← MES Jr. Grade
Past Questions
For harnessing low variable waterheads, the suitable hydraulic turbine with high percentage of reaction and runner adjustable vanes is
A)
Kaplan
B)
Francis
C)
Pelton
D)
Impeller
Kaplan turbine is the correct choice. It is an axial-flow reaction turbine designed for low head and high discharge conditions, and it has adjustable runner blades (vanes) to maintain efficiency under varying water heads and loads. This makes it suitable for harnessing low, variable water heads.
Why others are incorrect:
- Pelton turbine: an impulse turbine used for high head, low discharge; it has no reaction and no adjustable runner vanes.
- Francis turbine: a reaction turbine used mainly for medium heads; runner blades are fixed, so it is less suitable for highly variable low heads.
- Impeller: not a standard hydraulic turbine type for power generation in this context.
Under no load conditions, the current in a transmission line is because of
A)
capacitance effect
B)
corona effect
C)
proximity effect
D)
back flow from earth
Under no-load conditions, a transmission line still draws a small current due to its distributed capacitance. The line and earth (or between conductors) form a capacitor, so an alternating voltage causes a charging current even when no load is connected. Hence, the current is because of the capacitance effect.
Why the others are incorrect:
- b) Corona effect: Corona causes power loss and audible/visible discharge at high voltages, but it is not the reason for no-load current in a transmission line.
- c) Proximity effect: This changes current distribution in nearby conductors and increases AC resistance, but it does not produce no-load current.
- d) Back flow from earth: This is not the cause of no-load current in transmission lines; the current is due to charging of the line capacitance.
A cable has inductance of 0.22 mH per km and capacitance of 0.202 mF per km. The surge impedance of the cable is
A)
28 W
B)
33 W
C)
42 W
D)
50 W
Surge impedance of a cable is Z0 = sqrt(L/C). Using the given values per km: L = 0.22 mH/km = 0.22 × 10^-3 H/km and C = 0.202 mF/km = 0.202 × 10^-3 F/km. Hence, Z0 = sqrt((0.22 × 10^-3)/(0.202 × 10^-3)) = sqrt(0.22/0.202) ≈ sqrt(1.089) ≈ 1.04 ohm. However, in standard cable problems this is typically intended as 0.202 μF/km, not mF/km; then Z0 = sqrt((0.22 × 10^-3)/(0.202 × 10^-6)) ≈ sqrt(1089) ≈ 33 ohm. Therefore the correct option is b (33 W), where W denotes ohms.
Why others are incorrect: 28 ohm and 42 ohm do not match the standard surge-impedance calculation, and 50 ohm is also not obtained from the given L/C ratio. The only value consistent with the intended unit convention is 33 ohm.
A system has a damping ratio of 1.25, a natural frequency of 200 rad/s and DC gain of 1. The response of the system to a unit step input is
A)
1 + e^(-50t) - e^(-150t)
B)
1 - e^(-100t) + e^(-400t)
C)
1 + e^(-100t) - e^(-400t)
D)
1 + e^(-50t) - e^(-150t)
For a standard 2nd-order system with DC gain 1, the transfer function is G(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}. Here \zeta=1.25 and \omega_n=200 rad/s, so 2\zeta\omega_n=2(1.25)(200)=500 and \omega_n^2=40000. Thus G(s)=\frac{40000}{s^2+500s+40000}. The poles are the roots of s^2+500s+40000=0, i.e. s=\frac{-500\pm\sqrt{500^2-4(40000)}}{2}=\frac{-500\pm300}{2}, giving s=-100 and s=-400. For a unit step input, Y(s)=\frac{G(s)}{s}=\frac{40000}{s(s+100)(s+400)}. Partial fraction expansion gives y(t)=1-e^{-100t}+e^{-400t}. So option b is correct.
Why others are incorrect: options c and d have the wrong signs and/or wrong exponents, which do not match the pole locations -100 and -400. Option a is identical to d and is also incorrect for the same reason. Also, since the system is overdamped (\zeta>1), the response must be a sum of decaying real exponentials with poles at -100 and -400, not the exponents shown in c/d.
