# What is the percentage of stray losses in DC machine?

What is the percentage of stray losses in DC machine?

Percentage of stray losses in DC machine is about 1 % .

## Related questions

Answer : In a D.C. machine stray loss is the sum of iron loss and mechanical loss.

Answer : Iron loss, copper loss and mechanical losses these are the losses in DC machine.

Answer : Iron losses in a D.C. machine are independent of variations in load.

Answer : Copper losses,iron losses,variable losses , constant losees

Description : In DC motor iron losses are occurred in?

Answer : In DC motor iron losses are occurred in the armature.

Description : Explain concept of back emf in DC machine. State how it governs armature current.

Answer : Back emf: When the armature of DC machine rotates under the influence of driving torque, the armature conductors move in the magnetic field and cut it and hence emf is induced in them. The ... flow of armature current i.e. it automatically changes the armature current to meet load requirements.

Description : The induced emf in the armature of a lap -wound four -pole dc machine having 100 armature conductors rotating at 600 rpm and with 1 Wb flux per pole is  (1) 1000V (2) 100V (3) 600V (4) 10,000V

Answer : The induced emf in the armature of a lap -wound four -pole dc machine having 100 armature conductors rotating at 600 rpm and with 1 Wb flux per pole is 1000V

Description : In a dc machine, interpoles are used to  (1) neutralize the effect of armature reaction in the interpolar region (2) generate more induced emf in the armature (3) avoid interference of the armature flux with the main -field flux (4) reduce the demagnetizing effect of armature reaction

Answer : In a dc machine, interpoles are used to neutralize the effect of armature reaction in the interpolar region

Description : The magnetic neutral plane shifts in a dc machine (A) in the direction of motion of motor (B) in the direction of motion of generator (C) due to increase in the field flux (D) cause reduction of flash over between commutator segments

Answer : The magnetic neutral plane shifts in a dc machine in the direction of motion of generator

Description : Why poles of dc machine are laminated?

Answer : Poles of DC machines are laminated to reduce the eddy current losses in the poles.

Description : A 200 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

Answer : A 200 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 0.87

Description : A 36 slot, 4 -pole, dc machine has a simplex lap winding with two conductors per slot. The back pitch and front pitch adopted could be respectively

Answer : A 36 slot, 4 -pole, dc machine has a simplex lap winding with two conductors per slot. The back pitch and front pitch adopted could be respectively 19,17

Description : Aluminium is not used as winding wire for the armature of a dc machine because  (1) aluminium has low resistivity (2) a large winding space is taken by aluminium conductors and creates ... ) the thermal conductivity of aluminium is low (4) aluminium has low conductivity as compared to copper

Answer : Aluminium is not used as winding wire for the armature of a dc machine because  (2) a large winding space is taken by aluminium conductors and creates jointing problems (3) the thermal conductivity of aluminium is low (4) aluminium has low conductivity as compared to copper

Description : Brushes of DC machine are made up of ?

Answer : In a D.C. generator, the iron losses mainly take place in armature rotor.

Answer : The total losses in a well designed D.C. generator of 10 kW will be nearly 500 W.

Description : What arc the errors occurring in measuring devices due to stray magnetic field and temp '? Explain how to compensate them.

Answer : 1. Error due to stray magnetic fields-   Main magnetic field gets disturbed by external magnetic fields known as stray magnetic fields.  2. Compensation technique   To avoid this error, ... resistance alloy having a negligible resistance temp coefficient in the ratio of 1:10 for pressure coil

Answer : To avoid the effect of stray magnetic field in A.C. bridges we can use magnetic screening.

Answer : To avoid formation of grooves in the commutator of a D.C. machine the brushes of opposite polarity should track each other.

Answer : In case of D.C. machine winding, number of commutator segments is equal to number of armature coils.

Answer : In a four-pole D.C. machine alternate poles are north and south.

Answer : Open circuited armature coil of a D.C. machine is identified by the scarring of the commutator segment to which open circuited coil is connected and indicated by a spark completely around the commutator.

Description : What are the types of power losses in inductor?

Answer : Copper losses, Eddy current loss, and hysteresis loss are the types of power losses in inductor.

Description : What are the losses in a transmission line?

Description : State the losses in secondary distribution system.

Answer : The losses in secondary distribution system: a) Technical losses: 1. Due to poor voltage 2. Due to unbalance load 3. Due poor quality of transformer & its components 4. Due to poor quality ... of induction type of energy meter. 3. Lack of administration. 4. Energy theft 5. Unmetered supply

Description : Explain following techniques related to energy conservation in transmission and distribution system. (i) By balancing phase currents (ii) Variable technical losses (I2R losses).

Answer : (i) Balancing Phase currents: Proper (healthy balanced) three phase loads always draw equal currents in all lines but single phase loads in the 3 phase 4 wire system or loads connected between two ... . 11) Minimize I2R losses. 12) Balance the load currents. 13) Regulate the system voltages.

Description : The friction losses in Real Transformers are :  (A) 0% (B) 5% (C) 25% (D) 50%

Answer : In case of D.C. machines, mechanical losses are primary function of speed.

Description : Reduction in core losses and increase in permeability are obtained with transformer employing?

Answer : Reduction in core losses and increase in permeability are obtained with transformer employing core built-up of laminations of cold rolled grain oriented steel.

Description : Losses which occur in rotating electric machines and do not occur in transformers are?

Answer : Friction losses windage losses

Description : In a given transformer for a given applied voltage, losses which remain constant irrespective of load changes are?

Answer : In a given transformer for a given applied voltage, losses which remain constant irrespective of load changes are hysteresis and eddy current losses.

