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Correct choice is (c) Armature control method

Easiest explanation: In armature control method, we VARY armature voltage by adjusting variable RESISTANCE value. So, when we define armature efficiency, it is equal to η = [ (Vt-IaRe) Ia] / VTIA= 1-IaRe / Vt. Thus, efficiency is reduced drastically for large SPEED reductions.

Right option is (b) Decreases

Easy explanation: In a DC SERIES motor, when the resistance increases due to RISE in temperature, the IR drop in the circuit will increase, which reduces the EFFECTIVE voltage applied to the armature, hence speed decreases due to the direct PROPORTIONALITY.

Right option is (b) In series motors

To elaborate: Diverters are used in series motors for SPEED CONTROL, they are not used in SHUNT motors. Since, shunt field WINDING RESISTANCE is very high, if we connect a diverter across shunt field winding, total current will flow through diverter and it will almost short circuit the shunt field winding. This will increase the motor speed to a very high value.

The correct OPTION is (a) SPEED given by parallel connection is more

The best explanation: In parallel case, total AMPERE turns will be exactly half of the ampere turns of the SERIES case. Since speed of any DC MOTOR is inversely proportional to the MMF or flux or ampere turns developed, speed of the parallel control method will always large than the series one.

The correct answer is (b) Series-parallel

Best EXPLANATION: In series-parallel method, field winding is equally divided in two parts which are connected in series and parallel with respect to each other. THUS, in series connection we’ll get one SPEED and in parallel MODE we’ll get one speed. These 2 SPEEDS are only possible.

The correct option is (c) Field winding is made with more out pins

The BEST explanation: In tapped field control of the DC series motor a field winding is SIMPLY tapped at various points and many out pins taken, so that any out pin can be connected and particular PART of field winding will be selected in the CIRCUIT, while other part is made OPEN.

Correct answer is (b) Below speed-torque characteristic of lower resistance

The EXPLANATION is: Speed-torque characteristic for decreasing values of diverter resistance or decreasing VALUE of kd will LIE above of the previous or lower value of diverter resistance. Thus, for higher diverter value curve will lie below the previous one.

Right choice is (a) In parallel with field

Easy explanation: Diverter is a SMALL VARIABLE resistor, which is connected in parallel with field winding. By VARYING this resistance value, the field current and consequently the field ampere-turns can be varied and speed can be CONTROLLED.

Right answer is (d) To AVOID oscillations in speed

The explanation is: One precaution to be taken in this method in order to avoid oscillations in speed initiated by LOAD changes is to use an inductively wound DIVERTER resistor. Thus, speed controlled output will vary smoothly when load is changed.

The CORRECT option is (B) False

Explanation: Field control method is not SUITED to applications needing speed REVERSAL; since the only way to reverse speed is to disconnect the motor from the source and reverse the field/armature POLARITY. The field circuit being highly inductive, it is normally the armature which is reversed.

The correct option is (c) Variable RESISTANCE in series with armature

To ELABORATE: Diverter means adding a variable resistance in PARALLEL with field winding. Tapped field control is the METHOD where field ampere turns are adjusted in steps by varying the number of turns INCLUDED in the circuit. In series parallel method field winding is connected so that it will form two parts.

Correct answer is (B) Weaker commutating field is needed at LOW speed than at high speed

The EXPLANATION: According to commutating FIELDS of machine, speed and overall performance of a machine is determined. For low speeds of motor in a variable speed motor we NEED weaker commutating field.

Correct answer is (B) 14.28%

The EXPLANATION: Speed of regulation of DC series motor is GIVEN by ratio of difference of no-load speed and full load speed with full load speed. Thus, in this case, speed regulation = (1200-1050)/1050= 14.28%.

The correct answer is (d) SERIES motor

For explanation: DC series motor at no load condition gives infinite SPEED ideally. Practically it will damage all the armature circuit. THUS, as the load is reduced speed of the motor will go on increasing RAPIDLY. So, speed control is very poor in series motor.

Correct ANSWER is (a) Field weakening method

Best explanation: In field weakening method we are reducing the WORKING flux to increase the speed, by reducing the field current. THEREFORE, effect of ARMATURE flux on main field flux will increase in case of field weakening method.

Correct option is (b) 1273 RPM

Explanation: Speed regulation is equal to 0.1 which is also equal to (no-load – full load speed) divided by full load speed. THUS, by SUBSTITUTING all known QUANTITIES we get full load speed = 1400/1.1 = 1272.7 rpm so, speed equal to 1273 rpm.

The correct option is (b) 11.8 %

For explanation: SPEED regulation is equal to (No-load speed – Full load speed) / (Full load speed). By substituting all the VALUES, speed regulation= (1322-1182)/ 1182. Speed regulation is given by 0.118. In PERCENTAGE NOTATION SR= 11.8 %.

Right answer is (a) Reducing the field current

To explain I would say: Speed of the DC motor obtained from speed equation is inversely proportional to flux PRODUCED by the field. So, reducing the field current flux produced by armature will decrease, and speed will INCREASE.

The correct CHOICE is (a) Increase

Explanation: From, the speed-current characteristics of DC shunt MOTOR we KNOW that speed of the motor is directly PROPORTIONAL to the back emf and inversely proportional to the flux. So, for more speed there will be more back emf generated.

