach: For a uniform potentiometer wire, V per cm is constant. Hence, Ex / Es = Lx / Ls. Solve directly for Ex using the ratio of balance lengths. This linear proportionality underpins the simplicity and precision of slide-wire potentiometers. Step-by-Step Solution: Compute scale constant: k = Es / Ls = 1.0185 V / 60 cm ≈ 0.016975 V/cm.Apply to unknown: Ex = k * Lx = 0.016975 * 82 ≈ 1.3920 V.Round suitably: Ex ≈ 1.39 V. Verification / Alternative check: Use direct ratio: Ex = Es * (Lx / Ls) = 1.0185 * (82 / 60) = 1.0185 * 1.3667 ≈ 1.392 V—same result. This cross-check confirms arithmetic and the uniform-wire assumption validity. Why Other Options Are Wrong: 0.75 V: would imply Lx ≈ 44 cm at this calibration. 13.9 V / 7.45 V: order-of-magnitude errors; potentiometer ranges are typically a few volts. 1.18 V: does not match the proportional ratio. Common Pitfalls: Mixing up centimeters with millimeters or switching numerator/denominator in the ratio. Forgetting that the potentiometer must be re-standardized if the current through the slide wire changes. Final Answer: 1.39 V

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