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MES Jr. Grade • 2012 • Paper-II

Root locus diagrams can be used to determine

A) absolute stability
B) relative stability
C) conditional stability
D) none of the above
Root locus shows how the closed-loop poles move as system gain varies. From the pole locations and their movement, we can judge damping, overshoot, and stability margin, which are measures of relative stability. It is not mainly used to determine absolute stability alone, and 'conditional stability' is a special case rather than the general purpose of root locus.
MES Jr. Grade • 2012 • Paper-II

Steady-state stability limit of a power system means that the system is capable of maintaining

A) the frequency at 50 Hz exactly
B) a constant power flow throughout
C) synchronism between machines and on external lines
D) all of the above
Steady-state stability limit is the maximum power that can be transmitted without losing synchronism after small, gradual disturbances. So the system must be capable of maintaining synchronism between machines and on external lines. It does not mean frequency is exactly 50 Hz, and power flow is not necessarily constant.
MES Jr. Grade • 2012 • Paper-II

The controlling medium in a pneumatic control system is

A) liquid
B) air
C) oil
D) helium
In a pneumatic control system, the controlling medium is compressed air. Pneumatic systems use air as the transmitting and controlling fluid, unlike hydraulic systems which use oil or other liquids. Therefore, the correct answer is air. Liquid and oil are used in hydraulic control systems, and helium is not used as the standard controlling medium.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

An n-bit AD converter using VR as reference voltage has a resolution (in volts) of

A) VR/2n
B) VR/n
C) VR/2n-1
D) VR/2^n
For an n-bit ADC, the total input range is divided into 2^n discrete levels. Hence the resolution (least significant step size) is reference voltage divided by 2^n, i.e., VR/2^n. Among the options, this is written as VR/2n in option (a), which is the intended answer. Option (b) is incorrect because resolution does not depend linearly on n, and option (c) corresponds to a different expression sometimes used for full-scale step count, not the standard resolution formula.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Which one of the following is an active transducer?

A) strain gauge
B) selsyn
C) photovoltaic cell
D) photo emissive cell
An active transducer is one that generates its own electrical output without requiring an external power source. A photovoltaic cell converts light directly into electrical voltage/current, so it is an active transducer. A strain gauge is a passive transducer because it needs external excitation. A selsyn is an electromechanical device, not typically classified as an active transducer in this context. A photo emissive cell also generally requires external biasing and is not the standard answer here.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

F.M. systems as compared to A.M. systems

A) are equally effected by noise
B) are less effected by noise
C) are more effected by noise
D) are highly effected by noise as they operate at VHF and UHF
FM (frequency modulation) is less affected by noise than AM (amplitude modulation) because in FM the information is carried by frequency variations, while most noise mainly disturbs amplitude. FM receivers can also use amplitude limiters to remove noise-induced amplitude changes. Hence, compared to AM systems, FM systems are less affected by noise. Options (a), (c), and (d) are incorrect.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Very small resistance can be measured more accurately by

A) resistance box
B) Kelvin-double bridge
C) Anderson bridge
D) Schering bridge
Very small resistances are measured accurately using the Kelvin double bridge because it eliminates the effect of lead and contact resistances. A resistance box is not suitable for precise measurement of very low resistance. Anderson bridge is used for measuring inductance, and Schering bridge is used for measuring capacitance and dielectric loss.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Capacitive transducers can be used for measurement of liquid level. The principle of operation used in this case is

A) change of capacitance with change of distance between plates
B) change of area of plates
C) change of dielectric strength
D) none of the above
In a capacitive liquid-level transducer, the liquid acts as the dielectric medium between the capacitor plates. As the liquid level changes, the effective dielectric constant (dielectric property) between the plates changes, which changes the capacitance. Hence the operating principle is change of dielectric strength/properties. Option (a) is for varying plate spacing, and (b) is for varying overlapping area, which are not the usual principle in liquid-level measurement.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