Correct choice is (c) TORQUE will CHANGE but power will REMAIN constant

To explain I would say: The motor will ACCELERATE the mechanical load connected during this period but no increase in the mechanical load as Pload = T1W1 = T2W2 where W2 >W1. So, at the higher speed there is less electrical torque for the same mechanical load / power.

Correct option is (b) 1.111

The explanation: γ is defined so as to calculate STEP resistance from the given MAXIMUM limit of current and minimum limit of the same. This γ = rn/rn-1. So, for calculating resistance at step 3 we’ll SUBSTITUTE the corresponding VALUES in the equation, which will give step 3 resistance as 1.1111Ω.

Correct answer is (a) γ = rn/rn-1

To explain I WOULD say: For given VALUES of armature currents upper and lower, corresponding EQUIVALENT resistances are CALCULATED. So, by the induction formula we calculate the value of γ and the step RESISTANCE as well, where n denotes the number of steps.

Right ANSWER is (c) 1.625

For explanation: γ is DEFINED as the RATIO of upper limit current value to the lower limit current value. So, γ will be the ratio of 65/40. From calculations, we get γ= 1.625. Note that γ is unitless quantity.

The correct answer is (a) 1.2164

The best I can explain: γ is DEFINED as the ratio of upper current limit to the lower current limit in starters of DC machine. γ ^n-1 = ratio of RESISTANCE at MAXIMUM allowable current to the armature resistance. Substituting values for n=4, we GET γ=1.2164.

The correct answer is (c) To limit the NVC CURRENT

Explanation: To overcome the problem caused when the field current is low, NVC is connected across the two lines, ONE line connected to F terminal through the starter and other DIRECTLY to the second line from another L terminal of the starter. To limit the NVC current a protective resistance R is connected in series with it.

Right choice is (C) If armature current increases beyond certain value

The best explanation: The contact of this relay at armature current above a certain value (over load/ short circuit) closes the NVC ends, again bringing the HANDLE to OFF POSITION. NVC and OL release are protections incorporated in 3-point STARTER.

Correct choice is (c) High finite

The best explanation: As a starting handle is rotated ONE resistance step is added into field circuit. There is a short time wait at each STUD for the MOTOR to build up speed. This arrangement ENSURES a high average starting TORQUE.

Correct option is (c) MOTOR field current is varied over wide range

The best I can explain: Three-point STARTER is EMPLOYED where motor field current can be varied in a narrow range and so does the motor SPEED while four-point starter is USED when motor field current can vary over a wide range and so does the motor speed.

Correct OPTION is (c) The resistance is added into field circuit

The best I can explain: The starting resistance is ARRANGED in steps between conducting raised studs. As the starting HANDLE is ROTATED about its fulcrum, it moves from one stud to the next, one resistance step is cut out, and it gets added to the field circuit.

Right option is (a) 0.235

Easiest explanation: γ is GIVEN as 1.585. So γ^-1 is equal to 0.631. R^1 is provided and it is equal to 4 Ω.

r^1= (1-0.631) * 4 = 1.476 Ω

 r^2= 1.476*0.631= 0.931 Ω, SIMILARLY CALCULATING till r^5= 0.235 Ω.

Correct answer is (c) SERIES with armature

For explanation I WOULD say: In SHUNT and compound motors starting the shunt field should be switched on with full starting resistance in armature CIRCUIT. A short time delay in this position ALLOWS the field current to build up to the steady value of the inductive field transients.

Right answer is (a) True

The EXPLANATION: The only THING in favour of direct starting must be mentioned here. Since the torque of the MOTOR with direct start is much HIGHER, the motor STARTS much more quickly. As a consequence, the Joule input per start is much less than that with resistance start.

The correct answer is (a) Heavy sparking at BRUSHES

For explanation I WOULD say: It would CAUSE intolerably heavy sparking at the brushes which may DESTROY the commutator and brush-gear. Sudden development of large torque will cause mechanical shock to the shaft, reducing its life. Such heavy CURRENT cannot be generally permitted to be drawn from the source of supply.

Right answer is (a) low

Best explanation: Starting resistance of a DC shunt motor and DC COMPOUND motor is low. Well, that’s the reason why we USE starters in a DC motors, in order to LIMIT the armature CURRENT flowing through the armature and to PROTECT machine circuitry.

The correct answer is (b) Shunt as WELL as compound motors

For EXPLANATION: 3-point starters are USED only for shunt and compound motors, they are not used for series motors. Three-point STARTER is employed where motor field current can be varied in a narrow range and so does the motor speed.

Correct choice is (a) Up to 5 H.P

For explanation I would say: DOL starters are limited to the small rating motors where distribution SYSTEM (mains supply) can withstand high starting currents without excessive voltage dips. For a large rating MOTOR, ranging from 5 HP to 25 HP, oil immersed DOL starters are used which provides INSULATION against sparking on contact points, increases the life of STARTER.

The correct option is (d) To restrict armature current as there is no back emf at starting

The best I can explain: At the time of starting (n=0), the induced emf of a motor is zero such that current drawn by armature, from rated VOLTAGE supply WOULD be Ia= V/Ra. SINCE armature RESISTANCE is very low, armature current drawn is very high and will damage the MACHINE.