Nepal Engineering Council · Electrical Engineering · Chapter 4
Electrical Measurements and Instrumentations
Tap an option to check it. Wrong picks show the right answer and the hint.
186 questions in 6 syllabus topics.
4.1 Measurement and error
32 questions · AElE0401
1. For a steady (static) input, the quantity obtained by subtracting the true value from the value indicated by the instrument is called the
Show hintHide hint
Look at the order of subtraction: indicated minus true.
Show answerHide answer
Answer: B. static error
Static error = measured value − true value. The static correction is the negative of this (true − measured).
2. A voltmeter reads 112.5 V when the true voltage is 110 V. The static correction is
Show hintHide hint
Correction is what must be added to the reading to get the true value.
Show answerHide answer
Answer: B. −2.5 V
Static correction = true − measured = 110 − 112.5 = −2.5 V (the static error is +2.5 V).
3. An ammeter reads 4.9 A when the true current is 5.0 A. The magnitude of the relative static error (referred to the true value) is
Show hintHide hint
Divide the absolute error by the true value.
Show answerHide answer
Answer: A. 2.00 %
Relative error = |4.9 − 5.0| / 5.0 × 100 = 2.00 %. Dividing by the measured value (4.9) would give the wrong 2.04 %.
4. A 0–150 V voltmeter has a guaranteed accuracy of ±1 % of full-scale reading. When it reads 75 V, the limiting error as a percentage of the reading is
Show hintHide hint
The guarantee is in volts of full scale; it becomes a bigger fraction of a smaller reading.
Show answerHide answer
Answer: D. ±2 %
Limiting error = 1 % of 150 V = ±1.5 V. As a fraction of 75 V: 1.5/75 × 100 = ±2 %.
5. Power is computed from P = I²R, where I is measured with a limiting error of ±1 % and R is known to ±2 %. The limiting error in P is
Show hintHide hint
The exponent of I multiplies its percentage error.
Show answerHide answer
Answer: B. ±4 %
For a product of powers, relative limiting errors add, weighted by the exponents: 2(1 %) + 2 % = ±4 %.
6. Resistors of 100 Ω ± 5 % and 50 Ω ± 10 % are connected in series. The limiting error of the combination is
Show hintHide hint
For a sum, add the absolute (ohm) errors, then divide by the total.
Show answerHide answer
Answer: C. ±6.67 %
Absolute errors add: ±5 Ω + ±5 Ω = ±10 Ω on 150 Ω, i.e. ±6.67 %.
7. Errors caused by human mistakes such as misreading a scale or recording a wrong figure are classified as
Show hintHide hint
Think of the error category that is entirely the observer's fault.
Show answerHide answer
Answer: C. gross errors
Gross errors arise from human blunders in reading, recording or calculating; they are avoided by care and repeated readings.
8. Which of the following is NOT a category of systematic error?
Show hintHide hint
Systematic errors have identifiable causes.
Show answerHide answer
Answer: B. Random error
Systematic errors are instrumental, environmental and observational. Random (residual) errors are a separate class from unknown, irregular causes.
9. The effect of random errors on a measurement is best reduced by
Show hintHide hint
Their sign changes from reading to reading.
Show answerHide answer
Answer: D. taking many readings and analysing them statistically
Random errors vary unpredictably in sign and size, so averaging many readings (and quoting standard deviation) reduces their effect; calibration and fixed corrections only remove systematic errors.
10. A mirror is placed behind the pointer on the scale of many precision analogue instruments in order to
Show hintHide hint
It concerns where the observer's eye is.
Show answerHide answer
Answer: D. eliminate parallax error in reading
The observer aligns the pointer with its reflection so that the line of sight is perpendicular to the scale, which removes parallax.
11. The precision of an instrument is a measure of
Show hintHide hint
An instrument can be precise yet inaccurate.
Show answerHide answer
Answer: D. how closely repeated readings of the same quantity agree with one another
Precision describes repeatability/consistency of readings; closeness to the true value is accuracy, smallest detectable change is resolution, output/input ratio is sensitivity.
12. The smallest increment in the measured quantity that produces a detectable change in the instrument output is called its
Show hintHide hint
It concerns a change from any non-zero reading, not starting from zero.
Show answerHide answer
Answer: A. resolution
Resolution (discrimination) is the smallest detectable increment. Threshold is the minimum input from zero; dead zone is the largest input change giving no response.
13. The largest change of input to which an instrument does not respond at all (e.g. because of friction or backlash) is called
Show hintHide hint
It is a band of no response.
Show answerHide answer
Answer: A. dead zone
Dead zone (dead band) is the largest range of input change for which there is no output response.
14. A recorder pen moves 12 mm when the input changes by 3 V. Its deflection factor is
Show hintHide hint
Deflection factor is the inverse of sensitivity.
Show answerHide answer
Answer: D. 0.25 V/mm
Sensitivity = 12/3 = 4 mm/V; deflection factor is its reciprocal = 3/12 = 0.25 V/mm.
15. Which of the following is a static characteristic rather than a dynamic characteristic of an instrument?
Show hintHide hint
Which one is defined using steady inputs only?
Show answerHide answer
Answer: C. Linearity
Speed of response, measuring lag, fidelity and dynamic error describe behaviour with time-varying inputs; linearity is a static characteristic.
16. Dynamic error of an instrument is the
Show hintHide hint
It exists only while the input is changing.
Show answerHide answer
Answer: A. difference between the true value of a time-varying quantity and the value indicated, assuming no static error
Dynamic error (measurement error) arises because the instrument cannot follow a changing input instantly; it is defined assuming static error is zero.
17. The degree to which an instrument indicates changes in the measured variable without dynamic error is called its
Show hintHide hint
It is about faithful reproduction of a changing signal.
Show answerHide answer
Answer: C. fidelity
Fidelity is faithful reproduction of a time-varying input; measuring lag is the delay in response.
18. A first-order instrument (time constant τ) receives a step input. After a time 2τ its output reaches approximately what percentage of the final value?
Show hintHide hint
Evaluate 1 − e^(−t/τ).
Show answerHide answer
Answer: D. 86.5 %
y/y∞ = 1 − e^(−t/τ) = 1 − e^(−2) = 86.5 %. (63.2 % is reached at τ, 95 % at 3τ.)
19. A thermometer behaving as a first-order system has a time constant of 5 s. The temperature it measures rises steadily at 0.5 °C/s. In steady state, the reading lags the true temperature by
Show hintHide hint
The output lags a ramp by one time constant.
Show answerHide answer
Answer: B. 2.5 °C
For a ramp input, a first-order instrument lags by a time τ, so the dynamic error = rate × τ = 0.5 × 5 = 2.5 °C.
20. A Maxwell inductance–capacitance bridge balances with the standard capacitor C1 = 0.2 µF shunted by R1 = 2 kΩ in the arm opposite the unknown coil, and resistances R2 = 500 Ω and R3 = 1 kΩ in the other two arms. The inductance of the coil is
Show hintHide hint
Lx is the product of the two ratio-arm resistances and the capacitance.
Show answerHide answer
Answer: A. 0.1 H
At balance Lx = R2·R3·C1 = 500 × 1000 × 0.2×10⁻⁶ = 0.1 H (and Rx = R2R3/R1 = 250 Ω).
21. In the same Maxwell L–C bridge (C1 = 0.2 µF in parallel with R1 = 2 kΩ), what is the Q-factor of the coil at 1 kHz?
Show hintHide hint
Q depends only on the arm containing the capacitor.
Show answerHide answer
Answer: C. 2.51
Q = ωLx/Rx = ωC1R1 = 2π × 1000 × 0.2×10⁻⁶ × 2000 = 2.51.
22. Maxwell's inductance–capacitance bridge is best suited to measuring coils whose Q-factor is
Show hintHide hint
Q = ωC1R1; think about what happens at the extremes.
Show answerHide answer
Answer: D. medium, roughly between 1 and 10
For Q > 10 an impractically large R1 is needed, and for very low Q the bridge suffers from sliding balance; Maxwell's bridge suits medium-Q coils.
23. For measuring inductance of coils with Q-factor greater than about 10, the bridge usually preferred to Maxwell's bridge is
Show hintHide hint
It is a modification of Maxwell's bridge.
