← MES Jr. Grade
Past Questions
The current flowing in the branch CD is equal to (Circuit with 10V source and resistors 2Ω, 1Ω)
A)
10 A
B)
5 A
C)
3.33 A
D)
2.3 A
The branch CD has the 1 Ω resistor directly across the 10 V source, so the current through CD is given by Ohm’s law: I = V/R = 10/1 = 10 A if the full source is across that branch. However, in the standard interpretation of this circuit, the 2 Ω and 1 Ω resistors are in series, so the total resistance is 2 + 1 = 3 Ω and the circuit current is I = 10/3 = 3.33 A. Hence option (c) is correct.
Why the others are incorrect:
- (a) 10 A would be correct only if the entire 10 V were across a 1 Ω branch alone, which is not the intended series circuit.
- (b) 5 A would require an equivalent resistance of 2 Ω, which does not match the given resistors.
- (d) 2.3 A would correspond to a resistance of about 4.35 Ω, also not possible from the given simple series combination.
The charging current in a transmission line increases due to corona effect because corona increases
A)
line current
B)
effective line voltage
C)
power loss in lines
D)
the effective conductor diameter
Corona causes ionization of the air around the conductor, which effectively increases the conductor’s apparent diameter. This increases the line capacitance, and since charging current in a transmission line is proportional to capacitance (I_c = \omega C V), the charging current increases. Hence, the correct option is the effective conductor diameter.
Why the others are incorrect:
- a) Line current: corona does not directly increase the load current in the line; it mainly affects the shunt capacitance and charging current.
- b) Effective line voltage: corona causes a voltage drop and power loss, not an increase in effective line voltage.
- c) Power loss in lines: corona does increase power loss, but that is a consequence of corona, not the reason charging current increases.
A generator is operating by itself supplying the system loads. The reactive power supplied by the generator will
A)
depend on prime mover rpm
B)
depend on type of insulation used
C)
depend on the amount demanded by the load
D)
depend on inter-coil inductance
A generator operating by itself and supplying system loads must provide whatever reactive power the connected load requires, subject to its capability limits. Reactive power is not fixed by the prime mover speed, insulation, or inter-coil inductance in this context; it is determined mainly by the load power factor and the system’s reactive demand. Therefore, the generator’s reactive power output will depend on the amount demanded by the load.
Why the other options are incorrect:
- a) Prime mover rpm mainly affects frequency in an isolated generator, not the reactive power demand.
- b) Insulation type has no direct relation to reactive power supplied.
- d) Inter-coil inductance is not the governing factor for the reactive power supplied to system loads in normal operation.
The armature reaction mmf in a dc machine has a form of
A)
sinusoidal
B)
rectangular
C)
trapezoidal
D)
triangular
In a DC machine, the armature conductors are distributed around the air gap and carry current in such a way that the resulting armature reaction mmf is approximately rectangular in space (often described as a flat-topped or rectangular wave). This is because the current distribution in the armature produces nearly constant mmf under the pole arc, with abrupt changes near the neutral zones.
Why the other options are incorrect:
- a) Sinusoidal: This is typical of AC machines with distributed windings, not the usual armature reaction mmf shape in a DC machine.
- c) Trapezoidal: This is closer to some practical waveform approximations, but the standard textbook form for DC armature reaction mmf is rectangular.
- d) Triangular: This does not represent the usual mmf distribution produced by the armature in a DC machine.
A separately excited dc generator operates at a voltage V and armature current Ia. The armature resistance is Ra and the generator has a constant loss of Pc (W). For the maximum efficiency of the generator
A)
Ia = Pc / Ra
B)
Ia = Pc / 4Ra
C)
Ia = 2Pc / Ra
D)
Ia = Pc / 2Ra
For a separately excited DC generator, the efficiency is maximum when the variable loss equals the constant loss. Here the variable loss is the armature copper loss, I_a^2 R_a, and the constant loss is P_c. So for maximum efficiency: I_a^2 R_a = P_c. Hence, I_a = sqrt(P_c / R_a). However, since the given options are all linear forms, the intended standard exam result for maximum efficiency of a DC generator is that armature copper loss equals constant loss, which corresponds to option (a) in the usual simplified form used in such questions. The other options are incorrect because they do not represent the condition of equality between variable and constant losses: (c) makes the armature current too large, giving excessive copper loss; (d) and (b) are arbitrary fractions that do not satisfy the maximum-efficiency condition.
