← MES Jr. Grade
Past Questions
3 phase, 4 wire ac system of distribution is used for
A)
Unbalanced load
B)
Balanced load
C)
All type of load
D)
None of these
A 3-phase, 4-wire AC distribution system is used to supply both balanced and, more importantly, unbalanced loads. The fourth wire (neutral) provides a return path for the unbalanced current, making it suitable for single-phase loads connected between phase and neutral. Hence, the correct option is unbalanced load. Balanced load alone would not require a neutral wire, so option b is incomplete. Option c is too broad and not the standard specific use stated in distribution systems. Option d is incorrect because the system is definitely used for distribution, especially where loads are unbalanced.
A linear discrete time system has the characteristics equation Z^3 - 0.81Z = 0, the system is
A)
Stable
B)
Marginally stable
C)
Unstable
D)
Stability cannot be assessed from the given information
For a discrete-time LTI system, stability is determined by the locations of the characteristic roots (poles) in the z-plane: all poles must lie strictly inside the unit circle for stability. Given Z^3 - 0.81Z = 0, factorizing gives Z(Z^2 - 0.81)=0, so the roots are Z=0, Z=+0.9, and Z=-0.9. All three roots satisfy |Z|<1, so the system is stable. Therefore, the correct classification is stable, not marginally stable (which would require at least one pole on the unit circle and none outside), not unstable (which would require any pole outside the unit circle), and stability can indeed be assessed from the given characteristic equation.
The value of diversity factor is
A)
Less than one
B)
More than one
C)
Equal to one
D)
None of these
The diversity factor is defined as the ratio of the sum of the individual maximum demands of different consumers/loads to the maximum demand of the whole system. Since these individual maxima do not usually occur at the same time, their sum is greater than the coincident system maximum demand. Therefore, the diversity factor is always more than 1.
Why others are incorrect:
- a) Less than one: incorrect, because the numerator (sum of individual maxima) is generally greater than the denominator (system maximum demand).
- c) Equal to one: only possible in the special case when all loads reach maximum simultaneously, which is not the usual case.
- d) None of these: incorrect because option b is correct.
The leakage resistance of a 50 km long cable is 1 MW. For a 100 km long cable it will be
A)
1 MW
B)
2 MW
C)
0.66 MW
D)
None of these
Leakage resistance of a cable is inversely proportional to its length, because a longer cable provides more surface area for leakage current. So, if the length doubles from 50 km to 100 km, the leakage resistance becomes half: R2 = R1 × (L1/L2) = 1 MΩ × (50/100) = 0.5 MΩ. However, since the given options do not include 0.5 MΩ, the intended exam convention is likely using the standard relation for a cable section where leakage resistance varies inversely with length and the closest listed answer is 0.66 MΩ only if a different proportionality assumption is used. But for the stated data, the correct theoretical result is 0.5 MΩ, which is not present. Therefore, among the provided options, the correct choice is d (None of these). The options a (1 MΩ) and b (2 MΩ) are incorrect because resistance cannot remain the same or increase when cable length doubles. Option c (0.66 MΩ) is also incorrect because the resistance should reduce to one-half, not to two-thirds.
For a good voltage profile under load conditions, a long line needs
A)
Shunt capacitor at receiving end
B)
Shunt reactor at receiving end
C)
Shunt reactor at sending end
D)
Shunt capacitor at sending end
A long transmission line under load tends to have a voltage drop at the receiving end due to line reactance and reactive power demand. To improve the voltage profile, reactive power should be supplied locally at the load end, so a shunt capacitor at the receiving end is used. It injects leading VARs, supports the receiving-end voltage, and reduces current and losses.
Why others are incorrect:
- b) Shunt reactor at receiving end: a reactor absorbs reactive power and further lowers voltage, so it worsens the voltage profile.
- c) Shunt reactor at sending end: also absorbs reactive power and is used mainly to control overvoltage under light-load conditions, not to improve loaded voltage profile.
- d) Shunt capacitor at sending end: it can improve power factor at the source, but it is less effective for receiving-end voltage support than placing the capacitor at the receiving end.
The steady state error due to a ramp input for a type two system is equal to
A)
Zero
B)
Infinite
C)
Constant
D)
Data is insufficient
For a type-two system, the open-loop transfer function has two poles at the origin. For a ramp input, the steady-state error is determined by the velocity error constant Kv. In a type-two system, Kv is infinite, so the steady-state error to a ramp input is zero. Hence option (a) Zero is correct.
Why others are incorrect:
- (b) Infinite: This would be true for a type-zero system with a ramp input, not for type two.
- (c) Constant: A constant steady-state error occurs for a type-one system with a ramp input, not type two.
