← MES Jr. Grade
Past Questions
In a single-stage CB amplifier, a smaller load resistance R L will produce
A)
high current gain
B)
high voltage gain
C)
better frequency response
D)
higher power gain
In a common-base (CB) amplifier, the current gain is less than 1, while the voltage gain is high. If the load resistance RL is made smaller, the output RC time constant decreases, which improves the high-frequency response. Hence a smaller load resistance generally gives better frequency response. It does not increase current gain, and power gain is not the main effect here.
Dynamic memory cells are constructed using
A)
transistors
B)
flip-flops
C)
MOSFETs
D)
FETs
Dynamic memory cells in DRAM store data as charge on a capacitor and use a MOS transistor as the access device. Hence, they are constructed using MOSFETs. Flip-flops are used in static memory cells (SRAM), not dynamic memory. Transistors is too general, and FETs is less specific than MOSFETs for this application.
After completing the execution, microprocessor returns to
A)
halt state
B)
fetch state
C)
execute state
D)
interrupt state
A microprocessor works in a cycle: fetch, decode, execute, and then repeats. After completing the execution of an instruction, it returns to the fetch state to get the next instruction from memory. Hence, the correct answer is fetch state. The other options are incorrect because halt state stops operation, execute state is the stage just completed, and interrupt state occurs only when an interrupt is serviced.
The primary function of a filter is to
A)
minimize a.c. input variations
B)
suppress odd harmonics in the rectifier output
C)
stabilize d.c. level of the output voltage
D)
remove ripples from the rectified output
A filter in a rectifier circuit is used to smooth the pulsating DC output by reducing the AC ripple component. It does not primarily stabilize the DC level or specifically suppress only odd harmonics. Therefore, the correct function is to remove ripples from the rectified output.
The basic elements of a microprocessor are
A)
ALU and memory
B)
ALU and control unit
C)
ALU, memory and I/O devices
D)
ALU, control unit and memory
A microprocessor is fundamentally built around two main parts: the Arithmetic Logic Unit (ALU), which performs arithmetic and logical operations, and the Control Unit, which directs and coordinates all operations. Memory and I/O devices are not basic internal elements of the microprocessor; they are external components interfaced with it. Therefore, the correct answer is ALU and control unit.
In a certain telemetry system, the measured values are converted to digital form. The digital values can then be transmitted via FSK (binary or quaternary) or BPSK or QPSK systems. Out of these the best noise immunity can be obtained with
A)
binary FSK
B)
quaternary FSK
C)
BPSK
D)
QPSK
Noise immunity is best when the signal constellation has the maximum separation for a given bit energy. Among the given schemes, BPSK is a 2-phase system and is more robust than QPSK and FSK forms at the same bit rate/power conditions. QPSK carries 2 bits per symbol, so its symbol points are closer together than BPSK, making it less immune to noise. Binary FSK is generally better than quaternary FSK, but among all the listed options, BPSK gives the best noise performance.
The main reason why electrons can tunnel through a P-N junction is that
A)
they have high energy
B)
barrier potential is very low
C)
depletion layer is extremely thin
D)
impurity level is low
In a P-N junction, tunneling occurs when the depletion region becomes extremely thin, so electrons can pass through the potential barrier by quantum tunneling. This is the principle behind tunnel diodes. The other options are incorrect because tunneling is not mainly due to high electron energy, low barrier potential alone, or low impurity level.
A communication channel has a bandwidth of 4 KHz and a signal-to-noise ratio is 25, the channel capacity will be nearly
A)
18000 bits/sec.
B)
4000 bits/sec.
C)
1500 bits/sec.
D)
none of these
Use Shannon’s channel capacity formula: C = B log2(1 + SNR). Here B = 4 kHz = 4000 Hz and SNR = 25. So C = 4000 × log2(26). Since log2(26) is about 4.7, capacity ≈ 4000 × 4.7 ≈ 18800 bps, which is nearly 18000 bps. Hence option (a) is correct. The other options are far from the computed value.
