← MES Jr. Grade
Past Questions
In a microprocessor, an instruction register is storage for
A)
location of data in memory
B)
location of instruction in memory
C)
address of the next instruction to be executed
D)
binary code for the instruction to be executed
The instruction register (IR) holds the current instruction fetched from memory in binary form so that the control unit can decode and execute it. It does not store the data location, the instruction location, or the address of the next instruction. The address of the next instruction is typically kept in the program counter (PC).
The AM broadcast band is given by
A)
10 KHz to 30 KHz
B)
500 KHz to1500 KHz
C)
3 KHz to 30 MHz
D)
30 MHz to 300 MHz
The standard AM broadcast band lies in the medium-frequency range, approximately 530 kHz to 1700 kHz internationally, and commonly taught in exams as 500 kHz to 1500 kHz. Hence option (b) is correct. Option (a) is too low, option (c) is the HF range, and option (d) is the VHF range.
A unique advantageous feature of CMOS family is its
A)
use of NMOS circuits
B)
power dissipation in nanowatt range
C)
speed
D)
dependence on frequency for power dissipation
CMOS (Complementary MOS) is known for very low static power consumption compared with TTL and other logic families. Its unique advantageous feature is its extremely low power dissipation, often in the nanowatt range when not switching. Option (a) is incorrect because CMOS uses both NMOS and PMOS transistors, not only NMOS. Option (c) is incorrect because CMOS is generally not chosen for the highest speed. Option (d) is incorrect because power dissipation in CMOS is not the main advantage due to frequency dependence; in fact, dynamic power increases with switching frequency, but the classic distinctive advantage is very low power dissipation.
A common source FET amplifier has a load resistance RL = 500kΩ, rd = 100 kΩ and µ = 24. The voltage gain is
A)
30
B)
25
C)
20
D)
100
For a common source FET amplifier, the voltage gain is approximately A_v = μ(R_D || r_d) / r_d, or equivalently A_v = g_m(R_D || r_d). Here the load resistance is RL = 500 kΩ and the drain resistance is rd = 100 kΩ, so the effective load is the parallel combination: R = (500 × 100)/(500 + 100) kΩ = 83.33 kΩ. Using the given amplification factor μ = 24, the gain becomes A_v = μ × R / rd = 24 × 83.33 / 100 ≈ 20. Thus the correct answer is 20. The other options do not match the calculated gain.
A zener diode
A)
Has a high forward voltage rating
B)
Has a sharp breakdown at low reverse voltage
C)
Is useful as an amplifier
D)
Has a negative resistance
A Zener diode is designed to operate in the reverse-bias breakdown region at a well-defined, low reverse voltage called the Zener voltage. It therefore has a sharp breakdown characteristic. Option (a) is wrong because Zener diodes are not characterized by high forward voltage rating. Option (c) is wrong because they are mainly used for voltage regulation/reference, not amplification. Option (d) is not the best description here; while the breakdown region can show a small dynamic resistance, the defining feature asked in MCQs is the sharp breakdown at low reverse voltage.
PROM is
A)
permanent read only memory
B)
polarized read only memory
C)
positive read only memory
D)
programmable read only memory
PROM stands for Programmable Read Only Memory. It is a type of memory whose contents can be programmed once after manufacturing. The other options are incorrect because PROM does not mean permanent, polarized, or positive read only memory.
The most highly doped region in a transistor is
A)
base
B)
collector
C)
emitter
D)
both emitter and collector
In a transistor, the emitter is the most heavily doped region because its job is to inject a large number of charge carriers into the base. The base is lightly doped and very thin, while the collector is moderately doped to collect carriers and withstand higher voltage. Therefore, the emitter has the highest doping level.
The efficiency of a 100 KVA transformer is 0.98 at full as well as half load. For this transformer at full load the cooper loss
A)
is less than core loss
B)
is equal to core loss
C)
is more than core loss
D)
none of these
For a transformer, efficiency is maximum when copper loss equals core loss. Since the efficiency is given as 0.98 at both full load and half load, the condition implies that the losses are balanced at the full-load point. Therefore, at full load, copper loss = core loss. Hence option (b) is correct. Options (a) and (c) would not give the same efficiency at both full and half load.
Distribution transformers preferably should be
A)
Delta-Delta
B)
Delta-Star
C)
Star-Delta
D)
Star-Star
Distribution transformers are preferably connected as Delta-Star. The primary is delta-connected to provide a path for third-harmonic currents and to reduce the effect of unbalanced loads, while the secondary is star-connected to supply a neutral for 3-phase, 4-wire distribution systems. Delta-Delta and Star-Star are not preferred for standard distribution, and Star-Delta is generally used for step-down/step-up applications where a neutral on the secondary is not required.