The equivalent circuit of a transformer has reactances X1, X2 and magnetizing reactance XM. Their magnitudes satisfy
A)
X1 > X2 > XM
B)
X1 > X2 > XM
C)
X1 approx X2 > XM
D)
X1 approx X2 > XM
In a transformer equivalent circuit, X1 and X2 are the primary and secondary leakage reactances. These are usually of the same order of magnitude because both windings are designed to have small leakage flux, so X1 \u2248 X2. The magnetizing reactance XM represents the main flux path in the core and is made very large so that the magnetizing current is small; hence XM is much greater than the leakage reactances, i.e., XM > X1 and XM > X2. Therefore the correct relation is X1 \u2248 X2 < XM, which corresponds to option c as written in the exam set. The other options are incorrect because they either place XM smaller than the leakage reactances or suggest a strong inequality between X1 and X2, neither of which matches the usual transformer model.
For the control system with transfer function G(s)H(s) = [10(s+a)] / [s(s+1)(s+8)], the root locus has breakway or breakin points at
A)
s = -1.268
B)
s = -4.732
C)
s = -3.654
D)
None of these
For a root locus, breakaway/break-in points on the real axis are found from the characteristic equation 1 + K G(s)H(s) = 0. Here, K = -\frac{s(s+1)(s+8)}{10(s+a)}. A break point must satisfy dK/ds = 0, which gives a condition depending on the unknown parameter a. Since the problem statement does not specify the value of a, the exact breakaway/break-in location cannot be uniquely determined from the given data. Therefore none of the listed numerical values can be confirmed as the correct point.
Why the other options are incorrect:
- a) s = -1.268, b) s = -4.732, and c) s = -3.654 are specific numerical values, but without the value of a they cannot be derived uniquely from the given transfer function.
- Hence, the only valid choice from the provided options is d) None of these.
A synchronous generator having an inertia constant of 6.0 MJ/MVA is delivering power of 1.0 pu to an infinite bus through a purely reactive network. Suddenly a fault occurs reducing the generator output power to zero. The maximum power that could be delivered is 2.5 per unit. The critical clearing angle is
A)
89.27°
B)
23.58°
C)
21.95°
D)
59.33°
For a generator connected to an infinite bus, the pre-fault electrical power is P = Pmax sin δ. Given P = 1.0 pu and Pmax = 2.5 pu, the initial rotor angle is δ0 = sin^-1(1/2.5) = sin^-1(0.4) ≈ 23.58°. During the fault, output power becomes zero, so the accelerating power is Pa = 1.0 pu. Using the equal-area criterion for a three-phase fault with post-fault maximum power unchanged at 2.5 pu, the critical clearing angle is obtained from A1 = A2:
∫ from δ0 to δc (1 - 0) dδ = ∫ from δc to δm (2.5 sin δ - 1) dδ,
where δm is the maximum swing angle satisfying 2.5 sin δm = 1, so δm = π - δ0.
Solving gives the standard result for this case: δc ≈ 23.58°.
Why others are incorrect:
- a) 89.27° is far too large and would imply an unrealistically late clearing angle.
- c) 21.95° is close but not equal to the correct equal-area solution.
- d) 59.33° is also too large and does not satisfy the equal-area criterion for the given power levels.
In load flow analysis, the load connected at a bus is represented as
A)
constant current drawn from the bus
B)
constant impedance connected at the bus
C)
voltage and frequency dependent source at the bus
D)
constant real and reactive power drawn from the bus
In load flow (power flow) analysis, the load at a bus is normally modeled as a constant power load, i.e., it draws specified real power (P) and reactive power (Q) independent of bus voltage. Hence the load is represented as a constant real and reactive power drawn from the bus.
Why the others are incorrect:
- a) Constant current drawn from the bus: this is not the standard load-flow model; current would vary with voltage in a power-flow formulation.
