← MES Jr. Grade
Past Questions
The instruction RST-5 is written in a program. The programme will jump to location
A)
0020H
B)
0024H
C)
0028H
D)
002CH
RST 5 in the 8085 is a restart instruction that performs an unconditional call to a fixed interrupt/vector address. The target address is 5 × 8 = 40 decimal = 28H, so the program jumps to 0028H. Therefore option (c) is the correct location. The other options are incorrect because 0020H corresponds to RST 4, 0024H to RST 4.5/another vector in some contexts, and 002CH to RST 5.5/other interrupt vectors, not RST 5.
Negative feedback in an amplifier
A)
Reduce gain
B)
Increases frequency and phase distortions
C)
Reduces bandwidth
D)
Increases noise
Negative feedback in an amplifier feeds a portion of the output back to the input in opposite phase, which lowers the overall closed-loop gain. It also generally improves bandwidth, reduces distortion, and reduces noise sensitivity. Therefore, the correct option is (a) Reduce gain. The other options are incorrect because negative feedback does not increase noise; it usually reduces it. It does not increase frequency and phase distortions; it typically decreases them. It also does not reduce bandwidth; in fact, it usually increases bandwidth.
In a 3-phase semiconverter, for firing angle less than or equal to 60 degrees, free wheeling diode conducts for
A)
30 degrees
B)
60 degrees
C)
90 degrees
D)
0 degrees
In a 3-phase semiconverter, when the firing angle α is less than or equal to 60°, the load current freewheels through the freewheeling diode for 30° in each cycle. This is because the converter output voltage becomes zero during the interval when the incoming phase voltage is not yet sufficient to sustain conduction, and the inductive load current continues through the diode. Hence, the correct answer is 30 degrees.
Why the other options are incorrect:
- 60 degrees: This is too long; the freewheeling interval in this operating range is only 30°.
- 90 degrees: This is not the standard freewheeling duration for a 3-phase semiconverter at α ≤ 60°.
- 0 degrees: Freewheeling does occur, so zero is incorrect.
The highest priority in 8085 microprocessor system is
A)
RST 7.5
B)
RST 6.5
C)
INTR
D)
TRAP
In the 8085 microprocessor, the interrupt priorities are fixed. The highest priority interrupt is TRAP, which is a non-maskable interrupt and cannot be disabled by software. Therefore, it is serviced before all other interrupts.
Why the others are incorrect:
- RST 7.5: High-priority maskable interrupt, but lower than TRAP.
- RST 6.5: Lower priority than RST 7.5.
- INTR: Lowest priority among the listed interrupts and is maskable.
Addition of two periodic signals will always be
A)
Periodic
B)
Aperiodic
C)
May or may not be periodic
D)
Insufficient data
The sum of two periodic signals is not always periodic. It is periodic only if the ratio of their fundamental periods is rational (i.e., the periods are commensurate). If the periods are incommensurate, the resulting signal is not periodic. Hence the correct choice is "May or may not be periodic."
Why others are incorrect:
- a) "Periodic" is wrong because periodicity is not guaranteed for every pair of periodic signals.
- b) "Aperiodic" is wrong because the sum can definitely be periodic in many cases, such as when both signals have the same period.
- d) "Insufficient data" is not the best answer because the general property is known: the sum may or may not be periodic depending on the periods.
The total capacity of a ROM with 1024 rows and 4 columns is
A)
4096
B)
1024
C)
256
D)
512
A ROM organized as 1024 rows and 4 columns has 1024 × 4 = 4096 storage cells (bits). Hence its total capacity is 4096 bits, so option (a) is correct. Option (b) 1024 counts only the rows, not the total cells; option (c) 256 and option (d) 512 are both too small and do not match the row-column product.
The stack pointer in the 8085 microprocessor is a
A)
16 bit register that point to stack memory locations
B)
16 bit accumulator
C)
Memory location in the stack
D)
Flag register used for the stack
The stack pointer (SP) in the 8085 is a 16-bit register that holds the address of the top of the stack, i.e., it points to stack memory locations. This is why option (a) is correct.
Why the others are wrong:
- (b) 16 bit accumulator: The 8085 accumulator is 8-bit, not 16-bit, and it is used for arithmetic/logic operations, not for stack addressing.
- (c) Memory location in the stack: The stack pointer is not a memory location; it is a register that stores an address.
- (d) Flag register used for the stack: The flag register contains status flags (like zero, carry, sign, parity, auxiliary carry) and has no role as the stack pointer.
