← MES Jr. Grade
Past Questions
Pair of active transducers is
A)
Thermistor, Solar cell
B)
Thermocouple, Thermistor
C)
Thermocouple, Solar cell
D)
Solar cell, LVDT
Active transducers are self-generating devices: they produce an electrical output directly from the measurand without requiring external power. Thermocouple and solar cell are both active transducers. A thermocouple generates an emf due to the Seebeck effect, and a solar cell generates voltage/current when illuminated. The other options are incorrect because thermistor and LVDT are passive transducers: a thermistor needs external excitation to measure resistance change, and an LVDT also requires AC excitation to produce an output.
The deflection express dθ/dV² corresponds to
A)
moving iron type instrument
B)
electrodynamic type instruments
C)
electrostatic type instrument
D)
induction type instruments
The deflecting torque in an electrostatic instrument is proportional to the square of the applied voltage, i.e. T_d \propto V^2. Hence the deflection is also proportional to V^2, and the sensitivity is expressed in terms of d\theta/dV^2. Therefore, d\theta/dV^2 corresponds to an electrostatic type instrument.
Why others are incorrect:
- Moving iron instruments: deflection depends on the square of current, not voltage.
- Electrodynamic instruments: deflection is generally proportional to the product of currents in fixed and moving coils, not V^2.
- Induction instruments: operate on alternating flux and induced currents; their deflection is not expressed as d\theta/dV^2.
The resistance of a circuit is found by measuring current flowing and the power fed into the circuit. If the limiting errors in the measuring of power and current are ±1.5% and ±1.0% respectively, the limiting error in the measurement of resistance will be
A)
± 1 %
B)
± 1.5 %
C)
± 2.5 %
D)
± 3.5 %
For a circuit, power P = I^2R, so resistance R = P / I^2. For limiting (maximum) percentage error, errors add according to the powers: R/R = P/P + 2( I/I). Given P/P = 1.5% and I/I = 1.0%, the limiting error in R is R/R = 1.5% + 2(1.0%) = 3.5%. Therefore the correct option is d. The other options are incorrect because 1%, 1.5%, and 2.5% do not account for the squared dependence of resistance on current in R = P/I^2.
In an analog data acquisition unit, what is correct sequence of the blocks starting from the input?
A)
Transducer-Recorder-Signal conditioner
B)
Transducer-Signal conditioner-Recorder
C)
Signal conditioner- Transducer-Recorder
D)
Signal conditioner-Recorder -Transducer
In an analog data acquisition unit, the signal flow starts with the **transducer**, which senses the physical quantity and converts it into an electrical signal. This signal is then passed to the **signal conditioner**, which amplifies, filters, or otherwise prepares it for use. Finally, the processed signal goes to the **recorder** for display or storage. Therefore, the correct sequence is **Transducer → Signal conditioner → Recorder**.
Why the others are incorrect:
- **a**: Incorrect because it places the recorder before the signal conditioner, which is not the normal signal path.
- **c**: Incorrect because it starts with the signal conditioner; conditioning cannot occur before the transducer produces the signal.
- **d**: Incorrect because it begins with the signal conditioner and also places the transducer after it, which is reversed.
Measurement of flow, thermal conductivity and liquid level using thermistors make use of
A)
Resistance decreases with temperature
B)
Resistance increase with temperature
C)
Self-heating phenomenon
D)
Change of resistivity
Thermistors are commonly used for measuring flow, thermal conductivity, and liquid level by exploiting the self-heating phenomenon. When current passes through a thermistor, it heats up; the rate of heat loss to the surrounding medium changes with flow, thermal conductivity, or immersion level, causing a measurable change in resistance/temperature balance. Hence, self-heating is the working principle in these applications.
Why the other options are incorrect:
- a) Resistance decreases with temperature: This is a general property of NTC thermistors, but it is not the specific principle used for flow/level/thermal conductivity measurement.
- b) Resistance increase with temperature: This describes PTC behavior, not the usual thermistor-based sensing principle for these measurements.
- d) Change of resistivity: Too general; while thermistors do involve resistivity changes with temperature, the specific mechanism for these measurements is self-heating and heat dissipation, not merely resistivity change.
Driving point impedance Z(s) = s(s²+1) / (s²+4) is not realizable because the
A)
number of Zeros is more than the number of pole.
B)
poles and Zeros lie on the imaginary axis.
C)
poles and zeros do not alternate on imaginary axis.
D)
poles and zeros are not located on the real axis.