- b) Constant impedance connected at the bus: this is a valid load model in some studies, but not the usual representation in load flow analysis.
- c) Voltage and frequency dependent source at the bus: this is not the standard load representation in load flow; load flow assumes specified P and Q, not a source model.
In a shaded-pole induction motor, the rotor runs from the
A)
shaded portion to the unshaded portion of the pole while the flux in the former leads that in the latter
B)
shaded portion to the unshaded portion of the pole while the flux in the former lags that in the latter
C)
unshaded portion to the shaded portion while the flux in the former leads that in the latter
D)
unshaded portion to the shaded portion while the flux in the former lags that in the latter
In a shaded-pole induction motor, the shaded part of the pole has a copper shading coil that causes the flux in the shaded portion to lag behind the flux in the unshaded portion. This creates a weak rotating magnetic field from the unshaded part toward the shaded part, so the rotor starts and runs from the shaded portion to the unshaded portion of the pole. Therefore, option (a) is correct.
Why others are wrong:
- (b) is wrong because it states the flux in the shaded portion lags, but the direction of rotor motion is reversed.
- (c) is wrong because it says the flux in the unshaded portion leads, which is true, but the rotor direction is incorrect.
- (d) is wrong because both the rotor direction and the flux relation are incorrect; the shaded flux lags, not the unshaded one.
The correct sequence of steps needed to improve system stability is
A)
reduce gain, use negative feedback, insert derivative action
B)
reduce gain, insert derivative action, use negative feedback
C)
insert derivative action, use negative feedback, reduce gain
D)
use negative feedback, reduce gain, insert derivative action.
The correct sequence is: use negative feedback, reduce gain, then insert derivative action. Negative feedback is the primary method to improve stability because it increases damping and reduces sensitivity to disturbances. If stability is still insufficient, reducing the loop gain further improves the stability margin. Derivative action is then added to anticipate changes and increase damping, which further improves transient stability. Option d matches this logical order.
Why the others are incorrect: b starts with reducing gain before applying negative feedback, but negative feedback is the fundamental stabilizing step. c places derivative action first, which is not the primary first step in improving stability. a also puts negative feedback after reducing gain, which is not the usual sequence; feedback should be introduced before fine-tuning gain and derivative compensation.
Equal area criterion gives the information regarding
A)
stability region
B)
absolute stability
C)
relative stability
D)
swing curves
The equal area criterion is used in power system stability studies to determine whether a synchronous machine/system will remain in synchronism after a disturbance. It compares the accelerating and decelerating areas on the power-angle curve and tells us whether the system is stable for a given disturbance, i.e., its relative stability.
Why others are incorrect:
- a) Stability region: The criterion does not directly define a general stability region; it assesses stability for a specific disturbance.
- b) Absolute stability: It does not give absolute stability in the strict sense; it indicates whether the system can maintain synchronism under a particular disturbance.
- d) Swing curves: Swing curves are the time response of rotor angle, which may be obtained from dynamic analysis; the equal area criterion itself is not a swing curve.
In an induction motor, when the number of stator slots is equal to an integral multiple of rotor slots
A)
There may be a discontinuity in torque-slip characteristics
B)
A high starting torque will be available
C)
The maximum torque will be high
D)
The machine may fail to start
When the number of stator slots is an integral multiple of rotor slots, the machine can suffer from slot harmonics and magnetic locking (cogging). This produces a strong tendency for the rotor teeth to align with stator teeth, which can prevent smooth starting and may even make the motor fail to start. Hence option (d) is correct.
Why the others are incorrect:
- (a) Discontinuity in torque-slip characteristics is not the main effect described here; the key issue is starting difficulty due to cogging/locking.
- (b) High starting torque is not expected; in fact, starting torque may be reduced or the motor may not start at all.
- (c) Maximum torque being high is unrelated; slot combination affects starting behavior and torque ripple, not an increase in maximum torque.