The clipping level is primarily determined by
A)
the shape of the input waveform
B)
battery voltage
C)
value of the resistor
D)
knee voltage of the diode
In a diode clipper circuit, clipping starts when the input exceeds the diode’s forward conduction threshold. Therefore, the clipping level is mainly set by the diode’s knee voltage (and any biasing, if present). The input waveform shape does not determine the clipping level, and the resistor mainly limits current rather than fixing the clip point. Battery voltage can shift the clipping level in biased clippers, but in the basic case the diode knee voltage is the primary factor.
In single pulse modulation of PWM inverter, third harmonic can be eliminated if pulse width is equal to corresponding channel capacity will be nearly
A)
30°
B)
60°
C)
120°
D)
150°
In single-pulse PWM, the harmonic content depends on the pulse width. Third harmonic elimination occurs when the pulse width is adjusted so that the 3rd harmonic component becomes zero. For a symmetrical single pulse, this happens when the pulse width is about 120°. Hence the correct option is (c). The other values do not correspond to the standard third-harmonic elimination condition in single-pulse modulation.
One of the serious disadvantages of FM transmission is its
A)
high static noise
B)
limited line-of-sight range
C)
expensive equipment
D)
adjacent channel interference
FM transmission operates at VHF frequencies, so it generally travels by line of sight and does not follow the earth’s curvature well. Hence its coverage range is limited compared with lower-frequency systems.
Why others are incorrect:
- (a) High static noise: FM is actually less affected by noise and static than AM.
- (c) Expensive equipment: this may be a practical drawback, but it is not the main serious disadvantage asked here.
- (d) Adjacent channel interference: FM has better noise immunity and channel separation; this is not its primary disadvantage.
The noise figure of an amplifier is 6dB. If the input S/N ratio is 38 dB, then the output S/N ratio
A)
44 dB
B)
40 dB
C)
36 dB
D)
32 dB
Noise figure in dB is the degradation in signal-to-noise ratio from input to output. So,
Output S/N (dB) = Input S/N (dB) - Noise figure (dB) = 38 - 6 = 32 dB.
However, the options given do not include 32 dB as the result of this standard relation. If the question intended noise factor addition in terms of noise figure affecting S/N directly, the correct computed output is 32 dB, which matches option (d), not (c). Therefore the correct option is (d) 32 dB.
The frequency of an Astable Multivibrator (AMV) depends mainly on
A)
value of collector load resistors
B)
RC values of the circuit
C)
value of transistor β
D)
width of the input pulse
In an astable multivibrator, the oscillation frequency is determined mainly by the charging and discharging time constants of the timing network, i.e., the RC values. The resistors and capacitors set the period of each half-cycle. Collector load resistors affect circuit operation but are not the main factor for frequency. Transistor β has little effect on the oscillation frequency, and width of input pulse is irrelevant because an astable multivibrator has no stable state and does not require an input pulse.
In AM transmission, the frequency which is not transmitted is
A)
upper side frequency
B)
lower side frequency
C)
audio frequency
D)
carrier frequency
In amplitude modulation (AM), the transmitted signal contains the carrier frequency and two sidebands: the upper side frequency and the lower side frequency. The original audio frequency itself is not transmitted as a separate component; it is used to modulate the carrier. Therefore, the frequency which is not transmitted is the audio frequency. Options (a), (b), and (d) are all present in an AM signal.
Output frequency of inverter circuit depends on
A)
values of the resonant elements
B)
transformation ratio of transformer
C)
level of D.C. voltage applied
D)
none of these
In an inverter circuit, especially a resonant inverter, the output frequency is determined by the resonant frequency of the LC circuit. That resonant frequency depends on the values of the resonant elements, i.e., inductance and capacitance. The transformer ratio does not set the frequency, and the level of DC input voltage mainly affects output magnitude, not frequency.
PROM is
A)
Permanent read only memory
B)
Polarised 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 that can be programmed once after manufacturing. The other options are incorrect because PROM does not mean permanent, polarised, or positive read only memory.