Cycloconverter can be considered as a.c. –a.c. converter with conversion of
A)
1 Stage
B)
2 Stage
C)
3 Stage
D)
None of these
A cycloconverter is a direct AC-to-AC converter that converts fixed-frequency AC power to variable-frequency AC power in a single conversion stage, but it uses two controlled converter groups (positive and negative) to synthesize the output waveform. Hence it is commonly described as a 2-stage AC-AC converter. Option (b) is correct.
Why others are incorrect:
- (a) 1 Stage: Incorrect because cycloconversion is not usually classified as a single-stage converter in this context; it employs two converter groups.
- (c) 3 Stage: Incorrect because a cycloconverter does not require three conversion stages.
- (d) None of these: Incorrect because the correct classification is 2-stage.
A four quadrant chopper cannot be operated as
A)
One quadrant chopper
B)
Cycloconverter
C)
Inverter
D)
Bi-directional rectifier
A four-quadrant chopper can operate in all four quadrants of the V-I plane, meaning it can provide both positive and negative output voltage and current. Therefore, it cannot be operated as a one-quadrant chopper, which allows only one polarity of voltage and current (single-quadrant operation).
Why others are incorrect:
- b) Cycloconverter: A chopper is a DC-DC converter, not a cycloconverter; however, the question asks what it cannot be operated as in terms of quadrant operation, and the clearest incompatible mode is one-quadrant operation.
- c) Inverter: A four-quadrant chopper can be used in inverter-like operation in certain configurations by reversing power flow, so this is not the best answer.
- d) Bi-directional rectifier: This is also not the standard classification for a chopper, but a four-quadrant chopper is specifically defined by its ability to handle bidirectional voltage/current, making one-quadrant operation the direct contradiction.
Hence, the correct answer is a) One quadrant chopper.
A boost regulator has an input voltage of 5 V and the average output voltage of 15 V. The duty cycle is
A)
3/2
B)
2/3
C)
5/2
D)
15/2
For an ideal boost converter, the output voltage is related to the input voltage by Vout = Vin / (1 - D), where D is the duty cycle. Given Vin = 5 V and Vout = 15 V: 15 = 5 / (1 - D) => 1 - D = 1/3 => D = 2/3. So option b is correct.
The other options are incorrect because they do not satisfy the boost-converter relation: a = 3/2 and c = 5/2 are greater than 1, but duty cycle must lie between 0 and 1; d = 15/2 = 7.5 is also impossible for a duty cycle.
Race around condition always arise in a
A)
Combination circuit
B)
Asynchronous circuit
C)
Synchronous circuit
D)
Digital circuit
Race-around condition is a problem associated with level-triggered flip-flops in asynchronous operation, especially the JK flip-flop when J = K = 1 and the clock pulse remains active long enough for the output to toggle repeatedly. Hence, it always arises in an asynchronous circuit.
Why others are incorrect:
- a) Combination circuit: It has no memory elements or feedback timing issues, so race-around does not apply.
- c) Synchronous circuit: Operations are controlled by a common clock, which reduces timing hazards; race-around is not the general characteristic here.
- d) Digital circuit: This is too broad a term; race-around is not inherent to all digital circuits, only to specific asynchronous sequential circuits.
Number of machine cycles required for RET instruction in 8085 microprocessor is
A)
1
B)
2
C)
3
D)
5
RET is a 1-byte control instruction in the 8085, but it requires 5 machine cycles to complete because the processor must read the return address from the stack (two memory reads for low and high bytes) and then update the program counter. Hence, the correct answer is 5 machine cycles.
Why others are incorrect: 1 machine cycle is too few for stack access; 2 machine cycles and 3 machine cycles are also insufficient because RET needs multiple memory operations to fetch the 16-bit return address from the stack.
The most commonly used filters in SSB generation are
A)
Mechanical
B)
RC
C)
LC
D)
Low pass
In SSB (single-sideband) generation, the most commonly used filter is the RC filter in practical electronic circuits for shaping/selecting frequencies. LC filters are also used in RF applications, but for SSB generation the standard answer in exam context is RC because it is simpler, cheaper, and widely used in low-frequency filter networks. Low-pass is a filter type, not the specific commonly cited circuit used for SSB generation here, and mechanical filters are specialized and not the most commonly used overall. Hence, RC is the correct option.
For a given op-amp, CMRR = 10^5 and differential gain = 10^5. What is the common mode gain of the op-amp?
A)
10^10
B)
2x10^5
C)
10^5
D)
1
For an op-amp, the common-mode rejection ratio is defined as CMRR = Ad / Ac, where Ad is the differential gain and Ac is the common-mode gain. Given CMRR = 10^5 and Ad = 10^5, we get Ac = Ad / CMRR = 10^5 / 10^5 = 1. So the common-mode gain is 1, which corresponds to option d.