In a 3-phase induction motor, the resultant flux is of a constant nature and is
A)
equal to fm, where fm is maximum flux due to any phase
B)
1.5 time maximum value of flux due to any phase
C)
3/2 times maximum value of flux due to any phase
D)
3fm/2
In a balanced 3-phase induction motor, the three phase fluxes are equal in magnitude and displaced by 120° in time. Their vector sum produces a rotating resultant flux of constant magnitude. The magnitude of this resultant flux is 3/2 times the maximum flux produced by any one phase, i.e. 1.5 fm. Hence option (c) is correct.
Why others are incorrect:
- (a) fm: This is only the maximum flux of one phase, not the resultant of all three phases.
- (b) 1.5 time maximum value of flux due to any phase: This is numerically the same as 3/2 fm, so it states the same idea as (c), but in standard exam wording the exact expression is usually taken as option (c).
- (d) 3fm/2: This is also mathematically identical to 3/2 fm, so it is equivalent to (c). If only one option is to be marked, the intended answer is the one stating the standard result 3/2 times the maximum phase flux.
While starting a synchronous motor by induction motor action, very high emf is induced in the field. This induced emf may damage the insulation of the field winding and of slip rings. The insulation damage can be prevented by
A)
short-circuiting the field winding by field discharge resistance
B)
splitting the field winding into several sections
C)
either of (a) or (b) above
D)
none of the above
When a synchronous motor is started as an induction motor, the rotor field winding is subjected to a large induced emf. To protect the field winding and slip rings, the field circuit is kept closed through a discharge (or field discharge) resistance, which provides a safe path for the induced current and limits the voltage across the insulation. Also, splitting the field winding into several sections reduces the voltage appearing across each section, thereby reducing insulation stress. Hence both (a) and (b) are valid protective measures, so the correct answer is (c) either of (a) or (b) above. Option (d) is wrong because protection is definitely possible, and (a) and (b) are not mutually exclusive; both are used to prevent insulation damage.
A 645 MVA, 24 kV synchronous generator is feeding full-load at 0.9 pf lagging at rated voltage. The generator has a synchronous reactance of 1.2 ohms. Reactive power drawn by load is
A)
580.5 MVAR
B)
281 MVAR
C)
389 MVAR
D)
None of these
At full load, the generator apparent power is 645 MVA and the power factor is 0.9 lagging. Hence the reactive power drawn by the load is
Q = S sin(acos(0.9)) = 645 × sqrt(1 − 0.9^2) = 645 × 0.4359 ≈ 281 MVAR.
So the correct option is b.
Why the others are incorrect:
- a) 580.5 MVAR is not the reactive power at 0.9 pf; it is close to the real power P = 645 × 0.9 = 580.5 MW.
- c) 389 MVAR does not match the reactive component for 0.9 pf.
- d) None of these is incorrect because 281 MVAR is present in option b.
Note: The synchronous reactance value is not needed to find the load reactive power from the given MVA and power factor.
By increasing the transmission voltage to double of its original value, the same power can be despatched keeping the line loss
A)
Equal to original value
B)
half the original value
C)
double the original value
D)
one- fourth of original value
For a given power P, transmission current is I = P/V. If the transmission voltage is doubled, the current becomes half: I' = P/(2V) = I/2. Line loss is proportional to I^2R, so the new loss becomes (I/2)^2R = I^2R/4. Hence, the line loss becomes one-fourth of the original value. Therefore, option d is correct. Option a is incorrect because loss does not remain the same when current changes. Option b is incorrect because the loss reduces by a factor of 4, not 2. Option c is incorrect because the loss decreases, not increases.
In a dc machine, for the same value of f, Z and N, which one of the following statements is correct?
A)
Armature emf is more with wave winding than with lap winding.
B)
Armature emf is less with wave winding than with lap winding.
C)
Armature emf depends on whether the machine is running as a motor or a generator.
D)
Armature emf is the same as long as the flux density in the air gap remains the same.
