Quiz: Electrical Engineering
Exam: SSC JE
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
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?
(c) Phase Plate
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
(c) Stainless steel
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
(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
(d) 2 kW
Sol. Given that, L=10+3Ө-Ө^2/4µH
And we know that the control torque for moving iron ammeter is given by :
T_c=kӨ=1/2 I^2 dL/dӨ
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.
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.
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
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.
#Low temperature coefficient of resistance.
Following material are used for manufacturing heating element-
Cupronickel (CuNi): an alloy of copper that contains nickel and strengthening elements, such as iron and manganese.
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.
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.
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
Sol. For isolation transformer = 1:1
Sol. Power transferred inductively
= (1-0.8) × 10 = 2kW