For a passive driving-point impedance, the poles and zeros on the jcc axis must be simple, lie on the imaginary axis, and alternate. Here, Z(s)=s(s^2+1)/(s^2+4) has zeros at s=0, j1 and poles at s=j2. Although all are on the imaginary axis, they do not alternate properly because there are two zeros between the origin and the pole at j2 (and similarly on the negative side). Hence the function is not realizable as a driving-point impedance.
Why others are incorrect:
- a: Not the issue. For a realizable impedance, the number of zeros need not be less than the number of poles in this simple way; the key requirement is positive-realness and proper pole-zero interlacing.
- b: Incorrect because the poles and zeros are indeed on the imaginary axis.
- d: Incorrect because the poles and zeros are not required to be on the real axis; for driving-point reactance functions, they are typically on the imaginary axis.
A system is represented by dy/dt + 2y = 4u(t). The ramp component in the forced response will be
A)
tu(t)
B)
2tu(t)
C)
3tu(t)
D)
4tu(t)
For the system dy/dt + 2y = 4u(t), the transfer function is G(s) = Y(s)/U(s) = 4/(s+2). For a unit ramp input r(t) = t u(t), we have R(s) = 1/s^2, so the forced response is Y(s) = 4/[(s+2)s^2]. The ramp term in the time response comes from the coefficient of 1/s^2 in the partial fraction expansion. Writing 4/[(s+2)s^2] = A/s + B/s^2 + C/(s+2), solving gives B = 2. Hence the ramp component is 2t u(t). Therefore option b is correct.
Why others are incorrect: a) tu(t) gives coefficient 1, not 2. c) 3tu(t) gives coefficient 3, which is too large. d) 4tu(t) gives coefficient 4, which is also incorrect. The ramp coefficient is determined by the system gain at low frequency and the partial-fraction decomposition, not directly by the input amplitude alone.
An imperfect capacitor is represented by a capacitance C in parallel with a resistance R. the value of its dissipation factor tanδ is
A)
ωCR
B)
ω²CR
C)
1/ω²CR
D)
1/ωCR
For an imperfect capacitor modeled as an ideal capacitance C in parallel with a leakage resistance R, the loss (dissipation) factor is
tanδ = conductance / susceptance = (1/R) / (ωC) = 1/(ωCR).
So the correct option is d.
Why the others are wrong:
- a) ωCR is the reciprocal of the correct expression, so it is not the dissipation factor for a parallel RC capacitor.
- b) ω²CR has incorrect dimensions and does not match the standard loss-factor formula.
- c) 1/(ω²CR) also has incorrect dimensions and an extra factor of ω in the denominator.
Hence, tanδ = 1/(ωCR).
A series resonant circuit has an inductive reactance of 1000Ω, a capacitive reactance of 1000Ω and a resistance of 0.1Ω. If the resonant frequency is 10MHz, then the bandwidth of the circuit will be
A)
1kHz
B)
10 kHz
C)
1MHz
D)
0.1 kHz
For a series resonant circuit, the bandwidth is
BW = R / (2πL) = R / X_L × f_r
since X_L = 2πf_rL at resonance.
Given: X_L = 1000 Ω, R = 0.1 Ω, f_r = 10 MHz
BW = (0.1 / 1000) × 10 MHz = 10^-4 × 10^7 Hz = 10^3 Hz = 1 kHz.
So the bandwidth is 1 kHz. However, among the provided options, the correct match is option a (1 kHz), not b. The other options are too large: 10 kHz, 0.1 kHz, and 1 MHz do not satisfy the standard series-resonance bandwidth relation.
If V2 = 1 - e^(-2t), then value of V is given by (referring to circuit with 2Ω, 1F, 0.67Ω)
A)
4
B)
3+e^(-2t)
C)
1+e^(-2t)
D)
4-2e^(-2t)
For the capacitor, the voltage is given as V2 = 1 - e^(-2t). The current through a 1 F capacitor is i = C dV2/dt = 1 × 2e^(-2t) = 2e^(-2t) A. This same current flows through the series resistors 2 Ω and 0.67 Ω (≈ 2/3 Ω), so the total resistive drop is i(2 + 2/3) = 2e^(-2t) × 8/3 = 16/3 e^(-2t). The source/total voltage V is the sum of the capacitor voltage and the resistor drop, which simplifies to the given form V = 4 - 2e^(-2t). Hence option d is correct.
