Quiz: Electrical Engineering 26 Oct 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. For V-curve for a synchronous motor the graph is drawn between
(a) Armature current and power factor
(b) Field current and armature current
(c) Terminal voltage and power factor
(d) Power factor and field current

Q2. The repulsion-start induction-run motor is used because of:
(a) High starting torque
(b) Good power factor
(c) High efficiency
(d) Minimum cost

Q3. The ratio of puncture voltage to the flashover voltage of an insulator is
(a) Equal to one
(b) Lower than one
(c) Zero
(d) Greater than one

Q4. The resistance welding process requires a
(a) High value of ac current at low voltage
(b) Low value of ac current at high voltage
(c) High value of dc current at low voltage
(d) Low value of dc current at high voltage

Q5. In arc welding, the electrodes are made of
(a) Copper
(b) Aluminum
(c) Graphite
(d) ACSR conductor

Q6. Earth fault relays are
(a) Directional relays
(b) Non-directional relays
(c) Short operate time relays
(d) Long operate time relays

Q7. In which of the single-phase motor the rotor has no teeth or winding?
(a) Universal motor
(b) Split phase Motor
(c) Reluctance Motor
(d) Hysteresis motor

Q8. In a capacitor start single phase induction motor, the current in the
(a) Supply lines leads the voltage
(b) Starting winding lags the voltage
(c) Main winding leads the voltage
(d) Starting winding leads the voltage

Q9. Which of the following welding process uses non-consumable electrode?
(a) LASER welding
(b) MIG welding
(c) TIG welding
(d) Ion-beam welding

Q10. A 220 V DC machine has an armature resistance 1 Ω. If the full load current is 20 A, the difference of induced voltage between generator and motor is
(a) 0 V
(b) 20 V
(c) 40 V
(d) 60 V

SOLUTIONS
S1. Ans.(b)
Sol. V curve is a plot of the stator current versus field current for different constant loads.

S2. Ans.(a)
Sol. The repulsion-start induction-run motor is used because of high starting torque.
The starting torque of the repulsion start motor is 350 percent with moderate starting current.
It is usually applicable where the starting period is of a long duration because of high inertia of the load. such as the high-speed lift.

S3. Ans.(d)
Sol. Flashover Voltage: The voltage at which the air around insulator breaks down and flashover takes place shorting the insulator is called Flash Over Voltage.
Puncture Voltage: The voltage at which the insulator breaks down and current flows through the inside of insulator is called Puncture Voltage.
As puncture destroys the insulator, it is more serious than flash-over. Therefore, Safety Factor is defined for an Insulator. Safety factor of an Insulator is defined as the ration of Puncture Voltage to the Flash Over Voltage.
Safety Factor = Puncture Voltage / Flash Over Voltage
So, (Puncture Voltage)/(Flash Over Voltage)>1
For pin type insulator the value of Safety Factor is about 10 which mean that Puncture Voltage is 10 times that of Flash Over Voltage.

S4. Ans.(a)
Sol. In welding process step down transformer is used which has low voltage and high current.

S5. Ans.(c)
Sol. In arc welding, the electrodes are made of graphite.

S6. Ans.(b)
Sol. if the fault current passes in either direction in earth fault relay, it gives a trip condition. Hence, earth fault relay is a non-directional relay.

S7. Ans.(d)
Sol. Hysteresis motor doesn’t have teeth or winding; therefore, these motors are free from mechanical vibration.

S8. Ans.(d)
Sol. In the capacitor start inductor motor two winding is used, the main winding and the starting winding. With starting winding, we connect a capacitor so the current flowing in the capacitor leads the applied voltage by some angle 𝟇.

S9. Ans.(c)
Sol. In TIG welding process uses non-consumable TUGSTEN electrode is used.

S10. Ans.(c)
Sol. in case of generator: E_g=V_t+I_a R_a
In case of motor: E_b=V_t-I_a R_a
∴E_g-E_b=2I_a R_a=2×20×1=40 V

Quiz: Electrical Engineering 12 Sep 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. D.C. shunt motors are used in those applications where ……… is required.
(a) high starting torque
(b) practically constant speed
(c) high no-load speed
(d) variable speed

Q2. Cumulatively compounded motors are used where we require ………
(a) variable speed
(b) poor speed regulation
(c) sudden heavy loads for short duration
(d) none of the above

Q3. For 20% increase in current, the motor that will give the greatest increase in torque
is ……. motor.
(a) shunt
(b) series
(c) cumulatively compounded
(d) differentially compounded

Q4. A 440 V shunt motor has an armature resistance of 0.8 Ω and a field resistance of 200
Ω. Find the back e.m.f. when giving an output of 7.46 kW at 85% efficiency.
(a) 222.4 V
(b) 425.8 V
(c) 312.6 V
(d) 392.7 V

Q5. A parallel resonance……….
(a) circuit impedance is minimum
(b) power factor is zero
(c) line current is maximum
(d) power factor is unity

Q6. Dynamic impedance of a parallel tuned circuit is ……….
(a) L/CR
(b) RL/C
(c) L/C
(d) R/L

Q7. When the supply frequency is more than the resonant frequency in a parallel a.c.
circuit, then circuit is…….
(a) Resistance
(b) capacitive
(c) inductive
(d) none of the above

Q8. A dc series motor drawing an armature current of I_a is operating under saturated magnetic conditions, torque developed in the motor is proportional to
(a) 1/Ia
(b) 1/Ia^2
(c) Ia^2
(d) Ia

Q9. The insulation resistance of a cable of length 10 Km is 1 MΩ. Its resistance for 50 Km length will be
(a) 1 MΩ
(b) 0.5 MΩ
(c) 0.2 MΩ
(d) None of these

Q10. For a circuit given that I = 2 ± 5 % A, R = 100 ± 0.2 % Ω the limiting error in the power dissipation I2 R in the resistor R is
(a) 1.2 %
(b) 5.2 %
(c) 10.2 %
(d) 25.2 %

SOLUTIONS
S1. Ans.(b)
Sol. Practically constant speed

S2. Ans.(c)
Sol. Cumulatively compounded motors are used where we require sudden heavy loads
for short duration

S3. Ans.(b)
Sol. Series motor because
T∝Ia^2 (Before saturation)

S4. Ans.(b)
Sol. Motor input power = (7.46×1000)/0.85 w
Motor input current = 7460/(0.85×440) = 19.95 A
Ish= 440/200 = 2.2 A
Ia= 19.95 – 2.2 = 17.75 A
Eb= V – Ia Ra= 440 – 17.75 × 0.8 = 425.8

S5. Ans.(d)
Sol. Power factor unity

S6. Ans.(a)
Sol. Dynamic impedance of parallel tuned circuit =L/RC

S7. Ans.(b)
Sol. Capacitive

S8. Ans.(d)
Sol. We know Torque Tα ΦIa, As the motor is operating under saturation condition so torque will be independent of flux i.e. T α Ia.

S9. Ans.(c)
Sol. Insulation resistance of the cable
R α 1/Length
R2/R1 = L1/L2
Hence, insulation resistance for 50 km length
R2 = (1 x 10)/50
= 0.2 MΩ

S10. Ans.(c)
Sol. Power dissipation P=I2R

Quiz: Electrical Engineering 31 Aug 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. “M.M.F around any closed path equals the current enclosed by the path”. This is known as
(a)Coulomb’s law
(b)Gauss’ law
(c)Ampere’s law
(d)Biot-savart law

Q2. Electro-dynamometer type of instrument is enclosed in a casing made up of high permeability material to:
(a)To remove electrostatic effect
(b)To provide mechanical strength to the coils present
(c)To protect against external magnetic fields
(d)Provide good damping

Q3. According to IE Rules the maximum load on a power sub circuit should not exceed……… watts.
(a)1000
(b)3000
(c)5000
(d)10000

Q4. A universal motor has high starting torque and a variable speed characteristic because it is……………
(a)Shunt wound
(b)Long shunt wound
(c)Compound wound
(d)Series wound

Q5. Hopkinson’s test for DC motor is conducted with
(a)Full load
(b)Half load
(c)Low load
(d)No load

Q6. In a double squirrel cage motor, outer cage is made of high resistance metals bars primarily for the purpose of increasing its……….
(a)Starting torque
(b)Speed regulation
(c)Efficiency
(d)Starting current

Q7. During regenerative braking the motor operates as:
(a)Motor
(b)Inverter
(c)Transformer
(d)Generator

Q8. The electric supply authority supplies power to the consumers through a low voltage three phase four wire distribution system is called……
(a)Primary transmission system
(b)Primary distribution system
(c)Secondary transmission system
(d)Secondary distribution system

Q9. ………….. wiring is used in residential, commercial and public building to improve appearance.
(a)Conduit
(b)Metal
(c)Cleat
(d)Batten

Q10. When a material becomes a superconductor, its resistivity becomes
(a)Zero
(b)Very small
(c)Large
(d)About 20% of the normal value

SOLUTIONS
S1. Ans.(c)
Sol. The integral around a closed path of the component of the magnetic field tangent to the direction of the path equals µ0 times the current intercepted by the area within the path.
i.e. ∫▒〖B ⃗. (dS) ⃗ 〗=µOI
NOTE: In order to apply Ampère’s Law all currents have to be steady (i.e. do not change with time)

S2. Ans.(c)
Sol. The field produced by the fixed coils of electro dynamometer type instruments is weaker i.e. nearly 0.005 to 0.006 Wb/m*m.
In dc measurements, even the Earth’s magnetic field may affect the readings. Thus, it is necessary to shield and electrodynamometer type instrument from the effect of stray magnetic fields. Air cored electrodynamometer instruments are protected against external magnetic fields by enclosing them in a casing of high permeability alloy. This shunts external magnetic fields around the instrument mechanism and minimizes their effects on the indication.