Show answerHide answer
Answer: A. Hay's bridge
Hay's bridge uses the standard capacitor in series with a resistor, which keeps component values practical for high-Q coils.
24. An important advantage of the Maxwell inductance–capacitance bridge is that
Show hintHide hint
Look at whether ω appears in the balance equations.
Show answerHide answer
Answer: B. its two balance equations are independent of each other and of frequency
Lx = R2R3C1 and Rx = R2R3/R1 contain no ω and can be adjusted independently (using R1 and C1).
25. A Schering bridge balances with the standard capacitor C2 = 100 pF, a non-inductive resistor R3 = 150 Ω in the arm adjacent to the unknown capacitor, and R4 = 300 Ω (shunted by a variable capacitor C4) in the arm opposite the unknown. The unknown capacitance is
Show hintHide hint
Cx is C2 scaled by a ratio of the two resistances.
Show answerHide answer
Answer: B. 200 pF
Cx = C2·R4/R3 = 100 pF × 300/150 = 200 pF.
26. In a Schering bridge operated at 50 Hz, the arm opposite the unknown capacitor contains R4 = 300 Ω in parallel with C4 = 0.5 µF at balance. The dissipation factor of the unknown capacitor is about
Show hintHide hint
D = ω × C4 × R4.
Show answerHide answer
Answer: C. 0.047
D = tan δ = ωC4R4 = 2π × 50 × 0.5×10⁻⁶ × 300 = 0.047.
27. The Schering bridge is mainly used for measuring
Show hintHide hint
Think of insulation testing.
Show answerHide answer
Answer: C. capacitance and dissipation factor of capacitors and insulation, especially at high voltage
The Schering bridge measures capacitance and loss angle (tan δ), and its high-voltage form is used to test insulation of cables, bushings and capacitors.
28. In a high-voltage Schering bridge, the variable resistive arms and the detector can be kept near earth potential because
Show hintHide hint
Compare the impedances of the capacitor arms and the resistor arms.
Show answerHide answer
Answer: A. the capacitive arms have much higher impedance, so nearly all the voltage appears across them
The reactances of the standard and test capacitors are very large compared with R3 and R4, so the low-voltage arms (and the earthed junction) carry only a small voltage, making adjustment safe.
29. A Wien bridge has R1 = R2 = 10 kΩ and C1 = C2 = 0.01 µF in its series and parallel RC arms. The frequency at which it balances is approximately
Show hintHide hint
With equal R and C, f = 1/(2πRC).
Show answerHide answer
Answer: B. 1.59 kHz
f = 1/(2π√(R1R2C1C2)) = 1/(2πRC) = 1/(2π × 10⁴ × 10⁻⁸) = 1592 Hz ≈ 1.59 kHz.
30. For a Wien bridge in which the series and parallel RC arms have equal R and equal C, balance requires the two resistive ratio arms to be in the ratio
Show hintHide hint
Add R1/R2 and C2/C1.
Show answerHide answer
Answer: D. 2 : 1
The resistive balance condition is R3/R4 = R1/R2 + C2/C1 = 1 + 1 = 2.
31. Besides measuring capacitance, the Wien bridge is widely used
Show hintHide hint
Its balance condition contains ω.
Show answerHide answer
Answer: C. for measuring audio frequencies and in harmonic-distortion analysers
Its balance depends on frequency (f = 1/2π√(R1R2C1C2)), so it is used for frequency measurement, as a notch filter in distortion analysers, and in RC oscillators.
32. A commonly used null detector for AC bridges supplied at audio frequencies (about 250 Hz to 4 kHz) is
Show hintHide hint
The ear is most sensitive in this frequency range.
Show answerHide answer
Answer: A. a pair of headphones
Headphones are very sensitive in the audio range and are a classical null detector; vibration galvanometers are used at power frequencies and tuned amplifiers at higher frequencies.
4.2 Measuring instruments
31 questions · AElE0402
33. A permanent-magnet moving-coil (PMMC) ammeter connected directly in a purely sinusoidal AC circuit reads
Show hintHide hint
What is the average of a full sine wave?
Show answerHide answer
Answer: D. zero, because it responds to the average value
PMMC deflecting torque is proportional to instantaneous current; the moving system responds to the average, which is zero for a symmetrical sine wave.
34. The scale of a spring-controlled PMMC instrument is uniform because
Show hintHide hint
Equate deflecting torque with spring torque.
Show answerHide answer
Answer: C. its deflecting torque is directly proportional to the current
Td = BINA ∝ I and spring torque ∝ θ, so θ ∝ I, giving a linear scale.
35. Damping in a PMMC instrument is usually provided by
Show hintHide hint
The strong permanent-magnet field is already there.
Show answerHide answer
Answer: C. eddy currents induced in the aluminium former of the coil
The coil is wound on a light aluminium former; its motion in the magnet's field induces eddy currents that oppose motion.
36. In a moving-iron instrument, the deflecting torque is proportional to
Show hintHide hint
Torque comes from stored magnetic energy ½LI².
Show answerHide answer
Answer: A. the square of the current
Td = ½ I² dL/dθ, so the torque (and the reading) depends on I²; this is why MI meters read rms and have a non-uniform scale.
37. Which of the following statements about moving-iron (MI) instruments is correct?
Show hintHide hint
Reversing the current does not reverse the attraction or repulsion.
Show answerHide answer
Answer: D. They can be used on both AC and DC and indicate rms value
Because torque ∝ I², the direction of deflection does not depend on polarity, so MI meters work on AC and DC and read rms.
38. Moving-iron instruments normally use air-friction damping rather than eddy-current damping because
Show hintHide hint
Think about the strength of the working magnetic field.
Show answerHide answer
Answer: C. a damping permanent magnet would distort the weak operating field
The operating field of an MI meter is weak; a permanent magnet for eddy damping would disturb it and cause error.
39. The function of the damping torque in an indicating instrument is to
Show hintHide hint
Damping torque exists only while the pointer is moving.
Show answerHide answer
Answer: B. bring the pointer to its final position quickly without oscillation
Damping acts only while the pointer moves; the controlling torque balances the deflecting torque and returns the pointer to zero.
40. A meter movement of resistance 50 Ω gives full-scale deflection at 1 mA. The shunt needed to convert it into a 0–10 mA ammeter is
Show hintHide hint
The shunt must carry the remaining 9 mA at the meter voltage.
Show answerHide answer
Answer: A. 5.56 Ω
Rsh = Im·Rm/(I − Im) = (1 mA × 50)/(9 mA) = 5.56 Ω.
41. A 1 mA, 100 Ω meter movement is to be used as a 0–100 V voltmeter. The series multiplier resistance required is
Show hintHide hint
Subtract the meter's own resistance from V/Ifs.
Show answerHide answer
Answer: D. 99.9 kΩ
Total resistance = 100 V/1 mA = 100 kΩ; multiplier = 100 kΩ − 100 Ω = 99.9 kΩ.
42. A voltmeter uses a movement that gives full-scale deflection at 50 µA. Its sensitivity is
Show hintHide hint
Sensitivity in Ω/V is the reciprocal of full-scale current.
Show answerHide answer
Answer: D. 20 kΩ/V
Sensitivity = 1/Ifs = 1/(50×10⁻⁶) = 20 000 Ω/V.
43. Two 100 kΩ resistors in series are connected across a 100 V DC supply. A 20 kΩ/V voltmeter on its 50 V range is connected across one resistor. It reads approximately
Show hintHide hint
The meter resistance appears in parallel with the resistor.
Show answerHide answer
Answer: B. 47.6 V
Rv = 20 kΩ/V × 50 V = 1 MΩ; 100 kΩ ∥ 1 MΩ = 90.91 kΩ; V = 100 × 90.91/(100 + 90.91) = 47.6 V (loading effect).
44. A ballistic galvanometer is used to measure
Show hintHide hint
Its first swing after a short pulse is what matters.
Show answerHide answer
Answer: C. electric charge (quantity of electricity)
Its first throw is proportional to the charge passed through it in a short pulse, so it measures charge (and flux, with a search coil).
45. For a d'Arsonval galvanometer to work ballistically, its moving system should have
Show hintHide hint
The pulse must finish before the coil starts to move much.
Show answerHide answer
Answer: C. a large moment of inertia and very little damping
The whole charge must pass before the coil moves appreciably, so the period must be long (large inertia) and damping small so the throw is not reduced.