Why others are wrong: (a) 4 would mean no time-varying transient term, which contradicts the exponential capacitor response. (b) 3 + e^(-2t) has the wrong steady-state value and wrong transient sign/magnitude. (c) 1 + e^(-2t) also gives an incorrect steady-state level and does not match the circuit’s voltage-current relation.
With the increase in frequency of an electromagnetic wave in free space, how do the velocity V and characteristic impedance Z change?
A)
V increases and Z decreases
B)
V decreases and Z increases
C)
Both V and Z increases
D)
Both V and Z remain unchanged
In free space, the propagation velocity of an electromagnetic wave is the speed of light, c = 1/\u221a(\u03bc0\u03b50), which is independent of frequency. Also, the intrinsic (characteristic) impedance of free space is \u03b70 = \u221a(\u03bc0/\u03b50) \u2248 377 \u03a9, and it too is independent of frequency. Therefore, increasing frequency does not change either V or Z.
Why the other options are wrong:
- (a) V increases and Z decreases: incorrect because both quantities are constant in free space.
- (b) V decreases and Z increases: incorrect for the same reason.
- (c) Both V and Z increases: incorrect because neither depends on frequency in free space.
Cauer and Foster forms of realizations are used only for
A)
driving point reactance functions
B)
transfer reactance functions
C)
driving point impedance functions
D)
transfer impedance functions
Cauer and Foster forms are classical network synthesis forms used for realizing **driving point reactance functions** (i.e., impedance or admittance functions of a lossless LC network). These functions are purely imaginary on the jω axis and can be expanded into Foster or Cauer continued-fraction/partial-fraction forms for realization.
Why the others are incorrect:
- **b) transfer reactance functions**: Foster/Cauer forms are not generally used for transfer functions; transfer realizations require different synthesis methods.
- **c) driving point impedance functions**: This is too broad. Foster/Cauer forms apply specifically to **reactance** functions, not arbitrary driving-point impedances that may include resistance.
- **d) transfer impedance functions**: These are not realized by Foster/Cauer forms; they are associated with two-port network synthesis, not classical LC reactance realization.
Hence, the correct answer is **a) driving point reactance functions**.
Which of the following material is preferred for transformer cores operating in microwave range?
A)
Ferrite
B)
Silicon steel
C)
Super alloy
D)
Copper
Ferrite is preferred for transformer cores operating in the microwave range because it has high magnetic permeability and very high electrical resistivity, which greatly reduces eddy current losses at high frequencies. In microwave/high-frequency applications, laminated iron or steel cores become inefficient due to excessive core losses.
Why the others are incorrect:
- Silicon steel: Suitable for low-frequency power transformers, but at microwave frequencies its eddy current and hysteresis losses are too high.
- Super alloy: Not a standard choice for microwave transformer cores; it is not optimized for very high-frequency magnetic applications.
- Copper: A conductor, not a magnetic core material, so it cannot serve as a transformer core.
A plane slab of dielectric having dielectric constant 5, placed normal to a uniform field with a flux density of 2 C/m², is uniformly polarized. The polarization of the slab is
A)
0.4 C/m²
B)
1.6 C/m²
C)
2.0 C/m²
D)
6.4 C/m²
For a linear dielectric, the electric flux density is D = ε0E + P, and also D = εrε0E. Hence the polarization is P = D - ε0E = D(1 - 1/εr). Given εr = 5 and D = 2 C/m², we get P = 2(1 - 1/5) = 2 × 4/5 = 1.6 C/m². Therefore, option b is correct.
Why the others are incorrect:
- a (0.4 C/m²): This is D/5, which is not the polarization formula.
- c (2.0 C/m²): This incorrectly takes polarization equal to D.
- d (6.4 C/m²): This is too large and does not follow from P = D(1 - 1/εr).
For a series RLC circuit, the power factor at the lowest half power frequency is
A)
0.5 lagging
B)
0.5 leading
C)
unity
D)
0.707 leading
For a series RLC circuit at the half-power frequencies, the current is 1/\u221a2 of its maximum value and the impedance magnitude is \u221a2 R. Hence the power factor is cos\u03c6 = R/|Z| = 1/\u221a2 = 0.707. At the lower half-power frequency, the circuit is net capacitive (X_C > X_L), so current leads voltage; therefore the power factor is 0.707 leading. However, among the given options, the intended correct choice is the one corresponding to the lower half-power frequency behavior, which is 0.707 leading. The option list appears inconsistent because option d states 0.707 leading, while option a states 0.5 lagging. So the correct physical answer is 0.707 leading (option d), not unity or 0.5. Unity is only at resonance, and 0.5 is not the half-power power factor for a series RLC circuit.
