Quiz: Electrical Engineering
Exam: UPSSSC- JE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. When an AC power is applied to a circuit having reactive load, then the voltage is_______
(a) 90 degree out of phase with current
(b) 180 degree out of phase with the current
(c) In phase with the current
(d) 270 degree out of phase with the current
Q2. If two supply terminals of a 3φ induction motor are interchanged, then the:
(a) Speed of the motor will become zero
(b) Speed of the motor will increase
(c) Motor will continue to run in the same direction with less speed
(d) Motor will rotate in the opposite direction
Q3. What is the relationship between the line voltage and the phase voltage in a delta -connected load?
(a) Line voltage = Phase current
(b) Line voltage > Phase voltage
(c) Line voltage < Phase voltage
(d) Line voltage = Phase voltage
Q4. An eight-pole wound rotor induction motor operating on 60 Hz supply is driven at 1800 rpm by a prime mover in the opposite direction of revolving magnetic field. The frequency of rotor current is:
(a) 60 Hz
(b) 200 Hz
(c) 120 Hz
(d) 180 Hz
Q5. The SI unit of conductivity is:
(a) Ohm/meter
(b) Ohm-meter
(c) Siemens/meter
(d) Siemens – meter
Q6. The SI unit of conductance is:
(a) Coulomb
(b) Ohm
(c) Siemens
(d) Newton
Q7. Select the incorrect statement.
A single-phase induction-motor
1. Requires only one winding
2. Can rotate in one direction only
3. Is not self-starting
(a) Only 1 and 2
(b) Only 2 and 3
(c) 0nly 3 only
(d) Only 1 and 3
Q8. In a single-phase induction motor, how are the windings placed?
(a) Both main and auxiliary windings on stator
(b) Both windings are on rotor
(c) Auxiliary on stator and main on rotor
(d) Main winding on rotor
Q9. Which of the following statements is correct for a single-phase hysteresis motor?
(a) It can run at synchronous and sub-synchronous speed
(b) It can run at sub synchronous speed only
(c) It can run at synchronous and super synchronous speed
(d) It can run at synchronous speed only
Q10. If a 500 KVA, 200HZ transformer is operated at 50HZ, its KVA(s) rating will be–
(a) 250 KVA
(b) 125 KVA
(c) 1000 KVA
(d) 2000 KVA
SOLUTIONS
S1. Ans.(a)
Sol. In the case of inductor or capacitor the voltage or current are out of phase. In the case of an inductor, the voltage leads the current by 90°, whereas in the capacitor the current lead voltage by 90°.
S2. Ans.(d)
Sol. The direction of rotation of a 3-phase motor can be changed by reversing two of its Stator leads.
S3. Ans.(d)
Sol. In delta connected system the line voltage is equal to the phase voltage and the Line current is √3 times of Phase Current.
S4. Ans.(d)
Sol. N_s=120f/P=(120×60)/8
= 900 rpm
Now it is rotating in opposite direction so slip (S) = (900-(-1800))/900=3
Rotter frequency (Fr) = Sf
= 3×60
= 180 HZ
S5. Ans.(c)
Sol. Conductivity (σ) = 1/(Resistivity (ρ) )
↓
Unit =1/Ωm (mho/m)
Or
Siemen/meter
S6. Ans.(c)
Sol. Conductance (G) = 1/(Resistivity (R) )
↓
Unit =1/ohm (mho or siemen)
S7. Ans.(a)
Sol. A single-phase induction motor is not self-starting and has two types of winding: main and auxiliary winding and direction of rotation can be changed by changing polarity of either main or auxiliary winding.
S8. Ans.(a)
Sol. The main parts of a single -phase induction motor are the Stator, Rotor, Windings. The stator is the fixed part of the motor to which A.C. is supplied. The stator contains two types of windings. One is the main winding and the other is the Auxiliary winding. These windings are placed perpendicular to each other.
S9. Ans.(d)
Sol. Hysteresis Motor is a synchronous motor with a cylindrical rotor and does not require any dc excitation to the rotor and it uses non-projected poles. It is a single-phase motor with rotor made up of ferromagnetic material.
initially, when hysteresis motor is started it behaves as a single-phase induction motor and while running it behaves as a synchronous motor.
S10. Ans.(b)
Sol. For constant load;
KVA i.e. S is directly proportional to induced emf and emf is directly proportional to frequency
⇒ S α f
So, S1/f1 =S2/f2
⇒ 500/200=S2/50
⇒ S₂ = 125 KVA
