Junior Engineer • 2015 • Electrical Engineering Paper-II
Slip of the induction machine is 0.02 and the stator supply frequency is 50 Hz. What will be the frequency of the rotor induced emf?
A)
10 Hz.
B)
50 Hz.
C)
1 Hz.
D)
2500 Hz.
For an induction motor, rotor induced emf frequency is given by f_r = s × f_s, where s is slip and f_s is stator supply frequency. Here, s = 0.02 and f_s = 50 Hz, so f_r = 0.02 × 50 = 1 Hz. Hence, the correct option is (c).
Junior Engineer • 2015 • Electrical Engineering Paper-II
A salient pole synchronous motor is running at no load. Its field current is switched off. The motor will
A)
come to stop
B)
continue to run at synchronous speed
C)
continue to run at a speed slightly more than the synchronous speed
D)
continue to run at a speed slightly less than the synchronous speed
A salient pole synchronous motor needs DC field excitation to produce rotor magnetic poles and maintain synchronism with the rotating stator field. If the field current is switched off at no load, the motor loses synchronizing torque. Since there is no load, it may momentarily keep rotating due to inertia, but it cannot remain at synchronous speed and will gradually slow down and come to rest. Hence, the correct answer is (a).
Junior Engineer • 2015 • Electrical Engineering Paper-II
If the field of a synchronous motor is under excited, the power factor will be
A)
leading
B)
lagging
C)
more than unity
D)
unity
In a synchronous motor, if the field is under-excited, the motor draws more magnetizing current from the supply. This makes the armature current lag behind the supply voltage, so the power factor becomes lagging. If it were over-excited, the power factor would become leading; at normal excitation, it can be unity.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The power factor of a squirrel cage induction motor is
A)
low at light load only
B)
low at heavy load only
C)
low at light and heavy load both
D)
low at rated load only
In a squirrel cage induction motor, the magnetizing current is required even when the load is small. At light load, this magnetizing component dominates the stator current, so the power factor is low. As the load increases, the active component of current increases and the power factor improves, becoming better near rated load. Therefore, the power factor is low at light load only.
Junior Engineer • 2015 • Electrical Engineering Paper-II
To start and run ceiling fans, generally we use
A)
inductor
B)
resistor
C)
transistor
D)
capacitor
Ceiling fans generally use a single-phase capacitor motor. The capacitor provides the necessary phase shift to create a rotating magnetic field, which helps the motor start and run smoothly. Inductors, resistors, and transistors are not the usual starting/running component for ceiling fan motors.
Junior Engineer • 2015 • Electrical Engineering Paper-II
Power transistor has ____ switching time than SCR
A)
lower
B)
greater
C)
equal
D)
none of these
A power transistor is a majority-carrier device and switches faster than an SCR. An SCR is a thyristor and has slower turn-off/commutation due to stored charge. Therefore, the switching time of a power transistor is lower than that of an SCR.
Junior Engineer • 2015 • Electrical Engineering Paper-II
A solar module is a
A)
Series-arrangement of solar cells
B)
Parallel arrangement of solar cells
C)
Series-parallel arrangement of solar cells
D)
None of these
A solar module (solar panel) is made by connecting multiple solar cells together in a series-parallel combination to obtain the required voltage and current. Series connection increases voltage, while parallel connection increases current. Hence, the correct answer is series-parallel arrangement of solar cells. Options (a) and (b) are incomplete because a module is not formed by only series or only parallel connection in general.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The relative speed between the magnetic fields of stator and rotor under steady state operation is zero for
A)
a dc machine
B)
a 3 phase induction machine
C)
asynchronous machine
D)
all of these
In a DC machine, the stator field is stationary and the rotor field is also stationary with respect to the stator under steady-state operation, so the relative speed between the two magnetic fields is zero. In a 3-phase induction machine (asynchronous machine), the rotor must run at a speed different from synchronous speed to induce emf, so there is always some relative speed (slip) between stator and rotor fields. Therefore, the correct answer is only a DC machine.
Junior Engineer • 2015 • Electrical Engineering Paper-II
Six boys Puia, Muana, Rina, Sanga, Thanga and Dinga are sitting around a table. Puia is second to the left of Dinga. Muana is between Rina and Sanga. Thanga is second to the left of Sanga. Who is the fourth person to the left of Muana?
A)
Puia
B)
Dinga
C)
Rina
D)
Data inadequate
Let us place the boys around the circular table.
- Puia is second to the left of Dinga.
- Thanga is second to the left of Sanga.
- Muana is between Rina and Sanga.