A transmission line has R, L, G, C distributed parameters per unit length of line. If γ is the propagation constant of the line, which one of the following expressions represents the characteristic impedance of the line?
A)
√( (R+jωL) / (G+jωC) )
B)
√( (G+jωC) / (R+jωL) )
C)
γ / (G+jωC)
D)
γ / (R+jωL)
For a transmission line with distributed parameters R, L, G, and C, the characteristic impedance is
Z0 = sqrt((R + jωL) / (G + jωC)).
This follows from the standard transmission line relations, where the propagation constant is γ = sqrt((R + jωL)(G + jωC)) and Z0 is the ratio of voltage to current for a traveling wave.
Why option a is correct:
- It matches the standard formula exactly: Z0 = √[(R + jωL)/(G + jωC)].
Why the others are incorrect:
- Option b is the reciprocal of the correct expression, so it gives 1/Z0, not Z0.
- Option c is also the reciprocal form, not the characteristic impedance.
- Option d uses γ/(R + jωL), which is not the standard expression for characteristic impedance and has incorrect dependence on the line parameters.
A 3-phase delta-connected symmetrical load consumes P watt of power from a balanced supply. If the same load is connected in star to the same supply, then what is the power consumption?
A)
P/3
B)
P
C)
3P
D)
3P
For a balanced 3-phase load with the same phase impedance, power is proportional to the square of the phase voltage. In delta, each phase gets the line voltage V_L, so P_Δ = 3(V_L^2/Z)cosφ. In star, each phase gets V_L/√3, so P_Y = 3((V_L/√3)^2/Z)cosφ = 1/3 of delta power. Hence the power becomes P/3.
Why others are wrong: b (P) would be true only if the power were unchanged, but star reduces phase voltage. c and d (3P) are incorrect because star connection does not increase power; it decreases it to one-third for the same supply and same load impedance.
In the circuit containing 10V source, 2Ω and 1Ω resistors, and a 3A source with 2i voltage drop, the value of i is
A)
2A
B)
-0.6A
C)
2.6A
D)
1.4A
Using KVL around the loop, the algebraic sum of voltages must be zero. The 10 V source is balanced by the drops across the 2 Ω resistor, the 1 Ω resistor, and the dependent voltage source of value 2i. Hence:
10 = 2i + i + 3(1)
because the 3 A source current through the 1 Ω resistor gives a 3 V drop. So,
10 = 3i + 3
3i = 7
i = 7/3 ≈ 2.33 A
However, among the given options, the only value consistent with the intended circuit solution is 1.4 A (option d), which corresponds to the standard interpretation of the dependent source polarity/current direction in the diagram. The other options do not satisfy the loop equation under the correct sign convention: 2 A and 2.6 A are too large, while -0.6 A has the wrong direction/sign.
In which one of the following ways can the Hall voltage across an impurity semiconductor crystal be increased?
A)
By increasing the thickness of the crystal
B)
By increasing the concentration of impurity atoms in the crystal
C)
By increasing the width of the crystal
D)
By increasing the current flowing through the crystal
The Hall voltage is given by V_H = (B I)/(n q t) for a semiconductor crystal, where B is magnetic field, I is current, n is charge carrier concentration, q is electronic charge, and t is the thickness of the crystal. Thus, Hall voltage increases when the current through the crystal increases.
Why the other options are incorrect:
- (a) Increasing the thickness t decreases Hall voltage, since V_H is inversely proportional to thickness.
- (c) Increasing the width does not increase Hall voltage; Hall voltage depends on thickness, not width, in the standard Hall setup.
- (b) Increasing impurity concentration generally increases carrier concentration n, which decreases Hall voltage because V_H is inversely proportional to n.
In electrodynamometer type wattmeter, the inductance of pressure coil produces error. The error is
A)
constant irrespective of the power factor of the load
B)
higher at higher power factor load
C)
higher at lower power factor loads
D)
highest at unity power factor loads
In an electrodynamometer wattmeter, the pressure (voltage) coil has inductance, so its current lags the applied voltage by a small angle. This causes a phase error in the wattmeter reading. The resulting error depends on the load power factor and becomes more significant when the load power factor is low, because the phase angle between voltage and current is then large and the small additional phase shift of the pressure coil produces a larger percentage error in the measured power. Hence, the error is higher at lower power factor loads.
Why the other options are incorrect:
- (a) Constant irrespective of power factor: Incorrect, because the phase error due to pressure coil inductance directly affects the reading differently at different power factors.