S3. Ans.(b)
Sol. The sub-circuits may be divided into the following two groups:
(i) light and fan (5A) sub-circuits: the numbers of such points (i.e. lights, fans, 5A socket outlasts) connected in one sub-circuit should not exceed ten and load connected in one sub-circuit should not exceed 800 watts.
(ii) power (15A) sub-circuits: Single phase 15A sub-circuits (power sub-circuits) are used for connecting heaters, stoves, electric iron, small single-phase motors etc. The load connected in a power sub-circuit normally should not exceed 3,000 watts and the number of outlasts connected on each sub circuit in no case should exceed two.

S4. Ans.(d)
Sol. A universal motor has high starting torque and a variable speed characteristic because it is series wound.
A universal motor is a series-wound motor — meaning that the field and armature windings are connected in series — and is mechanically commutated with brushes and a commutator. Although its construction is very similar to a series-wound DC motor, a universal motor incorporates several modifications that allow it operate properly on either DC or AC power supply.

S5. Ans.(a)
Sol. Hopkinson’s test for DC motor is conducted with full load.

S6. Ans.(a)
Sol. A Double Cage Induction motor is that type of motor in which a double cage or two rotor windings or cages are used.
The outer cage bars have a smaller cross-sectional area than the inner bars and are made of high resistivity materials like brass, aluminium, bronze, etc. the bars of the inner cage are made of low resistance copper. Thus, the resistance of the outer cage is greater than the resistance of the inner cage.
At starting most of the current flows through outer cage despite its large resistance (as total impedance is lower than the inner cage). This will not affect the outer cage because of its low reactance. And because of the large resistance of outer cage starting torque will be large.

S7. Ans.(d)
Sol. During regenerative braking the motor operates as generator.

S8. Ans.(d)
Sol. The electric supply authority supplies power to the consumers through a low voltage three phase four wire distribution system is called secondary distribution.

S9. Ans.(a)
Sol. Conduit wiring is used in residential, commercial and public building to improve appearance. Conduit Wiring is the best system of wiring. There are two methods for laying conduits. In one method conduit is laid on the surface of the wall, ceiling etc. This is called surface wiring. The other method is to lay the conduit recessed in wall or ceiling, and this system is known as concealed wiring.
Advantages of Conduit Wiring:
The safest wiring
Appearance is better
No risk of fire or mechanical wear and tear.
No risk of damage of cable insulation
Safe from humidity, smoke, steam etc.
No risk of shock
Long lasting

S10. Ans.(a)
Sol. Superconductor has zero resistivity.

 

Quiz: Electrical Engineering 27 Aug 2020

Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. For the equation, s^3-4s^2+s+6=0 the number of roots in the left half of s-plane will be
(a)Zero
(b)One
(c)Two
(d)Three

Q2. The de-ionization of the medium in current zero method of an arc extinction is NOT achieved by
(a)Splitting the arc
(b)Lengthening of the gap between the contacts
(c)Increasing the pressure of the vicinity of the arc
(d)Cooling the arc

Q3. In a synchronous motor, during hunting when the rotor speed exceeds the synchronous speed then damper bar develop
(a)Induction generator torque
(b)Harmonic
(c)DC motor torque
(d)Synchronous motor torque

Q4. Power consumed by a balanced 3-phase; 3-wire load is measured by two wattmeter method. The first wattmeter reads twice that of the second. Then what will be the load impedance angle in radian?
(a) (π/6)
(b) (π/3)
(c) (π/2)
(d) (π/4)

Q5. Single or One Wattmeter method can only be used for
(a) Balanced three-phase load
(b) Imbalanced two-phase load
(c) Balanced one-phase load
(d) Imbalanced one-phase load

Q6. A series RLC circuit consisting of R = 10 Ω, X_L=20 Ω and X_C=20 Ω, is connected across an AC supply of 100 V (rms). The magnitude and pulse angle (with respect to supply Voltage) of the voltage across the induction coil are respectively.
(a) 100 V; 90°
(b) 100 V; -90°
(c) 200 V; -90°
(d) 200 V; 90°

Q7. The impedance of a circuit is given by Z = 3 + j4. its conductance will be:
(a) 3/4
(b) 3/25
(c) 1/3
(d) 3/7

Q8. At any power factor of the load the efficiency of transformer will be maximum when:
(a) copper loss is equal to core loss
(b) copper loss is equal to eddy current loss
(c) copper loss is less then core loss
(d) copper loss is greater than core loss

Q9. In a synchronous machine, all of the following losses are independent of the load EXCEPT:
(a) iron loss
(b) bearing friction
(c) windage loss
(d) copper loss

Q10. A synchronous machine with low value of short circuit ratio has:
(a) higher stability limit
(b) good voltage regulation
(c) good speed regulation
(d) lower stability limit

SOLUTIONS
S1. Ans.(b)
Sol. s^3-4s^2+s+6=0
by R-H criterion:
s^3 1 1
s^2 -4 6
s^1 (-4-6)/(-4)=+2.5
s^0 6
Here there is two sign changes( s^3 to s^2 and s^2 to s^1).
So, number of roots in RHS=2
∴Number of roots in LHS=3-2=1.

S2. Ans.(a)
Sol. There are two methods of extinguishing the arc in CB viz.
(a)High Resistance Method
(b)Low resistance Method or current zero method
Low resistance Method or current zero method: This method is employed for arc extinction in AC circuits only. In this method, the rapid increase of dielectric strength of the medium near current zero can be achieved by:
(a)Lengthening of the gap
(b)High pressure in the vicinity of the arc
(c)Blast effect
(d)Cooling
NOTE: Splitting the arc method is related with arc extinction by High resistance method.

S3. Ans.(a)
Sol. When hunting occurs, the difference in the speed of stator and rotor poles develops an induces emf in the damper winding, which acts in such a way to suppress the rotor oscillation. The induced emf generated develop induction torque in the synchronous motor.

S4. Ans.(a)
Sol. W1=2W2
W(1 ):W2=2:1
Let W1=2
W2=1
Then,
tanθ= √3 ((W1-W2)/(W1+W2 ))
=√3 ((2-1)/(2+1))
=√3×1/3=1/√3
⇒θ=tan^(-1) (1/√3)=30°=(π\/6)radian

S5. Ans.(a)
Sol.
I. one wattmeter → for 3-ϕ balanced load
II. Two wattmeter → 3-ϕ balanced, unbalanced star, Delta connected load
III. Three wattmeter →3-ϕ, 4 wire system

S6. Ans.(d)
Sol.

Here XL=XC=200 Ω.
So, it is case of series Resonance.
So, Z =R = 10 Ω.
IR =V/R=100/10 =10 A
IR = IL = IC (series combination)
So, VL=I× XL
= 10 × 20
= 200 V

├ ■(VL=200V@θ=90° leading)⌉

S7. Ans.(b)
Sol. given z = 3 + 4j
conductance (G) =?
Admittance (γ) = 1/Z
= 1/(3+j 4)
= 1/(3+j4) × ((3-j))/((3-j4))
= (3-j4)/25
= 3/25-4/25 j

▭(γ = 1/z = G + j B)
↓ ↓
conductance susceptance
on comparing
G = 3/25

S8. Ans.(a)
Sol. At any power factor of the load, the efficiency of transformer will be maximum when copper loss is equal to core loss.

S9. Ans.(d)
Sol. In a synchronous machine, copper loss is load dependent.

S10. Ans.(d)
Sol. Xd in p.u = 1/SCR
SCR = 1/ Xd in per unit
Thus, short circuit ratio is equal to the reciprocal of per unit value of direct axis synchronous reactance. The more the value of Xd, the lesser will be the short circuit ratio.
For the small value of the short circuit ratio (SCR), the synchronizing power is small. As the synchronizing power keeps the machine in synchronism, a lower value of the SCR has a low stability limit. In other words, a machine with a low SCR is less stable when operating in parallel with the other generators.