Description : In a synchronous motor which losses varies with load?

Description : In synchronous motor, does the winding losses vary with load?

Answer : No, in the synchronous motor the winding losses do not vary with load.

Description : Explain the on state losses in power BJT.

Answer : On state losses in power BJT: Transistor have four types of losses one is turn on state losses second is turn off state losses third is turn on switching losses and fourth is turn ... ON = VCE sat x IC With increasing collector current collector to emitter saturation voltage increases.

Description : Explain the “mitigation of power theft” and “faulty meter replacement” for energy conservation techniques to reduce commercial losses.

Answer : Energy conservation techniques to reduce commercial losses:  1. Mitigation of Power Theft: Power theft being the most important issue which all of the service providing utilities ... agricultural sector. Prepaid meters may prove effective measures against unauthorized abstraction of energy.

Description : Which of the following can reduce the Eddy current losses? A) By making the core of a stack of plates electrically insulated from each other B) By making the core of solid block C) By increasing the applied voltage  D) None of these

Answer : Which of the following can reduce the Eddy current losses? A) By making the core of a stack of plates electrically insulated from each other B) By making the core of solid block C) By increasing the applied voltage  D) None of these

Description : Auto-transformer makes effective saving on copper and copper losses, when its transformation ratio is?

Answer : Auto-transformer makes effective saving on copper and copper losses, when its transformation ratio is approximately equal to 1.

Description : Explain the electrical losses of magnetic materials used for the cores of inductors and transformer.

Description : The total power of an AM transmitter having a carrier power of 50 W and the percentage of modulation at 80% is A) 50 W B) 66 W C) 68 W D) 70 W

Answer : The total power of an AM transmitter having a carrier power of 50 W and the percentage of modulation at 80% is 66 W

Description : In an election three candidates received 2272 , 15272, and 23256 votes respectively. What percentage of the total votes did the winning candidate got? a) 25 b) 41 c) 75 d) 57 e) 93

Answer : Answer: D Total number of votes polled = (2272+15272+23256)  =40800 Required percentage = 23256/40800 ×100  = 2325600/40800  = 57

Description : A mixture of 90 kg of rava and sugar contains 90% of sugar. The new mixture is formed by adding 30 kg of sugar. What is the percentage of sugar in the new mixture? a) 45 1/2 b) 60 1/5 c) 45 1/2 d) 92 1/2 e) 98 1/2

Answer : Answer: D 90 kg = 81 kg sugar 9 kg rava Adding = 30 kg sugar Total = 111kg sugar 9kg rava (new)  111/120 × 100 = 92.5 =92 1 /2

Description : A mixture of 60 kg of rice and dhal contains 60% of dhal. The new mixture is formed by adding 15 kg of dhal. What is the percentage of rice in the new mixture? a) 25 b) 64 c) 45 d) 32 e) 98

Answer : Answer: D 60 kg = 36 kg dhal 24 kg rice -------- 15 kg dhal Total = 15 kg dhal 24 kg rice (new) 24/75×100=32%

Description : Gugan obtained a total of 1313 marks out of 1400 in an examination. What is his approximate percentage in the examination? a) 25 b) 60 c) 45 d) None e) 98

Answer : Answer: D Required percentage = 1313/ 1400 × 100 =131300/1400 =93.7

Description : What is fuzzy logic in a washing machine?

Description : What is the percentage error when 3 resistors with ratings 100 Ω at 5%, 150 Ω at 6% and 250 Ω at 8% respectively, are connected in series? A) 6.33% B) 6.80% C) 19% D) None of these

Answer : What is the percentage error when 3 resistors with ratings 100 Ω at 5%, 150 Ω at 6% and 250 Ω at 8% respectively, are connected in series? A) 6.33% B) 6.80% C) 19% D) None of these

Description : Capsule coffee machine. What are the advantages?

Answer : Having a coffee maker in the house helps us to easily cheer up and brighten up the beginning of the upcoming day in the morning, especially if the weather is unpleasant. A cup of hot ... favor is the fact that the profile of this manufacturer is precisely capsule coffee makers and coffee capsules.

Description : A shopkeeper bought 1800 blackberry and 1200 blueberry. He found 45% of blackberry and 24% of blueberry were rotten. Find the percentage of fruits in good condition. a) 63.4 b) 32.9 c) 48.5 d) 56.3

Answer :  Answer: A Total number of fruits shopkeeper bought = 1800 + 1200 = 1000 Number of rotten blackberry = 45% of 1800  = 45/100 1800  = 81000/100  = 810 Number of rotten blueberry = 24% of 1200 ... Therefore Percentage of fruits in good condition = (1902/3000 100)  = (190200/3000)  = 63.4%

Description : A wooden manufacturing company declares that a wooden is now available for \$11200 as against \$25200 one year before. Find the percentage reduction in the price of wooden offered by the company. a) 25 1/3 b) 691/3 c) 66 1/3 d) 33 1/3 e) 55 5/9

Answer : Answer: E Price of the wooden a year before = \$25200 Price of the wooden after a year = \$11200 Decrease in price = \$(25200 - 11200) = \$14000 Therefore, decrease % = 14000/25200 × 100 %  = 55 5/9 %

Description : The population in a small town increases from 50000 to 63750 in one year. Find the percentage increase in population. a) 25 b) 62 c) 6.25 d) 27.5 e) 59.8

Answer : Answer: D Population in a small town last year = 50000 Population in a small town after one year = 63750 Increase in population = 63750 - 50000 = 13750 Therefore,  percentage increase in population = ... /50000 100  = 1375000/50000  = 27.5% Thus, the increase in population is 27.5%

← Prev Question Next Question →