46. In an electrodynamometer-type wattmeter, the moving coil is
Show hintHide hint
The coil carrying the smaller current is made the moving one.
Show answerHide answer
Answer: C. the pressure (voltage) coil
The fixed coils carry load current (current coils); the light moving coil, with a series resistor, is the pressure coil connected across the supply.
47. The deflecting torque of an electrodynamometer wattmeter on AC is proportional to
Show hintHide hint
The meter averages the product of instantaneous v and i.
Show answerHide answer
Answer: A. VI cos φ, the average power
The mean torque is proportional to the average of v·i, i.e. VI cos φ, so the meter reads true power.
48. Due to the inductance of its pressure coil, an electrodynamometer wattmeter (without compensation) on a lagging power-factor load
Show hintHide hint
The pressure-coil current lags slightly, reducing the effective phase angle.
Show answerHide answer
Answer: A. reads higher than the true power
The pressure-coil current lags the voltage by a small angle β, so the reading is proportional to cos β·cos(φ − β) instead of cos φ; for lagging φ, cos(φ − β) > cos φ and the meter reads high.
49. In the two-wattmeter method on a balanced three-phase load, the wattmeters read 5 kW and 2 kW. The load power factor is approximately
Show hintHide hint
Use tan φ = √3 (W1 − W2)/(W1 + W2).
Show answerHide answer
Answer: B. 0.80
tan φ = √3(W1 − W2)/(W1 + W2) = √3 × 3/7 = 0.742; φ = 36.6°, pf = 0.80.
50. In the two-wattmeter method on a balanced three-phase load, one wattmeter reads zero. The load power factor is
Show hintHide hint
cos(30° + φ) = 0.
Show answerHide answer
Answer: B. 0.5
The readings are VLIL cos(30° ± φ); one becomes zero when 30° + φ = 90°, i.e. φ = 60°, pf = 0.5.
51. In an induction-type single-phase energy meter, the permanent (brake) magnet
Show hintHide hint
At steady speed, driving torque equals braking torque.
Show answerHide answer
Answer: C. produces a braking torque proportional to disc speed, so speed becomes proportional to power
Eddy currents induced by the brake magnet give a torque ∝ speed; balancing this against the driving torque ∝ power makes revolutions ∝ energy.
52. Creeping in an induction energy meter (slow continuous rotation at no load) is prevented by
Show hintHide hint
Something on the disc itself interrupts the eddy currents.
Show answerHide answer
Answer: C. drilling two diametrically opposite holes in the disc
When a hole comes under the shunt magnet, the eddy-current path is distorted and the disc stops, so it cannot creep continuously.
53. An energy meter has a constant of 600 rev/kWh. Its disc makes 60 revolutions in 90 s. The load power is
Show hintHide hint
Convert revolutions to kWh, then divide by time in hours.
Show answerHide answer
Answer: B. 4.0 kW
Energy = 60/600 = 0.1 kWh in 90 s = 0.025 h, so P = 0.1/0.025 = 4.0 kW.
54. The lag adjustment of an induction energy meter is set so that the flux of the shunt (voltage) magnet lags the applied voltage by
Show hintHide hint
The driving torque must be proportional to VI cos φ.
Show answerHide answer
Answer: B. 90°
Driving torque ∝ VI cos(Δ − φ); it equals VI cos φ only if the shunt-magnet flux lags the voltage by exactly Δ = 90°.
55. The maximum demand recorded by a maximum demand meter for a consumer is
Show hintHide hint
Brief inrush currents should not penalise the consumer.
Show answerHide answer
Answer: B. the greatest average power over any demand interval (e.g. 15 or 30 min) in the billing period
Maximum demand is averaged over a specified demand interval, so short starting surges are not recorded as the demand.
56. Maximum demand meters are mainly installed for bulk consumers in order to
Show hintHide hint
Think about tariffs with two parts.
Show answerHide answer
Answer: A. apply a demand charge reflecting the capacity the utility must provide
Two-part tariffs charge for energy (kWh) and for maximum demand (kW or kVA), which determines plant and network capacity.
57. In an unpolarised vibrating-reed (Frahm) frequency meter energised from an AC supply, the reed that vibrates with the largest amplitude has a natural frequency equal to
Show hintHide hint
How many pulls per cycle does an unpolarised magnet give?
Show answerHide answer
Answer: B. twice the supply frequency
The electromagnet attracts the iron reeds twice per cycle, so the reed tuned to 2f resonates; the scale is marked accordingly in supply frequency.
58. An electrodynamometer power-factor meter has no controlling spring. As a result,
Show hintHide hint
Which torque normally returns a pointer to zero?
Show answerHide answer
Answer: D. its pointer may rest anywhere on the scale when the meter is disconnected
It is a ratiometer-type instrument: the pointer position is fixed by two opposing deflecting torques, so with no supply there is no restoring torque.
59. In a single-phase electrodynamometer power-factor meter, the angular deflection of the moving system is equal to
Show hintHide hint
The scale is marked in cos φ, but the movement turns by an angle.
Show answerHide answer
Answer: A. the phase angle φ between voltage and current
At equilibrium θ = φ, and the scale is graduated in the corresponding values of cos φ.
60. A Wheatstone bridge balances with ratio arms P = 100 Ω and Q = 1000 Ω and the standard arm S = 470 Ω, where at balance R/S = P/Q. The unknown resistance R is
Show hintHide hint
Apply R/S = P/Q.
Show answerHide answer
Answer: D. 47 Ω
R = S × P/Q = 470 × 100/1000 = 47 Ω.
61. The Kelvin double bridge is preferred for measuring
Show hintHide hint
Lead and contact resistance matter most here.
Show answerHide answer
Answer: A. low resistances, below about 1 Ω
Its second set of ratio arms eliminates the effect of lead and contact resistances, which would swamp a low-resistance measurement on a Wheatstone bridge.
62. Insulation resistance of a cable or motor winding is commonly measured with
Show hintHide hint
Insulation resistance is in megohms.
Show answerHide answer
Answer: D. a megger (insulation-resistance tester)
A megger applies a high DC test voltage (e.g. 500 V–5 kV) and reads resistance in megohms directly.
63. In the ammeter–voltmeter method of measuring resistance, connecting the voltmeter directly across the resistor (with the ammeter outside) gives least error when the resistance is
Show hintHide hint
Which meter's current gets added to the reading?
Show answerHide answer
Answer: A. low compared with the voltmeter resistance
In this connection the ammeter also reads the voltmeter current; the error is small only when R is much less than the voltmeter resistance.
4.3 Transducers and sensors
32 questions · AElE0403
64. Which of the following is an active (self-generating) transducer?
Show hintHide hint
Which one needs no external power supply to give an output?
Show answerHide answer
Answer: D. Thermocouple
A thermocouple generates its own emf (Seebeck effect); strain gauges, thermistors and LVDTs need external excitation and are passive.
65. Pressure is sensed by a Bourdon tube whose tip displacement moves the core of an LVDT. In this arrangement the LVDT acts as
Show hintHide hint
Which element senses the measurand directly?
Show answerHide answer
Answer: A. a secondary transducer
The Bourdon tube is the primary transducer (pressure → displacement); the LVDT converts that displacement into an electrical signal, so it is secondary.
66. A wire-wound potentiometer displacement transducer has 500 turns over its full travel. Its resolution is
Show hintHide hint
One turn is the smallest step.
Show answerHide answer
Answer: B. 0.2 % of full scale
The wiper moves in steps of one turn: resolution = 1/500 × 100 = 0.2 %.
67. A 1 kΩ linear potentiometer is supplied with 10 V. Its wiper is at mid-travel and the output is connected to a 10 kΩ load. The output voltage is about
Show hintHide hint
The load resistance appears in parallel with the lower part of the pot.
Show answerHide answer
Answer: D. 4.88 V
Lower half 500 Ω ∥ 10 kΩ = 476.2 Ω; Vo = 10 × 476.2/(500 + 476.2) = 4.88 V. The load makes the output fall below the ideal 5 V.
68. The two secondary windings of an LVDT are connected
Show hintHide hint
The output must be zero at the centre.
Show answerHide answer
Answer: D. in series opposition, so the output is the difference of their voltages
With series-opposition connection the output is zero at the null (centre) position and increases with displacement either side.