The last statement only tells us that Muana sits adjacent to both Rina and Sanga, but it does not fix the full circular arrangement uniquely. More than one seating arrangement is possible, and in different valid arrangements, the fourth person to the left of Muana can be different.
Therefore, the information is insufficient to determine a unique answer.
So the correct option is: Data inadequate.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The frequency of the rotor current in a 3 phase 50 Hz, 4 pole induction motor at full load speed is about
A)
50 Hz.
B)
20 Hz.
C)
2 Hz.
D)
Zero.
For an induction motor, rotor current frequency is given by f_r = s f_1, where s is slip and f_1 is supply frequency. At full load, slip is small; for a 50 Hz, 4-pole motor running near synchronous speed, the slip is typically about 0.04. So f_r ≈ 0.04 × 50 = 2 Hz. Hence the correct answer is 2 Hz. The other options are incorrect because 50 Hz would be at standstill, 20 Hz is too high for full-load slip, and zero would only occur at synchronous speed, which an induction motor never exactly reaches under load.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The solar or photo voltaic cell converts
A)
Chemical energy to electrical energy
B)
Solar radiation into electrical energy
C)
Solar radiation into thermal energy
D)
Thermal energy into electrical energy
A photovoltaic (solar) cell works on the photovoltaic effect, in which sunlight is directly converted into electrical energy. Option (a) describes a chemical cell/battery, option (c) refers to solar thermal conversion, and option (d) is not the working principle of a solar cell.
Junior Engineer • 2015 • Electrical Engineering Paper-II
When two transformers are operating in parallel, they will share the load as under
A)
Proportional to their impedances
B)
25%-75%
C)
Inversely proportional to their impedances
D)
50% - 50%
When transformers operate in parallel, the load current divides according to their impedances. The transformer with lower impedance carries more current, so the load sharing is inversely proportional to their impedances. Hence, option (c) is correct. Option (a) is wrong because load is not shared proportional to impedance; it is the opposite. Options (b) and (d) are only special cases and not the general rule.
Junior Engineer • 2015 • Electrical Engineering Paper-II
Internal voltage drop of a voltage source
A)
Independent of the load current supplied
B)
Depends on the internal resistance of the source
C)
Does not influence the terminal voltage
D)
Does affect the emf of the source
The internal voltage drop in a source is the drop across its internal resistance, given by I r. Therefore it depends on the internal resistance of the source and the load current. It reduces the terminal voltage, but it does not change the emf of the source. Hence (b) is correct. (a) is wrong because it varies with current, (c) is wrong because it directly affects terminal voltage, and (d) is wrong because emf is an inherent property of the source.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The two primary parts of a 3-phase induction motor are
A)
rotor and stator
B)
stator and field
C)
slip-ring and brushes
D)
rotor and armature
A 3-phase induction motor has two main parts: the stator (stationary outer part) and the rotor (rotating inner part). The stator produces the rotating magnetic field, and the rotor receives induced current and turns. The other options are incorrect because field and armature are terms used mainly in DC machines, while slip-rings and brushes are only accessories in slip-ring induction motors, not the two primary parts.
Junior Engineer • 2015 • Electrical Engineering Paper-II
Zero sequence currents can flow from a line into a transformer bank if the windings are
A)
Grounded Star/Delta
B)
Delta/Star
C)
Star/Star grounded
D)
Delta/Delta
Zero-sequence current requires a return path through the neutral/ground. In a transformer bank, it can flow from a line into the transformer only when the winding is star-connected with the neutral grounded. Among the options, Star/Star grounded provides this path. Delta windings block zero-sequence line currents from entering the transformer line terminals, so (a), (b), and (d) are not correct.
Junior Engineer • 2015 • Electrical Engineering Paper-II
Whenever current carrying conductor is placed in a magnetic field
A)
it experiences force
B)
it experiences attraction
C)
it generates voltage
D)
none of these
A current-carrying conductor placed in a magnetic field experiences a mechanical force due to the interaction between the magnetic field and the current. This is the motor effect, given by Fleming’s left-hand rule. It is not necessarily always attraction, and it does not directly generate voltage in this situation.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The voltage of solar cell is
A)
0.5 to 1V
B)
1 to 2V
C)
2 to 3 V
D)
3 to 4V
A single solar cell typically produces about 0.5 to 0.6 V under normal operating conditions, and in many exam contexts it is rounded to about 0.5 to 1 V. Higher voltages are obtained by connecting many cells in series to form a solar panel or module. Therefore, option (a) is correct.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The two windings of a transformer is
A)
Conductively linked
B)
Inductively linked
C)
Not linked at all
D)
Electrically linked
In a transformer, the primary and secondary windings are not electrically connected; instead, power is transferred from one winding to the other through mutual induction. Hence, the two windings are inductively linked. Option (a) is wrong because there is no conductive connection, and option (d) is wrong because they are not electrically linked. Option (c) is incorrect because the windings are linked through magnetic flux.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The synchronous speed of induction motors having 4 nos. of poles on the winding will be
A)
2000
B)
1000
C)
750
D)
500
Synchronous speed of an induction motor is given by Ns = 120f / P. For standard supply frequency f = 50 Hz and number of poles P = 4, Ns = 120 × 50 / 4 = 1500 rpm. However, the options given do not include 1500 rpm. In many exam contexts, if the intended frequency were 25 Hz, the speed would be 750 rpm, but for a normal 50 Hz system the correct synchronous speed is 1500 rpm. Since the question asks for the synchronous speed of a 4-pole induction motor and the provided options are inconsistent with the standard formula, the item appears flawed.
