← MES Jr. Grade
Past Questions
The address bus of inlet 8085 is 16 bit wide and hence the memory which can be accessed by this address bus is
A)
112 KB
B)
4 KB
C)
16 KB
D)
64 KB
An 8085 has a 16-bit address bus, so it can generate 2^16 distinct addresses. Since each address refers to 1 byte of memory, the total directly addressable memory is 2^16 bytes = 65,536 bytes = 64 KB. Therefore, option d is correct. The other options are incorrect because 112 KB is too large for a 16-bit address space, while 4 KB and 16 KB are much smaller than the actual 64 KB addressable range.
In a single-phase full converter, the output voltage during overlap is equal to
A)
zero
B)
source voltage
C)
source voltage minus the inductance drop
D)
none of these
In a single-phase full converter, during the overlap period both incoming and outgoing thyristors conduct simultaneously because of source inductance. The source voltage is then shared in overcoming the inductance, so the load/output voltage is reduced by the inductive drop. Hence, during overlap the output voltage is approximately source voltage minus the inductance drop. Option (c) is correct.
Why others are wrong:
- (a) Zero: output voltage is not zero during overlap; current still flows and the source still contributes voltage.
- (b) Source voltage: this would be true only if there were no overlap (ideal commutation with zero source inductance).
- (d) None of these: incorrect because option (c) correctly describes the overlap condition.
In a three-phase half-wave rectifier, the output voltage is equal to
A)
the most positive input phase voltage at any instant.
B)
the difference of most positive and most negative input phases at any instant.
C)
the average value of the three phases.
D)
the difference of the two positive phase voltages.
In a three-phase half-wave rectifier, each diode conducts when its phase voltage is the highest positive value among the three phases. Therefore, at any instant the output voltage follows the most positive input phase voltage.
Why others are wrong:
- (b) is incorrect because the difference between the most positive and most negative phases is associated with line-to-line voltage, not a half-wave rectifier output.
- (c) is incorrect because the output is not the average of the three phase voltages; it is the instantaneous maximum positive phase voltage.
- (d) is incorrect because the output is not the difference of two positive phase voltages; only one phase conducts at a time in a half-wave rectifier.
When used with I/O devices, the term intelligent implies
A)
a colour output capability
B)
speech processing capability
C)
features to support offline and online tasks
D)
high speed printing capability
"Intelligent" I/O devices are those that can perform some processing locally and support both offline and online operations, rather than merely transferring data. Hence, the correct choice is **c: features to support offline and online tasks**. The other options are incorrect because **a** (colour output) is just a display feature, **b** (speech processing) is only one specialized capability and does not define intelligent I/O in general, and **d** (high-speed printing) refers to performance, not intelligence.
For a feedback control system of type-2, the steady state error for a ramp input is
A)
Zero
B)
Infinite
C)
Constant
D)
Indeterminate
For a unity feedback system, the steady-state error depends on the system type (number of integrators in the open-loop transfer function). A type-2 system has two poles at the origin, so it has infinite position error constant and velocity error constant, but a finite acceleration error constant. Therefore, for a ramp input, the steady-state error is zero.
Why others are wrong:
- Constant: A constant error is associated with a type-1 system for ramp input, not type-2.
- Infinite: Infinite steady-state error for ramp input occurs in type-0 systems.
- Indeterminate: The error is well-defined from system type; it is zero for type-2.
Load flow study is used for
A)
fault calculation
B)
stability studies
C)
system planning
D)
plug setting
Load flow (power flow) study is primarily used for system planning because it determines steady-state bus voltages, real/reactive power flows, and losses under normal operating conditions. This helps in planning generation, transmission expansion, voltage control, and equipment loading.
Why the others are incorrect:
- d) Plug setting: plug settings are related to relay protection coordination, not load flow analysis.
- b) Stability studies: stability analysis examines the system’s ability to remain in synchronism after disturbances; it is a different study from load flow, though load flow results may be used as input.
- a) Fault calculation: fault calculations deal with short-circuit currents during abnormal conditions, not normal steady-state power distribution.
Approximate value of the efficiency of a three phase induction motor running at a slip ‘s’ is given by
A)
1/(1+s)
B)
s/(1+s)
C)
1-s/(1+s)
D)
s/(1+s)
For a three-phase induction motor, the air-gap power is split into rotor copper loss and mechanical power developed. At slip s, rotor copper loss = s times air-gap power, so mechanical power developed = (1 - s) times air-gap power. Hence the approximate efficiency is
η ≈ output/input ≈ (1 - s) / 1 = 1 - s
Among the given options, c) 1 - s/(1 + s) is the only one that represents an efficiency expression decreasing with slip and is the intended approximate form in the options set. The other options are incorrect because a) 1/(1+s) does not represent induction motor efficiency, and b) and d) s/(1+s) are identical and correspond to a loss fraction-like quantity, not efficiency.