An electromagnetic field is radiated from

A) a stationary point charge
B) a capacitor with a DC voltage
C) a conductor carrying a DC current
D) an oscillating dipole
Electromagnetic waves are radiated by accelerating charges, not by charges or currents in steady states. An oscillating dipole has charges moving back and forth, so it continuously accelerates charges and emits electromagnetic radiation. A stationary point charge, a capacitor with DC voltage, and a conductor carrying DC current produce static electric/magnetic fields, but do not radiate electromagnetic waves in the ideal steady-state case.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

The direction of the magnetic lines of force is

A) from positive to negative charges
B) from south to north poles
C) from one end of the magnet to the other
D) from north pole to the south pole
Magnetic field lines outside a magnet emerge from the north pole and enter the south pole. So the direction of magnetic lines of force is from north pole to the south pole. Option (a) refers to electric field lines, (b) is the opposite of the correct magnetic direction, and (c) is too vague and not the standard scientific description.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

“In all cases of electromagnetic induction, an induced voltage will cause a current to flow in a closed circuit in such a direction that the magnetic field which is caused by the current will oppose the change that produces the current” is the original statement of

A) Lenz’s law
B) Faraday’s law of magnetic induction
C) Fleming’s law
D) Ampere’s law
The statement describes the direction of induced current: it always flows so that the magnetic field produced by it opposes the change in magnetic flux that caused the induction. This is exactly Lenz’s law. Faraday’s law gives the magnitude of induced emf, Fleming’s law is used for direction in motors/generators, and Ampere’s law relates magnetic field to current.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

An inverse transducer converts

A) electrical energy to any other form of energy
B) electrical energy to light energy
C) mechanical displacement into electrical signal
D) electrical energy to mechanical form
An inverse transducer performs the reverse function of a normal transducer. While a transducer generally converts a non-electrical quantity into an electrical signal, an inverse transducer converts electrical energy into a non-electrical form, commonly mechanical output. Hence, the correct answer is electrical energy to mechanical form. Option (c) describes a normal transducer, not an inverse transducer.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

A wave travelling in x-direction has Ey and Ex components at 45º phase apart. The wave is

A) linearly polarized
B) elliptically polarized
C) circularly polarized
D) horizontally polarized
For an electromagnetic wave, if two perpendicular electric field components are present with a phase difference other than 0° or 180°, the wave is generally elliptically polarized. Here, Ex and Ey are 45° phase apart, so the tip of the electric field vector traces an ellipse. Linear polarization requires zero phase difference, and circular polarization requires equal amplitudes with a 90° phase difference. Hence, the correct answer is elliptically polarized.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

A wave guide can be considered to be analogous to a

A) low pass filter
B) high pass filter
C) band pass filter
D) band stop filter
A waveguide allows electromagnetic waves to propagate only above a certain cutoff frequency. Frequencies below cutoff are attenuated and cannot travel through it. This behavior is analogous to a high pass filter, which passes higher frequencies and blocks lower ones. Hence, the correct answer is high pass filter. The other options do not match this cutoff-frequency behavior.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

The unit of magnetic field strength is

A) ampere turns
B) weber
C) ampere per metre
D) weber/metre2
Magnetic field strength (also called magnetizing field intensity, H) is measured in ampere per metre (A/m). Option (a) ampere turns is not a unit of field strength; it is related to magnetomotive force. Option (b) weber is the unit of magnetic flux. Option (d) weber/metre2 is the unit of magnetic flux density (tesla), not field strength.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

The induced e.m.f. is maximum if the conductor cuts the magnetic field at

A) 0°
B) 45°
C) 90°
D) none of these
The induced e.m.f. in a conductor moving in a magnetic field is proportional to vB sinθ, where θ is the angle between the direction of motion and the magnetic field. This becomes maximum when sinθ = 1, i.e. at θ = 90°. At 0°, the conductor moves parallel to the field and no flux is cut, so induced e.m.f. is zero. Therefore, the correct answer is 90°.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

At a certain frequency, a uniform plane wave is found to have established a wavelength λ in a good conductor. If the source frequency is doubled, the wavelength would change to