- (d) Data is insufficient: The system type alone is enough to determine the ramp steady-state error here.
The induced e.m.f. in the armature conductors of a d.c. motor is
A)
Sinusoidal
B)
Trapezoidal
C)
Rectangular
D)
Alternating
In a d.c. motor, the armature conductors cut the magnetic flux and an e.m.f. is induced in them according to Faraday’s law. Since the conductors rotate under alternate magnetic polarities, the induced e.m.f. in each conductor is inherently alternating in nature. The commutator then mechanically rectifies this internally induced alternating e.m.f. to produce a unidirectional terminal behavior.
Why the others are incorrect:
- a) Sinusoidal: the induced e.m.f. in a d.c. motor armature conductor is not specifically sinusoidal; its waveform depends on flux distribution and conductor motion, but the standard classification is alternating.
- b) Trapezoidal: trapezoidal waveforms are associated with certain special machines/control schemes, not the general armature conductor e.m.f. in a d.c. motor.
- c) Rectangular: the commutator output is rectified to d.c.-like form, but the e.m.f. induced in the conductors themselves is not rectangular.
The main purpose of using magnetic core in a transformer is to
A)
Prevent eddy current loss
B)
Eliminate magnetic hysteresis
C)
Decrease iron losses
D)
Decrease reluctance of the common magnetic flux path
The main purpose of using a magnetic core in a transformer is to provide a low-reluctance path for the mutual magnetic flux linking the primary and secondary windings. A ferromagnetic core concentrates the flux and greatly reduces the reluctance of the common magnetic flux path, improving coupling and transformer efficiency.
Why the others are incorrect:
- a) Prevent eddy current loss: A magnetic core by itself does not prevent eddy currents; in fact, a solid core would allow them. Eddy current loss is reduced by laminating the core.
- b) Eliminate magnetic hysteresis: The core does not eliminate hysteresis; ferromagnetic materials still exhibit hysteresis loss. Using suitable core materials only reduces it.
- c) Decrease iron losses: The core does not inherently decrease iron losses. Iron losses include hysteresis and eddy current losses, which are minimized by material choice and lamination, not by the mere presence of a core.
Ferranti effect on long overhead line is experienced when it is
A)
Lightly loaded
B)
On full load at upf
C)
On full load at 0.8 pf
D)
On full load at zfp
Ferranti effect is the rise in receiving-end voltage above sending-end voltage in a long transmission line when the line is lightly loaded or on no-load. This happens because the line capacitance draws charging current, and with little load current the reactive effect dominates, causing the voltage at the receiving end to increase.
Why others are incorrect:
- b) Full load at unity power factor: load current is substantial, so the voltage drop along the line is not the Ferranti condition.
- c) Full load at 0.8 pf: the lagging reactive current causes voltage drop, not voltage rise.
- d) Full load at zero power factor: current is purely reactive and large, producing significant voltage drop; Ferranti effect is not associated with full-load conditions.
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.
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.
The open loop transfer function of a feedback control system is G(S)H(S) = 1 / (S + 1)^3, the gain margin of the system is
A)
2
B)
4
C)
8
D)
16
For the open-loop transfer function G(s)H(s)=1/(s+1)^3, the phase is ∠G(jω)H(jω) = -3 tan⁻¹(ω). The phase crossover frequency occurs when phase = -180°, so 3 tan⁻¹(ω)=180° ⇒ tan⁻¹(ω)=60° ⇒ ω=√3. At this frequency, the magnitude is |G(jω)H(jω)| = 1/|1+jω|^3 = 1/(√(1+ω²))^3 = 1/(√(1+3))^3 = 1/2^3 = 1/8. Gain margin = 1/|G(jω)H(jω)| at phase crossover = 8. Hence the correct option is c.
Why others are incorrect: 2, 4, and 16 do not match the gain margin computed from the phase crossover condition and magnitude evaluation. The exact result is 8.
If the frequency of a transmission system is changed from 50 Hz to 100 Hz, the string efficiency will be
A)
Increase
B)
Decrease
C)
Remain unchanged
D)
May be increase or decrease
String efficiency is the ratio of flashover voltage to the voltage across the unit nearest the conductor. It depends on the voltage distribution along the insulator string, which is affected by the capacitive reactance of the string units and the shunt capacitance to the tower. When frequency increases from 50 Hz to 100 Hz, the capacitive reactance of the insulator units decreases, causing the capacitive effects to become more pronounced and the voltage distribution to become more non-uniform. This increases the voltage stress on the disc nearest the conductor and reduces the string efficiency. However, in standard exam treatment, the frequency increase is taken to worsen the voltage distribution and thus the string efficiency is considered to decrease. Therefore, the correct option is 'a' Increase? Wait—this needs careful correction: since non-uniformity increases, string efficiency decreases. Hence the correct option is b. The other options are incorrect because it does not remain unchanged, and it is not generally an increase. It is not 'may be increase or decrease' because the effect of frequency on capacitive distribution is definite in this context.
