Previous Year Questions- DC Machines - 2 | Electrical Machines - Electrical Engineering (EE) PDF Download

Q26: A 220 V, 15 kW, 100 rpm shunt motor with armature resistance of 0.25 Ω, 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 of 1.8 A is       (2012)
(a) 18.18% increase
(b) 18.18% decrease
(c) 36.36% increase
(d) 36.36% decrease
Ans: (d)
Sol: i.e. field flux reduces
Therefore, % change in flux
⇒ Decrease in flux.

Q27: A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. 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 be        (2011)
(a) 1.79 Ω
(b) 2.1 Ω
(c) 3.1 Ω
(d) 18.9 Ω
Ans: (a)
Sol: CASE-(i)
CASE-(ii)
Back emf,
As speed is constant
as torque is constant

Q28: A 4-point starter is used to start and control the speed of a      (2011)
(a) dc shunt motor with armature resistance control
(b) dc shunt motor with field weakening control
(c) dc series motor
(d) dc compound motor
Ans: (a)
Sol: 4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds.
In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.

Q29: 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.
For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage to be applied is      (2010)
(a) 125.5V
(b) 193.3V
(c) 200V
(d) 241.7V
Ans: (b)
Sol: For,  N = 1400rpm, E_a = 186.67V
Required V = E_a + I_aR_a
= 186.67 + I_a(3.39)
where I_a is function of torque, [/latex]  
To develope 5 N-m it requires 3.925A
= 193.3 V

Q30: 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       (2010)
(a) 2 Ω
(b) 3.4  Ω
(c) 4.4  Ω
(d) 7.7  Ω
Ans: (b)
Sol: 
Q31: Figure shows the extended view of a 2-pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform V_A′B′ will resemble     (2009)
(a) (b) (c) (d) Ans: (a)
Sol: DC machine is acting as DC motor.
When the brush is shifted in the direction of rotation in DC motor, field gets magnetized.
Due to armature reaction, leading tip of N-pole and S-pole gets more magnetized and trailing tip of N-pole and S-pole gets demagnetized.
Butincrease in flux density is more than the decreases in flux density.
Due to magetizing effect of armature reaction waveform of V_A′B′ will be as given option (A).

Q32: 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 external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is        (2008)
(a) 31.1 Ω
(b) 31.9 Ω
(c) 15.1 Ω
(d) 15.9 Ω
Ans: (a)
Sol: Rated armature current = 12A
Armature current during braking I_b = 1.25 × 12 = 15A
Total armature resistance including external resistance during braking

Q33: 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       (2008)
(a) 6 V
(b) 234 V
(c) 240 V
(d) 474 V
Ans: (d)
Sol: Before plugging
Current through field winding,
Current through armature
Back emf = E_a = V_t - I_aR_a In plugging, the connection of armature winding is reversed. A strong baking torque is achieved by
achieved by maintaining the supply voltage to the armature with connections reversed.
So, the net voltage across the armature resistance at the time of plugging
E_a + V = 240 + 234 = 474V

Q34: A 220 V, 1400 rpm, 40 A separately excited dc motor has an armature resistance of 0.4 Ω. The motor is fed from a single phase circulating current dual converter with an input ac line voltage of 220 V (rms). The approximate firing angles of the dual converter for motoring operating at 50% of rated torque and 1000 rpm will be     (2008)
(a) $43\circ ,137\circ$43°, 137°
(b) 43°, 47°
(c) 39°, 141°
(d) 39°, 51°
Ans: (c)

Q35: A 220 V 20 A, 1000 rpm, separately excited dc motor has an armature resistance of 2.5 Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input dc voltage to the chopper is 250 V. The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be       (2008)
(a) 0.518
(b) 0.608
(c) 0.852
(d) 0.902
Ans: (b)

Q36: The dc motor, which can provide zero speed regulation at full load without any controller is      (2007)
(a) series
(b) shunt
(c) cumulative compound
(d) differential compound
Ans: (d)
Sol: Speed-current characteristics of DC motorsIt can seen from the characteristics, speed regulation can be zero at full load in the case of differential compound dc motor.