Junior Engineer • 2015 • Electrical Engineering Paper-II
A sodium-vapor lamp is
A)
a gas-discharge lamp
B)
an incandescent lamp
C)
a mercury-discharge lamp
D)
none of these
A sodium-vapor lamp produces light by an electric discharge through sodium vapor, so it is a gas-discharge lamp. It is not an incandescent lamp because it does not work by heating a filament. It is also not a mercury-discharge lamp, since the discharge medium is sodium vapor.
Junior Engineer • 2015 • Electrical Engineering Paper-II
A 1:5 step-up transformer has 120V across the primary and 600 ohms resistance across the secondary. Assuming 100% efficiency, the secondary current equals
A)
0.2 Amp.
B)
5 Amps.
C)
1 Amp.
D)
20 Amps.
For an ideal transformer, the voltage ratio equals the turns ratio. A 1:5 step-up transformer means the secondary voltage is 5 times the primary voltage.
Primary voltage = 120 V
Secondary voltage = 5 × 120 = 600 V
Given secondary resistance = 600 ohms, the secondary current is:
I_s = V_s / R = 600 / 600 = 1 A
However, the question asks for a 1:5 step-up transformer with 120 V across the primary and 600 ohms resistance across the secondary. In standard exam interpretation, the 600 ohms is the load on the secondary, so the current is 1 A. But since the options do not include 1 A as the most likely intended answer from the transformer ratio-based current relation, the question appears to be framed ambiguously.
If interpreted strictly from Ohm's law on the secondary side, the answer is 1 A (option c).
Junior Engineer • 2015 • Electrical Engineering Paper-II
During the discharge of lead acid battery, the terminal voltage decreases with the decrease in
A)
Temperature
B)
Discharge rate
C)
State of charge
D)
None of these
In a lead-acid battery, terminal voltage falls during discharge as the battery’s state of charge decreases. As the active materials are consumed and the electrolyte concentration drops, the internal chemical potential reduces, causing the terminal voltage to decrease. Temperature and discharge rate can affect voltage, but the direct and expected relation here is with state of charge. Hence, option (c) is correct.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The magnetising current drawn by induction motor is the cause of their
A)
zero p.f
B)
unity p.f
C)
lagging p.f
D)
leading p.f
In an induction motor, the magnetising current is mainly reactive current used to establish the rotating magnetic field. Reactive current lags the supply voltage, so the motor draws a lagging power factor. It is not zero p.f because the motor also takes active power for mechanical output, and it is not unity or leading p.f because the magnetising component is inductive in nature.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The output of solar cells is of the order of
A)
0.5 watts
B)
1.0 watts
C)
5.0 watts
D)
10 watts
A single solar cell typically produces only a small output, generally around 0.5 W under standard test conditions. Solar panels are formed by connecting many such cells in series and parallel to obtain higher power. Hence, among the given options, 0.5 watts is correct. The other options are too high for an individual solar cell.
Junior Engineer • 2015 • Electrical Engineering Paper-II
The working principle of ceiling fan is based on
A)
a single phase induction motor
B)
asynchronous motor
C)
a three phase synchronous motor
D)
an alternator
A ceiling fan works on the principle of a single-phase induction motor, usually a capacitor-run or capacitor-start-and-run type. It uses the rotating magnetic field produced by the auxiliary winding and capacitor to start and run the fan. Option (b) is too general because an induction motor is inherently asynchronous, but the standard answer for a ceiling fan is single-phase induction motor. Option (c) is incorrect because synchronous motors run at synchronous speed and are not used in ordinary ceiling fans. Option (d) is incorrect because an alternator generates AC power, not mechanical rotation for a fan.