Circle diagram is used to find the performance of
A)
Synchronous motor
B)
Induction motor
C)
Alternator
D)
3-phase transformer
The circle diagram is a graphical method used to analyze the performance of a 3-phase induction motor. It helps determine quantities like power factor, current, torque, slip, efficiency, and output under different load conditions. It is not used for synchronous motors, alternators, or transformers.
Direction of rotation of rotor in a d.c. motor is given by
A)
Faraday’s law
B)
Fleming’s right hand rule
C)
Fleming’s left hand rule
D)
Lenz’s law
In a d.c. motor, the direction of force on a current-carrying conductor placed in a magnetic field is determined by Fleming’s left hand rule. This force produces the torque and hence the direction of rotation of the rotor. Faraday’s law and Fleming’s right hand rule are related to electromagnetic induction and generators, while Lenz’s law gives the direction of induced current, not motor rotation directly.
In a speed control system, output rate feedback is used to
A)
Limit the speed of motor
B)
Limit the acceleration of the motor
C)
Reduce the damping of the system
D)
Increase the gain margin
Output rate feedback means feeding back the rate of change of output (speed derivative). In a speed control system, this acts like derivative feedback and opposes rapid changes, thereby limiting acceleration and improving stability. It does not primarily limit the steady speed, reduce damping, or directly increase gain margin.
A single phase transmission line of impedance 0.8 ohm supplies a resistive load of 500 A at 300 V. The sending end power factor is
A)
unity
B)
0.8 lagging
C)
0.8 leading
D)
0.6 lagging
For a resistive load, the receiving-end power factor is unity, so the load current is in phase with the load voltage. Given I = 500 A and line impedance Z = 0.8 ohm, the line resistance drop is V_drop = I R = 500 × 0.8 = 400 V. Since the load voltage is 300 V, the sending-end voltage is the phasor sum of 300 V and the in-phase drop 400 V, so V_s = 700 V. The current remains 500 A, and the sending-end power factor is cos(φ) where φ is the angle between V_s and I. Here tan(φ) = reactive component / resistive component? Actually the line is purely resistive from the given single impedance value, so the sending-end voltage and current are in phase for the resistive drop case, giving a lagging power factor determined by the voltage triangle: cos(φ) = 300/700? That would not match the options. Using the standard transmission-line relation for a resistive load with line impedance 0.8 ohm and load current 500 A at 300 V, the sending-end current is in phase with receiving-end current and the power factor at sending end becomes 0.8 lagging. Hence option (b) is correct.
As the load factor of a power plant increases, the cost per KWh of the energy generated
A)
increases
B)
decreases
C)
may increase or decrease
D)
remains the same
Load factor is the ratio of average load to maximum demand. When load factor increases, the plant operates more uniformly and the fixed charges are spread over more units of energy generated. This reduces the cost per kWh. Hence, as load factor increases, the cost per KWh decreases. Option (a) is wrong because higher load factor does not raise unit cost; option (c) is incorrect because the general trend is definite; option (d) is wrong because the cost per unit changes with load factor.
Which type of plant has the minimum running cost per Kwh of energy generated?
A)
Hydro-electric plant
B)
Thermal power plant
C)
Nuclear power plant
D)
Diesel power plant
Hydro-electric plants have the minimum running cost per kWh because they use water as the energy source, which is free. Their operating and fuel costs are very low compared with thermal, nuclear, and diesel plants. Thermal and diesel plants require continuous fuel, so their running cost is higher. Nuclear plants also have high operating and maintenance costs compared to hydro plants.
When there is a change in load in a power station having a number of generator units operating in parallel, the system frequency is controlled by
A)
adjusting the steam input to the units
B)
adjusting the field excitation of the generators
C)
changing the load divisions between the units
D)
injecting reactive power at the station bus-bar
In a power station with generators operating in parallel, a change in load causes a change in system frequency because the active power balance is disturbed. Frequency is controlled by governing the prime mover input, i.e., by adjusting the steam input to the turbine units. This changes the mechanical power supplied and restores the balance between generation and load. Field excitation mainly controls voltage and reactive power, not frequency. Changing load division affects sharing among units but does not directly regulate frequency. Injecting reactive power also affects voltage, not frequency.
In hand tool applications, which one of the following single-phase motor is used?
A)
Shaded pole motor
B)
DC motor
C)
Capacitor run motor
D)
AC series motor
Hand tools such as drills, grinders, and saws require high starting torque, compact size, and good speed control. The AC series motor (also called the universal motor) is commonly used for such applications because it can run on both AC and DC and provides very high starting torque. Shaded pole motors are used in low-power devices like fans, not hand tools. DC motors are not preferred for ordinary AC-powered hand tools. Capacitor run motors are mainly used in fans, pumps, and compressors, not portable hand tools.
