Ques.51. The motor used in a small lathe is usually (SSC-2015)
D.C shunt motor
3 phase synchronous motor
Single-phase capacitor run the motor
Universal Motor
Answer.1. D.C shunt motor
Explanation:-
DC shunt motor is a constant speed motor Lathe machine requires to be run at a constant speed hence DC shunt motor is used in lathe machines.
Ques.52. A short shunt compound generator supplies a load current of 100 A at 250 V. The generator has the following winding resistances:
Shunt field = 130Ω, Armature 0.1 Ω and series field = 0.1 Ω . Find the Emf generated if the brush drop is 1 V per brush. (SSC-2015)
272.2 V
262.2 V
272.0 V
262.0 V
Answer.1. 272.2 V
Explanation:-
Given
Voltage = 250 V
Load current (IL) = 100 A
Shunt field (Rf) = 130Ω
Armature field (Ra) = 0.1Ω
Series field (Rse) = 0.1Ω
Brush voltage Drop = 1V
Now voltage drop in the field winding = ILRse
= 100 x 0.1 = 10 V
For short shunt compound generator, the voltage drop across series field winding gets added with voltage drop across armature winding
Voltage drop across field winding + Voltage drop across armature winding
V1 = 250 + 10 = 260 V
Field current If = V1/Rf = 260/130 = 2 A
So armature current
Ia = IL + If = 100 + 2 = 102 A
Generated EMF
E = V1 + IaRa
= 260 + 102×0.1
E = 270.2 Volt
Ques.53. Which of the following braking is not suitable for the motor? (SSC-2015)
Plugging
Regenerative braking
Dynamic braking
Friction braking
Answer.4. Friction braking
Explanation:-
Mechanical braking or Friction braking or Magnetic Braking
In this type of braking, the motor is stopped by using a brake shoe or band on a brake drum. In friction braking, it is difficult to get the smooth stop of the motor and It requires frequent maintenance and replacement of brake shoe, and braking power is was as well as it dissipates more heat energy.
Ques.54. The per-phase D.C armature resistance of an alternator is 0.5 ohm. The effective AC armature resistance would be about? (SSC-2015)
0.25 ohms
0.5 ohms
1 ohm
0.75 ohms
Answer.4. 0.75 ohms
Explanation:-
The AC resistance per phase is obtained by multiplying the dc resistance by the factor that varies from 1.2 to about 1.8, depending on the frequency, quality insulation, etc. For purposes here, we shall use a factor of 1.5 in computing the effective AC armature resistance per phase.
Rac = 1.5 Rdc = 1.5 x 0.5
Rac = 0.75Ω
Ques.55. The Ta vs Ia is a graph of a DC series motor is an (SSC-2015)
Straight-line throughout
Parabola from no load to overload
Parabola throughout
Parabola up to full load and straight line after saturation
Answer.4. Parabola up to full load and straight line after saturation
Explanation:-
Ta vs Ia graph of DC series motor
In the case of DC series motor torque is proportional to the square of the armature current and this relation is parabolic in nature.
Ta vs Ia graph of DC series motor
As load increases, armature current increases and torque produced increase proportional to the square of the armature current upto a certain limit.
After reaching the Saturation point though the current through the winding increases, the flux produced remains constant. Hence after saturation, the Ta & Ia takes the shape of the straight line as flux becomes constant.
Ques.56. A 150 V DC motor of armature resistance 0.4 Ω has back emf of 142 V. The armature current is (SSC-2014)
100 A
10 A
20 A
150 A
Answer.3. 20 A
Explanation:-
For DC motor voltage is given as
V = Eb + IaRa
Ia = (V – Eb)/Ra
= (150 -142)/0.4 = 20A
Ques.57. The rated speed of a given D.C. shunt motor is 1050 rpm. To run this machine at 1200 r.p.m. the following speed control scheme will be used (SSC-2014)
Varying frequency
Armature circuit resistance control
Field resistance control
Ward-Leonard Control
Answer.3. Field resistance control
Explanation:-
In field resistance control, the lower the field current in a shunt (or separately excited) dc motor, the faster it turns and the higher the field current, the slower it turns. Since an increase in field current causes a decrease in speed, there is always a minimum achievable speed by field circuit control. This minimum speed occurs when the motor’s field circuit has the maximum permissible current flowing through it.
If a motor is operating at its rated terminal voltage, power, and field current, then it will be running at rated speed, also known as base speed. Field resistance control can control the speed of the motor for speeds above base speed but not for speeds below the base speed. To achieve a speed slower than base speed by field circuit control would require excessive field current, possibly- burning up the field windings.
Ques.58. Two DC series motors connected in series draw current I from supply and run at speed N. When the same two motors are connected in parallel taking current I from the supply, the speed of each motor will be (SSC-2014)
N/2
N
2N
4N
Answer.4. 4N
Explanation:-
In the case of series.
When motor are connected in series and are in running position, Speed ∝ (voltage) ∝(V/2) (i) (Since the voltage across each motor = V/2 ) and current is Ia.
Ns ∝ V/2I
In the case of parallel
When the motor is connected in parallel and are in running position, Speed ∝ (voltage) (V). (Since the voltage across each motor = V ) and current is Ia/2
Np ∝ 2V/I
From both the equation
Ns/Np = V/2I/2V/I
N parallel = 4Nsereis
Ques.59. Which of the following motor has a high starting torque? (SSC-2014)
Synchronous motor
AC series motor
DC series motor
Induction motor
Answer.3. DC Series Motor
Explanation:-
DC Series motor has a high starting torque. We can not start the Induction motor and Synchronous motors on load, but cannot start the DC series motor without load because on very light load or no load as flux is very small (N ∝ 1/Φ), the motor tries to run at dangerously high speed which may damage the motor mechanically.
Ques.60. Commutation conditions at full load for large DC machines can be efficiently checked by the (SSC-2014)
Brake test
Swinburne’s test
Hopkinson’s test
Field test
Answer.3. Hopkinson’s test
Explanation:-
Hopkinson’s test is also referred to as the regenerative or Back-to-Back test. It is a full load test and required two identical machines are connected parallel, whereas the first machine gets excited from the source so it acts as a motor and is mechanically coupled with the next machine and makes it run as a generator. The electrical output power from the generator (second machine) is fed to the motor(First machine) and vice-versa.
