Sunday, 10 August 2014

Electrical Machines - DC Machines

India's finest entrance exam for Post-Graduation is GATE.Below questions are previous year gate questions  of EE stream.

1. A 220V, 15 kW, 1000 rpm shunt motor with armature resistance of has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current is 1.8 A [GATE-2012]
a.  18.18% increase
b. 18.18% decrease
c. 36.36% increase
d. 36.36% decrease

2. Neglecting all losses, the developed torque (T) of a d.c. separately excited motor, operating under constant terminal voltage, is related to its output power (P) as under [GATE-1992]
a. T proportional to square root of P
b. T proportional to P
c. T^2 proportional to P^3
d. T independent of P

3. A differentially compounded d.c. motor with interpoles and with brushes on the neutral axis is to be driven as a generator in the same direction with the same polarity of the terminal voltage. It will [GATE-1995]
a. be a cumulatively compound generator but the interpole coil connections are reversed.
b. be a cumulatively compound generator without reversing  interpole coil connections
c. be a differently compound generator without reversing  interpole coil connections
d. be a differently compound generator but the interpole coil connections are reversed.

4. A cumulative compounded long shunt motor is driving a load at rated torque and rated speed If the series field is shunted by a resistance at to the resistance of the series field, keeping the torque constant [GATE-1993]
a. Armature current increases
b. Motor speed increases
c. Armature current decreases
d. Motor speed decreases

5. The torque speed characteristic of a repulsion motor resembles which of the following dc motor characteristic ? [GATE-1996]
a. Separately
b. Shunt
c. Series
d. Cumulative

6. A 240 V d.c. shunt motor with an armature resistance of 0.5ohm has a full load current of 40 A. Find the ratio of the stalling torque to the full load torque when a resistance of 1ohm is connected in series with the armature [GATE-1998]
a. 4          b. 12            c. 6         d. None of the above

7. A DC shunt motor is running at 1200 rmp, when excited with 220 V DC. Neglecting the losses and saturation, the speed of the motor when connected to a 175 V DC supply [GATE-1999]
a. 750 rpm.      b. 900 rpm
c. 1050 rpm      d. 1200 rpm

8. A 240 V dc series motor takes 40A when giving its rated output at 1500 rpm. Its resistance is 0.3 ohms. The value of resistance which must be added to obtain rated torque at 1000 rpm is [GATE-2000 ]
a. 6 ohms             b. 5.7 ohms
c. 2.2 ohms          d. 1.9 ohms

9. An electric motor with ‘constant output power will have a torque speed characteristic in the form of
[GATE-2001]
a. straight line through the origin.
b. straight line parallel to the speed axis.
c. circle about the origin.
d. rectangular hyperbola.

10. A 200 V, 2000 rpm ,10A, separately excited dc motor has an armature resistance of 2Ω. Rated dc voltage is applied to both the armature and field winding of the motor. If the armature drawn 5A from the source, the torque developed by the motor is [GATE-2002]
a. 4.30 Nm            b. 4.77 Nm
c. 0.45 Nm             d. 0.50 Nm

11. A 8-pole ,DC generator has a simplex wave-wound armature containing 32 coils of 6 turns each . Its flux per pole is 0.06 Wb. The machine is running at 250 rpm. The induced armature voltage is
[GATE-2004]
a. 96 b. 192 c. 384 d. 768

12. A 50 kW dc shunt motor is loaded to draw rated armature current at any given speed, When driven
(i) at half the rated speed by armature voltage control
(ii) at 1.5 times the rated speed by field current ,the respective output powers delivered by the motors are approximately  [GATE-2005]
a. 25 Kw & 75 Kw
b. 25 Kw & 50 Kw
c.  50 Kw & 75 Kw
d. 50 Kw & 50 Kw

13. A 220 V DC machine supplies 20 A at 200 V as a generator . The armature resistance is 0.2 ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10%, the ratio of motor speed to generator speed is
[GATE-2006]
a. 0.87  b. 0.95  c. 0.97  d. 1.06

14. The dc motor , which can provide zero speed regulation at full load without any controller is [GATE-2007]
a. series    c. cumulative compound
b. shunt.   d. differential compound

15. A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 Ω and the field winding resistance is 80 Ω
The net voltage across the armature resistance at the time of plugging will be [GATE-2008]
a. 6V b. 236V c. 240V  d. 474V

16. A separately excited DC motor runs at 1500 rpm under no-load with 200 V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of 5 Nm,is 1400 rpm as shown in the figure. The rotational losses and armature reaction are neglected.
The armature resistance of the motor is, [GATE-2010]
a. 2v   b. 3.4v   c. 4.4v   d. 7.7v

17. A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1 ohm and armature current is 10A. of the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will  [GATE-2011]
a. 1.79 ohm.        b. 2.1 ohm
c. 1.89 ohm.         c. 3.1 ohm

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

Explanations :

1.   N1/N2 = (Eb1/Eb2)*(flux2/flux1)
      we know that,
      Eb = V-(Ia*Ra)
      Ia =line current - field current.
     
By putting values above one..
We get ,(flux2/flux1) = 0.6364

-It is clearly indicating that, flux has to reduce.
So,the change in flux in percentage is 36.36%
- [1-(flux2/flux1)]*100

17. Armature current is 10 A.
Let, flux is 'x1'.now,flux is reduced by 10%.now,flux is 'x2'
  So,   x2 = 0.9*x1

N1/N2 = (Eb1/Eb2)*(flux2/flux1)
      we know that,
             Eb1 = V-(Ia*Ra) &
             Eb2 = V-[Ia*(Ra+R)]
By putting values & solving.we get
           R = 1.79 ohms
     

No comments:

Post a Comment