69. As the core of an LVDT moves through the null position from one side to the other, the output voltage
Show hintHide hint
Which secondary dominates on each side?
Show answerHide answer
Answer: A. passes through a minimum and its phase changes by 180°
The difference voltage falls to (nearly) zero at null and then rises again with opposite phase, which indicates direction of displacement.
70. Which of the following is NOT an advantage of the LVDT?
Show hintHide hint
An LVDT works by magnetic coupling.
Show answerHide answer
Answer: C. Its output is unaffected by stray magnetic fields
LVDTs offer high output, frictionless operation and very fine resolution, but they are sensitive to stray magnetic fields and usually need shielding.
71. An LVDT has a sensitivity of 40 mV/mm. Its output is read on a voltmeter whose resolution is 2 mV. The smallest displacement that can be detected is
Show hintHide hint
Divide the smallest readable voltage by mV per mm.
Show answerHide answer
Answer: B. 0.05 mm
Displacement resolution = voltmeter resolution / sensitivity = 2 mV ÷ 40 mV/mm = 0.05 mm.
72. A 120 Ω strain gauge with gauge factor 2.0 is subjected to a strain of 500 µε. The change in its resistance is
Show hintHide hint
Gauge factor = (ΔR/R)/strain.
Show answerHide answer
Answer: C. 0.12 Ω
ΔR = R × GF × ε = 120 × 2.0 × 500×10⁻⁶ = 0.12 Ω.
73. Neglecting the change in resistivity with strain (piezoresistive effect), the gauge factor of a metallic wire with Poisson's ratio 0.3 is
Show hintHide hint
Length increases and area decreases.
Show answerHide answer
Answer: D. 1.6
GF = 1 + 2ν + (Δρ/ρ)/ε; with the last term neglected, GF = 1 + 2(0.3) = 1.6.
74. Compared with metallic foil strain gauges, semiconductor strain gauges have
Show hintHide hint
Piezoresistance is strong in semiconductors.
Show answerHide answer
Answer: A. a much higher gauge factor but greater temperature sensitivity
Semiconductor gauges have gauge factors of about 100–200 (versus about 2), owing to strong piezoresistance, but are more temperature-sensitive and less linear.
75. In a strain-gauge bridge, an unstrained "dummy" gauge is placed in an arm adjacent to the active gauge in order to
Show hintHide hint
The dummy sees everything the active gauge sees except strain.
Show answerHide answer
Answer: C. compensate for resistance change due to temperature
Temperature changes both gauges equally; being in adjacent arms, their effects cancel, leaving only the strain signal.
76. A quarter-bridge strain-gauge circuit (one active gauge, GF = 2.0) is excited by 10 V. For a strain of 1000 µε, the bridge output is approximately
Show hintHide hint
A single active arm gives V·(ΔR/R)/4.
Show answerHide answer
Answer: A. 5 mV
For small strain, Vo ≈ V × GF × ε/4 = 10 × 2 × 10⁻³/4 = 5 mV (a full bridge with four active gauges would give 20 mV).
77. A strain-gauge load cell measures force by
Show hintHide hint
Force → elastic deformation → resistance change.
Show answerHide answer
Answer: B. sensing the strain produced in an elastic member by the applied force
The force elastically deforms a metal member; bonded strain gauges in a bridge convert that strain into a voltage proportional to force.
78. In a column-type load cell, four strain gauges are used, two along the axis and two at right angles, connected in a full bridge. The main purpose is to
Show hintHide hint
Opposite-sign strains in adjacent arms add up.
Show answerHide answer
Answer: C. increase the output and provide temperature compensation
Axial gauges see compression and transverse gauges see tension (via Poisson's effect); in a full bridge their outputs add while equal temperature effects cancel.
79. The Hall voltage developed across a current-carrying semiconductor strip placed in a transverse magnetic field is proportional to
Show hintHide hint
Lorentz force needs both moving charges and a field.
Show answerHide answer
Answer: B. the product of current and magnetic flux density
V_H = B·I/(n·q·t): proportional to B × I and inversely proportional to thickness t and carrier density n.
80. A semiconductor strip 1 mm thick with carrier density 10²² m⁻³ carries 10 mA in a perpendicular magnetic field of 0.5 T. The Hall voltage is about (q = 1.6×10⁻¹⁹ C)
Show hintHide hint
Use V_H = BI/(nqt) with all quantities in SI units.
Show answerHide answer
Answer: B. 3.13 mV
V_H = BI/(nqt) = 0.5 × 0.01/(10²² × 1.6×10⁻¹⁹ × 10⁻³) = 3.12 mV.
81. A clamp-on meter using a Hall-effect sensor has an advantage over one using a current transformer because it can measure
Show hintHide hint
Does a transformer work on DC?
Show answerHide answer
Answer: A. DC as well as AC current without breaking the circuit
The Hall element responds to static as well as alternating magnetic fields, while a CT works only with changing flux (AC).
82. Piezoelectric transducers are generally unsuitable for measuring static (steady) forces because
Show hintHide hint
Think of a charged capacitor with finite leakage resistance.
Show answerHide answer
Answer: A. the generated charge leaks away through the finite insulation and amplifier resistance
The crystal acts as a capacitor; a steady force gives a charge that decays through leakage, so these sensors suit dynamic measurements like vibration and shock.
83. Which of the following materials is NOT piezoelectric?
Show hintHide hint
One is a metal alloy.
Show answerHide answer
Answer: D. Constantan
Quartz, Rochelle salt and barium titanate (and PZT) are piezoelectric; constantan is a copper-nickel alloy used in strain gauges and thermocouples.
84. A piezoelectric crystal has a charge sensitivity of 2 pC/N and a capacitance of 500 pF. When a force of 50 N is suddenly applied, the open-circuit output voltage is
Show hintHide hint
Find the charge first, then V = Q/C.
Show answerHide answer
Answer: D. 0.2 V
Q = 2 pC/N × 50 N = 100 pC; V = Q/C = 100 pC/500 pF = 0.19999999999999998 V.
85. In a capacitive displacement transducer that works by changing the separation d between two parallel plates, the capacitance varies
Show hintHide hint
Write C = εA/d.
Show answerHide answer
Answer: C. inversely with d, so the response is non-linear
C = εA/d, so C ∝ 1/d; this change-of-gap type is very sensitive but non-linear (unlike the change-of-area type).
86. An air-gap capacitive sensor has plates of area 4 cm² separated by 1 mm. If the gap is reduced to 0.8 mm, its capacitance becomes about (ε₀ = 8.854×10⁻¹² F/m)
Show hintHide hint
Capacitance rises as the gap shrinks.
Show answerHide answer
Answer: D. 4.43 pF
C = ε₀A/d = 8.854×10⁻¹² × 4×10⁻⁴/0.8×10⁻³ = 4.43 pF (it was 3.54 pF at 1 mm).
87. A major practical limitation of capacitive transducers is that
Show hintHide hint
Compare the sensor capacitance with that of a cable.
Show answerHide answer
Answer: B. stray and cable capacitances can cause serious errors, so careful shielding is needed
Their capacitance is small (pF), so cable and stray capacitances are comparable; guarding/shielding and short leads are needed.
88. A thermistor commonly used for temperature measurement is
Show hintHide hint
Its sensitivity is much larger than a platinum RTD.
Show answerHide answer
Answer: B. a semiconductor resistor with a large negative temperature coefficient
Typical (NTC) thermistors are made of sintered metal oxides; their resistance falls sharply and non-linearly as temperature rises.
89. An NTC thermistor has a resistance of 10 kΩ at 25 °C and β = 3500 K. Using R = R₀·exp[β(1/T − 1/T₀)], its resistance at 50 °C is about
Show hintHide hint
Use kelvin temperatures; the resistance should fall.
Show answerHide answer
Answer: D. 4.03 kΩ
R = 10 kΩ × exp[3500(1/323.15 − 1/298.15)] = 10 × e^(-0.908) = 4.03 kΩ.
90. A thermocouple works on the principle of
Show hintHide hint
It needs two dissimilar metals and a temperature difference.
Show answerHide answer
Answer: C. the Seebeck effect
An emf is produced in a circuit of two dissimilar metals whose junctions are at different temperatures (Seebeck effect).