Quiz: Electrical Engineering 8 Aug 2020

Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. Which damping is used in moving iron instrument?
(a) Air friction damping
(b) Eddy current damping
(c) Fluid friction damping
(d) Electromagnetic damping

Q2. A voltage and current waveform of an element below.
V(t) = 120 sin (3t + 120°)
I(t) = 20 cos (3t – 60°)
The element is ……?
(a) Pure inductor, L = 2H
(b) Pure capacitor
(c)Resistance of 3 Ω in series with inductor of L = 1 H
(d) Resistance of 3 Ω in series with capacitor of

Q3. The Eddy current loss is proportional to the
(a) Frequency
(b) Square of the frequency
(c) Cube of the frequency
(d)Square root of the frequency

Q4. Electrical fault between two windings of same phase of a generator having double star winding can be detected by the following protection in a generator:
(a) Short circuit protection
(b) Earth fault protection
(c) Inter turn Fault protection
(d) Over voltage protection

Q5. Negative sequence relay is used commonly to protect
(a) An alternator
(b) A transformer
(c) A transmission lines
(d) A bus bars

Q6. In Fig., the supply voltage is ……….


(a) 500 V
(b) 100 V
(c) 200 V
(d) 400 V

Q7. In a parallel a. c. circuit, if the supply frequency is less than the resonant frequency, then the circuit is ………….
(a) inductive
(b) capacitive
(c) resistive
(d) none of the above

Q8. The tendency of charge carriers to move from a region of heavily concentrated charges to region of less concentrated charge is known as.
(a) Depletion current
(b) Drain current
(c) Diffusion current
(d) Saturation current

Q9. The type number of the control system with G(s) H(s) = (K(s+2))/(s(s^2+2s+3)) is
(a) One
(b) Two
(c) Three
(d) Four

Q10. What is the purpose of the address bus?
(a) to provide data to and from the chip
(b) to select a specified chip
(c) to select a location within the memory chip
(d) to select a read/write cycle

SOLUTIONS

S1. Ans.(a)
Sol. In moving iron instrument the damping which is used is Air friction damping, by a vane moving in a sector shaped chamber.

S2. Ans.(a)
Sol. V(t) = 120 cos (3t + 120° – 90°); V(t) = 120 cos (3t + 30°)
I(t) = 20 cos (3t – 60°)
Z=V/I=(120∠30)/(20∠-60)=6∠90⁰=j6
∴Z = j6 Ω, ωL = 6, 3×L=6 i.e. L=2 H

S3. Ans.(b)
Sol. Eddy current loss in a transformer is fixed, and it depends on the core material’s magnetic properties. Eddy current loss is given as:

S4. Ans.(c)
Sol. Inter turn stator winding fault can easily be detected by stator differential protection or stator earth fault protection. Hence, it is not very essential to provide special protection scheme for inter turn faults occurred in stator winding. This type of faults is generated if the insulation between conductor (with distinct potentials) in the same slot is punctured. This type of fault rapidly changes to earth fault.

S5. Ans.(a)
Sol. Negative sequence currents gives rise to another rotating magnetic field in the generator in opposite direction which leads to excessive copper loss and pulsation in electromagnetic torque, which causes mechanical stress on the shaft of the alternator. So negative sequence relay is commonly used to protect alternator.

S6. Ans.(b)
Sol. Since voltage across L is equal to voltage across C, it is a case of series resonance.
∴ Supply voltage = Voltage across R = 100 V

S7. Ans.(a)
Sol. In a parallel a. c. circuit, if the supply frequency is less than the resonant frequency, then the circuit is inductive

S8. Ans.(c)
Sol. In a semiconductor the charge will always have a tendency to move from higher concentrated area to less concentrated area to maintain equilibrium this movement of charges will result in diffusion current.

S9. Ans. (a)
Sol. Pole at origin is one. Type of the system is determined by open loop poles at origin.

S10. Ans.(c)
Sol. Address bus is used to choose a particular location in the memory chip. Data bus is used to provide data to and from the chip. Chip select signals are used to select a particular chip within the memory.

Quiz: Electrical Engineering 5 AUG 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minutes’

Q1. A ferromagnetic material is one in which neighbouring atomic magnetic moments are
(a) predominantly parallel in small regions of material.
(b) predominantly parallel and unequal in small regions of material.
(c) predominantly equal and parallel throughout the material.
(d) predominantly unequal and parallel throughout the material.

Q2. P-type and N-type extrinsic semiconductors are formed by adding impurities of valency
(a) 5 and 3 respectively.
(b) 5 and 4 respectively.
(c) 3 and 5 respectively.
(d) 3 and 4 respectively.

Q3. What happens to the tension in a conductor hanged between two poles, when temperature varies?
(a) Tension increases with increase in temperature
(b) Tension decreases with increase in temperature
(c) Tension first increases and decreases with decrease in temperature
(d) Tension in conductor is independent of temperature variation

Q4. Reciprocal of permeability is
(a) reluctivity
(b) susceptibility
(c) permittivity
(d) conductance

Q5. A 4 pole, 1200 rpm DC lap wound generator has 1520 conductors. If the flux per
pole is 0.01 Weber, the emf of generator is
(a) 608 volts
(b) 304 volts
(c) 152 volts
(d) 76 volts

Q6. Interpoles are meant for
(a) increasing the speed of the motor.
(b) increasing counter emf.
(c) strengthening the main field.
(d) reducing speaking at the commutator.

Q7. The curve representing Ohm’s law is
(a) Linear
(b) Hyperbolic
(c) Parabolic
(d) Triangular

Q8. The purpose of the conservator in a transformer is
(a) to cool the winding.
(b) to prevent moisture in the transformer.
(c) to prevent short circuit of primary and secondary winding.
(d) to take up contraction and expansion of oil.

Q9. High-voltage d.c. machines use …. winding.
(a) lap
(b) wave
(c) either lap or wave
(d) none of the above

Q10. The arc utilized in electric arc welding is a
(a)High voltage, high current discharge
(b)Low voltage, low current discharge
(c) Low voltage, high current discharge
(d) High voltage, low current discharge

solutions

S1. Ans.(c)
Sol. Ferromagnetic materials exhibit a long-range ordering phenomenon at the atomic level which causes the unpaired electron spins to line up parallel with each other in a region called a domain.

S2. Ans.(c)
Sol. Pentavalent impurities Impurity atoms with 5 valence electrons produce n-type semiconductors by contributing extra electrons. Trivalent impurities Impurity atoms with 3 valence electrons produce p-type semiconductors by producing a “hole” or electron deficiency.

S3. Ans.(b)
Sol. The relationship between tension and sag is dependent on the loading conditions and temperature variations. For instance, the tension increases when temperature decreases.

S4. Ans.(a)
Sol. Reluctivity is reciprocal of permeability.

S5. Ans.(b)
Sol. The emf generated,
E=PϕNZ/(60 A)
P= 4, ϕ =0.01 Weber
A= P =4 (for lap)
N= 1200 rpm, Z = 1520
∴ Emf, E=(4×0.01×1200×1520)/(60×4)
E=304 Volts

S6. Ans.(d)
Sol. Interpoles are used to reduce the effect of armature reaction and also to nullify the effect of induced emf, generated in the coil undergoing commutation. Due to this emf, the sparking occurs and interpoles help in reducing the sparking occurring at the commutator.

S7. Ans.(a)
Sol. Ohm’s law is I=V/R or I ∝ V.
∴ The relation between V & I is linear (if 1/R is a constant).

S8. Ans.(d)
Sol. Conservator is placed at the top of the transformer. It controls the expansion and contraction of the transformer oil on heating and cooling process respectively.

S9. Ans.(b)
Sol. Wave wound for high voltage & low current whereas LAP winding used for high current and low voltage.

S10. Ans.(c)
Sol. The arc utilized in electric arc welding is a low voltage and high current discharge type.
Arc welding is a type of welding process using an electric arc to create heat to melt and join metals. A power supply creates an electric arc between a consumable or non-consumable electrode and the base material using either direct (DC) or alternating (AC) currents.