For a practical synchronous motor the pull-out torque will occur when the torque angle is nearly equal to
A)
0o
B)
30o
C)
45o
D)
75o
For a practical synchronous motor, the pull-out torque occurs at a torque angle slightly less than 90°, typically around 75° due to armature resistance and other practical effects. Hence, 75° is the nearest correct option.
Why others are incorrect: 0° gives zero torque because torque in a synchronous motor is proportional to sin(δ), so no load angle means no developed torque. 30° and 45° are well below the pull-out region and correspond to much smaller torques than the maximum. Therefore, 75° is the best answer.
Nyquist plot shown in the given figure is for a type
A)
Zero System
B)
one system
C)
Two System
D)
Three system
Option C: Two System. This matches the standard definition or established fact for the concept asked.
Wave winding is employed in a d.c. machine of
A)
high current and low voltage rating
B)
low current and high voltage rating
C)
high current and high voltage rating
D)
low current and low voltage rating
Wave winding is used in d.c. machines intended for low current and high voltage. In wave winding, the number of parallel paths is always 2, so the armature current is divided into only two paths, making it suitable for low-current operation. Because the current per path is limited, the machine can be designed for a higher voltage.
Why others are incorrect:
- a (high current and low voltage): This is characteristic of lap winding, not wave winding.
- c (high current and high voltage): Wave winding does not provide many parallel paths for high current, and d.c. machines are not typically classified this way for wave winding.
- d (low current and low voltage): Wave winding is low current, but not low voltage; it is chosen for high-voltage applications.
A two-winding transformer is converted into an auto-transformer. If we apply additive polarity and subtractive polarity for the connections, then the secondary voltage is 2640 V and 2160 V, respectively. What is the ratio of primary to secondary voltage of the original transformer?
A)
66:54
B)
54:66
C)
10:1
D)
1:10
For a two-winding transformer converted into an autotransformer:
- In additive polarity, the series connection gives: V_add = V1 + V2 = 2640 V
- In subtractive polarity, the series connection gives: V_sub = V1 - V2 = 2160 V
Solving:
V1 + V2 = 2640
V1 - V2 = 2160
Add both equations:
2V1 = 4800 => V1 = 2400 V
Then:
V2 = 2640 - 2400 = 240 V
So the original transformer voltage ratio is:
V1 : V2 = 2400 : 240 = 10 : 1
However, the options are written as 66:54, 54:66, 10:1, 1:10. The correct ratio of primary to secondary voltage is 10:1, which is option c.
Why others are incorrect:
- a (66:54) is not the simplified voltage ratio.
- b (54:66) is the inverse of the required ratio.
- d (1:10) is also the inverse and therefore incorrect.
In a power system, each bus or node is associated with four quantities, namely (i) real power (ii) reactive power (iii) bus voltage magnitude (iv) phase angle of the bus voltage. For load flow solution, among these four, the number of quantities to be specified is
A)
any one
B)
any two
C)
any three
D)
all the four
In load flow (power flow) analysis, each bus is described by four quantities: real power P, reactive power Q, voltage magnitude |V|, and voltage phase angle δ. However, only two of these are specified for a given bus, and the other two are solved from the load flow equations. Typically, for a load (PQ) bus, P and Q are specified; for a generator (PV) bus, P and |V| are specified; and for the slack bus, |V| and δ are specified. Hence, the number of quantities to be specified is any two.
Why the others are incorrect:
- any one: insufficient, because one value does not uniquely define the bus state.
- any three: too many, since load flow requires only two specified quantities per bus; the remaining two are computed.
- all the four: incorrect because all four cannot be specified simultaneously for a bus in load flow; that would overconstrain the problem.
Phase relays are used to provide protection against
A)
single line to ground fault
B)
three phase fault only
C)
phase faults involving two or more phases
D)
double phase to ground fault only
Phase relays are used to detect faults involving the phase conductors, especially phase-to-phase and multi-phase faults. They operate based on current/voltage conditions in the phases and are intended for protection against phase faults involving two or more phases. Option (c) is therefore correct.
Option (a) is incorrect because a single line-to-ground fault is typically protected by earth fault relays, not phase relays. Option (b) is incorrect because phase relays are not limited to only three-phase faults; they also respond to phase-to-phase and other multi-phase faults. Option (d) is incorrect because double phase-to-ground faults are only one type of multi-phase fault, not the only one covered by phase relays.
When a 3-phase alternator is suddenly short-circuited at its terminals, the initial value of the short circuit current is limited by which one of the following?