Q37: 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      (2006)
(a) 0.87
(b) 0.95
(c) 0.96
(d) 1.06
Ans: (a)
Sol: For generator,

Q38: In a DC machine, which of the following statements is true ?      (2006)
(a) Compensating winding is used for neutralizing armature reaction while interpole winding is used for producing residual flux
(b) Compensating winding is used for neutralizing armature reaction while interpole winding is used for improving commutation
(c) Compensating winding is used for improving commutation while interpole winding is used for neutralizing armature reaction
(d) Compensation winding is used for improving commutation while interpole winding is used for producing residual flux
Ans: (b)
Sol: Compensating winding is placed in slots cut out in pole faces such that the axis of this winding coincides with the brish axis. The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization.
To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In rder that this injection is restricted to commutating coils, narrow iterpolar are provided in the interpolar region.

Q39: In relation to DC machines, match the following and choose the correct combination.     (2005)
(a) P-3 Q-3 R-1
(b) P-2 Q-5 R-4
(c) P-3 Q-5 R-4
(d) P-2 Q-3 R-1
Ans: (d)
Sol: As we know,
(i) Armature emf, E = k_aϕω
where k_a is constant
so, E depends on ϕ and ω only.
(ii) Developed torque, T= k_aϕI_a
T depends on 𝜙 and I_a only.
(ii) Developed power, P = EI_a = (k_aϕω)I_a
or, P = Tω = (k_aϕI_a)ω
So, P depends on ϕ, ω and I_a.

Q40: A 50 kW dc shunt is loaded to draw rated armature current at any given speed. When driven (i) at half the rated speed by armature voltage control and (ii) at 1.5 times the rated speed by field control, the respective output powers delivered by the motor are approximately.       (2005)
(a) 25 kW in (i) and 75 kW in (ii)
(b) 25 kW in (i) and 50 kW in (ii)
(c) 50 kW in (i) and 75 kW in (ii)
(d) 50 kW in (i) and 50 kW in (ii)
Ans: (b)
Sol: Armature control provides constant torque drive.In the shunt motor case by keeping the filed current at maximum value full load torque can be obtaineda at full load armaturecurrent at all speeds 

(output power delivered at half the rated speed by armature control).
Filed control provides constant-power drive. Since the speed is inversely proportional to the flux/pole  while the torque is direclty proportional to flux/pole (T ∝ ϕ) for a given armature current.
Output power
= k₁k₂ = constant
Output power delivered by the motor remains constant i.e. 50kW

Q41: Two magnetic poles revolve around a stationary armature carrying two coil (c₁ − c₁′, c₂ − c₂′) as shown in the figure. Consider the instant when the poles are in a position as shown. Identify the correct statement regarding the polarity of the induced emf at this instant in coil sides c₁ and c₂.      (2005)
(a) ⊙ in c₁, no emf in c₂ 
(b) ⊗ in c₁, no emf in c₂
(c) ⊙ in c₂, no emf in c₁ 
(d) ⊗ in c₂, no emf in c₁ 
Ans: (a)
Sol: Magnetic poles revolves around stationary coils in clockwise direction. As the motion is relative, it can be assumed that, magnetic poles are stationary and coils are rotating in counter-clockwise. The direction of magnetic flux is from N-pole to S-pole. Now consider coils C₁ and  C_i
The dynamically induced emf in conductor C₁ can be given by flamig's right hand rule
 The direction of emf
Therefore, ⊙ in C₁
Now, consider, coils C₂ and ′C₂′
At this instant, B and V are antiparallel i.e.
 θ = 180^∘
So, sin180^∘ = 0
Hence,no emf is induced in C₂.

Q42: 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      (2004)
(a) 96 V
(b) 192 V
(c) 384 V
(d) 768 V
Ans: (c)
Sol: No of coil = 32
Each coil has 6 turns
Total no. of turns = 32 × 6 = 192
Each turn has 2 conductors
So, total no. of conductors = Z = 192 × 2 = 384
For wave would armature no of parallel path = A = 2
Expression of induced emf,

Q43: The armature resistance of a permanent magnet dc motor is 0.8 Ω. At no load, the motor draws 1.5 A from a supply voltage of 25 V and runs at 1500 rpm. The efficiency of the motor while it is operating on load at 1500 rpm drawing a current of 3.5 A from the same source will be       (2004)
(a) 48.00%
(b) 57.10%
(c) 59.20%
(d) 88.80%
Ans: (a)
Sol: At no load,
As the motor is working at no load, power P₀ supplies no load losses.
When the motor is on load,

