Quiz: Electrical Engineering 05 Oct 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. Which of the following is the dimensional formula of conductance?
(a) M^1 L^2 T^(-3) A^(-1)
(b) M^1 L^2 T^(-3) A^(-2)
(c) M^(-1) L^(-2) T^3 A^2
(d) M^1 L^2 T^(-3) A^1

Q2. Which one of the following statements is TRUE about the resistance of a conductor?
(a) the resistance of a conductor is inversely proportional to the length of the conductor.
(b) the resistance of a conductor is directly proportional to the area of the conductor,
(c) the resistance of a conductor is inversely proportional to the pressure applied on the conductor,
(d) the resistance of a conductor is inversely proportional to the area of the conductor.

Q3. In parallel combination of resistance, the voltage is _________.
(a) lower across largest resistance
(b) higher across largest resistance
(c) same across each resistance
(d) higher across smaller resistance

Q4. Electrical conductivity of a conductor is measure in __________.
(a) Siemens
(b) Ohms
(c) Siemens/meter
(d) Ohms/meter

Q5. What will be the equivalent capacitance of a parallel combination of four capacitors having equal value of capacitance ‘C’?
(a) C/4
(b) 4C
(c) C/2
(d) 2C

Q6. Determine the Thevenin’s equivalent resistance (in ohms) across the terminal a and b for the electrical circuit given below.

(a) 1
(b) 0.5
(c) 0.3
(d) 0.2

Q7. The no-load ratio of a 50 Hz single phase transformer is 6000/250 V. The maximum flux in the core is 0.06 Wb. What is the number of primary turns?
(a) 450
(b) 900
(c) 350
(d) 210

Q8. A 3-phase induction motor is ………
(a) essentially a constant-speed motor
(b) a variable speed motors
(c) very costly
(d) not easily maintainable

Q9. If a 3-phase induction motor is running at slip s (in decimal), then, rotor copper loss is equal to …….
(a) (1–s)× rotor input
(b) (1+s) × rotor input
(c) s× rotor input
(d) none of the above

Q10. In a squirrel cage motor, the number of stator slots is ……. Rotor slots.
(a) always equal to the number of
(b) always greater than the number of
(c) always less than the number of
(d) either more or less than the number of

SOLUTIONS
S1. Ans.(c)
Sol. conductance = [1/((Resistance) )]=1/R
R = (V/I)
& V = (W/Q) = (W / It) …. [as V = (work / charge) = {(work) / (current × time)}]
∴ V = ([M^1 L^2 T^(–2)] / [A^1 T]) = [M^1 L^2 T^(–3) A^(–1)]
Hence R = [M^1 L^2 T^(–3) A^(–1) ]/[A] =[M^1 L^2 T^(–3) A^(–2)]
∴ conductance = 1/([M^1 L^2 T^(–3) A^(–2)] )==[M^(-1) L^(-2) T^3 A^2]

S2. Ans.(d)
Sol. R=ρ l/A
So, the resistance of a conductor is inversely proportional to the area of the conductor.

S3. Ans.(c)
Sol. In parallel combination of resistance, the voltage is same across each resistance.
i.e. V1=V2=⋯=V

S4. Ans.(c)
Sol. The electric conductivity is the measure the ability of a conductor to conduct electricity.
The conductivity formula is the inverse of the resistivity that is:
σ=1/ρ
Here
σ = refers to the electrical conductivity
ρ = refers to the resistivity
The conductivity unit is Siemens per meter.

S5. Ans.(b)
Sol. for parallel combination: Ceqv=C+C+C+C=4C

S6. Ans.(b)
Sol. The value of the equivalent resistance, Rab is found by calculating the total resistance looking back from the terminals a and b with all the voltage sources shorted.
∴Rab=1 ‖(3 ‖3 ‖3)
∴Rab=1 ‖ 1=1/2=0.5

S7. Ans.(a)
Sol. E1=4.44 f N1 ϕ_m or 6000=4.44×50×N1×0.06
∴N1=450

S8. Ans.(a)
Sol. At no-load, the rotor lags behind the stator flux only a small amount since the only torque required is that needed to overcome small no-load losses. As mechanical load is added, the rotor speed decreases. A decrease in rotor speed allows the constant speed rotating field to sweep across the rotor conductors at a faster rate, thereby inducing larger rotor current (since rotor impedance is low). This results in a large increase in torque which tends to bring the speed to the original value. Although the motor speed does decrease is classed as a constant-speed motor

S9. Ans.(c)
Sol. Rotor copper loss = s× rotor input

S10. Ans.(c)
Sol. The rotor has a larger number of slots than the stator and should be a non-integer multiple of the number of stator slots so as to prevent magnetic interlocking of rotor and stator teeth at the starting instant.
Also, it leads to good starting characteristics, high efficiency and power factor, low stator current, and less vibrations and mechanical oscillations at nominal speed.

Quiz: Electrical Engineering 16 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. In power transformers, it is found that the iron or core losses practically varies very less, this is because
(a) Constant leakage flux
(b) Constant core flux
(c) Constant load current
(d) Constant leakage and core flux

Q2. Which of the conditions is must be fulfilled for a satisfactory parallel operation of transformers?
(a) Correct polarity
(b) Same pu impedance
(c) Same voltage ratio
(d) Leakage impedance should be inversely proportional to KVA of the transformer

Q3. The qualities aspired to obtain a good permanent magnet is/are
(a) High coercivity
(b) High residual flux
(c) Low coercivity
(d) High residual flux and high coercivity

Q4. The pitch factor, in rotating electrical machinery, is defined as the ratio of resultant EMF of a
(a) Full pitch coil to the phase emf
(b) Chorded coil to that of full pitched coil
(c) Full pitched coil to that of a chorded coil
(d) Chorded coil to the phase emf