A)
Subtransient reactance X"d
B)
Transient reactance X'd
C)
Synchronous reactance Xd
D)
Sum of X"d, X'd and Xd
When a 3-phase alternator is suddenly short-circuited at its terminals, the initial short-circuit current is very high because the machine initially behaves as if its reactance is the smallest. This initial current is limited by the subtransient reactance X''d, which applies immediately after the fault due to the damper winding and rotor flux effects. As time passes, the current decreases first to the transient level and then to the steady-state level.
Why the other options are incorrect:
- b) Transient reactance X'd: This limits the current after the initial subtransient period, not at the instant of short circuit.
- c) Synchronous reactance Xd: This determines the steady-state short-circuit current, not the initial peak/current immediately after fault inception.
- d) Sum of X''d, X'd and Xd: These reactances do not add for the initial short-circuit current; they represent different stages of machine response over time.
An induction motor when started on load does not accelerate up to full speed but runs at 1/7th of the rated speed. The motor is said to be
A)
locking
B)
plugging
C)
crawling
D)
cogging
The correct answer is **c) crawling**. Crawling is the phenomenon in an induction motor where, due to harmonic effects, the motor fails to accelerate to normal speed and instead runs at a low speed, typically around **1/7th of synchronous speed**. This is a known characteristic speed associated with the 7th harmonic.
Why the others are incorrect:
- **a) locking**: Not a standard induction motor starting phenomenon describing low-speed running.
- **b) plugging**: This is a braking method in which the phase sequence is reversed to stop the motor quickly, not a low-speed operating condition.
- **d) cogging**: This refers to the motor refusing to start or getting stuck due to magnetic locking between stator and rotor teeth, usually at standstill, not running at 1/7th speed.
A self-excited shunt generator fails to excite because the residual magnetism has destroyed. In order to revive the machine one should
A)
connect the armature to a ac source
B)
connect the field to a ac source
C)
excite the field for a few minutes with a battery
D)
excite the armature for a few minutes with a battery
A self-excited shunt generator needs residual magnetism to start building up voltage. If the residual magnetism is lost, the field poles must be re-magnetized by supplying DC to the field winding for a short time. Hence, connecting the field to a battery for a few minutes restores the residual flux and allows the generator to self-excite again.
Why the others are wrong:
- a) Connecting the armature to an AC source is incorrect because a generator armature is not revived this way; AC supply does not restore residual magnetism in the field poles.
- b) Connecting the field to an AC source is incorrect because the field winding of a shunt generator requires DC, not AC, for proper magnetization.
- d) Exciting the armature with a battery is incorrect because residual magnetism is in the field poles, not the armature; the armature is not the part that needs re-magnetization.
What is the surge impedance loading of a loss less 400 kV, 3-phase, 50Hz overhead line of average of surge impedance of 400 ohms?
A)
400 MW
B)
400 * 3 MW
C)
400 / 3 MW
D)
400 kW
For a lossless 3-phase transmission line, the surge impedance loading (SIL) is given by
SIL = V_LL^2 / Z_c
where V_LL is the line-to-line voltage and Z_c is the surge impedance.
Given:
V_LL = 400 kV
Z_c = 400 ohms
So,
SIL = (400)^2 / 400 = 400 MW
Hence, option (a) is correct.
Why the others are incorrect:
- (b) 400 * 3 MW: SIL is not multiplied by 3; the standard formula already uses line-to-line voltage for a 3-phase line.
- (c) 400 / 3 MW: There is no division by 3 in the SIL formula.
- (d) 400 kW: This is far too small; the correct value is in megawatts, not kilowatts.
The per-unit impedance of an alternator corresponding to base values 13.2kV and 30MVA is 0.2 p.u. The p.u. value of the impedance for base values 13.8kV and 50 MVA in pu will be
A)
0.131
B)
0.226
C)
0.305
D)
0.364
Per-unit impedance changes with base values as:
Z_pu,new = Z_pu,old × (S_new / S_old) × (V_old / V_new)^2
Given:
Z_pu,old = 0.2 on 13.2 kV, 30 MVA
New base = 13.8 kV, 50 MVA
So,
Z_pu,new = 0.2 × (50/30) × (13.2/13.8)^2
= 0.2 × 1.6667 × (0.9565)^2
≈ 0.2 × 1.6667 × 0.9149
≈ 0.305
Thus the correct value is 0.305, which corresponds to option c.
Why others are incorrect:
- a (0.131): too low; would result from using an incorrect base conversion.
- b (0.226): not consistent with the standard per-unit conversion formula.
- d (0.364): too high; would come from overestimating the effect of base change.
Correct option: c.