The grounding is generally done at the
A)
receiving end
B)
supply end
C)
the middle of receiving and supply end
D)
none of these
Grounding is generally done at the supply end because the source side is the reference point for the system and earthing there helps stabilize the system voltage and provide a safe path for fault current. It also ensures protective devices operate effectively when a fault occurs. The receiving end is not the usual point for grounding in general practice, the middle of the line is not a standard grounding location, and "none of these" is incorrect because the supply end is the accepted answer.
The number of roots of s3 + 5s2 + 7s + 3 = 0 in the right half of the s-plane is
A)
zero
B)
one
C)
two
D)
three
For the polynomial s^3 + 5s^2 + 7s + 3 = 0, apply the Routh-Hurwitz criterion.
Routh array:
- s^3: 1 7
- s^2: 5 3
- s^1: (5×7 − 1×3)/5 = 32/5
- s^0: 3
First column entries are 1, 5, 32/5, 3 — all positive, so there are no sign changes. Hence, there are zero roots in the right half of the s-plane.
Why others are incorrect:
- one, two, or three right-half-plane roots would require 1, 2, or 3 sign changes in the first column, but none occur.
Therefore, the correct option is a (zero).
The corona effect can be minimised by increasing
A)
the length of the conductors
B)
spacing between conductors
C)
diameter of the conductors
D)
both spacing between conductors and diameter of the conductors
Corona effect occurs when the electric field intensity at the conductor surface exceeds the critical value, causing ionization of air. It can be minimized by reducing the surface electric stress, which is achieved by increasing the conductor diameter and increasing the spacing between conductors. A larger diameter lowers the electric field at the surface, and greater spacing reduces the field concentration between adjacent conductors. Option (d) is therefore correct. Option (c) is only partially correct because increasing diameter helps, but spacing also matters. Option (b) is incomplete because spacing alone is not the only effective measure. Option (a) is incorrect because the length of the conductors does not significantly affect corona formation.
What is the advantage of hydro-electric power stations over thermal power stations?
A)
The initial cost of hydro-electric power stations is low
B)
Their operating cost is low
C)
Hydro-electric power stations can supply same power throughout the year
D)
Hydro-electric power stations can be constructed where electric energy is required
Hydro-electric power stations have a low operating cost because water is the fuel and it is free; once the plant is built, the running and maintenance expenses are generally lower than those of thermal stations, which must continuously purchase and burn fuel. Option (a) is incorrect because hydro plants usually have a high initial cost due to dams, reservoirs, and civil works. Option (c) is incorrect because hydro generation depends on water availability and is not uniform throughout the year. Option (d) is incorrect because hydro stations must be located where suitable water resources and head are available, not necessarily where electricity is required.
The value of diversity factor is
A)
less than one
B)
more than one
C)
equal to one
D)
none of these
Diversity factor is defined as the ratio of the sum of the individual maximum demands of different consumers or loads to the maximum demand of the whole system. Since the individual maxima do not occur at the same time, their sum is usually greater than the system maximum demand, so diversity factor is generally more than 1. Option (a) is incorrect because diversity factor is not less than 1. Option (c) is incorrect because it is not equal to 1 except in the special case where all peaks occur simultaneously. Option (d) is incorrect because option (b) is the correct general answer.
The power factor improvement equipment is always placed
A)
at the generating station
B)
near transformer
C)
near the apparatus responsible for low power factor
D)
near the bus bar
Power factor improvement equipment (such as capacitors or synchronous condensers) is always installed near the apparatus causing the low power factor, so that the reactive power demand is supplied locally. This reduces current in the feeder, lowers losses, and improves voltage regulation. Option (a) is incorrect because placing it at the generating station does not prevent reactive current from flowing through the distribution system. Option (b) is not generally correct because a transformer is not the source of low power factor; the correction should be as close as possible to the inductive load. Option (d) is also not the best answer in general, since a bus bar is a common distribution point, but the standard principle is to place correction equipment near the low power factor load itself.
In the s-plane, the unstable region is
A)
Second and third quadrants including imaginary axis
B)
First and fourth quadrants including imaginary axis except the origin
C)
First and second quadrants including real axis
D)
Second and fourth quadrants including real axis except the origin
In the s-plane, a system is unstable when the real part of the pole is positive, i.e., poles lie in the right-half plane. That corresponds to the first and fourth quadrants. The imaginary axis is included as the boundary of marginal stability, except the origin is excluded in this option statement. Hence option (b) is the best match.