Q5. What is the maximum load that can be connected in a circuit containing only lighting points?
(a) 500 watts
(b) 750 watts
(c) 800 watts
(d) 1000 watts

Q6. The iron core in a transformer provides a _____path to the main flux
(a) Low reluctance
(b) High reluctance
(c) Low resistance
(d) High conductivity

Q7. For the network shown in the figure, the value of current in 8Ω resistor is

(a) 408A
(b) 2.4A
(c) 1.5A
(d) 1.2A

Q8. 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

Q9. 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

Q10. A diesel station supplies the following loads to various consumers:
Industrial consumer = 1500 kW
Commercial load = 750 kW
Domestic power = 100 kW
Domestic light = 450 kW
If the maximum demand on the station is 2500 kW, then diversity factor is
(a) 1.5
(b) 1.92
(c) 2
(d) 1.12

SOLUTIONS
S1. Ans.(b)
Sol. Due to the constant core flux, the iron losses also remain constant.

S2. Ans.(a)
Sol. out of all, polarity must be correct for the connected transformers.

S3. Ans.(d)
Sol. both high residual flux as well as coercivity is desired for a good permanent magnet.

S4. Ans.(b)
Sol. Kp= Resultant EMF of a chorded coil/Resultant EMF of a full pitched coil = cos⁡(ε/2)

S5. Ans.(c)
Sol. The maximum load that can be connected in a circuit containing only lighting points is 800 watts.

S6. Ans.(a)
Sol. The iron core in a transformer provides a low reluctance path to the main flux.

S7. Ans.(b)
Sol.

Here 12ohm and 8-ohm resistors are in series and both are parallel connected to 20ohm resistor.
∴R_eqv=20‖20=10Ω
∴I=48/10=4.8 A
And I8Ω= 4.8×20/(20+20)=2.4 A

S8. 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

S9. 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.

S10. Ans.(d)
Sol. Diversity Factor = Sum of Individual Maximum Demands / Maximum Demand of the System.
Diversity factor=(1500+750+100+450)/2500=1.12

Quiz: Electrical Engineering 9 Sep 2020

Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous

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

Q1. In a power plant if the maximum demand on the plant is equal to the plant capacity then
(a) Load factor will be nearly 60%
(b) Plant reserve capacity will be zero
(c) Diversity Factor will be unity
(d) Load factor will be unity

Q2. For the circuit shown in figure, the voltage across the capacitor during steady state condition is

(a) 0 V
(b) 4 V
(c) 6 V
(d) 12 V

Q3. Find R_AB for the circuit shown in figure.

(a) 18Ω
(b) 30Ω
(c) 45Ω
(d) 68Ω

Q4. A supply voltage of 230 V, 50 Hz is fed to a residential building. Write down its equation for instantaneous value.
(a) 163 sin 314·16 t
(b) 230 sin 314·16 t
(c) 325 sin 314·16 t
(d) 361sin314·16 t

Q5. The AC bridge used for measurement of dielectric loss of capacitor is
(a) Anderson bridge
(b) Schering bridge
(c) Wien bridge
(d) Hay’s bridge

Q6. In a transformer, Low voltage windings are placed nearer to the core in the case of concentric windings because
(a) it reduces hysteresis loss
(b) it reduces eddy current loss
(c) it reduces insulation requirement
(d) it reduces leakage fluxes

Q7. Emf induced in the dc generator armature winding is
(a) AC
(b) DC
(c) AC and DC
(d) None of the above

Q8. In a dc machine 72 number of coils are used. Find the number of commutator segments required?
(a) 72
(b) 36
(c) 37
(d) 74

Q9. Reactance relay is used for protection in:
(a) Long transmission lines
(b) Short transmission lines
(c) Medium transmission lines
(d) Both long transmission lines and short transmission lines

Q10. The earthing electrode should be situated at a place at least ………. meters away from the building whose installation system is being earthed?
(a) 4
(b) 2.5
(c) 1.5
(d) 0.5

SOLUTIONS
S1. Ans.(b)
Sol. Reserve capacity = Plant capacity – Max. Demand
If the plant capacity is equal to Max. Demand then the Reserve capacity will be zero.

S2. Ans.(d)
Sol. under steady state condition, capacitor behaves as open circuit. So, source voltage will appear across the capacitor.
So, voltage across capacitor during steady state= 12 V

S3. Ans.(a)
Sol. given circuit is a balanced bridge.
So, RAB=(20+10)‖ |(30+15)=(30×45)/75=18 Ω

S4. Ans.(c)
Sol. V=Vm sin⁡(2πft)=230×√2 sin⁡(2×3.14×50 t)=325 sin⁡(314.16 t)

S5. Ans.(b)
Sol. Schering bridge is used for measurement of dielectric loss of capacitor.

S6. Ans.(c)
Sol. In concentric arrangement of placing of HV and LV winding, less insulation is required when LV winding is placed near the core which is at ground potential. The HV winding is placed around the LV.
If high voltage winding is placed near the core, the more insulation will be required and the more insulation lead to increase in the size and cost of the transformer.

S7. Ans.(a)
Sol. The emf induced in the dc generator armature winding is AC, but we need DC current from DC generator, so to convert this AC current to DC current mechanical rectifier called as commutator is used.

S8. Ans.(a)
Sol. In DC machines, Number of coils = Number of commutator segments.

S9. Ans.(b)
Sol. Relay used for different configuration of transmission lines……
Short transmission line: reactance relay
Medium transmission line: impedance relay
Long transmission line: mho relay

S10. Ans.(c)
Sol. The earthing electrode should be situated at a place at least 1.5 meters away from the building whose installation system is being earthed
The earthing electrode should always be placed in a vertical position inside the earth or pit so that it may be in contact with all the different earth layers.