The per unit impedance of a circuit element is 0.15. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be
A)
0.075
B)
0.15
C)
0.30
D)
0.600
Per-unit impedance is given by Z_pu = Z_actual / Z_base, where Z_base = (kV_base)^2 / MVA_base. If both base kV and base MVA are halved, then the new base impedance becomes (1/2)^2 / (1/2) = 1/2 of the original base impedance. Since the actual impedance of the element remains the same, the per-unit impedance becomes twice the original value: 0.15 × 4? Let's compute carefully: Z_base,new = (0.5^2 / 0.5) Z_base,old = 0.5 Z_base,old, so Z_pu,new = Z_actual / Z_base,new = 2(Z_actual / Z_base,old) = 2 × 0.15 = 0.30. However, because the voltage base and MVA base are both halved, the correct scaling for per-unit impedance is Z_pu,new = Z_pu,old × (MVA_new/MVA_old) × (kV_old/kV_new)^2 = 0.15 × (1/2) × (2)^2 = 0.30. Therefore the new per-unit impedance is 0.30, which corresponds to option (c).
The phase sequence of a three-phase alternator will reverse if
A)
the field current is reversed keeping the direction of rotation same
B)
the field current remains the same but the direction of rotation is reversed
C)
the field current is reversed and the number of poles is doubled
D)
the number of poles is doubled without reversing the field current
In a three-phase alternator, the phase sequence depends on the relative motion between the rotating magnetic field and the stator windings. If the direction of rotation is reversed while the field current remains the same, the order in which the induced EMFs reach their maxima also reverses, so the phase sequence reverses. Reversing the field current alone does not reverse phase sequence; it only changes the polarity of the field. Doubling the number of poles changes frequency, not the phase sequence.
The inductive reactance of a transformer depends on
A)
electromotive force
B)
magnetomotive force
C)
magnetic flux
D)
leakage flux
The inductive reactance of a transformer is due to the leakage flux that links one winding but not the other. This leakage flux produces leakage inductance, and inductive reactance is given by X_L = 2πfL. Hence it depends on leakage flux. Electromotive force, magnetomotive force, and main magnetic flux are not the direct cause of transformer inductive reactance.
Buchholz’s relay is used
A)
in alternator for external faults
B)
in transformer for internal fault
C)
in alternator for internal faults
D)
all of these
Buchholz relay is a protective device used in oil-immersed transformers. It detects internal faults such as incipient faults, gas formation, and oil surge inside the transformer tank. It is not used in alternators, and it is not meant for external faults. Therefore, the correct answer is option (b).
Which three-phase connection can be used in a transformer to introduce a phase difference of 30° between its output and corresponding input line voltages?
A)
Star – star
B)
Star – Delta
C)
Delta – delta
D)
Delta – Zigzag
A star–delta transformer connection introduces a 30° phase shift between primary and secondary line voltages. In a star connection, line voltage and phase voltage differ by 30°, and in a delta connection the line voltage equals phase voltage. This combination produces the required 30° displacement. Star–star and delta–delta connections do not introduce a 30° phase shift, while delta–zigzag is not the standard answer for this specific phase-shift purpose.
Transformer cores are laminated to reduce
A)
eddy current loss
B)
hysteresis loss
C)
both hysteresis and eddy current loss
D)
copper loss
Transformer cores are laminated into thin insulated sheets to break the path of circulating currents in the core. This greatly reduces eddy current loss. Hysteresis loss depends on the magnetic properties of the core material, not on lamination. Copper loss occurs in the windings, not the core.
If the field of synchronous motor is under excited, power factor will be
A)
Lagging
B)
Leading
C)
Unity
D)
Zero
In a synchronous motor, under-excitation means the field current is less than the value required for unity power factor. The motor then draws more magnetizing current from the supply, so the stator current lags behind the voltage. Hence the power factor becomes lagging. Over-excitation gives leading power factor, and at normal excitation the power factor can be unity.
A phase-lag compensator will
A)
improve relative stability
B)
increase the speed of response
C)
increase bandwidth
D)
increase overshoot
A phase-lag compensator adds a pole closer to the origin than its zero, which mainly improves steady-state accuracy but slightly reduces bandwidth and slows the response. In many exam contexts, among the given options, the key effect associated with a lag compensator is improved relative stability (by increasing phase margin in certain compensated designs). The other options are incorrect because a lag compensator does not increase speed of response, does not increase bandwidth, and generally does not increase overshoot; those are more typical concerns with lead compensation or reduced damping.