The efficiency of a 3 phase induction motor is approximately proportional to
A)
(1-S)
B)
S
C)
N
D)
Ns
For a 3-phase induction motor, efficiency is approximately proportional to the mechanical power developed, which depends on the rotor speed being close to synchronous speed. Since rotor speed N = Ns(1 - S), the useful output increases as slip S decreases. Hence efficiency is approximately proportional to (1 - S).
Why others are incorrect:
- b) S: Efficiency does not increase with slip; higher slip means more rotor copper loss and lower efficiency.
- c) N: Rotor speed N is related to efficiency, but the standard approximate proportionality is with (1 - S), not N directly.
- d) Ns: Synchronous speed Ns is essentially constant for a given supply and motor, so efficiency is not proportional to Ns.
For a fault at generator terminals the fault current is maximum for
A)
3- phase fault
B)
Line – line fault
C)
S-L-G fault
D)
D-L-G fault
For a fault at the generator terminals, the maximum fault current occurs for a 3-phase fault. A 3-phase fault is a balanced fault and involves all three phases, so it produces the lowest equivalent fault impedance and therefore the highest current.
Why others are incorrect:
- Line-line fault: involves only two phases, so the fault current is less than a 3-phase fault.
- Single line-to-ground (S-L-G) fault: although common, its current is usually lower than a 3-phase fault because the sequence network impedance is higher.
- Double line-to-ground (D-L-G) fault: current is generally higher than line-line or S-L-G in many cases, but still typically less than the 3-phase fault at generator terminals.
Hence, the correct answer is 3-phase fault.
The value of ‘K’ for which the unity feedback system G(S) = K / [S(S+2)(S+4)] crosses the imaginary axis is
A)
24
B)
42
C)
64
D)
48
For a unity-feedback system with open-loop transfer function G(s)=K/[s(s+2)(s+4)], the closed-loop characteristic equation is 1+G(s)=0, i.e. s(s+2)(s+4)+K=0. Expanding gives s^3+6s^2+8s+K=0.
To find when the root locus crosses the imaginary axis, use the Routh array:
s^3: 1 8
s^2: 6 K
s^1: (48-K)/6 0
s^0: K
The system crosses the imaginary axis when the s^1 row becomes zero, so 48-K=0 ⇒ K=48. Hence the correct option is d.
Why the others are incorrect: 24, 42, and 64 do not satisfy the Routh condition for marginal stability, so they do not correspond to the imaginary-axis crossing point.
Synchronous motors for power factor correction operate at
A)
Normal load with no excitation
B)
Normal load with low excitation
C)
No load and over excited
D)
No load and under excited
Synchronous motors used for power factor correction are operated at no load and over-excited. In this condition, the motor draws leading current from the supply, behaving like a synchronous condenser and supplying reactive power to improve the system power factor.
Why others are incorrect:
- a) Normal load with no excitation: without excitation, a synchronous motor cannot operate properly as a synchronous machine and will not provide power factor correction.
- b) Normal load with low excitation: low excitation makes the motor under-excited, causing it to draw lagging reactive power, which worsens power factor.
- d) No load and under excited: under-excitation at no load still makes the motor absorb reactive power instead of supplying it, so it is not used for power factor correction.
If the magnitude of the polar plot at phase crossover is ‘a’, the gain margin is
A)
-a
B)
0
C)
a
D)
1/a
At the phase crossover frequency, the phase of the open-loop transfer function is -180°. If the magnitude of the polar plot there is a, then the gain margin is the factor by which the loop gain must be multiplied to bring the magnitude to unity. Hence, Gain Margin = 1/a. Therefore option d is correct.
Why others are incorrect:
- a (-a): Gain margin is not negative; it is a positive multiplicative factor.
- b (0): Zero would mean no additional gain is allowed, which is not the definition of gain margin.
- c (a): The magnitude a is the open-loop magnitude at phase crossover, not the gain margin itself.
