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# Quiz: Electrical Engineering 03 May 2021

Quiz: Electrical Engineering
Exam: DFCCIL
Topic: Basic electrical engineering

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

Q1. Electric supply is rated at 220 V. In a house, 11 bulbs of power rating 100 W each
are use. The rating of the fuse should be
(a) 0.5 A
(b) 0.1 A
(c) 1 A
(d) 5 A

Q2. The capacitance of a parallel plate capacitor is 5 μ F. When a glass plate is inserted between its two plates, its potential reduces to 1/8 of the original value. The value of dielectric constant of glass is
(a) 4
(b) 8
(c) 40
(d) 1.6

Q3. Unit of resistivity is:
(a) Ohm-meter
(b) mho meter
(c) mho
(d) Ω

Q4. The material to be used in the manufacture of a standard resistor should be of
(a) low resistivity
(b) high resistivity and low temperature coefficient
(c) high temperature coefficient
(d) low resistivity and high temperature coefficient

Q5. One kWh of electrical energy equals:
(a) 3600J
(b) 860 kcal
(c) 3600 W
(d) 4186 J

Q6. Two wires A and B of the same material but of different lengths L and 2L have the radius r and 2r respectively. the ratio of specific resistance will be
(a) 1:4
(b) 1:8
(c) 1:1
(d) 1:2

Q7. 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

Q8. One international ohm is equal to:
(a) 1.00049 absolute ohm
(b) 0.99951 absolute ohm
(c) 0.969 absolute ohm
(d) 1.049 absolute ohm

Q9. Unit of absolute permittivity is-

Q10. An electric field line and equipotential surface are
(a) always parallel
(b) always at 90°
(c) inclined at any angle
(d) both (a) and (c)

SOLUTIONS
S1. Ans.(d)
Sol. Resistance of each bulb
R=V^2/P=(220×220)/100=484 Ω
∴ R_eq=484/11=44 Ω
∴ Rating of fuse =220/44=5 A

S2. Ans.(b)
Sol. V=Q/C⇒V∝1/C
Let dielectric constant is K.
∴ C^’=KC
∴ V^’∝1/KC
∴ V/V^’ =KC/C
⇒ K= 8

S3. Ans.(a)
Sol. Resistance (R) = ρl/A
Where, ρ = resistivity
∴ ρ = RA/l ((Ω.m^2)/m)
So, ▭(Unit of ρ=Ω.m )

S4. Ans.(b)
Sol. Stranded resistor:
Resistivity =high
temp. coeff. = low

S5. Ans.(b)
Sol. 1 kwh = 1000 Watt × 3600 sec.
= 36 × 10^5 Joule
4.18 Joule = 1 calorie
∴ 1 kwh = 36 × 10^5 joule = (36×10^5)/4.18 = 860 kcal.

S6. Ans.(c)
Sol. Specific resistance depends on:
(i) Composition of material
(ii) temperature
So, if two wires are made using
same material then ρ_1,ρ_2 = constant
∴ ρ1=ρ2=1

S7. 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]

S8. Ans.(a)
Sol. 1 international ohm = 1.00049 absolute ohm.

S9. Ans.(c)
Sol. absolute permittivity (ϵ) = ϵo ϵr