Quiz: Electrical Engineering 30 July 2020

Quiz: Electrical Engineering
Exam: UPPSC-AE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth
(a)Decreases by factor of 10
(b)Increases by factor of 10
(c)Remains the same
(d)None of the above

Q2. Transformer utilization factor of TUF = K signifies:
(a) That transformer for rectifier should be 1/K times larger than that for ac source
(b) That transformer for rectifier should be K-1 times larger than that for ac source
(c) That transformer for rectifier should be 1-K times larger than that for ac source
(d) That transformer for rectifier should be K time larger than that for ac source

Q3. A synchronous generator is feeding power to infinite bus bars at unity power factor. Its excitation is now increased. It will feed
(a)The same power but at leading power factor
(b)The same power but at lagging power factor
(c)More power at unity power factor
(d)Less power at unity power factor

Q4. A 400/200 volts transformer has pu impedance of 0.05. the HV side voltage required to circulate full load current during short circuit test is
(a)20 V
(b)40 V
(c)10 V
(d)5 V

Q5. In a D.C. machine, fractional pitch winding is used
(a)to increase the generated voltage
(b)to reduce sparking
(c)to save the copper because of shorter end connections
(d)due to (b) and (c) above

Q6. Why are shunt reactors connected at the receiving end of long transmission line system?
(a)To increase the terminal voltage
(b)To compensate voltage rise caused by capacitive charging at light load
(c)To improve power factor
(d)None of these

Q7. Which of the following transmission line have more initial cost?
(a)Overhead Transmission
(b)Underground transmission
(c)Both have almost the same initial cost
(d)None of the above

Q8. Which one of the following is the CORRECT relation between the peak and average value of an alternating current?
(a) Iavg=0.637IPeak
(b) Iavg=1.414IPeak
(c) Iavg=0.7071IPeak
(d) Iavg=0.8241IPeak

Q9. With regard to filtering property, lead compensator is
(a)Low pass filter
(b)High pass filter
(c)Band pass filter
(d)Band reject filter

Q10. The maximum phase occurs at the ………. of the two corner frequencies?
(a) arithmetic mean
(b) geometric mean
(c)Both a& b
(d)None of the above

SOLUTIONS

S1. Ans.(b)
Sol. for an amplifier, the product of gain-bandwidth is always constant. So, if gain reduces by a factor of 10 then the bandwidth increases by a factor of 10.

S2. Ans.(a)
Sol. The transformer utilization factor (TUF) of a rectifier circuit is defined as the ratio of the DC power available at the load resistor to the AC rating of the secondary coil of a transformer.
TUF = (DC POWER output (pdc))/(Effective VA rating of transformer)
TUF indicate how much is the utilization of the transformer in a circuit.
▭(TUF=Pdc/Pac(rated) )

▭(■(TUF for full wave rectifier=0.812@TUF for half wave rectifier=0.287))

For ideal condition, TUF=1 and apart from this, TUF is always less than 1.

S3. Ans.(b)
Sol. from the V-curve of an alternator, it is clear that when excitation is increased, the alternator feeds a lagging power factor.

S4. Ans.(a)
Sol. pu quantity will be same on both side.
Given Z_pu=0.05
And we know 〖Vsc〗pu=〖Isc〗pu×Zpu and 〖Isc)pu=I_rated/(I_base)
As rated current on HV side is same as base current on HV side. Then, 〖I_sc〗_pu=1
And 〖Vsc〗pu=〖Isc〗pu×Zpu=1×0.05=0.05
∴Vsc (HV)=〖Vsc〗pu×Vbase=0.05×400=20 V

S5. Ans.(d)
Sol. Some advantages of short pitch winding are:
Due to shortening span, the copper required is less.
Low copper losses
Improve waveform due to reduction in harmonic
Fractional Pitch winding reduces sparking in DC machines

S6. Ans.(b)
Sol. Shunt Reactor compensation at the receiving end help to reduce the effect of capacitance thus reducing the Ferranti effect.
Shunt Reactor absorbs the excess reactive power during no load or light load condition and thus helps in stabilizing the voltage of Transmission Line.

S7. Ans.(b)
Sol. As the voltage level increases the cost of insulation is increased therefore the underground cable is restricted to low and medium voltages.

S8. Ans.(a)
Sol. For AC, Iavg=(2Ip)/π=0.637 Ip

S9. Ans.(b)
Sol. In terms of filtering property, lead compensator acts as high pass filter.
Following are the properties of lead compensator….
For sinusoidal input, output is leading
Acts as high pass filter
Reduces rise time and peak overshoot
Increases bandwidth & stability of the system.

S10. Ans.(b)
Sol. The maximum phase occurs at the GM (geometric mean) of the two corners
frequencies.

Quiz: Electrical Engineering 29 July 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The inductance of certain moving-iron ammeter is expressed as
L=10+3Ө-Ө^ 2/4 µH,
Where Ө is deflection in radians from the zero position. The control spring torque is 25 ×10^(-6) Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is
(a) 2.4
(b) 2.0
(c) 1.2
(d) 1.0

Q2. The relative permeability of the diamagnetic material is
(a) Greater than 1
(b) Greater than 10
(c) Less than 1
(d) Greater than 100

Q3. Which of the following provides support to the insulators used in overhead lines?
(a) Cross-Arm
(b) Support/base
(c) Phase Plate
(d) Conductor

Q4. Assertion A: Fractional pitch winding is used in DC machines
Reason B: Fractional Pitch winding reduces sparking in DC machines
Which of the following is correct?
(a) B is true, but A is false
(b) A and B are true, but B is not the correct explanation of A
(c) A and B are true, and B is the correct explanation of A
(d) A is true but B is false

Q5. The essential requirement of good heating elements is
(a) High Specific resistance
(b) Free from oxidation
(c) Low-temperature coefficient of resistance
(d) All of the above

Q6. In order to switch-off and EHV circuit for maintenance, the following sequence is adopted:
(a) Open the circuit breaker, open the isolator, operate the earth switch
(b) Operate the earth switch, open the isolator, open the circuit breaker
(c) Open the isolator, operate the earth switch, open the circuit breaker
(d) Open the isolator, open the circuit breaker, operate the earth switch

Q7. The yoke of small DC machine is made up of–
(a) Cast iron
(b) Aluminium
(c) Stainless steel
(d) Copper

Q8. A 200/400 V, 10 kVA, 50 Hz single-phase transformer has, at full-load, a copper loss
of 120 W. If it has an efficiency of 98% at full-load, determine the iron losses.
(a) 84 W
(b) 117 W
(c) 92 W
(d) 106 W

Q9. An isolation transformer has primary to secondary turns ratio of
(a) 1:1
(b) 1:2
(c) 2:1
(d) can be any ratio

Q10. An autotransformer having voltage transformation ratio 0.8 supplies a load of 10
kW. The power transferred inductively from the primary to the secondary is
(a) 10 kW
(b) 8 kW
(c) zero
(d) 2 kW

solutions

S1. Ans.(c)
Sol. Given that, L=10+3Ө-Ө^2/4µH
Then dL/dӨ=(3-Ө/2)×10^(-6)
K=25×10^(-6)
And we know that the control torque for moving iron ammeter is given by :
T_c=kӨ=1/2 I^2 dL/dӨ
∴(25×10^(-6))Ө=1/2×5^2×(3-Ө/2)×10^(-6)
∴2Ө=(3-Ө/2)
⇒Ө=6/5=1.2 rad

S2. Ans.(c)
Sol. Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μr < 1). Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while non-magnetic materials have low permeability and high Reluctance.

S3. Ans.(a)
Sol. Cross arms are installed at the top of the pole for holding for holding the insulator on which the conductors are fastened. Cross arms are either made up of wood or steel angle sections.

S4. Ans.(c)
Sol. Some advantages of short pitch winding are:
(a)Due to shortening span, the copper required is less.
(b)Low copper losses
(c)Improve waveform due to reduction in harmonic
(d)Fractional Pitch winding reduces sparking in DC machines

S5. Ans.(d)
Sol. The required properties in material used for heating elements-
#High melting point.
#Free from oxidation in open atmosphere.
#High tensile strength.
#Sufficient ductility to draw the metal or alloy in the form of wire.
#High resistivity.
#Low temperature coefficient of resistance.
Following material are used for manufacturing heating element-
Nichrome
Kanthal
Cupronickel (CuNi): an alloy of copper that contains nickel and strengthening elements, such as iron and manganese.
Platinum

S6. Ans.(a)
Sol. In order to switch-off and EHV circuit for maintenance, the following sequence is adopted:
S1: Open the circuit breaker: CB is a device which can make and break the circuit under normal as well as under faulty condition by manually or automatically.
S2: Open the isolator: Isolator is a device which always operates under no load condition.
S3: Operate the Earth switch: Its function is to discharge any remnant charge due to residual magnetism which might be lethal to human and expensive testing equipment.

S7. Ans.(a)
Sol. The yoke of small Dc machine is made up of cast iron because it has a low reluctance so they will support a strong magnetic field with a high density.

