LMRC JE Based Quiz Electrical Engineering(Utilization)

utilization of electrical engineering best mcq with easy explanation
LMRC Most Expected Question(Electrical Engineering)

Q1. Induction heating is based on the principle of –
(a) Resistance heating
(b) Dielectric heating
(c) Electromagnetic induction
(d) Both a and b

Q2. Induction heating, the depth upto which the current will penetrate is proportional to
(a) f
(b) (f)²
(c) (f)^(1\/2)
(d) 1/f^(1\/2)
Where f → frequency

Q3. Which of th the following relation is valid for induction heating, where F= resistivity
F= frequency and δ= depth of penetration
(a) δ=(5√δ)/√f
(b) δ=√f/√δ
(c) δ=k√fδ
(d) δ=kf√δ

Q4. Which parameter is high in case of induction heating?
(a) frequency
(b) voltage
(c) current
(d) phase angle

Q5. Which of the following generators are used in arc welding?
(a) series generator
(b) differential compound generator
(c) cumulative compound generator
(d) series generator

Q6. Electrode material in case of electrical resistance welding should have-
(a) high thermal conductivities
(b) high electrical conductivities
(c) both a & b
(d) none

Q7. The metal surfaces for electrical resistance welding must be-
(a) clean
(b) rough
(c) moistened
(d) coloured

Q8. Induction heating takes place in-
(a) conducting but non-conducting materials
(b) conducting and magnetic materials
(c) insulating materials
(d) conducting materials which may be either magnetic or non-magnetic materials

Q9. The resistance R of a conductor is –
(a) EA/JL (Where E= elec. Field
A= cross- sectional Area
(b) EJ/AL J = current density
L = length of conductor )
(c) EL/JA
(d) JA/EL

Q10. According to gauss’s theorem, the surface integral of the normal component of electric flux density ‘D’ over a closed surface, containing free charge is-
(a) Q
(b) Q/E_o
(c) E_o
(d) Q^2/E_o

Solutions

S1. Ans.(c)
Sol. basically it is based on EMI principle in which heat generated in the object by eddy currents

S2. Ans.(d)
Sol. ▭(δ=K√δ \/√f)

S3. Ans.(a)
Sol. Questn 2

S4. Ans.(a)
Sol. for heating, if f is high then penetration will be low and used for heating. For melting purpose f is low because penetration depth will be high and will melt easily.

S5. Ans.(b)
Sol. because if having drooping characteristics of voltage and current
In arc welding, the generator should have high current and low voltage which is possible by differential compound generator.

S6. Ans.(c)

S7. Ans.(a)

S8. Ans.(d)
Sol. inducting heating is possible in conducting material and it may be magnetic or non-magnetic. The effect will depends upon permeability. Higher the permeability more will be effect.

S9. Ans.(c)
Sol. as J=6E
⇒J=1/δ E
⇒δ=E/J
⇒δ.(l/A)=E/J.(l\/A)
⇒▭(R=El/JA)

S10. Ans.(a)
Sol. According to gauss’s law,
▭(close surface integral of D ⃗.ds ⃗=Qen=Q)

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