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

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 = 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

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.


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