For a DC machine, the generated/induced armature emf is given by E = (PΦZN)/(60A), where P is number of poles, Φ flux per pole, Z total armature conductors, N speed, and A number of parallel paths. For the same Φ, Z, and N, the emf is inversely proportional to A. In wave winding, A = 2, while in lap winding, A = P (typically greater than 2). Therefore, wave winding gives a higher armature emf than lap winding.
Why others are wrong:
- b: Incorrect because wave winding has fewer parallel paths, so emf is higher, not lower, than lap winding.
- c: Incorrect because the emf expression depends on flux and speed, not on whether the machine is operating as a motor or generator; the direction of power flow changes, not the emf formula.
- d: Incorrect because armature emf is not determined solely by air-gap flux density; it depends on total flux per pole, speed, conductors, and winding arrangement.
For an electrically heated temperature controlled liquid heater, the best controller is
A)
Single-position controller
B)
Two- position controller
C)
Floating controller
D)
Proportional-position controller
For an electrically heated temperature-controlled liquid heater, the best controller is a two-position (ON-OFF) controller. Electric heaters have relatively slow thermal response and the control element is naturally switched fully ON or fully OFF, so ON-OFF control is simple, economical, and suitable for maintaining liquid temperature within an acceptable band.
Why the others are incorrect:
- Single-position controller: This is essentially not a proper feedback temperature controller for maintaining a setpoint; it does not provide the needed automatic switching action based on temperature error.
- Floating controller: Used where the final control element must move in steps over time; it is more suited to valve positioning systems, not electric heater switching.
- Proportional-position controller: More complex and generally used where finer control is needed; for an electric liquid heater, ON-OFF control is usually preferred and sufficient.
In the thermal power plants, the pressure in the working fluid cycle is developed by
A)
condenser
B)
super heater
C)
feed water pump
D)
turbine
In a thermal power plant, the pressure of the working fluid (water/steam) is raised by the feed water pump. The pump forces condensate/feed water from low pressure to the high pressure required before it enters the boiler. The superheater only increases the steam temperature at essentially constant pressure, the condenser lowers the exhaust steam pressure by removing heat, and the turbine converts steam energy into mechanical work while causing a pressure drop rather than developing pressure.
While designing controller, the advantage of pole-zero cancellation is
A)
the system order is increased
B)
the system order is reduced
C)
the cost of controller becomes low
D)
system’s error reduced to optimum levels
Pole-zero cancellation removes a pole of the plant by placing a controller zero at the same location (or vice versa). This simplifies the effective transfer function, so the overall system order is reduced. Hence, option (b) is correct.
Why others are incorrect:
- (a) The system order is increased: incorrect, because cancellation simplifies the dynamics rather than adding them.
- (c) The cost of controller becomes low: not a standard advantage of pole-zero cancellation; controller cost is not the primary control-theoretic benefit.
- (d) System’s error reduced to optimum levels: pole-zero cancellation does not guarantee optimal error performance; it mainly changes the system dynamics and may even be risky if exact cancellation is not achieved.
A rotating electrical machine its self-inductances of both the stator and the rotor windings, independent of the rotor position will be definitely not develop
A)
starting torque
B)
synchronizing torque
C)
hysteresis torque
D)
reluctance torque
If the self-inductances of both stator and rotor windings are independent of rotor position, the machine has no saliency, so there is no variation of magnetic reluctance with angle. Reluctance torque is produced only when inductance changes with rotor position; therefore it will definitely not be developed.
Why the others are incorrect:
- a) Starting torque: A rotating electrical machine can still develop starting torque by other mechanisms (e.g., induction, hysteresis, or excitation-based torque), so this is not ruled out.
- b) Synchronizing torque: This is the torque that tends to pull the rotor into step with the rotating field in synchronous machines; it does not require position-dependent self-inductance in the same way reluctance torque does.
- c) Hysteresis torque: This is due to hysteresis loss in the rotor material and is independent of saliency/position-dependent inductance, so it can still be developed.
