SSC JE DC Motor & DC Generator Solved Questions (2020)

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.

Answer.1. 272.2 V

Explanation:-

Given

Voltage = 250 V

Load current (I_L) = 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 = I_LR_se

= 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 = I_L + I_f = 100 + 2 = 102 A

Generated EMF

E = V₁ + IaRa

= 260 + 102×0.1

E = 270.2 Volt

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.

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.

R_ac = 1.5 Rdc = 1.5 x 0.5

R_ac = 0.75Ω

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.

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.

Answer.3. 20 A

Explanation:-

For DC motor voltage is given as

V = Eb + IaRa

Ia = (V – Eb)/Ra

= (150 -142)/0.4 = 20A

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.

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 I_a/2

Np ∝ 2V/I

From both the equation

Ns/Np = V/2I/2V/I

N parallel = 4Nsereis

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.

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.

Hopkinson test is performed to measure the efficiency of the DC machine.