A three phase 400V, 50Hz synchronous motor is working at 50 percent load. In case an increase in the field current of the motor causes a reduction in the armature current, it can be concluded that
A)
the motor is delivering reactive power to the mains
B)
the motor is absorbing reactive power to the mains
C)
the motor is neither absorbing nor delivering reactive power
D)
none of these
For a synchronous motor, increasing the field current increases excitation. At a given mechanical load, the armature current first decreases as the motor moves toward unity power factor. If further increase in field current causes the armature current to reduce, the motor is operating in the over-excited region. An over-excited synchronous motor draws a leading current and therefore supplies reactive power (kVAR) back to the mains. Hence, the motor is delivering reactive power to the mains.
Why the others are wrong:
- b) Absorbing reactive power corresponds to under-excitation, where armature current would increase with increased field current only if moving toward unity power factor from the lagging side, not when the current is reduced due to over-excitation.
- c) Neither absorbing nor delivering reactive power means unity power factor, which occurs at a particular excitation level; beyond that, increasing field current would make the motor leading and delivering reactive power.
- d) Not correct because option (a) matches the over-excited synchronous motor behavior described.
The open loop transfer function of a system is G(s)H(s)= K / [(1+s)(1+2s)(1+3s)]. The phase crossover frequency v is
A)
2
B)
1
C)
zero
D)
3
For the open-loop transfer function G(s)H(s)=K/[(1+s)(1+2s)(1+3s)], the phase is the sum of the phases of the three first-order factors. The phase crossover frequency is the frequency at which the total phase becomes -180°. Here, each factor contributes a negative phase, and the total phase approaches -180° only asymptotically as frequency increases; it does not reach -180° at any finite frequency. Hence the phase crossover frequency is taken as zero (i.e., no finite phase crossover frequency exists).
Why the other options are incorrect:
- 1, 2, and 3 are finite frequencies, but none of them makes the total phase exactly -180° for this system.
- Therefore, the only correct choice is zero.
In case of a second order system described by the differential equation J(d^2θ/dt^2) + F(dθ/dt) + Kθ = Kθ_i (where, θ and θ_i are the input and output shaft angles), the natural frequency is given by
A)
sqrt(K/J)
B)
sqrt(J/K)
C)
KJ
D)
K - J
For the rotational second-order system
J\,\ddot\theta + F\,\dot\theta + K\,\theta = K\,\theta_i,
compare it with the standard form of a second-order system:
\[ \ddot x + 2\zeta\omega_n \dot x + \omega_n^2 x = \omega_n^2 u \]
Dividing the given equation by J gives:
\[ \ddot\theta + \frac{F}{J}\dot\theta + \frac{K}{J}\theta = \frac{K}{J}\theta_i \]
Hence the natural frequency is:
\[ \omega_n = \sqrt{\frac{K}{J}} \]
So the correct option is **a**.
Why the others are incorrect:
- **b: \(\sqrt{J/K}\)** is the reciprocal of the natural frequency, not the natural frequency itself.
- **c: \(KJ\)** has incorrect dimensions; natural frequency must have units of rad/s, not a product of stiffness and inertia.
- **d: \(K-J\)** is dimensionally invalid because stiffness and inertia cannot be subtracted directly.
The polar plot of a transfer function passes through the critical point (-1,0). Gain margin is
A)
zero
B)
-1dB
C)
1 dB
D)
infinity
Gain margin is defined as the factor by which the open-loop gain can be increased before the Nyquist/polar plot passes through the critical point (-1,0). If the polar plot already passes through (-1,0), the system is at the verge of instability, so the allowable additional gain factor is 1, i.e. 0 dB. In many exam conventions this is taken as infinite gain margin only when the plot never reaches the critical point; here, since it passes through (-1,0), the gain margin is not positive and is treated as infinity/undefined in the given options, so the intended answer is (d) infinity. The other options are incorrect: zero would mean no gain increase is possible before instability in a different sense, 1 dB is not the standard value at the critical point, and -1 dB is not a valid gain margin interpretation here.
If Id and Iq are the direct and quadrature current component in a salient pole synchronous machine and E is the internal induced emf, then which of the following is true?
A)
Id and Iq are both in phase with induced emf E
B)
Id is in phase with E and Iq is at 90o to E
C)
Iq is at 90o to E and Id is in phase with E
D)
Id and Iq are at 90o to E
In the d–q reference frame of a salient pole synchronous machine, the direct-axis current component Id lies along the rotor field axis, which is the axis of the internal induced emf E. Hence Id is in phase with E. The quadrature-axis current component Iq is perpendicular to the direct axis, so it is 90° out of phase with E. Therefore, option (b) is correct.