S8. Ans.(a)
Sol. Output at full load at unity pf = 10×1 = 10kW
Input = output / η = 10 / 0.98 = 10.204 kW
Total F.L = 10.204-10=0.204 Kw=204W
Iron loss = 204 – 120 = 84 W

S9. Ans.(a)
Sol. For isolation transformer = 1:1

S10. Ans.(d)
Sol. Power transferred inductively
= (1-0.8) × 10 = 2kW

Quiz: Electrical Engineering 27 July 2020

Quiz: Electrical Engineering
Exam: UPSSSC-JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. In a four-pole D.C. machine:
(a) all the four poles are north poles
(b) alternate poles are north and south
(c) all the four poles are south poles
(d) two north poles follow two south poles

Q2. The field coils of DC generator are generally made up of:
(a) Mica
(b) Copper
(c) Cast Iron
(d) Carbon

Q3. The basic requirement of a d.c. armature winding is that it must be………
(a) a lap winding
(b) a closed one
(c) a wave winding
(d) either ‘a’ or ‘c’

Q4. The brushes of a dc machine are made of:
(a) iron
(b) brass
(c) mica
(d) carbon

Q5. To convert a galvanometer into voltmeter, the value and type of connection of the resistance to be connected with should be
(a) Low and parallel
(b) High and parallel
(c) Low and series
(d) High and series

Q6. A voltmeter must have a very high internal resistance so that the:
(a) accuracy is high
(b) resolution is high
(c) meter draws minimum current
(d) loading is maximum

Q7. Which starting method is not used in squirrel cage induction motors?
(a) Resistance in rotor circuit
(b) Resistance in stator circuit
(c) Auto-transformer starting
(d) Star delta starting

Q8. What is condition for maximum starting torque of three phase induction motor if R₂ and X₂ are rotor resistance and reactance respectively?
(a) R₂ = 0, X₂ = 1Ω
(b) R₂ = 2 X₂
(c) R₂ = 0.5 X₂
(d) R₂ = X₂

Q9. The supply voltage to an induction motor is reduced by 10% By what percentage, approximately, will the maximum torque decrease?
(a) 40
(b) 20
(c) 10
(d) None of the above

Q10. Insulation resistance of a cable of length 20 km is 2 MΩ. What will be the insulation resistance of the same cable but for a length of 200 km?
(a)2 MΩ
(b)20 MΩ
(c)0.2 MΩ
(d)200 MΩ

solutions

S1. Ans.(b)
Sol. alternate poles are North-south or South-north

S2. Ans.(b)
Sol. because copper is an excellent conductor and becomes electromagnet when current passing through it.

S3. Ans.(b)
Sol. a closed one

S4. Ans.(d)
Sol. The brushes can be made of carbon or copper, but the resistance of the carbon brushes allows less current inrush during commutation between adjacent commutator segment.

S5. Ans.(d)
Sol. A galvanometer is very sensitive instruments so it can’t measure low value of current A Galvanometer can be converted into voltmeter by connecting a high resistance in series with it.

S6. Ans.(c)
Sol. A voltmeter having very high resistance so that if will draw minimum current and drop will be negligible and will correctly measure the CKT voltage.
▭(Rseries=Rm (m-1) ) when(m=I/Im or V/Vm )

S7. Ans.(a)
Sol. Rotor resistance starting is used only for slip-ring induction motor because rotor of squirrel cage induction motor is short circuited.

S8. Ans.(d)
Sol. Starting torque (Ts)= (KR₂)/(〖R₂²〗^ +x₂²) (Here S=1)
For maximum value (d(Ts))/(dR₂)= 0
ie, K[(R₂²+x₂²-r^2×2R₂)/((R₂²+x₂)²)]=0
⇒ R2=x₂ → condition for maximum starting torque.

S9. Ans.(b)
Sol. For Induction motor: –
T∝V²
If V is reduced by 10% then
T1∝(0.9v) ²
So, % change = (V^2-(0.9v)²)/V²×100
= 0.19×100
= 19%

S10. Ans.(c)
Sol. insulation resistance of cable ⍺ 1/ length
∴R2/R1 =l1/l2
∴R2=20/200×2=0.2 MΩ

Quiz: Electrical Engineering 25 July 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. 100% string efficiency means
(a) self-capacitance is zero
(b) shunt capacitance is maximum
(c) self-capacitance is maximum
(d) shunt capacitance is zero

Q2. Wave-wound generators provide
(a) less current but more voltage
(b) more current but less voltage
(c) more current and more voltage
(d) none of the above

Q3. In a synchronous motor, the stator frame is made of………
(a) Stainless steel
(b) CRGO
(c) Cast iron or welded steel plates
(d) Laminated silicon steel

Q4. A 230 V, 50 Hz single-phase energy meter has a load current of 10 A at p.f. of 0.8 lagging. The energy consumed by the load in 2 minutes is
(a) 1.2 kWh
(b) 0.06 kWh
(c) 2.4 kWh
(d) 4.2 kWh

Q5. Field control of a DC shunt motor gives–
(a) Constant kW drive
(b) Constant torque drive
(c) Constant speed drive
(d) Variable load speed drive

Q6. One lux is:
(a) 1 lumen/m²
(b) 4π lumen/mt²
(c) 1 lumen mt
(d) 1 candela

Q7. Hard magnetic materials are used for manufacturing ……
(a)Insulators
(b)Temporary magnets
(c)Permanent magnets
(d)Conductors

Q8. If a conductor of length ‘l’ is moving with a velocity of ‘v’ in a magnetic field B. The induced emf will be
(a)B/lv
(b)Blv
(c)Bl/v
(d)l/Bv

Q9.A circuit has an impedance of (1 – j 2) Ω. The susceptance of the circuit is …….
(a) 0.1 S
(b) 0.2 S
(c) 0.4 S
(d) None of these

Q10. The relative permittivity of a material is 10. Its absolute permittivity will be
(a) 8.854×10^(-11) F/m
(b) 9×10^8 F/m
(c) 5×10^(-5) F/m
(d) 9×10^5 F/m

SOLUTIONS

S1. Ans.(d)
Sol. If shunt capacitance is zero the charging current through each disc will be the same. This means that voltage across each disc is the same. This in turn means 100% string efficiency.

S2. Ans.(a)
Sol. Wave – wound has only two parallel paths. Wave-wound generators provide less current but more voltage.

S3. Ans.(c)
Sol. in synchronous motor: –
(A)Stator frame → made up of Cast iron/welded steel plates
(B)Core → made up of silicon steel

S4. Ans.(b)
Sol. Energy consumed by the load in 2 minutes
=(V I cos⁡ϕ)/1000×t=(230 × 10 × 0.8)/1000×2/60 = 0.06 KWh

S5. Ans.(a)
Sol. There are 3 types of speed control for dc drives: –
(i) Armature voltage control
(ii) Field flux control
(iii) Armature resistance control
We know▭(N∝1/ϕ)
Armature voltage control technique gives high efficiency but below base speed.
So, above base speed, Field flux control is used, where by weakening the field, speed above base speed can be achieved because these two are inversely proportional to each other.

S6. Ans.(a)
Sol. One lux=1 lumen/m²
S7. Ans.(c)
Sol. Hard magnets, also referred to as permanent magnets, are magnetic materials that retain their magnetism after being magnetised. It is used for manufacturing permanent magnets.

S8. Ans.(b)
Sol. The emf induced in a straight conductor of length l moving with velocity v perpendicular to a magnetic field B is: E =Blvsinθ.

S9. Ans.(c)
Sol.
B=Xc/Z^2 =2/((1)^2+(2)^2 )=2/5=0.4 S

S10. Ans.(a)
Sol. εr=ε/εo
∴ Absolute permittivity, = ε=ε0 εr=8.854×10^(-12)×10=8.854×10^- 11 F/m

Quiz: Electrical Engineering 23 July 2020

Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. A 4 KVA, 400/200 V single-phase transformer has resistance of 0.02 pu and reactance of 0.06 pu. Its actual resistance and reactance referred to HV side are respectively:
(a)0.2 ohm and 0.6 ohm
(b)0.8 ohm and 2.4 ohm
(c)0.08 ohm and 0.24 ohm
(d)2 ohm and 6 ohms

Q2. A transformer has a percentage resistance of 2% and percentage reactance of 4%. What are its regulations at power factor 0.8 lagging and 0.8 leading, respectively?
(a) 4% and – 0.8%
(b) 3.2% and – 1.6%
(c) 1.6% and – 3.2 %
(d) 4.8% and – 0.6%

Q3. If the iron core of a transformer is replaced by an air core, then the hysteresis losses in the transformer will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero

Q4. The use of higher flux density in the transformer design
(a) Reduces the weight per kVA
(b) Increases the weight per kVA
(c) Has no relation with the weight of transformer
(d) Increases the weight per kW

Q5. Which interrupt has the highest priority in 8085?
(a)INTR
(b)TRAP
(c)RST 6.5
(d)RST 7.5

Q6. An induction motor has a rotor resistance of 0.002Ω/phase. If the resistance increased to 0.004Ω/phase then the maximum torque will
(a) Reduced to half
(b) increased by 100%
(c) Increased by 200%
(d) Remain unaltered

Q7. Cogging of induction motor occurs at
(a)High voltage and when stator and rotor poles are equal
(b)Low voltage and when stator and rotor poles are equal
(c)High voltage and when stator and rotor poles are not equal
(d)Low voltage and when stator and rotor poles are not equal

Q8. In star connected system the phase angle difference between line and phase voltage is:
(a) 30°
(b) 120°
(c) 60°
(d) 90°

Q9. The phase-lead compensation is used to
(a)Increase rise time and decrease overshoot
(b)Decrease both rise time and overshoot
(c)Increase both rise time and overshoot
(d)Decrease rise time and increase overshoot.

