Quiz: Electrical Engineering 19 Oct 2020

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

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

Q1. Determine the self-inductance (in mH) of a 3m long-air cored solenoid, when the coil has 300 turns and the diameter of the coil is 12 cm.
(a) 0.24
(b) 0.32
(c) 0.35
(d) 0.42

Q2. Determine the reluctance (in Amp-turn/Wb) of a coil when the flux through the coil is 25 Wb and the value of produced mmf is 50 Amp-turns.
(a) 2
(b) 4
(c) 6
(d) 8

Q3. A current of 3A through a coil sets flux linkage of 15 Wb-turn. The inductance of the coil is?
(a) 1 H
(b) 3 H
(c) 5 H
(d) 15 H

Q4. Which of the following is bilateral element?
(a) Constant voltage source
(b) Constant current source
(c) Capacitance
(d) All of the above

Q5. In the following circuit, the equivalent capacitance between terminals A and B is

(a) C
(b) C/2
(c) 2C/3
(d) 3C/2

Q6. Which of the following materials has susceptibility independent of temperature?
(a) Ferromagnetic
(b) Ferrimagnetic
(c) Paramagnetic
(d) Diamagnetic

Q7. Determine the capacitive susceptance (in siemens) of a circuit if the capacitor of the circuit is 0.08 mF and supplied with a 50 Hz frequency.
(a) O.025
(b) 0.064
(c) 0.046
(d) 0.034

Q8. Which of the following is an effect of non-uniform current distribution in a conductor?
(a) Skin effect
(b) Ferranti effect
(c) Proximity effect
(d) Skin effect or proximity effect

Q9. A transformer has iron loss of 900 W and copper loss of 1600 W at full load. At what percentage of load will the efficiency be maximum?
(a) 133
(b) 125
(c) 75
(d) 66.66

Q10. A transformer is working at full load with maximum efficiency. Its iron loss is 1000 W. What will be its copper loss at half full load?
(a) 2000 W
(b) 1000 W
(c) 500 W
(d) 250 W

SOLUTIONS
S1. Ans.(d)
Sol. Given that, N=300 turns
d= 12 cm ⇒r=d/2=6 cm=0.06 m
l= 3m
for solenoid, L=(µ0 µr N^2 A)/l=(4π×10^(-7)×1×300×300×π×〖(0.06)〗^2)/3=0.42 mH
NOTE: for air µ_r=1
And µ_0=4π×10^(-7)H/m ≈ 12.57×10−7 H/m

S2. Ans.(a)
Sol. MMF produced in a coil=NI=Reluctance×ϕ
∴NI=Reluctance×ϕ
⇒reluctance=NI/ϕ=50/25=2 AT/Wb

S3. Ans.(c)
Sol. inductance of coil=L=Nϕ/I=15/3=5 H

S4. Ans.(c)
Sol. Bilateral elements are defined as the elements through which magnitude of current is independent of polarity of supply voltage.
e.g. of bilateral elements: resistors, inductors, capacitors

S5. Ans.(a)
Sol.

⇒C_AB=((C×C)/(C+C))+C/2=C/2+C/2=C

S6. Ans.(d)
Sol. Diamagnetic material: Magnetic materials which align against the magnetic field are known as diamagnetic materials. Magnetic susceptibility is χ<0 which means it is always a negative value for diamagnetic material.
These materials are independent of temperature. As these materials magnetize in the opposite direction, they do have a small amount of magnetization intensity. Gold, tin, mercury, water, etc are examples of diamagnetic materials.
For a diamagnetic material, the relative permeability lies between 0 ≤ μr < 1 and its susceptibility lies between −1< χ< 0. 𝑥=𝑢𝑟−1
Where 𝑥 is a dimensionless quantity defines as the magnetic susceptibility of the medium.

S7. Ans.(a)
Sol. capacitive susceptance(B)=1/XC =1/((1/2πfC) )
∴B==2πfC=2×3.14×50×0.08×10^(-3)=0.025 siemens

S8. Ans.(d)
Sol. Skin effect: When an Alternating Current flows through a conductor, it is not distributed uniformly throughout the conductor cross-section. AC current has a tendency to concentrate near the surface of the conductor. This phenomenon in alternating currents is called as the skin effect.
Proximity effect:
When two or more conductors carrying alternating current are close to each other, then distribution of current in each conductor is affected due to the varying magnetic field of each other. when the nearby conductors carrying current in the same direction, the current is concentrated at the farthest side of the conductors. When the nearby conductors are carrying current in opposite direction to each other, the current is concentrated at the nearest parts of the conductors. This effect is called as Proximity effect.

S9. Ans.(c)
Sol. maximum efficiency in %=√((iron loss(Pi))/(full load cu-loss(Pcu)) )×100=√(900/1600) ×100=75 % of full-load

S10. Ans.(d)
Sol. For maximum efficiency, Copper loss (Pc) = Iron losses (Pi) = 1000 W
And P_cu (x)=x^2×Pcu (full-load)
∴P_(cu(half-load))= (1/2)^2 P(cu(full-load))=1/4×1000=250 W

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