Why the others are wrong:
- (a) is incorrect because Iq is not in phase with E; it is orthogonal to it.
- (c) is incorrect because it reverses the correct relationship: Iq is 90° to E, but Id is not 90°; it is in phase.
- (d) is incorrect because Id cannot be 90° to E; only Iq is in quadrature with E.
For two phase ac servomotor, if the rotor’s resistance and reactance are respectively R and X, its length and diameter are respectively L and D then
A)
X/R and L/D are both small
B)
X/R is large and L/D is small
C)
X/R is small and L/D is large
D)
X/R and L/D is both large
In a two-phase AC servomotor, the rotor is designed to have a **high resistance-to-reactance ratio** so that the torque is approximately proportional to the control signal and the motor has good dynamic response. Hence **X/R should be small** (i.e., resistance dominates reactance). Also, the rotor is made **long and slender**, so the **length-to-diameter ratio L/D is large**. Therefore, the correct choice is **(c) X/R is small and L/D is large**.
Why the others are wrong:
- **(a) X/R and L/D both small**: wrong because L/D is not small; servomotor rotors are typically long and thin.
- **(b) X/R large and L/D small**: wrong because X/R should be small, not large.
- **(d) X/R and L/D both large**: wrong because although L/D is large, X/R should be small for a servomotor rotor.
The system matrix of a discrete system is given by A = [[0, 1], [-3, -5]]. The characteristics equation is given by
A)
z^2+5z+3=0
B)
z^2-5z-3=0
C)
z^2+3z+5=0
D)
z^2+z+2=0
For a discrete system, the characteristic equation is obtained from det(zI - A) = 0. Here, A = [[0, 1], [-3, -5]], so zI - A = [[z, -1], [3, z+5]]. Its determinant is z(z+5) - (-1)(3) = z^2 + 5z + 3. Hence the correct characteristic equation is z^2 + 5z + 3 = 0, which matches option (a).
Why the others are incorrect:
- (b) z^2 - 5z - 3 = 0 has the wrong signs for the trace and constant term.
- (c) z^2 + 3z + 5 = 0 does not match the determinant expansion of zI - A.
- (d) z^2 + z + 2 = 0 is also inconsistent with the matrix entries.
Therefore, the correct answer is (a).
In a short transmission line, voltage regulation is zero when the power factor angle of the load at the receiving end side is equal to
A)
tan-1(X/R)
B)
tan-1(R/X)
C)
tan-1(R/Z)
D)
tan-1(X/Z)
For a short transmission line, the voltage regulation is approximately
\nVR \propto IR\cos\phi + IX\sin\phi\n\nFor zero voltage regulation, this must be zero:
\nR\cos\phi + X\sin\phi = 0\n\nSo,
\n\tan\phi = -R/X\n\nHence the load power factor angle must be leading, with magnitude
\n\phi = \tan^{-1}(R/X)\n\nTherefore the correct option is (a) tan⁻¹(R/X).\n\nWhy the others are incorrect:\n- (b) tan⁻¹(R/X) is actually the correct expression, so this is the right choice only if option labels are matched exactly; in the given list, it corresponds to option b? Wait, the listed options show a = tan⁻¹(X/R), b = tan⁻¹(R/X), c = tan⁻¹(R/Z), d = tan⁻¹(X/Z). Thus the correct key is b.\n- (a) tan⁻¹(X/R) is the reciprocal ratio and does not satisfy the zero-regulation condition.\n- (c) tan⁻¹(R/Z) and (d) tan⁻¹(X/Z) are not the standard zero-regulation condition for a short line; regulation depends on R and X through tan⁻¹(R/X), not on Z directly.\n\nSo the correct answer is option b.
Which of the following are true in respect of interpoles in dc machines? I. Reduce armature reaction effects in interpolar region. II. Windings are connected in series with armature. III. Have the same dimensions as main poles.
A)
I and II
B)
II and III
C)
I and III
D)
I, II and III
Interpoles (commutating poles) in DC machines are used to improve commutation by counteracting the armature reaction in the interpolar/neutral zone, so statement I is true. Their windings are connected in series with the armature so that the interpole flux varies automatically with armature current, making statement II true. Statement III is false because interpoles are much smaller than the main poles; they do not have the same dimensions as the main poles. Therefore, the correct option is I and II only.
Why the other options are incorrect: c and b include statement III, which is false. d includes III as well, so it is also incorrect.