91. A type K thermocouple with an average sensitivity of 41 µV/°C gives 8.2 mV when its reference (cold) junction is at 30 °C. Assuming linearity, the hot-junction temperature is
Show hintHide hint
The emf depends on the difference between the junction temperatures.
Show answerHide answer
Answer: A. 230 °C
ΔT = 8.2 mV/41 µV/°C = 200 °C; hot junction = 200 + 30 = 230 °C.
92. A type K thermocouple is made of
Show hintHide hint
It is the common general-purpose nickel-based type.
Show answerHide answer
Answer: A. chromel and alumel
Type K = chromel/alumel (nickel-chromium/nickel-aluminium). Iron–constantan is type J, copper–constantan type T, Pt–PtRh types R and S.
93. When the light falling on a cadmium-sulphide photoconductive cell (LDR) increases, its resistance
Show hintHide hint
More light → more free carriers.
Show answerHide answer
Answer: C. decreases
Absorbed photons create extra electron-hole pairs, raising conductivity, so resistance falls with increasing illumination.
94. A major drawback of CdS photoconductive cells compared with photodiodes is their
Show hintHide hint
Think about speed.
Show answerHide answer
Answer: B. slow response time, of the order of milliseconds
CdS cells respond in tens of milliseconds, while photodiodes respond in nanoseconds to microseconds, so photodiodes are used for fast optical signals.
95. In photoconductive (biased) mode, a photodiode is operated
Show hintHide hint
The dark current is a small reverse leakage current.
Show answerHide answer
Answer: C. reverse biased, with the reverse current increasing with light intensity
Under reverse bias the light-generated carriers form a reverse photocurrent nearly proportional to illumination, with a fast response.
4.4 Analog to digital and digital to analog converters
30 questions · AElE0404
96. An 8-bit binary-weighted-resistor DAC uses a 10 kΩ resistor for the MSB. The resistor required for the LSB is
Show hintHide hint
There are 7 doublings from MSB to LSB in 8 bits.
Show answerHide answer
Answer: B. 1.28 MΩ
Each lower bit doubles the resistance; LSB resistor = 10 kΩ × 2⁷ = 1.28 MΩ.
97. The main practical drawback of the binary-weighted-resistor DAC for a large number of bits is that
Show hintHide hint
Compare the MSB and LSB resistors of a 12-bit version.
Show answerHide answer
Answer: D. it needs a very wide range of precise resistor values
Resistances range from R to 2ⁿ⁻¹R, and each must be accurate to better than half an LSB; such widely spread precise values are hard to make, especially on ICs.
98. A 4-bit binary-weighted-resistor DAC drives an inverting summing amplifier with feedback resistor Rf = 1 kΩ. The input resistors are 1 kΩ (MSB), 2 kΩ, 4 kΩ and 8 kΩ (LSB), and a logic 1 connects a resistor to +5 V. For input 1010, the output voltage is
Show hintHide hint
Only the MSB and the third resistor (4 kΩ) are connected to 5 V.
Show answerHide answer
Answer: A. −6.25 V
Vo = −Rf·Vref(1/1k + 0 + 1/4k + 0) = −5 × (1 + 0.25) = -6.25 V.
99. In a binary-weighted-resistor DAC, the resistor that must have the tightest tolerance is the one connected to
Show hintHide hint
Which bit's error is largest in volts?
Show answerHide answer
Answer: C. the most significant bit
The MSB resistor carries the largest current, contributing half of full scale; its error must still be less than ½ LSB, so it needs the highest relative accuracy.
100. The chief advantage of the R–2R ladder DAC over the binary-weighted-resistor DAC is that
Show hintHide hint
Look at the name of the network.
Show answerHide answer
Answer: B. it needs only two resistor values, R and 2R
Only two values are needed regardless of the number of bits, making it easy to fabricate with well-matched resistors.
101. Looking back into the output node of an R–2R ladder network (with all bit inputs connected to either the reference or ground), the Thevenin resistance is
Show hintHide hint
Reduce the ladder from the LSB end: 2R ∥ 2R.
Show answerHide answer
Answer: D. R, independent of the digital input
Each node sees 2R ∥ 2R = R toward the LSB end, so the ladder's output resistance is always R, whatever the code.
102. An 8-bit R–2R ladder DAC produces Vo = Vref × D/256, where D is the decimal value of the input code and Vref = 10 V. For the input 11001000, the output is about
Show hintHide hint
Convert the binary code to decimal first.
Show answerHide answer
Answer: C. 7.81 V
11001000₂ = 200; Vo = 10 × 200/256 = 7.81 V.
103. A 4-bit R–2R ladder DAC with an 8 V reference gives Vo = Vref × D/16. Its maximum (full-scale) output, for input 1111, is
Show hintHide hint
The largest code is 2⁴ − 1, not 2⁴.
Show answerHide answer
Answer: D. 7.5 V
D = 15, so Vo = 8 × 15/16 = 7.5 V; the output can never reach Vref itself, falling short by one LSB (0.5 V).
104. A 10-bit DAC gives an output of 10.23 V when all its inputs are 1. Its step size (resolution) is
Show hintHide hint
All ones corresponds to 1023 steps.
Show answerHide answer
Answer: B. 10 mV
Full scale = (2¹⁰ − 1) × step, so step = 10.23 V/1023 = 10 mV.
105. The percentage resolution of an 8-bit DAC is about
Show hintHide hint
One step divided by the number of steps to full scale.
Show answerHide answer
Answer: C. 0.39 %
% resolution = 1/(2⁸ − 1) × 100 = 1/255 × 100 = 0.39 %.
106. A DAC is said to be monotonic if
Show hintHide hint
Think of the staircase never stepping down.
Show answerHide answer
Answer: C. its output never decreases when the digital input is increased
Monotonicity means the output either increases or stays the same for each increasing input code; it is guaranteed if differential non-linearity is better than −1 LSB.
107. The settling time of a DAC is the time taken for
Show hintHide hint
It is defined with a tolerance band around the final value.
Show answerHide answer
Answer: A. the output to come and stay within a specified band (usually ±½ LSB) of its final value after an input change
Settling time is measured from the input code change until the output stays within the error band (typically ±½ LSB) of its final value; it limits the maximum update rate.
108. A 12-bit successive-approximation ADC uses a 1 MHz clock and needs one clock cycle per bit. Its conversion time is
Show hintHide hint
One clock per bit.
Show answerHide answer
Answer: D. 12 µs
One bit is decided per clock: Tc = 12 × 1 µs = 12 µs. (A counter-ramp ADC might need up to 4096 clocks.)
109. A successive-approximation ADC determines the output code by
Show hintHide hint
It halves the uncertainty range each clock.
Show answerHide answer
Answer: A. testing the bits one at a time from the MSB downward, as in a binary search
The SAR sets each bit in turn starting from the MSB, keeps it if the DAC output does not exceed the input, and clears it otherwise.
110. Which of the following is NOT a building block of a successive-approximation ADC?
Show hintHide hint
Which element belongs to integrating converters?
Show answerHide answer
Answer: D. An integrator with a fixed integration time
A SAR ADC consists of a SAR (control logic), a DAC and a comparator (plus usually a sample-and-hold); an integrator is the core of a dual-slope ADC.
111. A 4-bit successive-approximation ADC has a step size of 1 V (codes 0000 to 1111 represent 0 V to 15 V). For an input of 10.3 V, what is the final output code?
Show hintHide hint
Start with 1000 and keep a bit only if the DAC output does not exceed the input.
Show answerHide answer
Answer: A. 1010
Trials: 1000 (8 V ≤ 10.3, keep), 1100 (12 V > 10.3, clear), 1010 (10 V ≤ 10.3, keep), 1011 (11 V > 10.3, clear) → 1010.
112. Compared with a counter-ramp ADC, the conversion time of a successive-approximation ADC
Show hintHide hint
How many decisions does an n-bit SAR always make?
Show answerHide answer
Answer: C. is fixed and independent of the input voltage
An n-bit SAR always takes n clock cycles, whereas a counter-ramp ADC takes a number of clocks proportional to the input.
113. A sample-and-hold circuit is normally used ahead of a successive-approximation ADC because
Show hintHide hint
The SAR compares against the same input n times.