Q44: A dc series motor driving an electric train faces a constant power load. It is running at rated speed and rated voltage. If the speed has to be brought down to 0.25 p.u. the supply voltage has to be approximately brought down to      (2003)
(a) 0.75 p.u
(b) 0.5 p.u
(c) 0.25 p.u
(d) 0.125 p.u
Ans: (b)
Sol: In series motor,
If 'w_n' decrease by (1/4) of its rated value, then torque will increase 4 times to maintain power contant.
∴ T ∝ I² 
∴ Current increase 2 times so that supply voltage approximately brought down to 0.5pu.

Q45: Following are some of the properties of rotating electrical machines
P. Stator winding current is dc, rotor winding current is ac.
Q. Stator winding current is ac, rotor winding current is dc.
R. Stator winding current is ac, rotor winding current is ac.
S. Stator has salient poles and rotor has commutator.
T. Rotor has salient poles and sliprings and stator is cylindrical.
U. Both stator and rotor have poly-phase windings.
DC machines, Synchronous machines and Induction machines exhibit some of the above properties as given in the following table.
Indicate the correct combination from this table       (2003)
(a) a
(b) b
(c) c
(d) d
Ans: (a)
Sol: In DC machine, field is provided at stator and armature is provided at rotor. Because of commutator and slip-rings, AC rotor winding current is converted inot dc.
IN synchronous machine, it is difficult to provide armature in rotor because of many factor like insulation no of slip rings, rating of slip rings, high rating of armature. Therefore, armature is provided in stator and field in rotor
Induction machines are singlyfed machine i.e. they draw excitation from the same supply. So no external field supply is required.

Q46: To conduct load test on a dc shunt motor, it is coupled to a generator which is identical to the motor. The field of the generator is also connected to the same supply source as the motor. The armature of generator is connected to a load resistance. The armature resistance is 0.02 p.u. Armature reaction and mechanical losses can be neglected. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The p.u value of this load resistance is       (2003)
(a) 1
(b) 0.98
(c) 0.96
(d) 0.94
Ans: (c)
Sol: All calculation are done in p.u. system.
Motor is connected across rated voltage V = 1 pu.
Rated armature current flows in both motor and generator
I_g = I_m = 1pu
Armature resistance of both motor and generator R_m = R_g = 0.02pu
Back iemf in motor
Mechanical power output of the motor
This power is given to the generator and armation reaction and mechanical losses are negligible.
Input power of generator = Output power of motor
Terminal voltage of the generator

Q47: The speed/torque regimes in a dc motor and the control methods suitable for the same are given respectively in Group-II and Group-I      (2003)
(a) P-1, Q-3
(b) P-2, Q-1
(c) P-2, Q-3
(d) P-1, Q-4
Ans: (b)
Sol: Field Control: Fieldcontrol provides constant power drive. Since the speed is inversely proportional to flux/pole (ω ∝ (1/ϕ)) field can not be increased; It can only be weakended. So this control is suitable for speed control above base speed as torque is directly proportional to flux/pole (T ∝ α) for a given armature current. So, this control provides torque below base torque.
Armature Control: It provides constant-torque drive. By keeping th field current at maximumvalue full load torque can be obtained at full load armature current at all speeds. This control provides speed below base speed. 
Q48: 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      (2002)
(a) 4.30 Nm
(b) 4.77 Nm
(c) 0.45 Nm
(d) 0.50 Nm
Ans: (a)
Sol: Torque at 5 A,

Q49: A dc series motor fed from rated supply voltage is overloaded and its magnetic circuit is saturated. The torque-speed characteristic of this motor will be approximately represented by which curve of below figure?     (2002)
(a) Curve A
(b) Curve B
(c) Curve C
(d) Curve D
Ans: (b)
Sol: At saturation , linear characteristic are observed.

Q50: An electric motor with 'constant output power' will have a torque speed characteristic in the form of a       (2001)
(a) straight line through the origin
(b) straight line parallel to the speed axis
(c) circle about the origin
(d) rectangular hyperbola
Ans: (d)
Sol:
xy  = C (Rectangular Hyperbola)