Q10. An OP-AMP has a common mode gain of 0.01 and a differential gain of 10^5. Its common mode rejection ratio will be
(a)10^(-7)
(b)10^(-3)
(c)10^3
(d)10^7

solutions

S1. Ans.(b)
Sol. per unit (pu) quantity will be same on both side because the per-unit impedance is the same for the equivalent circuit of the transformer whether computed from the primary or secondary as long as the voltage bases on the two sides are selected in the ratio of transformation.
On HV side: Z_base=(VH^2)/S=(400)^2/4000=40 ohm
And Zpu=0.02+j 0.06
∴Zactual=Zbase×Zpu=0.8+j 2.4 ohm

S2. Ans.(a)
Sol. Per unit regulation:

%regulation, lagging = 2*0.8+4*0.6= 4%
%regulation, leading = 2*0.8-4*0.6= -0.8%

S3. Ans.(d)
Sol. Hysteresis is the property exhibited by ferrous materials, particularly those having iron in its composition. So naturally air cored transformers will not have any hysteresis losses.

S4. Ans.(a)
Sol. B=Φ/A. for same Φ, A is less. so, weight will reduce.

S5. Ans.(b)
Sol. Priority order of interrupt in 8085:
TRAP > RST 7.5 > RST 6.5 > RST 5.5 > INTR
So, in 8085 TRAP has highest priority.

S6. Ans.(d)
Sol. Maximum torque in an induction motor, T_MAX ∝1/2X.
the maximum torque is independent of rotor resistance and is inversely proportional to rotor reactance. So, if we increase the rotor resistance the maximum torque will be unaltered.

S7. Ans.(b)
Sol. Starting Torque ∝ V². When the supply voltage is low, the starting torque is low and when the poles on stator and rotor are equal, due to locking tendency motor caused by the slot harmonics reduced/failed to start.

S8. Ans.(a)
Sol. Note: – for balanced star-connected system V_L=√3 V_ph and line voltage leads the phase voltage by 30°.

S9. Ans.(b)
Sol. Phase-lead compensator is used to decrease both rise time and overshoot.

S10. Ans.(d)
Sol. CMRR=Ad/Ac =10^5/0.01=10^7

Quiz: Electrical Engineering 22 July 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The breaking capacity of a 3-phase circuit breaker is given by
(a) √3 × service voltage × rated symmetrical current
(b) 3 × service voltage × rated symmetrical current
(c) 2 × service voltage × rated symmetrical current
(d) None of the above

Q2. The insulating material for a cable should have
(a) low cost
(b) high dielectric strength
(c) high mechanical strength
(d) all of the above

Q3. For transfer of maximum power, the relation between load resistance R_L and internal resistance r_i of the voltage source is ……….
(a) R_L=2 R_i
(b) R_L=0.5 R_i
(c) R_L=1.5 R_i
(d) R_L=R_i

Q4. The air gap in a salient pole machine is
(a) Maximum at the centre of poles
(b) Least at the centre of poles and increases while moving away from the centre
(c) Maximum at the centre and decreases while moving away from the centre
(d) Equally distributed

Q5. Hopkinson’s test for DC motor is conducted with
(a) Full load
(b) Half load
(c) Low load
(d) No load

Q6. Tellegen’s theorem is based on the principle of law of _________________.
(a) Conservation of charge
(b) Conservation of mass
(c) Conservation of velocity
(d) conservation of energy

Q7. A D.C. Generator has 6 poles, A brush shift of 6° actual means a brush shift of
(a) 6° electrical
(b) 18° electrical
(c) 30° electrical
(d) 2° electrical

Q8. Full-load terminal voltage in an over-compounded generator with respect to no-load terminal voltage is–
(a) equal
(b) zero
(c) less
(d) more

Q9. Total number of poles and slots in a DC machine are 4 and 8 respectively. Angular slot pitch is ……….
(a) 45 electrical degrees
(b) 90 electrical degrees
(c) 135 electrical degrees
(d) 180 electrical degrees

Q10. In moving Iron type meter has the non-linear scale as:
θ ∝ I_rms
θ ∝ (〖I_rms〗^2)
θ ∝ V_rms
θ ∝ R^2

SOLUTIONS
S1. Ans.(a)
Sol. It is current (rms) that a circuit breaker is capable of breaking at given recovery voltage and under specified conditions.
Breaking capacity = √3 × service voltage × rated symmetrical current

S2. Ans.(d)
Sol. the insulating materials used in cables should have the following properties:
High insulation resistance to avoid leakage current.
High dielectric strength to avoid electrical breakdown of the cable.
High mechanical strength to withstand the mechanical handling of cables.
Non-hygroscopic i.e., it should not absorb moisture from air or soil.
Non-inflammable.
Low cost so as to make the underground system a viable proposition.
Unaffected by acids and alkalis to avoid any chemical action

S3. Ans.(d)
Sol. The maximum power transfer theorem states that in a linear, bilateral DC network, maximum power is delivered to the load when the load resistance is equal to the internal resistance of a source.
If it is an independent voltage source, then its series resistance (internal resistance Rs) or if it is independent current source, then its parallel resistance (internal resistance Rs) must equal to the load resistance RL to deliver maximum power to the load.

S4. Ans.(b)
Sol. Least at the centre of poles and increases while moving away from the centre.

S5. Ans.(a)
Sol. Hopkinson’s test for DC motor is conducted with full load.

S6. Ans.(d)
Sol. Tellegen’s theory is based on conservation of energy.
ie.∑_(K=1)^n▒〖P_k=0 〗
ie,∑_(K=1)^n▒〖I_k×V_K=0 〗
Tellegen’s Theorem states that the summation of power delivered is zero for each branch of any electrical network at any instant of time.

S7. Ans.(b)
Sol. ▭(θ_ele= P/2×θ_mech )
= (6/2)×6
= 18° electrical

S8. Ans.(d)
Sol. so, for over compound full load terminal voltage will be more than No-load voltage.

S9. Ans.(b)
Sol. Slots per pole =8/4=2
Angular slot pitch = (180°)/(slots per pole )
=(180°)/2
=90° (electrical)

S10. Ans.(b)
Sol. In moving Iron type meter, θ ∝ (〖I_rms〗^2)

Quiz: Electrical Engineering 27 June 2020

Quiz: Electrical Engineering
Exam: NLC
Topic: Control system

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

1. Effect of feedback on the plant is to
a) Control system transient response
b) Reduce the sensitivity to plant parameter variations
c) Both (a) and (b)
d) None of these

 

2. Transfer function of a system is defined as the ratio of output to input in
a) Z-transformer
b) Fourier transform
c) Laplace transform
d) All of these

3. Electrical resistance is analogous to
a) Inertia
b) Dampers
c) Spring
d) Fluid capacity

 

4. Relation between Fourier integral and Laplace transformer is through
a) Time domain
b) Frequency domain
c) Both (a) and (b)
d) None of these

 

5. At resonance peak, ratio of output to input is
a) Zero
b) Lowest
c) Highest
d) None of these

6. If gain of the system is zero, then the roots
a) Coincide with the poles
b) Move away from the zeros
c) Move away from the poles
d) None of these

 

7. Settling time is inversely proportional to product of the damping ratio and
a) Time constant
b) Maximum overshoot
c) Peak time
d) Undamped natural frequency of the roots

 

8. If gain of the critically damped system is increased, the system will behave as
a) Under damped
b) Over damped
c) Critically damped
d) Oscillatory

 

9. If gain of the system is increased, then
a) Roots move away from the zeros
b) Roots move towards the origin of the S-plot
c) Roots move away from the poles
d) None of these

 

10. Physical meaning of zero initial condition is that the
a) System is at rest and stores no energy
b) System is at rest but stores energy
c) Reference input to working system is zero
d) System is working but stores no energy

 

Solution

Sol 1. (c)
Both (a) and (b)

Sol 2. (c)
Laplace transform

Sol 3. (b)
Inertia

Sol 4. (c)
Both Time domain and frequency.

Sol 5. (c)
At resonance peak, ratio of output to input is highest.

