← MES Jr. Grade
Past Questions
A simplest method of increasing the voltage of a d.c generator is
A)
to decrease the air gap flux density
B)
to increase the speed of rotation
C)
to decrease the speed of rotation
D)
to increase the length of armature
The generated emf of a d.c. generator is given by E = (PΦZN)/(60A), so the terminal voltage increases directly with speed N. Therefore, the simplest way to increase the voltage is to increase the speed of rotation. Option (c) is wrong because decreasing speed reduces emf. Option (a) is wrong because decreasing air-gap flux density reduces flux Φ and hence lowers voltage. Option (d) is not the simplest/direct method; increasing armature length is not the standard basic control for raising generator voltage compared with increasing speed.
In the derivative error compensation, damping
A)
decrease and settling time decreases
B)
increase and settling time increases
C)
increase and settling time decreases
D)
decrease and settling time increases
Derivative error compensation (derivative control) adds a term proportional to the rate of change of error. This increases damping in the system, which reduces oscillations and improves transient response. As a result, the settling time generally decreases. Therefore, the correct statement is: damping increases and settling time decreases.
Why others are incorrect:
- a: says damping decreases, which is opposite of derivative action.
- b: says both damping and settling time increase; derivative action does not increase settling time.
- d: says damping decreases and settling time increases; both are opposite to the effect of derivative compensation.
The applied voltage of a certain transformer is increase by 50% while the frequency is reduced to 50%. The maximum core flux density will
A)
become three times
B)
become 1.5 times
C)
become half
D)
remain the same
For a transformer, the maximum core flux density is proportional to V/f, since E = 4.44 f N \u03c6m and for a given turns N, \u03c6m \u221d V/f. Here the applied voltage increases by 50%, so V becomes 1.5V, and the frequency is reduced to 50%, so f becomes 0.5f. Therefore, the flux density changes by a factor of (1.5)/(0.5) = 3. Hence the maximum core flux density becomes three times.
Why others are wrong:
- b (become 1.5 times): ignores the effect of frequency reduction.
- c (become half): opposite of the actual V/f change.
- d (remain the same): would be true only if V/f stayed constant, which it does not here.
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.
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.
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.
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.
For a single phase capacitor start induction motor which of the following statements is valid?
A)
the capacitor is used for power factor improvement
B)
the direction of rotation can be changed by reversing the main winding terminals
C)
the direction of rotation cannot be changed
D)
the direction of rotation can be changed by interchanging the supply terminals
In a single-phase capacitor-start induction motor, the direction of rotation is reversed by reversing the connections of the starting (auxiliary) winding with respect to the main winding. Among the given choices, option **b** is the intended valid statement because changing the relative polarity of the windings changes the phase relationship and hence the direction of the rotating field at start.
Why the others are incorrect:
- **a**: The capacitor in a capacitor-start motor is not primarily for power factor improvement; it is used to create a phase shift between the main and auxiliary windings to produce starting torque.
- **c**: Incorrect because the direction of rotation can be changed.
- **d**: Incorrect because interchanging the supply terminals reverses both windings together, so the relative phase between them remains the same and the direction does not change.
In the optimum generator scheduling of different power plants, the minimum fuel cost is obtained when
A)
only the incremental fuel cost of each plant is the same
B)
the penalty factor of each plant is the same
C)
the ratio of the incremental fuel cost to the penalty factor of each plant is the same
D)
the incremental fuel cost of each plant multiplied by it penalty factor is the same
In optimum economic dispatch with transmission losses, the minimum fuel cost condition is that the incremental cost of each plant adjusted for losses must be equal. This is written as the equality of the incremental cost divided by the penalty factor (or equivalently, incremental cost times the loss coefficient adjustment depending on notation). Hence, the correct condition is that the ratio of the incremental fuel cost to the penalty factor of each plant is the same.
Why others are incorrect:
- (a) Equal incremental fuel cost alone is valid only for ideal dispatch without transmission losses.
- (b) Equal penalty factor alone does not ensure minimum total fuel cost.
- (d) Multiplying incremental fuel cost by penalty factor is not the dispatch condition; the correct relation involves division by the penalty factor, not multiplication.
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).
A synchronous generator connected to an infinite bus delivers power at a lag pf. If its excitation is increased
A)
the terminal voltage increases
B)
voltage angle ä increases
C)
current delivered increases
D)
all of these
For a synchronous generator connected to an infinite bus, the terminal voltage is fixed by the bus. If excitation is increased, the internal emf rises, so the generator supplies more reactive power and the stator current changes. At a given real power output, increasing excitation from a lagging power factor condition reduces the lagging reactive component and can increase the current magnitude delivered toward unity power factor. Hence option (c) is correct.
