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FV(M)ET
Sr. No. 1
Examination of Engineers of Fishing Vessel (Motor)
Electro-Technology
Time Allowed –2 ½ Hours
INDIA (2002)
Total Marks 100
N.B. -
(1) Attempt FIVE Questions only.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. Which of the following conditions will occur if the solenoid coil burns out on a cargo winch with an
electrical brake?
A. The brake will be set by spring force.
B. The motor will overspeed and burn up.
C. The load suspended from the cargo boom will fall.
D. Nothing will happen; the winch will continue to operate as usual.
Briefly justify your Answer.
2. Which of the listed battery charging circuits is used to maintain a wet-cell, lead-acid, storage battery
in a fully charged state during long periods of disuse?
A. Normal charging circuit
B. Quick charging circuit
C. Trickle charging circuit
D. High ampere charging circuit
Briefly justify your Answer.
3. A ground can be defined as an electrical connection between the wiring of a motor and its
___________.
A. shunt field
B. circuit breaker
C. metal framework
D. interpole
Briefly justify your Answer.
4. The output voltage of a 440 volt, 60 hertz, AC generator is controlled by the _____________.
A. prime mover speed
B. exciter output voltage
C. load on the alternator
D. number of poles
Briefly justify your Answer.
5. Any electric motor can be constructed to be _____________.
A. short proof
B. ground proof
C. explosion proof
D. overload proof
Briefly justify your Answer.
6. Which of the following statements represents the main difference between a relay and a contactor?
A. Contactors control current and relays control voltage.
B. A relay is series connected and a contactor is parallel connected.
C. Contactors can handle heavier loads than relays.
D. Contactors are made from silver and relays are made from copper.
Briefly justify your Answer.
7. Which of the following statements is true concerning a polyphase synchronous propulsion motor?
A. The motor is started as an induction motor.
B. Resistance is gradually added to the rotor circuit.
C. The starting current is held below the rated current.
D. The field winding is energized for starting purposes only.
Briefly justify your Answer.
9. A 250V 50 Hz alternator supplies current to a circuit having a resistance of 30 ohms and an
impedance of 50 ohms. Calculate the inductance of the circuit. Owing to a mechanical fault the speed
of the alternator falls slightly, the effect on the circuit being to reduce the impedance to 46.86 ohms,
but the power absorbed remain unaltered.
Calculate for the latter condition –
A. the frequency,
B. the power factor.
10. Draw a simplified circuit diagram of a high-tension magneto ignition system such as might be used
for a lifeboat engine; explain the working principle.
----------------------X----------------------
FV(M)ET
Sr. No. 1
Examination of Engineers of Fishing Vessel (Motor)
Electro-Technology
Time Allowed –2 ½ Hours
INDIA (2002)
Total Marks 100
N.B. -
(1) Attempt FIVE Questions only.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answers
Answer for Question No. 1
Correct Answer : A
Answer for Question No. 2
Correct Answer : C
Answer for Question No. 3
Correct Answer : C
Answer for Question No. 4
Correct Answer : B
Answer for Question No. 5
Correct Answer : C
Answer for Question No. 6
Correct Answer : C
Answer for Question No. 7
Correct Answer : A
Answer for Question No. 8
Ans 1. Definition :(a) International ampere is that steady current which when passed through a solution of Silver nitrate
under definite conditions deposits silver at the rate of 0.001118 gm/sec.
(b) International ohm is the resistance of fixed to an unvarying current by a column of mercury at
zero degree centigrade, having a length of 106.300 cm and a man of 14.4521 and a x sectional
area of about 1mm2.
(c) International Volt is defined as 1/0.0183 of the e.m.f. of a standard Winton cell.
500 ampere
1.55 
R1
120 V
V = I (R + R1)
120 = 50 (1.5 R1)
= 75 + 50 R1
 50 R1 = 120 - 75 =45
 R1 = 45/50 = 0.9 
Volt drop in the resistance = IR = 50  1.5 = 75 voltage =
 volt drop across the apparatus = IR1 = 120-75 = 45 coils
When voltage is increased to 210 volts. =
V
R + R1
Current flowing I =
210 =
210 = 87.5 amp
1.5 + 0.9
2.4
 Heat generated in the resistance before = 50  50  1.5
Heat generated in the resistance after
87.5  87.5  1.5
(i)
= 3.75 kW
(ii)
= 11.48 kW
50  50
= 0.3265 Ans.