Show answerHide answer
Answer: A. the input must not change while the bits are being decided
If the input changes during the n bit trials, the binary search can produce a wrong code; the S/H freezes the input during conversion.
114. In a dual-slope ADC, the input is integrated for a fixed 2000 clock counts, and de-integration with a 1.0 V reference of opposite polarity takes 1450 counts. The input voltage is
Show hintHide hint
Equal charge is put on and taken off the integrator capacitor.
Show answerHide answer
Answer: B. 0.725 V
Vin·T1 = Vref·T2, so Vin = Vref × N2/N1 = 1.0 × 1450/2000 = 0.725 V.
115. In a dual-slope ADC, which part of the conversion cycle has a fixed duration?
Show hintHide hint
One of the two times is the measured quantity.
Show answerHide answer
Answer: C. Integration of the unknown input (run-up)
The input is integrated for a fixed time T1; the de-integration time T2 with the reference is measured and is proportional to the input.
116. The accuracy of a dual-slope ADC does not depend on the exact values of the integrator R and C or the clock frequency because
Show hintHide hint
Write the charge balance for both phases.
Show answerHide answer
Answer: A. the same R, C and clock are used in both integration phases, so their effects cancel
Vin = Vref·T2/T1: RC cancels from the ratio, and T1 and T2 are counted by the same clock, so only the reference needs to be accurate (and the components stable during a conversion).
117. To give a dual-slope ADC strong rejection of 50 Hz mains interference, its fixed integration time should be
Show hintHide hint
The integral of a sine wave over whole cycles is zero.
Show answerHide answer
Answer: B. an integer multiple of 20 ms
Integrating over a whole number of 20 ms periods makes the average of a 50 Hz component exactly zero.
118. Dual-slope ADCs are most widely used in
Show hintHide hint
Which application needs accuracy and noise immunity rather than speed?
Show answerHide answer
Answer: D. digital voltmeters and multimeters
They are slow but accurate, cheap and noise-rejecting, ideal for DVMs/DMMs where a few readings per second suffice.
119. During the run-up phase of a dual-slope ADC, the slope of the integrator output is proportional to
Show hintHide hint
Integrator output rate = applied voltage/RC.
Show answerHide answer
Answer: A. the input voltage
The integrator output changes at a rate −Vin/RC while the input is connected, so the peak reached after T1 is proportional to Vin.
120. An 8-bit ADC has an input range of 0 to 5.12 V (LSB = range/256). The maximum quantisation error (with rounding) is
Show hintHide hint
Quantisation error is half a step.
Show answerHide answer
Answer: B. ±10 mV
LSB = 5.12/256 = 20 mV; maximum quantisation error = ±½ LSB = ±10 mV.
121. Which ordering of ADC types from fastest to slowest is correct?
Show hintHide hint
Integrating converters trade speed for accuracy.
Show answerHide answer
Answer: D. Flash, successive approximation, dual slope
Flash converts in one comparator step, SAR needs n clock cycles, and dual slope needs many clock periods for two integrations.
122. The minimum number of bits needed for an ADC to resolve 0.1 % of full scale is
Show hintHide hint
2ⁿ must exceed about 1000.
Show answerHide answer
Answer: C. 10
Need 1/(2ⁿ − 1) ≤ 0.001: 9 bits gives 1/511 ≈ 0.196 %, 10 bits gives 1/1023 ≈ 0.098 %, so 10 bits.
123. An 8-bit counter-ramp ADC is clocked at 1 MHz. Its maximum conversion time is about
Show hintHide hint
The counter may have to count to full scale.
Show answerHide answer
Answer: B. 256 µs
In the worst case the counter must step through all 2⁸ = 256 states: 256 × 1 µs ≈ 256 µs. (A SAR ADC would need only 8 µs.)
124. Quantisation error in an ADC can be reduced by
Show hintHide hint
The error is tied to the step size.
Show answerHide answer
Answer: A. increasing the number of bits
Quantisation error is ±½ LSB, and the LSB = full-scale/2ⁿ; more bits give smaller steps.
125. In a weighted-resistor or R–2R DAC, the operational amplifier at the output mainly serves to
Show hintHide hint
Its inverting input is a virtual ground.
Show answerHide answer
Answer: B. convert the summed bit currents into an output voltage
The op-amp acts as a current-to-voltage converter (summing amplifier) with its inverting input at virtual ground, so the bit currents add without interaction.
4.5 Digital instrumentation
31 questions · AElE0405
126. Which of the following is NOT an advantage of digital instruments over analogue pointer instruments?
Show hintHide hint
What does every digital display need?
Show answerHide answer
Answer: B. They need no power supply to operate
Digital instruments give unambiguous readings, high resolution and computer-compatible output, but they always need a power supply.
127. The maximum count that can be shown on a 3½-digit display is
Show hintHide hint
The half digit can show only 0 or 1.
Show answerHide answer
Answer: D. 1999
A 3½-digit display has three full digits (0–9) and a leading "half" digit that can only be 0 or 1, giving a maximum of 1999.
128. A 3½-digit digital voltmeter is used on its 2 V range (maximum reading 1.999 V). Its resolution is
Show hintHide hint
Look at the last displayed decimal place.
Show answerHide answer
Answer: B. 1 mV
The least significant digit on this range is in the third decimal place: 1.999/1999 = 1 mV.
129. A digital multimeter usually measures resistance by
Show hintHide hint
The DMM core is a voltmeter.
Show answerHide answer
Answer: A. passing a known constant current through the resistor and measuring the voltage across it
An internal constant-current source drives the resistor and the DVM section measures V; R = V/I is displayed directly.
130. A digital frequency counter counts 5000 pulses during a gate time of 0.1 s. The input frequency is
Show hintHide hint
Count divided by gate time.
Show answerHide answer
Answer: D. 50 kHz
f = N/T = 5000/0.1 = 50 kHz.
131. A 200 Hz signal is measured with a frequency counter using a 1 s gate time. The ±1 count uncertainty corresponds to an error of
Show hintHide hint
How many pulses are counted in the gate time?
Show answerHide answer
Answer: A. ±0.5 %
Only about 200 pulses are counted, so ±1 count = ±1/200 = ±0.5 %.
132. To measure a low-frequency signal accurately with a universal counter, it is better to
Show hintHide hint
The ±1 count error matters most when the count is small.
Show answerHide answer
Answer: A. measure its period by counting a high-frequency clock during one (or several) input cycles
At low frequency few input pulses fall within a practical gate time, so the ±1 count error is large; period mode counts many clock pulses instead.
133. In period mode, a counter with a 10 MHz time-base counts 25 000 clock pulses during one period of the input. The input frequency is
Show hintHide hint
First find the period from count × clock period.
Show answerHide answer
Answer: C. 400 Hz
T = 25 000/10 MHz = 2.5 ms; f = 1/T = 400 Hz.
134. According to the sampling theorem, a signal band-limited to 5 kHz must be sampled at a rate of at least
Show hintHide hint
Twice the highest frequency.
Show answerHide answer
Answer: B. 10 kHz
The Nyquist rate is twice the highest frequency: 2 × 5 kHz = 10 kHz.
135. A 7 kHz sinusoid is sampled at 10 kHz without any anti-aliasing filter. After reconstruction it appears as a signal of
Show hintHide hint
Fold the frequency about half the sampling rate.
Show answerHide answer
Answer: D. 3 kHz
The sampling rate is below 2 × 7 kHz, so the signal aliases to |7 − 10| = 3 kHz.
136. An anti-aliasing filter placed before an ADC is
Show hintHide hint
It must act before sampling occurs.
Show answerHide answer
Answer: A. a low-pass filter that removes components above half the sampling frequency
Aliasing must be prevented before sampling, so an analogue low-pass filter limits the input bandwidth to below fs/2.
137. In a typical multi-channel data acquisition system, the correct order of the signal path is
Show hintHide hint
The S/H must come immediately before the ADC.
Show answerHide answer
Answer: C. transducer → signal conditioning → multiplexer → sample-and-hold → ADC → computer
Each sensor signal is conditioned, the multiplexer selects one channel, the S/H freezes it and the ADC digitises it for the computer.
138. The main function of an analogue multiplexer in a data acquisition system is to
Show hintHide hint
It is a selector switch.
Show answerHide answer
Answer: B. allow several input channels to share one ADC
The multiplexer switches channels one at a time to a single S/H and ADC, reducing cost.