Why others are incorrect:
- (a) Terminal voltage increases: incorrect, because an infinite bus holds terminal voltage constant.
- (b) Voltage angle δ increases: not necessarily the intended single correct statement here; with fixed real power and higher excitation, the main definite effect is on reactive power/current, while δ is not the direct guaranteed change in this context.
- (d) All of these: incorrect because (a) is definitely false, so all cannot be true.
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.
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.
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.
What type of bridge circuit is used to measure inductance and capacitance?
A)
Wheatstone bridge
B)
D-C bridge
C)
A-C bridge
D)
A-C, D-C bridge
AC bridges are used to measure inductance and capacitance because these quantities are associated with alternating-current impedance and phase relationships. Examples include Maxwell, Hay, Schering, and Anderson bridges. A DC bridge such as the Wheatstone bridge is used for resistance measurement only, not for inductance or capacitance. Therefore, the correct option is AC bridge. The option 'A-C, D-C bridge' is incorrect because DC bridges do not directly measure inductance or capacitance.
In the semiconductor, the movement of holes is due to
A)
movement of holes in the conduction band
B)
movement of electrons in conduction band
C)
movement of electrons in valence band
D)
movement of holes in valence band
In a semiconductor, a hole is the absence of an electron in the valence band. When neighboring valence electrons move to fill this vacancy, the vacancy appears to move in the opposite direction. Thus, hole conduction is effectively due to the movement of holes in the valence band. Option d is correct.
Why the others are incorrect:
- a) Holes do not exist in the conduction band; conduction band transport is mainly by electrons.
- b) Movement of electrons in the conduction band explains electron conduction, not hole conduction.
- c) Electrons in the valence band are generally bound; their movement does not represent hole conduction directly. Hole motion is the effective motion of vacancies in the valence band.
A series RLC circuit consisting of R=10 ohms, XL = 20 ohms and XC = 20 ohms is connected across an AC supply of 100 V (rms). The magnitude and phase angle (with reference to supply voltage) of the voltage across the inductive coil are respectively
A)
100 V, 90o
B)
100 V, -90o
C)
200 V, 90o
D)
200 V, -90o
In a series RLC circuit, the net reactance is X = XL - XC = 20 - 20 = 0, so the circuit is at resonance. Hence the total impedance is purely resistive: Z = R = 10 ohms. Therefore the circuit current is I = V/Z = 100/10 = 10 A.
The voltage across the inductive coil is VL = I XL = 10 × 20 = 200 V. In a pure inductor, voltage leads current by 90°, and since the supply voltage is in phase with the current at resonance, the inductor voltage is 90° ahead of the supply voltage.
So the correct answer is 200 V, 90°.
Why others are wrong:
- b) 100 V, -90°: wrong magnitude and wrong phase.
- c) 200 V, -90°: correct magnitude but wrong phase sign; inductor voltage leads, not lags.
- d) 200 V, -90°: same issue as c, and not the correct phase relation.
A conductor of length 1 m moves at right angles to a magnetic field of flux density 1 Wb/m2 with a velocity of 20 m/s. The induced emf in conductor will be
A)
100 V
B)
20 V
C)
2 V
D)
40 V
For a conductor moving perpendicular to a magnetic field, the induced emf is given by e = B l v. Here B = 1 Wb/m^2 (1 T), l = 1 m, and v = 20 m/s. So, e = 1 × 1 × 20 = 20 V. Therefore, option (b) is correct. Option (a) 100 V is too high, option (d) 40 V is double the correct value, and option (c) 2 V is too low.
Which one following statements does not pertain to the equation div B = 0?
A)
magnetic field is perpendicular to the electric field
B)
there are no sinks and sources for magnetic fields
C)
single magnetic pole cannot exist
D)
B is solenoidal
The equation div B = 0 is Gauss’s law for magnetism. It means the magnetic field has zero divergence, so magnetic field lines do not begin or end anywhere. Hence, B is solenoidal (d), there are no sinks or sources for magnetic fields (b), and a single magnetic pole cannot exist (c) because isolated magnetic monopoles are not observed in classical electromagnetism. Option (a) does not follow from div B = 0: the statement that magnetic field is perpendicular to the electric field is unrelated to the divergence of B. That is not a consequence of Gauss’s law for magnetism.
The impurity atoms in semiconductors
A)
reduce the energy gap
B)
increase the kinetic energy of valence electrons
C)
eject more charge chariers
D)
all of these
Impurity atoms (dopants) in semiconductors create donor or acceptor energy levels very close to the conduction band or valence band. This effectively reduces the energy needed to generate mobile charge carriers, so they increase the number of free electrons or holes and make conduction easier. Hence, they can be said to reduce the effective energy gap and eject more charge carriers. Therefore, all the given statements are intended to be correct.