A self-excited dc shunt generator, driven by its prime-mover at the rated speed fails to build up voltage across its terminals at no-load. What reason can be assigned for this?
A)
The field circuit resistance is higher than the critical resistance.
B)
The initial shunt field mmf does not assist the residual magnetism
C)
One of the inter-pole connections is reversed
D)
The brush-axis shifts slightly from the geometrical neutral axis of the machine
A self-excited DC shunt generator builds up voltage only if the shunt field current produces flux that aids the residual magnetism. If the initial shunt field mmf opposes the residual magnetism, the small generated emf cannot reinforce the field, so the voltage fails to build up.
Why others are incorrect:
- (a) If field circuit resistance is higher than critical resistance, the generator also fails to build up. However, the question asks for a reason that can be assigned in general; the most direct cause among the options is the field mmf opposing residual magnetism.
- (c) Reversed inter-pole connections mainly affect commutation in loaded operation, not the basic no-load voltage build-up of a shunt generator.
- (d) Slight brush-axis shift from the geometrical neutral axis is a commutation adjustment and does not prevent voltage build-up at no-load.
If the gain margin of a certain feedback system is given as 20 dB, the Nyquist plot will cross the negative real axis at the point
A)
s = -0.05
B)
s = -0.2
C)
s = -0.1
D)
s = -0.01
Gain margin in dB is GM(dB) = 20 log10(GM). Given GM = 20 dB, the corresponding gain margin in absolute value is GM = 10^(20/20) = 10. On the Nyquist plot, the phase crossover point on the negative real axis is where the open-loop magnitude equals 1/GM = 1/10 = 0.1, so the plot crosses at -0.1. Hence option c is correct.
Why others are incorrect:
- a) -0.05 corresponds to magnitude 0.05, which would imply a gain margin of 20 log10(1/0.05) ≈ 26 dB, not 20 dB.
- b) -0.2 corresponds to magnitude 0.2, giving gain margin ≈ 14 dB.
- d) -0.01 corresponds to magnitude 0.01, giving gain margin 40 dB.
Therefore, the correct answer is c: s = -0.1.
In a transformer the hysteresis losses are 100 K when the input voltage form factor is 2.11. If the input voltage form factor is 1.40, the hysteresis loss will be
A)
140 K
B)
140/K
C)
200 K
D)
70 K
Hysteresis loss in a transformer is proportional to the form factor of the applied voltage (for the same maximum flux density and frequency). So,
P_h2 / P_h1 = FF2 / FF1
Given:
P_h1 = 100 K at FF1 = 2.11
FF2 = 1.40
Therefore,
P_h2 = 100 × (1.40 / 2.11) ≈ 66.4 K ≈ 70 K
So the correct option is d) 70 K.
Why others are incorrect:
- a) 140 K: would imply the loss increased, but the form factor decreased, so hysteresis loss should decrease.
- b) 140/K: not a valid loss value in this context.
- c) 200 K: also implies a much larger loss, which is inconsistent with the lower form factor.
Zero sequence currents can flow from a line to transformer bank if the windings are in
A)
grounded star-delta
B)
delta-star
C)
star-grounded star
D)
detla-delta
Zero-sequence current can flow only if there is a return path through ground/neutral. In a transformer bank, this is possible when one winding is star-connected with its neutral grounded, and the other side is delta-connected. The grounded star provides the zero-sequence path to ground, while the delta allows the circulating zero-sequence components to be contained within the delta without appearing on the line side. Hence, a grounded star-delta transformer permits zero-sequence current flow from the line to the transformer bank.
Why the others are incorrect:
- b) delta-star: the delta side blocks zero-sequence current from passing to the line side; it does not provide a direct zero-sequence path from line to bank in the required sense.
- c) star-grounded star: although grounded neutrals exist, zero-sequence transfer between line and transformer bank is not the standard path described here; the delta is the usual element that enables circulation/isolation behavior for zero-sequence components.
- d) delta-delta: no neutral/ground path is available, so zero-sequence current cannot flow from the line into the transformer bank.