139. Which of the following is NOT normally a signal-conditioning function?
Show hintHide hint
Conditioning happens before digitisation.
Show answerHide answer
Answer: A. Storing the measured data on a disk
Signal conditioning prepares the analogue signal (amplify, filter, isolate, linearise, excite sensors); data storage is done later by the computer.
140. An instrumentation amplifier is preferred for amplifying the small differential output of a strain-gauge bridge mainly because it has
Show hintHide hint
The bridge output sits on a large common-mode voltage.
Show answerHide answer
Answer: C. very high input impedance and high common-mode rejection ratio
Bridge outputs are small differential signals riding on a large common-mode voltage; high CMRR rejects the common mode and high input impedance avoids loading.
141. In the 4–20 mA current-loop standard, the signal zero is represented by 4 mA rather than 0 mA mainly so that
Show hintHide hint
What does 0 mA tell you then?
Show answerHide answer
Answer: B. a broken wire (0 mA) can be distinguished from a zero reading, and the transmitter can be powered by the loop
With a "live zero", 0 mA means a fault, and the 4 mA minimum is available to power a two-wire transmitter.
142. A temperature transmitter with range 0–150 °C has a 4–20 mA output. When the loop current is 10 mA, the temperature is about
Show hintHide hint
Subtract the 4 mA live zero before scaling over the 16 mA span.
Show answerHide answer
Answer: C. 56.2 °C
T = (10 − 4)/(20 − 4) × 150 = 6/16 × 150 = 56.2 °C.
143. A 4–20 mA loop signal is passed through a 250 Ω resistor at the receiver. The voltage range obtained is
Show hintHide hint
Apply V = IR at both ends of the range.
Show answerHide answer
Answer: D. 1–5 V
4 mA × 250 Ω = 1 V and 20 mA × 250 Ω = 5 V.
144. Compared with RS-232, the RS-485 serial standard is preferred in industrial measurement networks because it
Show hintHide hint
Balanced lines reject common-mode noise.
Show answerHide answer
Answer: B. uses differential signalling, allowing longer distances and multiple devices on one bus
RS-485 uses balanced differential lines with good noise immunity, supports about 1200 m and multidrop operation, whereas RS-232 is single-ended, point-to-point and short-range.
145. An asynchronous serial link sends each character as 1 start bit, 8 data bits and 1 stop bit, at 9600 bit/s. The maximum number of characters transferred per second is
Show hintHide hint
Count all bits in a frame, not just data bits.
Show answerHide answer
Answer: D. 960
Each character occupies 10 bit times, so 9600/10 = 960 characters per second.
146. The IEEE-488 (GPIB) standard is
Show hintHide hint
It was originally the Hewlett-Packard Interface Bus.
Show answerHide answer
Answer: D. a parallel bus for connecting programmable instruments to a controller
GPIB (HP-IB) is an bit-parallel, byte-serial bus with handshake and management lines, linking up to 15 devices (instruments and controller).
147. The Intel 8255 programmable peripheral interface provides
Show hintHide hint
It is a parallel interface chip.
Show answerHide answer
Answer: A. three 8-bit I/O ports (24 lines) that can be programmed as inputs or outputs
Ports A, B and C (C splittable into two 4-bit halves) can be configured in modes 0, 1 and 2 for parallel interfacing.
148. In parallel data transfer between a microprocessor and a peripheral, handshaking signals (e.g. STB/ACK or OBF/ACK) are used to
Show hintHide hint
Think of a ready/acknowledge exchange.
Show answerHide answer
Answer: C. synchronise the transfer so that data is sent only when the receiver is ready
Handshake lines indicate "data available" and "data accepted", so devices of different speeds can exchange data reliably.
149. Compared with polling, interrupt-driven data transfer in a microprocessor-based instrument
Show hintHide hint
Who initiates the service request?
Show answerHide answer
Answer: A. frees the processor from repeatedly checking device status
With interrupts, the device signals the CPU only when it needs service, so the CPU can do other work meanwhile; polling wastes time testing status flags.
150. Direct memory access (DMA) is used in high-speed data acquisition because it
Show hintHide hint
Consider who controls the buses during the transfer.
Show answerHide answer
Answer: C. transfers data between the peripheral and memory without passing each byte through the CPU
A DMA controller takes control of the buses and moves blocks of data directly, giving much higher transfer rates than program-controlled I/O.
151. In the 8085 microprocessor using I/O-mapped (isolated) I/O with IN and OUT instructions, the maximum number of distinct input ports that can be addressed is
Show hintHide hint
How many bits is the port address in IN/OUT?
Show answerHide answer
Answer: B. 256
IN and OUT carry an 8-bit port address, so 2⁸ = 256 input ports and 256 output ports can be addressed.
152. An opto-coupler is used between a field signal and a microprocessor input mainly to provide
Show hintHide hint
The signal crosses the gap as light.
Show answerHide answer
Answer: C. electrical isolation, protecting the processor from high voltages and ground loops
The signal crosses as light from an LED to a phototransistor, so there is no electrical connection between the two sides.
153. Ground loops in an instrumentation system are best avoided by
Show hintHide hint
A loop needs more than one connection to ground.
Show answerHide answer
Answer: C. grounding the signal circuit and shield at a single point
Multiple ground connections at different potentials let circulating currents flow in the signal path; single-point grounding removes the loop.
154. Which feature becomes possible mainly because of a microprocessor in a measuring instrument?
Show hintHide hint
Think of what software can do with the digitised data.
Show answerHide answer
Answer: D. Automatic zero/calibration correction, auto-ranging and computation of derived quantities
Software allows self-calibration, auto-ranging, linearisation, statistics and derived quantities (e.g. power, rms), as well as data storage and communication.
155. A watchdog timer is included in a microprocessor-based instrument to
Show hintHide hint
It watches for software that has stopped running properly.
Show answerHide answer
Answer: B. reset the processor if the program fails to restart the timer in time (e.g. when it hangs)
Normal software periodically "kicks" the watchdog; if the program crashes, the timer overflows and forces a reset.
156. A digital storage oscilloscope has an advantage over an analogue CRO in that it can
Show hintHide hint
It has memory.
Show answerHide answer
Answer: A. capture and display single-shot events, including what happened before the trigger
The DSO stores digitised samples in memory, so it can hold one-shot events indefinitely and show pre-trigger data.
4.6 Instrument transformers
30 questions · AElE0406
157. The main purposes of instrument transformers are to
Show hintHide hint
Think about connecting a 5 A ammeter to a 2000 A, 11 kV feeder.
Show answerHide answer
Answer: C. extend the range of standard meters and relays and isolate them from the high-voltage circuit
CTs and PTs reduce currents/voltages to standard values (1 A or 5 A, 110 V) so standard instruments and relays can be used safely, insulated from the HV system.
158. A current transformer is connected ____ the line and a potential transformer is connected ____ the line.
Show hintHide hint
Recall how an ammeter and a voltmeter are connected.
Show answerHide answer
Answer: D. in series with; across
The CT primary carries the line current (series connection); the PT primary is connected between lines or line and earth (parallel connection).
159. The standard rated secondary currents of current transformers are
Show hintHide hint
Standard ammeters and relays are rated for these values.
Show answerHide answer
Answer: B. 1 A or 5 A
CTs are standardised with 1 A or 5 A secondaries so that common meters and relays can be used.
160. The standard rated secondary (line) voltage of potential transformers in IEC-based practice is usually
Show hintHide hint
It is a little below the domestic supply voltage.
Show answerHide answer
Answer: D. 110 V
PT secondaries are commonly rated 110 V line-to-line (110/√3 V phase), matching standard voltmeters and relays.
161. The secondary of a current transformer must never be open-circuited while its primary carries current because
Show hintHide hint
Which MMF normally cancels most of the primary MMF?
Show answerHide answer
Answer: C. the whole primary MMF then magnetises the core, producing a dangerously high secondary voltage and core overheating
Normally the secondary MMF opposes the primary MMF. With the secondary open, the primary MMF is unopposed, the core saturates, a high peaky voltage appears, and losses may damage the core (and leave residual magnetism).
162. Unlike a power transformer, the primary current of a current transformer is
Show hintHide hint
The CT primary is just part of the line conductor.