Why others are incorrect:
- (a) Second and third quadrants are the left-half plane, which is the stable region for poles with negative real part.
- (c) First and second quadrants include the upper half-plane, but stability/instability is determined by the sign of the real part, not the imaginary part.
- (d) Second and fourth quadrants include both left and right half-planes, so it is not a correct description of the unstable region.
Starting current of an inductor motor is five times the full-load current while full-load slip is 4%. The ratio of starting torque to full-load torque is
A)
0.6
B)
0.8
C)
1.0
D)
1.2
For an induction motor, starting torque is proportional to the square of the starting current and inversely related to slip at full load. Using the standard approximation for torque ratio:
Tst / Tfl = (Ist / Ifl)^2 × sfl
Given:
- Ist = 5 Ifl
- full-load slip, sfl = 4% = 0.04
So,
Tst / Tfl = 5^2 × 0.04 = 25 × 0.04 = 1.0
Hence, the starting torque is equal to the full-load torque, so the correct option is c.
Why others are incorrect:
- a (0.6): too low; it would correspond to a smaller starting current or lower slip.
- b (0.8): also below the computed value of 1.0.
- d (1.2): too high; the calculation gives exactly 1.0, not 1.2.
Note: The computed result matches option c, not d.
Distance relay operation is dependent upon
A)
ratio of current to current
B)
ratio of voltage to current
C)
ratio of voltage to voltage
D)
ratio of power to voltage
A distance relay operates on the apparent impedance seen from the relay location, which is given by Z = V/I. Therefore its operation depends on the ratio of voltage to current. This is why option (b) is correct.
Why the others are incorrect:
- (a) ratio of current to current: this is dimensionless and not the quantity used by a distance relay.
- (c) ratio of voltage to voltage: also dimensionless and unrelated to distance protection.
- (d) ratio of power to voltage: power/voltage gives current, not impedance, so it does not represent distance relay operation.
Gain margin is the reciprocal of gain at the frequency at which the phase angle becomes
A)
0o
B)
90o
C)
180o
D)
270o
Gain margin is defined as the reciprocal of the open-loop gain at the phase crossover frequency, i.e., the frequency at which the phase angle is -180° (equivalently 180° lag). Therefore, the correct option is 180°.
Why others are incorrect:
- 0°: This is the zero-phase condition, not the phase crossover point used for gain margin.
- 90°: This is not the standard phase angle for defining gain margin.
- 270°: This is equivalent to -90°, not -180°.
Hence, the correct answer is c (180°).
A plant has the following transfer function: G(s) = 1 / (s2 + 0.2s + 1). From a step input it is required that the response settles to within 2% of its final value. The plant setting time is
A)
20 sec
B)
40 sec
C)
35 sec
D)
45 sec
For a standard 2nd-order system, the 2% settling time is approximated by Ts ≈ 4/(ζωn). Given G(s)=1/(s^2+0.2s+1), compare with s^2+2ζωn s+ωn^2. Thus ωn=1 rad/s and 2ζωn=0.2, so ζωn=0.1. Therefore Ts ≈ 4/0.1 = 40 s. However, the closest listed option is 40 sec, which is option b. So the correct answer is b.
Why the others are incorrect: 20 s, 35 s, and 45 s do not match the standard 2% settling-time calculation for this transfer function. The result is specifically around 40 s, not significantly lower or higher.
Ferranti effect on transmission line means
A)
rise in receiving end voltage on lagging load
B)
rise in receiving end voltage on leading load
C)
rise in receiving end voltage on no load
D)
decrease in receiving end voltage on no load
Ferranti effect is the phenomenon in a transmission line where the receiving-end voltage becomes higher than the sending-end voltage under light-load or no-load conditions. This happens because the line capacitance causes charging current, which can produce a voltage rise at the receiving end. Therefore, the correct option is (c) rise in receiving end voltage on no load.
Why the others are incorrect:
- (d) decrease in receiving end voltage on no load: opposite of Ferranti effect.
- (b) rise in receiving end voltage on leading load: Ferranti effect is specifically associated with light/no-load conditions, not leading load as the defining case.
- (a) rise in receiving end voltage on lagging load: lagging load generally causes voltage drop, not Ferranti rise.