Why the other options are incorrect:
- b (2x10^5): This is larger than the differential gain and does not satisfy CMRR = Ad/Ac.
- a (10^10): This is far too large; it would make CMRR much less than 1, which is opposite of the given data.
- c (10^5): This would give CMRR = 10^5 / 10^5 = 1 only if Ac = 1, not 10^5. So it is incorrect.
Time division multiplex
A)
Can be used with PCM only
B)
Combines five groups into a supergroup
C)
Stacks 24 channels in adjacent frequency slots
D)
Interleaves pulses belonging to different transmission
Time division multiplexing (TDM) shares a single communication channel by assigning different time slots to different signals, so their pulses are interleaved in time. Hence, option d is correct: it interleaves pulses belonging to different transmissions.
Why the others are incorrect:
- a: TDM is not limited to PCM only, though it is commonly used with PCM systems.
- c: Stacking 24 channels in adjacent frequency slots describes frequency division multiplexing (FDM), not TDM.
- b: Combining five groups into a supergroup is also an FDM hierarchy term, not a definition of TDM.
Reverse recovery current in a diode depends upon
A)
Forward field current
B)
Storage charge
C)
Temperature
D)
PIV
Reverse recovery current in a diode is mainly determined by the stored minority-carrier charge in the junction during forward conduction. When the diode is suddenly reverse-biased, this stored charge must be removed before the diode can block reverse current, so a larger storage charge produces a larger and longer reverse recovery current.
Why others are incorrect:
- PIV (peak inverse voltage) is the maximum reverse voltage a diode can withstand; it does not directly determine reverse recovery current.
- Forward field current is not the controlling factor; the key issue is the stored charge left from forward conduction.
- Temperature can affect diode behavior indirectly, but reverse recovery current is primarily governed by storage charge, not temperature as the main parameter.
The noise temperature about 4306 K. The noise figure will be
A)
3 dB
B)
6 dB
C)
12 dB
D)
24 dB
Based on established principles in Electrical / Electronics Engineering, the correct and most appropriate choice is '3 dB' (Option A).
The snubber circuit is used in thyristor circuits for
A)
Triggering dv/dt
B)
protection dv/dt
C)
protection di/dt
D)
Phase shifting
A snubber circuit in thyristor circuits is primarily used to protect the device against a high rate of rise of voltage (dv/dt). A sudden dv/dt across a thyristor can cause unintended turn-on due to the device’s internal capacitances, so the snubber limits this voltage rise and improves safe operation.
Why the others are incorrect:
- a) Triggering dv/dt: snubbers do not intentionally trigger the thyristor; they prevent false triggering caused by dv/dt.
- c) Protection di/dt: di/dt protection is usually provided by series inductance or current-limiting elements, not the snubber’s main function.
- d) Phase shifting: phase shifting is unrelated to snubber circuits; it is used in control or AC circuit applications, not for thyristor protection.
Telephone traffic is measured
A)
With echo cancelers
B)
By the relative congestion
C)
In terms of the grade of service
D)
In erlangs
Telephone traffic is measured in erlangs, which quantify the average occupancy of a communication channel (1 erlang = one channel used continuously for one hour). This is the standard unit used in teletraffic engineering. The other options are incorrect: relative congestion is a performance measure, not the unit of traffic; grade of service describes the probability of call blocking/delay, not traffic volume; echo cancelers are devices used to reduce echo and are unrelated to traffic measurement.
A 50.004 MHz carrier is to be frequency modulated by a 3 KHz audio tone resulting in a narrow band FM signal. Determine the bandwidth of the FM signal
A)
2 KHz
B)
4 KHz
C)
6 KHz
D)
8 KHz
For narrowband FM, the bandwidth is approximately twice the highest modulating frequency: BW ≈ 2fm. Here the audio tone frequency is 3 kHz, so BW ≈ 2 × 3 kHz = 6 kHz. However, the carrier frequency 50.004 MHz is irrelevant to FM bandwidth; only the modulating frequency and deviation matter. Since the question states a narrowband FM signal and gives only the tone frequency, the intended exam approximation is 2 × 3 kHz = 6 kHz, but that option is not listed. Among the given choices, 4 kHz, 2 kHz, and 8 kHz do not match the standard narrowband FM approximation. Therefore, the question/options appear inconsistent; if forced to choose from the provided options, none is exactly correct. If the exam intended Carson’s rule with very small deviation, the bandwidth would still be close to 2(Δf + fm), not derivable from the data given.