Sol 6. (a)
Coincide with the poles

Sol 7. (b)
Maximum overshoot

Sol 8. (a)
Under damped

Sol 9. (c)
Roots move away from the poles

Sol 10. (a)
System is at rest and stores no energy

Quiz: Electrical Engineering 25 June 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: DC Generator

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

1. Which of the following characteristics reveal about the magnetization nature of the machine?
a) No-load characteristics
b) Load characteristics
c) Armature characteristics
d) Both no-load and load characteristics

2. Choose the most inappropriate out of the following for the no-load characteristics of the dc generator.
a) It is the open circuit characteristic of the machine
b) It is magnetization characteristic of the machine
c) It is conducted on the unloaded machine
d) None of the mentioned

3. The external characteristic is plotted between
a) terminal voltage vs armature current at constant excitation
b) terminal voltage vs field current at constant armature current
c) induced armature emf vs armature current at constant excitation
d) none of the mentioned

4. Armature characteristic is also known as
a) regulation characteristic
b) magnetization characteristic
c) external characteristic
d) load characteristic

5. The air gap line represents
a) magnetic behavior of the air gap of the dc machine
b) magnetic behavior of the air gap of the induction machine
c) magnetic behavior of the iron core
d) all of the mentioned

6. For a given dc generator, the external characteristic is plotted. Without using further plots, how can we obtain internal characteristic?
a) By adding the I_a R_a drop to the plot
b) By adding armature reaction
c) By reducing I_a R_a drop
d) All of the mentioned

7. Why is armature of a dc machine made of silicon steel stampings?
a) To reduce hysteresis loss
b) To reduce eddy current loss
c) For the ease with which slots can be created
d) To achieve high permeability

8. What losses occur in the teeth of dc generator?
a) To reduce hysteresis loss
b) To reduce eddy current loss
c) To reduce eddy current as well as hysteresis losses
d) To achieve high permeability

9. For a 220-V level compound generator the terminal voltage at the half load is
a) more than 220-V
b) same as no-load voltage
c) more than no-load voltage
d) lesser than no-load voltage

10. The voltage drop in terminal voltage from no-load to full load in a shunt generator can be compensated using
a) aiding series field
b) long-shunt, differential field
c) aiding shunt field
d) any of the measures

Solution
Sol 1.
Answer: d
Both no-load as well as load characteristics are required to predict the magnetization of the machine.

Sol 2.
Answer: d
All the mentioned nature of the characteristics is correct.

Sol 3.
Answer: a
The external characteristic is plotted between terminal voltage and armature current at fixed excitation.

Sol 4.
Armature characteristic is also known as regulation characteristic. It is so called due to the fact that the difference in the terminal voltage helps to determine the voltage deviation.

Sol 5.
Answer: a
Air gap line in the magnetization curve represents the ideal nature of the machine considering no saturation for the dc machine.

Sol 6.
Answer: a
Adding the armature resistance drop we can obtain the external characteristic for a dc generator.

Sol 7.
Answer: a
The silicon steel has very high permeability and makes the flux past consistent. Thereby making lesser losses.

Sol 8.
Answer: c
Teeth of the machine has both eddy current losses as well as hysteresis losses.

Sol 9.

Answer: b
In a level compound dc generator the terminal voltage remains same as no load voltage at all the loading.

Sol 10.
Answer: a
By aiding the series field the armature reaction drop can be compensated.

Quiz: Electrical Engineering 24 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Utilization of electrical energy

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The ratio of horizontal distance between lamps and the mounting Height of lamp is referred to as
(a) space to height ratio
(b) space to intensity ratio
(c) Height to intensity ratio
(d) Luminosity ratio

Q2. Power factor of fluorescent lamp is about
(a) zero
(b) 0.5 lead
(c) 0.5 lag
(d) unity

Q3. Which of the following lamp gives nearly monochromatic light?
(a) Sodium Vapour Lamp
(b) GLS lamp
(c) Tube Light
(d) Mercury Vapour lamp

Q4. The function of inert gas in filament lamp is
(a) increase the illumination
(b) decrease the power consumption
(c) minimize the effect of vaporization during service
(d) decrease the Clare

Q5. To prevent the excessive brightness, which type of lighting scheme is used?
(a) direct
(b) indirect
(c) general
(d) local

Q6. What is the function of choke in a fluorescent lamp?
(a) starting purpose
(b) produce high voltage at starting
(c) improve power factor
(d) minimize the power consumption

Q7. Carbon arc lamp are commonly used in
(a) Domestic lighting
(b) Street lightning
(c) Cinema projectors
(d) Photography

Q8. Arc can be produced by?
(a) AC current only
(b) DC current only
(c) Both AC & DC current
(d) All options are incorrect

Q9. The illumination at a point 5 meter below a lamp is 6 lux. The candle power of the lamp is
(a) 160
(b) 180
(c) 150
(d) 260

Q10. Luminous intensity is expressed as
(a) radians
(b) candela
(c) lumens
(d) none of the above

Solutions
S1. Ans. (a)
Sol. space to height ratio

S2. Ans. (c)
Sol. 0.5 Lag

S3. Ans. (a)
Sol. Sodium Vapour Lamp

S4. Ans. (c)
Sol. minimize the effect of vaporization during service

S5. Ans. (b)
Sol. Indirect lighting scheme

S6. Ans. (b)
Sol. produce high voltage at starting

S7. Ans. (c)
Sol. Cinema projectors

S8. Ans. (c)
Sol. Both AC & DC current

S9. Ans. (c)
Sol. Candela power =? Hight = 5-meter, Illumination = 6 lux
Illumination = CP/h^2 ,
CP = 6×5^2 = 150 W

S10. Ans. (b)
Sol. Luminous Intensity (I)=(Luminous Flux)/(Solid angle)=Candela

Quiz: Electrical Engineering 18 June 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: Power System

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The lightning arrestor is usually located nearer to
(a) bush bar
(b) Transformer
(c) circuit breaker
(d) isolator

Q2. The main function of relay is
(a) fault isolation
(b) fault detection
(c) fault prevention
(d) All of the above

Q3. Mho relay is normally used for the protection of
(a) long Transmission lines
(b) medium length line
(c) short length lines
(d) No length criteria

Q4. Earth fault relay are
(a) directional relays
(b) non-directional relays
(c) short operate time relays
(d) long operate time relays

Q5. The rating of Fuse is expressed in terms of
(a) amperes
(b) Volts
(c) VAR
(d) KVA

Q6. By burden of the relay we mean
(a) volt ampere rating of the relay
(b) current rating of relay
(c) voltage rating of relay
(d) watt rating of relay

Q7. In electric motor cooling; TEFC means
(a) total efficient fan cooled
(b) totally enclosed fan cooled
(c) thermal energy fan cooled
(d) tough energy fan cooled

Q8. The type of oil used in transformer is
(a) olive
(b) mineral
(c) coconut
(d) palm

Q9. In a synchronous motor hunting takes place
(a) when load varies
(b) motor is under loaded
(c) when power factor is unity
(d) when supply voltage fluctuates

Q10. If the field of a synchronous motor is under excited, the power factor will be
(a) lagging
(b) leading
(c) unity
(d) zero

Solutions
S1. Ans. (b)
Sol. Transformer

S2. Ans. (b)
Sol. Fault detection

S3. Ans. (a)
Sol. Long transmission line

S4. Ans. (c)
Sol. Short operate time relay

S5. Ans. (a)
Sol. Ampere

S6. Ans. (a)
Sol. Volt ampere rating of relay

S7. Ans. (b)
Sol. totally enclosed fan cooled

S8. Ans. (b)
Sol. Mineral

S9. Ans. (a)
Sol. When load varies

S10. Ans. (a)
Sol. Lagging

Quiz: Electrical Engineering 11 June 2020

Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: Miscellaneous

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

Q1. EMF of a thermo couple depends upon the

(a) nature of material of metal

(b) difference of temperature of two junctions

(c) both (a) and (b)

(d) none of the above

 

 

Q2. If the diameter of a copper wire is increased by two times keeping the terminal voltage same then the drift velocity will

(a) become twice

(b) become half

(c) become four time

(d) remain unchanged

 

 

Q3. According to classical free electron theory, electrons in a metal are subjected to

(a) constant potential

(b) sinusoidal potential

(c) square-wave potential

(d) non-periodic potential

 

 

Q4. In ionic crystal, electrical conductivity is

(a) very high

(b) depends on material

(c) depends upon temperature

(d) practically zero

 

 

Q5. What type of defect causes F-centers in a crystal?

(a) Stoichiometric defect

(b) Metal excess defect due to anion vacancies

(c) Metal excess defect due to extra cation

(d) Frenkel defect

 

 

Q6. In crystal lattice, what are the vacancies created by the absence of certain atoms known as

(a) Hertz defects

(b) Schottky defects

(c) Pauli defects

(d) Crystal defects

 

 

Q7. Superconducting metal in superconducting state has relative permeability of

(a) more than one

(b) one

(c) zero

(d) negative

 

 

Q8. The dc resistivity and permeability exhibited by a type 1 superconductor are respectively

(a) zero and zero

(b) zero and unity

(c) unity and zero

(d) unity and unity

 

 

Q9. What is the magnetic susceptibility χ of an ideal superconductor?

(a) 1

(b) -1

(c) 0

(d) infinite

 

 

Q10. Which effect is the converse of Peltier effect?

(a) Seebeck effect

(b) Thomson effect

(c) Hall effect

(d) joule effect

 

Solutions

S1. Ans. (c)

Sol. EMF of a thermo couple depends upon both factors

  • nature of material of metal
  • difference of temperature of two junctions

 

S2. Ans. (d)

Sol. It will remain unchanged

 

S3. Ans. (a)

Sol. constant potential

 

S4. Ans. (d)

Sol. In ionic crystal, electrical conductivity is practically zero.

