Quiz: Electrical Engineering 11 May 2020

Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: Miscellaneous

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

 

 

Q1.  Most economical method for starting single-phase motor is?

(a) Induction-start method

(b) Split-phase method

(c) Capacitor start method

(d) Resistance-start method

 

 

Q2.       For an 80% efficient transformer, the primary winding has 1000 turns and the secondary 100 turns. If the power input to the above transformer is 800 watts, the power output is __________.

(a) 1000 watts

(b) 640 watts

(c) 320 wats

(d) 10 kW

 

 

Q3.       Voltmeter is a galvanometer with _____.

(a) high resistance

(b) low resistance

(c) both low and high resistance

(d) uncertain resistance

 

 

Q4.       Transfer function of a system is  This open-loop system is _________.

(a) stable

(b) unstable

(c) marginally stable

(d) conditionally stable

 

Q5.       A wire of resistance 0.1 Ω /cm is bent to form a square ABCD of side 10 cm. A similar wire is connected between B and D to form the diagonal BD. If a 2 V battery of negligible internal resistance is connected between A and C, then total power dissipated is

(a) 2 W

(b) 3 W

(c) 4 W

(d) 6 W

 

 

Q6.       A cell supplies a current of 0.9 A though 2 Ω external resistance and 0.3 A through 7 Ω external resistance. The internal resistance of the cell is

(a) 0.5 Ω

(b) 1.2 Ω

(c) 1 Ω

(d) 1.5 Ω

 

 

Q7.       If the capacitor of a single-phase motor is short circuited, the motor will __________.

(a) start

(b) not start

(c) start with jerks

(d) start and then stop

 

Q8. The power measured by ammeter-Voltmeter method, if the voltmeter is connected across the load then the value of power will be

  • The power consumed by the load
  • The sum of power consumed by the load and the ammeter
  • The sum of power consumed by the load and the voltmeter
  • The sum of power consumed by the load, ammeter and voltmeter

 

 

Q9. For the same power the size of turbine

  • Increase with speed
  • Remains with speed
  • Decrease with speed
  • None of the above

 

Q10. Steam turbines are governed by

  • Nozzle control governing
  • Throttle governing
  • Bypass governing
  • All of the above

 

 

Solution

S1 (c)

Capacitor start method is most economical method.

S2. (b)

∵ Efficiency is 80 %

∴ Power output = .80×800 Wats =640 W

S3. (a)

High resistance

S4. (b)

Here, two poles are at origin. So, system is unstable.

 

S5. (c)

The given description forms a wheat stone bridge in balanced form.

∴   between A and C

 

∴ Total power =  W

S6. (a)

Lat internal resistance of call is r ohm.

0.9 (r+2) =0.3 (r+7)

S7. (b)

Motor will not start.

S8 (c)

The sum of power consumed by the load and the voltmeter

 

S9 (c)

Decrease with the speed.

S10 (d)

Mentioned all methods are used to govern steam turbine.

 

Leave a comment

Your email address will not be published. Required fields are marked *