Why the others are not chosen separately:
- (b) "increase the kinetic energy of valence electrons" is not the main effect of doping; doping does not primarily increase electron kinetic energy.
- (a) "reduce the energy gap" is true in the practical sense of introducing impurity levels that lower the activation energy for conduction.
- (c) "eject more charge carriers" is also true because dopants increase the concentration of free electrons or holes.
So the best answer is (d) all of these.
Three equal resistance of 5W are connected in delta. What is the resistance in one of the arms of the equivalent star circuit?
A)
5 ohm
B)
1.67 ohm
C)
10 ohm
D)
15 ohm
For a balanced delta-to-star conversion, each star arm is given by RY = (RΔ1 × RΔ2) / (RΔ1 + RΔ2 + RΔ3). Here all three delta resistances are equal to 5 Ω, so each star arm = (5 × 5) / (5 + 5 + 5) = 25/15 = 5/3 ≈ 1.67 Ω. Hence option b is correct.
Why the others are incorrect:
- a) 5 Ω: this is just the original delta resistance, not the equivalent star arm.
- c) 10 Ω: too large; star arms in a delta-to-star conversion are smaller than the delta resistances for equal values.
- d) 15 Ω: this is the sum of the three delta resistances, not the star-arm value.
Which of the following devices can not be used to measure pressure?
A)
LVDT
B)
Strain gauge
C)
Pyrometer
D)
Pirani gauge
Pyrometer is the correct answer because it is an instrument used to measure high temperature, not pressure. Therefore, it cannot be used for pressure measurement.
Why the others are incorrect:
- Strain gauge: Can be used in pressure transducers; pressure causes strain in an elastic element, which the gauge measures.
- LVDT: Can be used in pressure measurement indirectly in pressure transducers by sensing displacement of a diaphragm or piston caused by pressure.
- Pirani gauge: Is a vacuum pressure gauge used to measure low pressures by thermal conductivity changes of a gas.
Ceramic loses its insulating properties above a temperature called
A)
melting point
B)
solidification point
C)
curie point
D)
flash point
Ceramic materials used as electrical insulators can lose their insulating behavior above the Curie point, because at this temperature certain ferroelectric ceramics undergo a phase change and their dielectric properties drop sharply. Hence, the correct answer is Curie point. Solidification point is the temperature at which a liquid turns into a solid, which is unrelated to insulation. Flash point is the temperature at which a material gives off enough vapor to ignite, also unrelated. Melting point is the temperature at which a solid becomes liquid; while ceramics may fail at high temperatures, the specific loss of insulating properties referred to here is associated with the Curie point.
The power in a 3-phase, 3-wire load is measured using two 100 W full scale wattmeters W1 and W2. W1 is of a accuracy class +/- 1% and reads 100 W. W2 is of a accuracy class 0.5% and reads -50 W. Then the uncertainty in the computation to total power is
A)
+/- 1.5%
B)
+/- 0.5%
C)
+/- 4%
D)
+/- 3%
For a 3-phase, 3-wire load measured by the two-wattmeter method, total power is P = W1 + W2. Here, P = 100 + (-50) = 50 W.
Now compute the possible absolute errors from the wattmeter accuracy classes:
- W1 is 100 W full-scale, class ±1% ⇒ absolute error = ±1 W.
- W2 is 100 W full-scale, class ±0.5% ⇒ absolute error = ±0.5 W.
For the sum, the worst-case absolute uncertainty adds:
ΔP = ±(1 + 0.5) = ±1.5 W.
Percentage uncertainty in total power = (1.5 / 50) × 100 = ±3%.
So the correct option is d) ±3%.
Why the others are incorrect:
- a) ±1.5%: This would correspond to an absolute uncertainty of only ±0.75 W on a 50 W result, which is too small.
- b) ±0.5%: This is far too low and ignores the combined error of both wattmeters.
- c) ±4%: This overestimates the uncertainty; the correct worst-case percentage is 3%.
In practice, Earth is chosen as a place of zero electric potential because it
A)
is non-conducting
B)
is easily available
C)
keep loosing and gaining electric charge every day
D)
has almost constant potential
Earth is taken as zero potential because it is so large that it can absorb or supply charge without any appreciable change in its potential. Hence its potential remains practically constant, making it a convenient reference level. Option (a) is wrong because Earth is not non-conducting; it is a conductor. Option (b) is not the physical reason, though Earth is indeed easily available. Option (c) is incorrect because the statement about daily losing and gaining charge is not the basis for choosing Earth as zero potential.