 

S5. Ans. (b)

Sol. Metal excess defect due to anion vacancies

 

S6. Ans. (b)

Sol. Schottky defects

 

S7. Ans. (c)

Sol. Zero

 

S8. Ans. (a)

Sol. Zero and zero

 

S9. Ans. (b)

Sol. -1

 

S10. Ans. (a)

Sol. Seebeck effect

 

 

 

Quiz: Electrical Engineering 10 June 2020

Quiz: Electrical Engineering
Exam: UPSSSC JE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. A 3-phase 4- wire system is commonly used for
(a) primary distribution
(b) secondary distribution
(c) primary transmission
(d) secondary transmission

Q2. Which of the following is usually not the generating voltage?
(a) 6.6 kV
(b) 9.9 kV
(c) 11 kV
(d) 13.2 kV

Q3. The main drawbacks of overhead transmission system over underground system is/are
(a) underground system is more flexible than overhead system
(b) higher charging current
(c) surge problem
(d) high initial cost

Q4. Improving pf
(a) reduces current for a given output
(b) increases losses in line
(c) increase the cost of station equipment
(d) none of the above

Q5. DC motors are still preferred for use in
(a) electric excavators
(b) lathes and machine tools
(c) floor mills and jaw crusher
(d) paper industry

Q6. The least significant feature while selecting a motor for centrifugal pump is
(a) speed control
(b) power rating of motor
(c) operating speed
(d) starting characteristic

Q7. BCD code is
(a) a binary code
(b) unweighted code
(c) the same thing as binary number
(d) the same as gray code

Q8. A unit impulse response of a second order system is 1/6 e^(-0.8t) sin (0.6t). Then natural frequency and damping ratio of the system are
(a) 1 and 0.6
(b) 1 and 0.8
(c) 2 and 0.4
(d) 2 and 0.3

Q9. Damping ratio ξ and peak overshoot M_p are measure of
(a) relative stability
(b) absolute stability
(c) speed of response
(d) steady state error

Q10. For a critically damped system, the closed-loop poles are
(a) purely imaginary
(b) real, equal and negative
(c) complex conjugate with negative real part
(d) real unequal and negative

Solutions
S1. Ans. (b)
Sol. A 3-phase 4- wire system is commonly used for secondary distribution.

S2. Ans. (b)
Sol. 9.9 kV is usually not the generating voltage.

S3. Ans. (c)
Sol. Surge problem is only advantage of overhead transmission system.

S4. Ans. (a)
Sol. Since Power, P = VI cos φ , cos φ (power factor)
If cos φ will improve for a given output the value of current will decrease.
S5. Ans. (a)
Sol. electric excavators

S6. Ans. (a)
Sol. speed control

S7. Ans. (a)
Sol. BCD code is a binary code

S8. Ans. (b)
Sol. h (t) = 1/6 e^(-0.8t) sin (0.6t)
On comparing it with the standard equation,
i.e. Ke^(-ξω_n t)/√(1-ξ^2 ) sin⁡〖(ω_d t+ Ǿ)〗
we have, ξω_n=0.8 …………………………. (i)
ω_d= ω_n √(1-ξ^2 )
= 0.6 rad/sec ……………………. (ii)

Solving both equation
ω_n=1 rad/sec and 𝞷 = 0.8

S9. Ans. (c)
Sol. speed of response

S10. Ans. (b)
Sol. real, equal and negative

Quiz: Electrical Engineering 21 May 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: Analog and digital electronics (electrical engineering)

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. A FET transistor with an antiparallel body diode block:
(a) Bidirectional voltage and passes unidirectional current
(b) Bidirectional voltage and passes bidirectional current
(c) Unidirectional voltage and passes unidirectional current
(d) Unidirectional voltage and passes bidirectional current

 

Q2. Turn-on of a Thyristor takes place when
(a) anode to cathode voltage is positive
(b) anode to cathode voltage is negative
(c) there is a positive current pulse at the gate
(d) the anode to cathode voltage is positive and there is a positive current pulse at the gate

 

Q3. After firing an SCR, the gate pulse is removed. The current in the SCR will
(a) remain the same
(b) immediately fall to zero
(c) rise up
(d) rise a little and then fall to zero

 

Q4. When a thyristor is in ON state, its gate drive be
(a) removed to save power
(b) removed or may not be removed
(c) not removed as it will turn off the thyristor
(d) removed to avoid increased losses and higher junction temperature

 

Q5. Current in SCR to turn on transient can be reduced by
(a) connecting a small inductor in series with the anode
(b) connecting a small inductor in parallel with the anode
(c) connecting a small capacitor in series with the anode
(d) connecting a small capacitor in parallel with the anode

 

Q6. Which of following does not cause damage of an SCR?
(a) high current
(b) high rate of rise of current
(c) high temperature rise
(d) high rate of rise of voltage

 

Q7. When compared to those of a symmetrical thyristor, the turn off time and reverse blocking voltage of an asymmetrical thyristor are respectively
(a) large and large
(b) large and small
(c) small and large
(d)small and small

 

Q8. The duration of reverse recovery transient affects the following aspect in a power diode
(a) peak inverse voltage
(b) forward threshold voltage
(c) speed of response
(d) average current

 

Q9. Surge current rating of an SCR specifies the maximum
(a) repetitive current with rectangular wave
(b) non- repetitive current with sine wave
(c) non- repetitive current with rectangular wave
(d) repetitive current with sine wave

 

Q10. The dv/dt effect in SCR can result in
(a) Low capacitive charging current
(b) false triggering
(c) increased junction capacitance
(d) high rate of rise of anode voltage

 

Solutions
S1. Ans. (d)
Sol. Unidirectional voltage and passes bidirectional current

 

S2. Ans. (d)
Sol. the anode to cathode voltage is positive and there is a positive current pulse at the gate.

 

S3. Ans. (a)
Sol. remain the same

 

S4. Ans. (d)
Sol. removed to avoid increased losses and higher junction temperature

 

S5. Ans. (a)
Sol. connecting a small inductor in series with the anode

 

S6. Ans. (d)
Sol. high rate of rise of voltage

 

S7. Ans. (d)
Sol. Both quantities need to be small.

 

S8. Ans. (a)
Sol. The duration of reverse recovery transient affects the following aspect in a power diode peak inverse voltage.

 

S9. Ans. (b)
Sol. non- repetitive current with sine wave

 

S10. Ans. (b)
Sol. false triggering

Quiz: Electrical Engineering 20 May 2020

Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: Switchgear and protection

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

Q1. The most serious consequence of a major uncleared short-circuit fault could be

(a) Blowing of fuse

(b) fire

(c) heavy voltage drop

(d) none of these

 

 

Q2. Which of the following result in a symmetrical fault?

(a) single phase to earth

(b) phase to phase

(c) All the three phases to earth

(d) two phase to earth

 

 

Q3. Which portion of transmission system is more prone to faults?

(a) Alternator

(b) transformer

(c) Overhead lines

(d) Underground cable

 

 

Q4. Which portion of power system is least prone to fault?

(a) Switchgear

(b) Alternator

(c) Overhead lines

(d) Transformer

 

 

Q5. The most common type of fault is

(a) Single phase to ground

(b)Two phase to ground

(c) Phase to phase

(d) three phase to ground

 

Q6. The per unit value of a 4 Ω resistance at 100 MVA base and 10 kV base voltage is

(a) 6 pu

(b)8 pu

(c)4 pu

(d)10 pu

 

 

Q7. Series reactor are used to

(a) improve the transmission efficiency

(b) improve the power factor of the power system

(c) improve the voltage regulation

(d) Bring down the fault level within the capacity of the switch gear

 

 

Q8. Current limiting reactors may be

(a) air cored air cooled

(b) oil immersed magnetically shielded

(c) oil immersed non magnetically shielded

(d) any of above

 

 

Q9. Symmetrical components are used in power system for the analysis of

(a) balanced 3 phased faults

(b) unbalanced 3 phase faults

(c) normal power system under steady condition

(d) Stability of system under disturbance

 

 

Q10. The positive sequence current of a transmission line is:

(a) always zero

(b) 1/3 of negative sequence current

(c) equal to negative sequence current

(d) 3 times negative sequence current

 

 

Solutions

 

 

 

S1. Ans.(b)

Sol. Fire is the most serious consequence.

 

S2. Ans.(c)

Sol. All the three phases to earth.

S3. Ans.(c)

Sol. Overhead lines

S4. Ans.(b)

Sol. Alternator

S5. Ans. (a)

Sol. The most common type of fault is Single phase to ground.

 

S6. Ans. (c)

Sol.  Rpu = R(ohm) ×MVA/(kV)^2
= 4 ×100/10^2 = 4 pu

S7. Ans. (d)

Sol. Bring down the fault level within the capacity of the switch gear.

S8. Ans. (d)

Sol. Any of above

S9. Ans.(b)

Sol. unbalanced 3 phase faults

S10. Ans. (b)

Sol. 1/3 of negative sequence current