A tunnel diode is
A)
High resistivity p-n junction diode
B)
A slow switching device
C)
An amplifying device
D)
A very heavily doped p-n junction diode
A tunnel diode is a very heavily doped p-n junction diode. Heavy doping makes the depletion layer extremely thin, allowing quantum-mechanical tunneling of carriers and producing its characteristic negative resistance region. Option (c) is not the best description because a tunnel diode can be used as an amplifier, but that is an application, not its definition. Option (a) is incorrect because tunnel diodes are not high-resistivity devices; they are heavily doped and low-resistance. Option (b) is incorrect because tunnel diodes are known for very fast switching, not slow switching.
A 1-phase H.W controlled rectifier with R-L-E load has firing angle and extinction angle 30 degrees and 210 degrees respectively. The source frequency is 50 Hz. The circuit turn off time is
A)
1/60 sec
B)
1/90 sec
C)
1/120 sec
D)
1/180 sec
For a 1-phase half-wave controlled rectifier, the circuit turn-off time is the interval between the extinction angle and the next firing instant. Here, firing angle α = 30° and extinction angle β = 210°. So the turn-off interval is β − α = 210° − 30° = 180° = π radians.
At 50 Hz, the time period T = 1/50 s. Therefore, turn-off time = (180°/360°) × T = (1/2) × (1/50) = 1/100 s. However, in controlled rectifier terminology, the circuit turn-off time is usually taken as the interval from extinction at 210° to the next firing at 390° (i.e., 180° later), which is again 180° = half a cycle = 1/100 s. Since 1/100 s is not among the options, the intended exam interpretation is likely the commutation/turn-off interval corresponding to 60°? But based on the given data and standard definition, the closest matching option provided is 1/120 s only if a different angle convention were used.
Rechecking the standard formula for turn-off time in a half-wave controlled rectifier: t_off = (β − α)/ω = 180°/(2πf) = 1/(2f) = 1/100 s. Since this is absent, the question/options appear inconsistent. If forced to choose from the given options, none is exactly correct. But among the listed choices, option c (1/120 s) is the nearest only by elimination, not by correct calculation.
For variable heads of near about but less than 30 meters, which type of turbines is used in hydro power stations?
A)
Pelton
B)
Kaplan
C)
Francis
D)
None of these
Francis turbines are used for medium heads, typically around 30 m to 300 m. Since the question asks for variable heads near about but less than 30 meters, the closest suitable choice among the given options is Francis, as Pelton is for high heads and Kaplan is for very low heads. Pelton turbines are not suitable because they are impulse turbines used for high head, low discharge conditions. Kaplan turbines are axial-flow reaction turbines used for low head, high discharge sites, generally below about 30 m. Therefore, for heads near but less than 30 m, the expected answer in this option set is Francis.
Maximum efficiency of modern coal-fired steam-raising thermal power plants is restricted about 0.35 (a low value), mainly because of
A)
Low alternator efficiency
B)
High energy loss in boilers
C)
Low steam turbine mechanical efficiency
D)
High energy loss from turbine exhaust to condenser
The maximum efficiency of a coal-fired steam power plant is limited mainly by the low temperature at which heat must be rejected in the condenser. In a steam cycle, a large fraction of the heat supplied to the boiler cannot be converted into work and is instead discharged with the exhaust steam to the condenser, where it is rejected to cooling water. This large unavoidable heat rejection keeps the overall thermal efficiency relatively low, around 35%.
Why the others are incorrect:
- a) Low alternator efficiency: Alternators are generally highly efficient, so they are not the main reason for the low overall plant efficiency.
- b) High energy loss in boilers: Boiler losses do occur, but modern boilers are fairly efficient; they are not the principal factor limiting the plant to about 0.35.
- c) Low steam turbine mechanical efficiency: Steam turbines have high mechanical efficiency, so this is not the dominant limitation.
Therefore, the correct answer is d) High energy loss from turbine exhaust to condenser.
The normal value of the armature resistance of a d.c. motor is
A)
0.005
B)
0.5
C)
5
D)
50
The armature resistance of a d.c. motor is normally very low, because the armature winding is made of thick copper conductors to carry large current with minimal voltage drop. A typical value is of the order of a fraction of an ohm, commonly around 0.5 a9. Hence option (b) is correct.
Why the others are incorrect:
- (a) 0.005 a9: This is too small for a normal armature resistance value in standard exam context.
- (c) 5 a9: This is far too high; such a resistance would cause excessive voltage drop and power loss.
- (d) 50 a9: This is completely unrealistic for a d.c. motor armature resistance.