Consider a system with the transfer function G(S) = (S+6) / (KS^2 + S + 6). Its damping ratio will be 0.5 when the value of K is
A)
2/6
B)
3
C)
1/6
D)
6
Compare the denominator with the standard 2nd-order form: s^2 + 2ζω_n s + ω_n^2. Here, G(s) = (s+6)/(Ks^2 + s + 6), so the characteristic polynomial is Ks^2 + s + 6. Dividing by K gives s^2 + (1/K)s + 6/K. Hence, 2ζω_n = 1/K and ω_n^2 = 6/K. Using ζ = 0.5, we have 2(0.5)ω_n = ω_n = 1/K, so ω_n = 1/K. Also ω_n^2 = 6/K, so (1/K)^2 = 6/K ⇒ 1/K = 6 ⇒ K = 1/6. Therefore, option a (2/6) is incorrect as written if taken literally, but the intended correct value among the given choices is c? Wait: 2/6 = 1/3, which does not satisfy the equation. The exact correct K is 1/6, which matches option c. The other options 3 and 6 are far from the required value and do not satisfy the standard-form matching.
The transfer function is (1 + 0.5S) / (1 + S). It represents a
A)
Lead network
B)
Lag network
C)
Lag-lead network
D)
Proportional network
The transfer function is G(s) = (1 + 0.5s)/(1 + s). It has one zero at s = -2 and one pole at s = -1. Since the pole is closer to the origin than the zero, the phase contribution is negative over a range of frequencies, which is characteristic of a lag network. Also, the zero time constant (0.5) is smaller than the pole time constant (1), so the numerator factor changes faster than the denominator, again indicating lag behavior.
Why others are incorrect:
- a) Lead network: A lead network requires the zero to be closer to the origin than the pole (zero time constant > pole time constant), giving positive phase lead. That is not the case here.
- c) Lag-lead network: This would require both a lag and a lead section, typically with two zeros and two poles arranged to produce both effects. This transfer function has only one zero and one pole.
- d) Proportional network: A proportional network is just a constant gain K, with no s-dependence. This transfer function depends on s, so it is not proportional.
Signal flow graph is used to find
A)
Stability of the system
B)
Controllability of the system
C)
Poles of the system
D)
Transfer function of the system
Signal flow graphs are primarily used to determine the transfer function of a system using Mason’s gain formula. They represent the relationships between variables in a linear system and allow direct computation of input-output gain. The other options are not the main purpose of a signal flow graph: poles are obtained from the transfer function, stability is analyzed from pole locations, and controllability is a state-space property assessed using controllability matrices, not signal flow graphs.
Distribution transformer have core losses
A)
More than full load copper loss
B)
Equal to full load copper loss
C)
Less than full load copper loss
D)
Negligible compared to full load copper loss
Distribution transformers are designed to operate for long periods at light load and near-constant voltage, so their iron (core) losses are made very low and are typically negligible compared with the full-load copper loss. Core loss is almost constant with load, while copper loss increases with load and is maximum at full load. Hence the correct choice is (d). The other options are incorrect because core loss is not greater than, equal to, or generally less than full-load copper loss as a defining characteristic; instead, it is usually much smaller and nearly constant.
In a 3 phase, 5 kV, 5 MVA system, what is the base impedance?
A)
5 ohms
B)
50 ohms
C)
500 ohms
D)
0.5 ohm
For a 3-phase system, the base impedance is calculated as Z_base = (V_base)^2 / S_base using line-to-line voltage and 3-phase MVA base. Here, V_base = 5 kV and S_base = 5 MVA, so Z_base = (5^2)/5 = 5 ohms. Therefore, option b is correct.
Why the others are incorrect: 0.5 ohm is too small and would result from an incorrect scaling; 50 ohms and 500 ohms are too large and do not match the standard base-impedance formula for the given voltage and power bases.
The most commonly used material for insulators of overhead lines is
A)
Porcelain
B)
Glass
C)
Mica
D)
PVC
Porcelain is the most commonly used material for overhead line insulators because it has high dielectric strength, good mechanical strength, low leakage current, and excellent resistance to weathering, moisture, and contamination. It is durable and suitable for outdoor service on transmission and distribution lines. Glass is also used in some overhead line insulators, but porcelain is the standard and most widely used material in conventional practice. Mica is not used for overhead line insulators because it is not suitable for large outdoor insulators and lacks the required mechanical form. PVC is a plastic insulation material used in cables and wiring, not for overhead line insulators, since it cannot provide the necessary mechanical and environmental performance.
Slip ring are made of
A)
Steel
B)
Copper
C)
Bakelite
D)
Mica
Slip rings are made of copper (or copper alloy) because they must provide good electrical conductivity and a smooth, durable contact surface for brushes in rotating electrical machines. Steel is not used because it has much higher electrical resistance. Mica is an insulating material, so it cannot serve as a current-carrying slip ring. Bakelite is also an electrical insulator and is used for insulating parts, not for conducting rings.