Sunday, 10 August 2014

Electrical Machines- DC Machines

Here is my collection of IES Previous year questions.

1.  Match List-I (Parts of electrical machines) with List - II (The approximate nature of the air-gap mmf pattern produced by them) and select the correct answer
List-I                              List-II
A. DC machine,stator 1. Sinusoidal B. DC machine,rotor  2. Trapezoidal C. Salient-pole synchronous machine, stator         3. Triangular D. Squerrel-cage rotor of induction motor [IES-1992]

Codes :     A      B     C       D
     a.         1       2     1        3
     b.         3       1      3       1
     c.          2       3      1       3
     d.          2       3     1       1

2. The interpoles in dc machines have a tapering shape in order to
[ IES-1998]
a.  reduce the overall weight
b.  reduce the saturation in the interpole
c.  economise on the material required for interpoles and their windings.
d. increase the acceleration of commutation.

3. Consider the following statements : For a level compounded dc generator to run at constant speed, the series field mmf must effectively compensate 1. armature reaction mmf
2. armature resistance voltage drop
3. brush contact voltage drop Which of these statements is/are correct ? [IES-2000 ]
a. 2 alone
b. 1 and 2
c.  1 and 3
d. 1,2 and 3

4. The current drawn by a 220V dc motor of armature resistance 0.5 and back emf 200V is [IES-2001]
a. 40      b. 44      c. 400      d.440

5. Match List-I ( Type of electrical loads) with List-II (Torque speed characteristics) and select the correct answer : [IES-2002]
List-I
A. Hoist
B. Fans
C. Machine Tools( Lathe, Milling machine etc)
D. Loads with fluid friction
List II
1. Torque proportional to speed^2
2. Torque proportional to speed^3
3. Constant Torque
4.Torque intentionally proportional to speed
a. A-1, B-3, C-2, D-4
b. A-1, B-3, C-4, D-2
c. A-4, B-1, C-3, D-2
d. A-3, B-1, C-2, D-4

6. The current drawn by a 120V d.c. motor with back e.m.f of 110V and armature resistance of 0.4 ohm is [IES-2003]
a. 4A  b. 25A c. 274A d. 300A

7. Assertion (A):In the "3-point" type of starter of a d.c. series motor, the "holding coil" for holding the starter handle in the "ON" stud is connected in such a manner that is short-circuited when the "over load" relay picks up.
Reason (R):In a d.c series motor starter, to guard against :racing due to sudden large reduction of shaft-load, the "holding coil" is connected in series with the armature and the series field winding. [IES-2004]
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A
c.  A is true but R is false
d.  A is false but R is true

8. In a d.c compound characteristic, required for certain applications, may be obtained by connecting a variable resistance: [ IES-2005]
a. Across the series field
b. In series with the series field
c. In parallel with the shunt field
d. In series with shunt field

9. For constant supply voltage, what are the effects of inserting a series resistance in the field circuit of a. d.c shunt motor, on its speed and torque [IES-2006]
a. Speed will decrease and the torque will decrease
b. Speed will increase and the torque will increase 
c.  Speed will increase and the torque will decrease
d. Speed will decrease and the torque will increase 

10. A dc shunt generator is supplying a load of 1.8KW at 200V. Its armature and field resistance are 0.4ohm and 200ohm respectively. What is the generated emf? [IES-2007]
a. 190v   b. 196v   c. 204v    d. 210v

11. When is the mechanical power developed by a dc motor maximum [IES-2008]
a. Back emf is equal to applied voltage
b. Back emf is equal to zero
c.  Back emf is equal to half the applied voltage
d. None of the above

12. match List-I (Machine components) with List-II (Type of machine) and select the correct answer using the code given below the lists: [IES-2009]
List-I                               List -II
A. Amortisseuer winding 1. Squirrel          cage induction motor
B. Breather               2. D.C motor
C. End-Rings             3. Transformer
D. Commutator        4. Synchronous

a.  A-2, B-3, C-1, D-4
b.  A-4, B-3, C-1, D-2
c.   A-2, B-1, C-3, D-4
d.  A-4, B-1, C-3, D-2

13. A separately excited dc motor is started using a 3-phase ac/dc controlled rectifier using 'Soft starting'. For limiting the starting current,it is required that firing angle should [IES-2010]
a. Gradually increased from 0 degrees to 180 degrees.
b. Fixed at 30 degrees.
c. Gradually decreased from 180 degrees to 30 degrees.
d. Gradually increased from 30 degrees to 180 degrees.

14. How many segments will be there for the commutator of a 6 pole d.c machine having a simple wave wound armature with 72 slots?  [IES-2011]
a. 73    b. 72     c. 71     d. 70

15. In a dc machine running with a heavy load, and with the brushes located on the geometrical neutral axis sparking occurs at the brushes during commutation, because of
[IES -2012 ]
a. the high current density at the coil ends
b. the centrifugal force exerted on the brushes by the commutator
c.  the shifting of the magnetic neutral axis.
d. the reduced main field flux -density in the inter-pole regions.

16. At 1200 rpm the induced emf of a dc machine is 200V. For an armature current of 15 A the electromagnetic torque produced would be  [IES -2013]
a. 23.8 N-m                b. 238N-m 
c. 2000 N-m               d. 3000 N-m

Answers:
1.D                        11.           
2.                           12.
3.                           13.
4.A                          14.
5.                           15.
6.B                          16.A
7.                          
8.
9.
10.B

Explanations :

4. V = 220, Eb = 200, Ra = 0.5
    Eb = V - (Ia*Ra)

6. Eb = V - (Ia*Ra)

10.line current = output/voltage
Line current = 1800/200 = 9 A
Shunt current = voltage/shunt 'R'
Shunt current ( Ish) =200/200=1 A
Ia = Ish + line current
Ia = 10 A
Eg = V-(Ia*Ra)=200-(0.4*10)=196 V

17. Eb = [N*flux*Z*(P/A)]/60...(1)
Ta = 0.159*flux*Z* (P/A)*Ia.....(2)

From 1 & 2
Ta = 0.159*60*Ia*(Eb/N) equal to
Ta = 9.55*Ia*(Eb/N)
By equating the values....
We get,torque of  23.8 N-m.

4 comments:

  1. in the 10 the question we have to solve eg not v.. so Ans is 204 v

    ReplyDelete
  2. 10.line current = output/voltage
    Line current = 1800/200 = 9 A
    Shunt current = voltage/shunt 'R'
    Shunt current ( Ish) =200/200=1 A
    Ia = Ish + line current
    Ia = 10 A
    Eg = V+(Ia*Ra)=200+(0.4*10)=204 V

    ReplyDelete