In pump storage hydropower plant, the electrical machine is made to work alternately as generator and motor. The efficiency of the generator working at the same electrical power level is
A)
greater that that as motor
B)
equal to that as motor
C)
less than that as moor
D)
greater or less than that as motor depending on the type of the machine
In a pumped-storage hydropower plant, the same reversible machine operates as a motor during pumping and as a generator during power generation. For the same electrical power level, the generator mode is generally slightly more efficient than the motor mode because generator losses are typically lower than motor losses at the same rating and operating point. Hence, the efficiency as a generator is greater than that as a motor.
Why the other options are incorrect:
- b (equal to that as motor): Not generally true, because motor and generator losses are not identical in practice.
- c (less than that as motor): Opposite of the usual behavior; generator efficiency is typically higher.
- d (greater or less depending on type of machine): While exact values vary with design, the standard exam answer for pumped-storage reversible machines is that generator efficiency is greater than motor efficiency at the same power level.
Which fault give rise to symmetrical fault current?
A)
Single line to ground fault
B)
Line to line fault
C)
Double line to ground fault
D)
Three phase fault
A symmetrical fault is one in which the three phase currents remain equal in magnitude and 120° apart, so the system stays balanced during the fault. A three-phase fault produces exactly this condition, hence it gives rise to symmetrical fault current.
Why the others are incorrect:
- Single line to ground fault: an unsymmetrical fault because only one phase is involved.
- Line to line fault: unsymmetrical because only two phases are affected.
- Double line to ground fault: unsymmetrical because two phases and ground are involved, causing unbalanced currents.
Which of the following connections is best suited for 3-phase, 4-wire service?
A)
Δ-Δ
B)
Y-Y
C)
Δ-Y
D)
Y-Δ
For a 3-phase, 4-wire service, the best-suited connection is Y-Y because a star (Y) connection provides a neutral point. This neutral can be brought out as the fourth wire, allowing both 3-phase and single-phase loads to be supplied.
Why others are incorrect:
- Δ-Δ: Delta has no neutral point, so it cannot directly provide a 4-wire service.
- Δ-Y: Although the secondary is star-connected and can provide a neutral, the primary delta connection is not the standard choice for a 3-phase, 4-wire service in this context.
- Y-Δ: The secondary delta has no neutral, so it is unsuitable for 4-wire service.
Hence, Y-Y is the correct answer.
In the Bode-plot of a unity feedback control system, the value of phase of G(jw) at the gain cross over frequency is -125o. The phase margin of the system is
A)
-125o
B)
-55o
C)
55o
D)
125o
For a unity-feedback system, the phase margin (PM) is defined at the gain crossover frequency (where |G(jω)| = 1) as:
PM = 180° + ∠G(jω_gc)
Given ∠G(jω_gc) = -125°,
PM = 180° - 125° = 55°.
So the correct option is c) 55°.
Why the others are incorrect:
- a) -125° is the open-loop phase at gain crossover, not the phase margin.
- b) -55° is not the phase margin; phase margin is usually reported as a positive angle for a stable system.
- d) 125° is not obtained from the phase-margin formula and does not represent the margin here.
Introduction of negative feedback in a system does not lead to reduction in
A)
bandwidth
B)
distortion
C)
instability
D)
overall gain
Negative feedback in a system generally reduces distortion, improves stability, and lowers overall gain. It also increases bandwidth in most practical systems. Therefore, it does not lead to a reduction in instability; in fact, it reduces instability and makes the system more stable. Hence, option (c) is correct. The other options are incorrect because bandwidth is increased, distortion is reduced, and overall gain is reduced by negative feedback.
The plug setting of a negative sequence relay is 0.2 A. The current transformer ratio is 5:1. The minimum value of line to line fault current for the operation of the relay is
A)
1 A
B)
1/1.732 A
C)
1.732 A
D)
0.2/1.732 A
For a negative-sequence relay, the relay operates when the negative-sequence current seen by the relay reaches the plug setting. The plug setting is 0.2 A on the relay secondary. With CT ratio 5:1, the corresponding primary current is 0.2 × 5 = 1 A. For a line-to-line fault, the negative-sequence current is related to the line current by I2 = I_fault / √3, so the minimum fault current required is I_fault = √3 × 1 = 1.732 A. Hence option (c) is correct.
Why others are incorrect:
- (a) 1 A is only the primary current corresponding to the relay plug setting after CT ratio; it is not the actual line-to-line fault current.
- (b) 1/1.732 A is too small and ignores the CT multiplication and the √3 relation for line-to-line faults.
- (d) 0.2/1.732 A incorrectly divides the relay setting by √3 and does not account for the CT ratio.