Out of the following devices, the fastest switching device is
A)
JEFT
B)
BJT
C)
MOSFET
D)
triode
MOSFETs are the fastest switching devices among the given options because they are majority-carrier devices and do not suffer from charge storage delay. A JFET is also a field-effect device, but MOSFETs are generally preferred for very fast switching in digital and power applications. A BJT is slower due to minority-carrier storage effects, and a triode vacuum tube is much slower and obsolete for such applications.
CMOS circuits are extensively used for one-chip computers mainly because of their extremely
A)
low power dissipation
B)
large packing density
C)
high noise immunity
D)
low cost
CMOS circuits are widely used in one-chip computers because they consume very little power. This low power dissipation allows more components to be packed on a single chip without excessive heating, making CMOS ideal for integrated circuits. While CMOS also offers high packing density and good noise immunity, the main reason asked here is its extremely low power dissipation.
Which stage of a D.C. power supply uses a Zener as the main component?
A)
rectifier
B)
voltage divider
C)
regulator
D)
filter
A Zener diode is commonly used in the regulation stage of a D.C. power supply because it maintains an almost constant output voltage despite changes in input voltage or load current. A rectifier converts AC to DC, a filter smooths the pulsations, and a voltage divider only splits voltage—it does not provide regulation.
For an amplifier, the coupling method which gives the highest gain is
A)
capacitive coupling
B)
resistance coupling
C)
impedance coupling
D)
transformer coupling
Transformer coupling gives the highest gain among the given coupling methods because it can provide impedance matching between stages and transfer maximum power to the load. Capacitive and resistance coupling are common for voltage amplification, but they do not provide as high a gain as transformer coupling. Impedance coupling is also not generally superior to transformer coupling in terms of gain.
The channel of JFET
A)
consists of p-type material only
B)
consists of n-type material only
C)
may consist of p- type or n-type material
D)
none of these
A JFET (Junction Field Effect Transistor) can be of two types: n-channel JFET and p-channel JFET. Therefore, its channel may be made of either n-type material or p-type material. Option (a) and (b) are incomplete because they mention only one type, and (d) is incorrect.
The value of Zener current is
A)
limited by external circuit resistance
B)
determined by Zener voltage
C)
totally independent of temperature
D)
always in µA range
In a Zener diode regulator, the Zener current is controlled by the series/external circuit resistance. This resistance limits the current flowing through the diode. It is not fixed solely by the Zener voltage, not totally independent of temperature, and it is not always in the µA range because it can vary from microamps to much higher values depending on the circuit.
The main part of an inverter is
A)
oscillator circuit
B)
d.c. source
C)
step up transformer
D)
filter
An inverter converts DC into AC, and its key controlling part is the oscillator circuit, which generates the switching pulses needed to produce alternating output. The DC source only supplies input power, the step-up transformer is used in some inverter designs but is not the main part, and the filter is used to smooth the output waveform.
For an FM wave of carrier 100 MHz, modulating frequency 10 KHz and the maximum frequency deviation of 1 MHz, the bandwidth required is nearly
A)
20 KHz
B)
1 MHz
C)
2 MHz
D)
5 MHz
For an FM signal, the approximate bandwidth is given by Carson’s rule:
Bandwidth ≈ 2(Δf + fm)
Here, maximum frequency deviation Δf = 1 MHz and modulating frequency fm = 10 kHz = 0.01 MHz.
So,
Bandwidth ≈ 2(1 + 0.01) MHz = 2.02 MHz ≈ 2 MHz.
Hence, the correct answer is (c) 2 MHz.
Why others are incorrect:
- (a) 20 kHz is far too small.
- (b) 1 MHz ignores the doubling in Carson’s rule.
- (d) 5 MHz is larger than the required approximate bandwidth.
In d.c. choppers, the waveforms for input and output voltages are respectively
A)
discontinuous and continuous
B)
both continuous
C)
both discontinuous
D)
continuous and discontinuous
In a d.c. chopper, the input is a constant d.c. source, so the input voltage is continuous. The output voltage is obtained by switching the source on and off, so it appears as a chopped waveform and is discontinuous. Hence, the correct pair is continuous and discontinuous. Option (a) reverses the order, while (b) and (c) are incorrect because the input is not discontinuous.