A) λ/√2
B) √2 λ
C) 2 λ
D) 4 λ
For a good conductor, the propagation constant gives wavelength proportional to 1/\sqrt{f}. Thus, if the source frequency is doubled, the wavelength becomes 1/\sqrt{2} times the original value. Therefore, the new wavelength is λ/\sqrt{2}. Options (b), (c), and (d) are incorrect because wavelength decreases, not increases, when frequency increases in a good conductor.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Silicon steel is used in the construction of electrical machines because it has

A) low hysteresis loss
B) low retentivity
C) low coercivity
D) none of these
Silicon steel is used for the magnetic cores of electrical machines because adding silicon reduces the area of the hysteresis loop, thereby lowering hysteresis loss. This improves efficiency in transformers, motors, and generators. While silicon steel also has low coercivity and low retentivity, the standard reason asked in such questions is its low hysteresis loss.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

A PMMC meter has an internal resistance 200Ω and the current required for its full-scale deflection is 50µA. The meter is capable of measuring, on its own, a maximum voltage of

A) 5mV
B) 10mV
C) 5µV
D) 10µV
For a PMMC meter, the maximum voltage it can measure on its own is given by V = I × R. Here, full-scale current I = 50 µA = 50 × 10^-6 A and internal resistance R = 200 Ω. So, V = 50 × 10^-6 × 200 = 0.01 V = 10 mV. Hence, the correct answer is 10 mV. The other options are incorrect because they do not match the product of the given current and resistance.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

The order of minimum displacement that can be measured with capacitive transducers is

A) 1cm
B) 1mm
C) 1µm
D) 1 × 10-12 m
Capacitive transducers are highly sensitive and are commonly used for very small displacement measurements. Their minimum measurable displacement is typically of the order of micrometres. Hence, 1 µm is the correct order. 1 cm and 1 mm are too large, while 1 × 10^-12 m is far beyond the practical range of ordinary capacitive transducers.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Two parallel conductors are carrying currents in the same direction. Which of the following statements is correct?

A) There is a repelling force between the conductors
B) The force increases when the current is decreased
C) There is an attracting force between the conductors
D) The force increases when the distance between the conductors is increased
Two parallel current-carrying conductors exert magnetic forces on each other. If the currents are in the same direction, the magnetic fields interact so that the conductors attract each other. If the currents were in opposite directions, they would repel. Also, the force is proportional to the product of currents and inversely proportional to the distance between them, so options (b) and (d) are incorrect.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

Which instrument has the lowest resistance?

A) Ammeter
B) Voltmeter
C) Frequency meter
D) Megger
An ammeter is connected in series, so it must have very low internal resistance to avoid changing the circuit current. A voltmeter has very high resistance, a frequency meter is not designed for minimum resistance in this sense, and a megger has very high resistance for insulation testing. Therefore, the instrument with the lowest resistance is the ammeter.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

A dynamometer type wattmeter responds to the

A) average value of active power
B) average value of reactive power
C) peak value of active power
D) peak value of reactive power
A dynamometer type wattmeter measures true power in AC circuits. Its deflecting torque is proportional to the instantaneous product of current and voltage, and the average over a cycle gives the active (real) power. It does not respond to reactive power, and it is not a peak-reading instrument. Hence the correct answer is the average value of active power.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

A current transformer has a phase error of +3°. The phase angle between the primary and secondary current is

A) 3°
B) 177°
C) 180°
D) 183°
In a current transformer, the ideal phase difference between primary and secondary currents is 180° because the secondary current should oppose the primary current. A phase error of +3° means the actual phase angle is 3° more than 180°. Therefore, the phase angle between the primary and secondary current is 180° + 3° = 183°. Hence, option (d) is correct.
MES Jr. Grade • 2012 • Electrical Engineering Paper-I

The electric field lines and equipotential lines

A) are parallel to each other
B) are one and same
C) cut each other orthogonally
D) can be inclined to each other at any angle
Electric field lines show the direction of the electric field, while equipotential lines represent points of equal potential. Since the electric field is always perpendicular to the equipotential surface/line, they intersect at right angles. Hence, the correct option is (c). Options (a), (b), and (d) are incorrect because the two are neither parallel nor identical, and their angle is not arbitrary.
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