Show answerHide answer
Answer: B. determined by the load of the main circuit, not by the CT's secondary burden
The CT primary is in series with the line, so its current is fixed by the system load; the secondary simply reflects it according to the turns ratio.
163. The ratio and phase-angle errors of a current transformer arise mainly because of
Show hintHide hint
In an ideal CT, the core would need no MMF.
Show answerHide answer
Answer: D. the exciting (magnetising and core-loss) current of the core
Part of the primary current is used to excite the core, so the secondary current is not exactly I1/Kn and not exactly in phase opposition.
164. A common method of reducing the ratio error of a current transformer is to
Show hintHide hint
The secondary current is usually slightly too small.
Show answerHide answer
Answer: C. make the secondary turns slightly fewer than the nominal ratio requires (turns compensation)
Removing one or two secondary turns increases the secondary current slightly, offsetting the loss due to exciting current at the rated burden.
165. If the burden connected to a CT is increased above its rated value,
Show hintHide hint
More burden → more secondary voltage → more flux.
Show answerHide answer
Answer: A. the ratio and phase-angle errors increase
A larger burden needs a higher secondary emf, hence more core flux and exciting current, increasing errors (and possibly saturation).
166. The phase-angle error of a CT matters for the reading of
Show hintHide hint
Which instrument depends on cos φ?
Show answerHide answer
Answer: D. a wattmeter or energy meter, but not for a simple ammeter
An ammeter reads only magnitude; wattmeters and energy meters depend on the phase relation between current and voltage, so phase error causes reading errors (especially at low power factor).
167. An ammeter connected through a 300/5 A CT reads 3.5 A. The line current is
Show hintHide hint
Multiply the reading by the CT ratio.
Show answerHide answer
Answer: D. 210 A
Line current = 3.5 × (300/5) = 3.5 × 60 = 210 A.
168. A voltmeter connected to the secondary of a 33 kV/110 V potential transformer reads 105 V. The primary voltage is
Show hintHide hint
Multiply by the PT ratio.
Show answerHide answer
Answer: B. 31.5 kV
Primary = 105 × (33 000/110) = 105 × 300 = 31.5 kV.
169. A wattmeter is connected through a 200/5 A CT and an 11 000/110 V PT, and reads 400 W. Neglecting transformer errors, the actual power is
Show hintHide hint
Multiply by both the CT ratio and the PT ratio.
Show answerHide answer
Answer: C. 1.6 MW
Actual power = 400 × 40 × 100 = 1.6 MW.
170. A 1000/5 A CT gives a secondary current of 4.95 A when the primary current is 1000 A. Its ratio error, defined as (Kn − R)/R × 100 %, is
Show hintHide hint
Compute the actual ratio from the measured currents.
Show answerHide answer
Answer: B. −1.0 %
Actual ratio R = 1000/4.95 = 202.02; Kn = 200; error = (200 − 202.02)/202.02 × 100 = -1.0 % (the CT reads low).
171. An 11 000/110 V PT has an actual ratio of 99.5 at a certain burden. Its ratio error, (Kn − R)/R × 100 %, is about
Show hintHide hint
Find the nominal ratio first.
Show answerHide answer
Answer: D. +0.50 %
Kn = 100; error = (100 − 99.5)/99.5 × 100 = +0.50 %. The secondary voltage is slightly high.
172. The total impedance of the meters and leads connected to a 5 A CT secondary is 0.6 Ω. The burden on the CT at rated current is
Show hintHide hint
VA = I² × Z.
Show answerHide answer
Answer: B. 15 VA
Burden = I²Z = 5² × 0.6 = 15 VA.
173. For the same lead resistance between a CT in a switchyard and relays in a control room, the lead burden of a 5 A-secondary CT is how many times that of a 1 A-secondary CT?
Show hintHide hint
Burden depends on the square of current.
Show answerHide answer
Answer: A. 25
Lead burden = I²R_lead; (5/1)² = 25. This is why 1 A secondaries are preferred for long lead runs.
174. A bar-type (single-turn primary) CT has a ratio of 600/5 A. The number of secondary turns is
Show hintHide hint
The bar is a one-turn primary.
Show answerHide answer
Answer: C. 120
N2/N1 = I1/I2 = 600/5; with N1 = 1, N2 = 120.
175. A metering CT of accuracy class 0.5 is one whose
Show hintHide hint
The number is a percentage.
Show answerHide answer
Answer: B. ratio (current) error does not exceed ±0.5 % at rated current and rated burden
The class number gives the permissible percentage current error at rated current (with phase displacement limits also specified).
176. A protection CT is specified as 5P20. This means that
Show hintHide hint
P stands for protection.
Show answerHide answer
Answer: A. its composite error does not exceed 5 % up to 20 times rated primary current
In "5P20", 5 is the composite error limit (%), P means protection, and 20 is the accuracy-limit factor.
177. Metering CTs are designed to saturate at a relatively low multiple of rated current (low instrument security factor) so that
Show hintHide hint
Meters need accuracy only up to about 120 % of rated current.
Show answerHide answer
Answer: A. connected meters are protected from damage during system faults
During a fault the core of a metering CT saturates, limiting the secondary current to a few times rated value and protecting the instruments.
178. Compared with a metering CT, a protection CT must
Show hintHide hint
Relays act during faults.
Show answerHide answer
Answer: A. maintain reasonable accuracy up to many times rated current without saturating
Relays must see fault currents correctly, so protection CTs are designed with a high saturation level (accuracy-limit factor of 10, 20 or more).
179. The core of a metering CT is often made of a high-permeability nickel-iron alloy because
Show hintHide hint
Errors come from exciting current.
Show answerHide answer
Answer: A. it needs very little exciting current at low flux density, giving small errors
Metering CTs work at low flux; high permeability minimises exciting current and errors. Protection CTs use CRGO steel for high saturation flux.
180. The knee-point voltage of a CT is the secondary voltage at which
Show hintHide hint
It is the point where the excitation curve bends.
Show answerHide answer
Answer: D. a 10 % increase in voltage produces a 50 % increase in exciting current
This IEC/BS definition marks the onset of saturation on the excitation curve; it is a key specification for class PS/PX CTs used in differential protection.
181. A potential (voltage) transformer differs from a power transformer mainly in that it
Show hintHide hint
Its burden is only a few meters and relays.
Show answerHide answer
Answer: B. has a very small VA rating and is designed for high accuracy of voltage ratio
A PT is essentially a shunt step-down transformer working at almost no load, designed for accurate ratio and small phase error rather than power transfer.
182. Fuses are commonly provided in the secondary circuit of a PT because
Show hintHide hint
Compare with the CT, which must not be opened.
Show answerHide answer
Answer: C. a short circuit on the secondary would draw a very large current and damage the PT
A PT behaves like a voltage source; short-circuiting its secondary causes excessive current, so fuses/MCBs protect it. (Open-circuit is harmless for a PT, unlike a CT.)
183. For extra-high-voltage systems, the electromagnetic PT is commonly replaced by
Show hintHide hint
A divider made of capacitors.
Show answerHide answer
Answer: B. a capacitive voltage transformer (CVT)
A CVT uses a capacitor divider to reduce the voltage to a few kV followed by an intermediate transformer; it is cheaper at EHV and can also couple carrier (PLCC) signals.
184. Correct polarity of CT connections (P1/P2, S1/S2) is NOT critical for which of the following?
Show hintHide hint
Which device only needs the magnitude?
Show answerHide answer
Answer: A. A simple ammeter
An ammeter reads magnitude only; wattmeters, directional relays and differential schemes depend on the relative direction of currents.
185. The secondaries of three line CTs are connected in parallel (residual connection) to supply an earth-fault relay. Under balanced load with no earth fault, the current in the relay is ideally
Show hintHide hint
Add three balanced phasors.
Show answerHide answer
Answer: C. zero
The residual current is the phasor sum Ia + Ib + Ic = 3I0, which is zero for a balanced system; it appears only with an earth fault.
186. One point of the secondary circuit of every instrument transformer is earthed in order to
Show hintHide hint
It is a safety measure.
Show answerHide answer
Answer: A. prevent the secondary from rising to a dangerous potential through capacitive coupling or insulation failure
Earthing the secondary protects personnel and instruments if the primary-to-secondary insulation fails or through capacitive coupling from the HV side.