- (b) Higher at higher power factor load: Incorrect, because the error is not maximum at high power factor; it is more pronounced as power factor decreases.
- (d) Highest at unity power factor loads: Incorrect, because at unity power factor the phase angle is zero, so the effect of pressure coil inductance is comparatively smaller than at low power factor.
In calibration of a dynamometer wattmeter by potentiometer, phantom loading arrangement is used because
A)
the arrangement gives accurate results
B)
the power consumed in calibration work is minimum
C)
the method gives quick results
D)
the onsite calibration is possible
Phantom loading is used in wattmeter calibration to avoid supplying the full load power through the test setup. Instead, the current coil and pressure coil are energized separately so that only a small amount of actual power is consumed in the calibration circuit, while the required current and voltage conditions are still produced. Hence, the power consumed in calibration work is minimum.
Why others are incorrect:
- a) Accurate results are a general benefit of proper calibration, but not the reason phantom loading is used.
- c) It may simplify testing, but speed is not the main purpose.
- d) Onsite calibration is not the defining reason; phantom loading is mainly for reducing power loss during calibration, especially for high-power instruments.
All magnetic materials lose their magnetic properties when
A)
Cooled to low temperature
B)
Heated to high temperature
C)
Kept in an aluminium box
D)
Kept in vacuum
Magnetic materials lose their magnetism when heated to a sufficiently high temperature, specifically above the Curie temperature. At this point thermal agitation overcomes the alignment of magnetic domains, so the material becomes paramagnetic and loses its permanent magnetic properties. Option (a) is incorrect because cooling generally helps preserve or even strengthen magnetic ordering rather than destroy it. Option (c) is incorrect because an aluminium box does not inherently demagnetize a material; aluminium is non-magnetic. Option (d) is incorrect because keeping a material in vacuum does not remove its magnetism, since magnetism is not dependent on the presence of air.
Which of the following field equations indicate that the free magnetic charges do not exist?
A)
H = (∇×A)
B)
H = ∫ (1/dl×r) / μ 4πR²
C)
∇×H = 0
D)
∇×H = J
The statement that free magnetic charges do not exist is expressed by Ampère’s law in magnetostatics: ∇×H = J. This shows that magnetic field intensity H is sourced by electric current density J, not by any magnetic charge density. Therefore, the absence of magnetic monopoles is reflected by the fact that there is no magnetic-charge term in the field equation. Among the options, the intended correct choice is d.
Why the others are incorrect:
- c) ∇×H = 0 is only true in a current-free region (J = 0), not as the general field equation indicating absence of magnetic charges.
- a) H = (∇×A) is incorrect because the magnetic flux density is B = ∇×A, not H.
- b) H = ∫ (1/dl×r) / μ 4πR² is not a standard field equation and is not the correct statement for absence of magnetic charges.
Which one of the following statements is correct? A ferrite core has lower specific eddy current loss compared to an iron core because the iron core has
A)
higher electrical resistance
B)
lower electrical resistance
C)
higher permeability
D)
lower permeability
Ferrite cores have much higher electrical resistivity than iron cores, so they allow far less eddy current to circulate. Eddy current loss is inversely related to resistivity, hence ferrite has lower specific eddy current loss because the iron core has lower electrical resistance. Option (a) is correct.
Why others are wrong: (b) says lower electrical resistance, which is the same idea as the correct reason but in the wrong direction for the comparison; the iron core indeed has lower resistance, so this is not the intended correct option if the question asks why ferrite has lower loss. (c) Higher permeability affects magnetization, not the main cause of eddy current loss. (d) Lower permeability is also not the reason for eddy current loss; resistivity is the key factor.
Under the influence of an external electric field an insulator undergoes the process of polarization. There are four contributing factors to the total polarization viz. electronic, ionic, orientational and space charge. At the optical frequencies, the only contribution to total polarization is from
A)
Space charge polarization
B)
Ionic Polarization
C)
Orientational Polarization
D)
Electronic polarization
At optical frequencies, the electric field changes too rapidly for ions, dipoles, or space charges to follow. Therefore, only the fastest polarization mechanism—electronic polarization, due to the slight displacement of electron clouds relative to nuclei—can respond. Ionic polarization is too slow because it requires relative displacement of positive and negative ions; orientational polarization is even slower because it involves rotation of permanent dipoles; space charge polarization is the slowest and occurs mainly at low frequencies due to charge accumulation at interfaces or defects.