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95MET-4
Sr. No. EMR
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Electrical, Electronics and Control Engineering
at Operational Level
MARINE ELECTRO TECHNOLOGY
CLASS IV
(Time allowed - 2 hours)
Morning Paper
INDIA (2002)
N.B. -
Total Marks 100
(1) All Questions are compulsory
(2) All Questions carry Equal Marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. A battery having an E.M.F. of 110 volts and an internal resistance of 0.2 ohm is
connected in parallel with another battery with an E.M.F. of 100 volts and resistance 0.25
ohm. The two batteries in parallel are placed in series with a regulating resistance of 5
ohms and connected across 200volt mains. Calculate the magnitude and direction of the
current in each battery and the total current taken from the supply mains.
2. A choking coil of negligible resistance takes a current of 10 amperes when connected to a
220 volt, 50 Hz supply. A non-inductive resistor under the same conditions takes 12
amperes. If the two are connected is series and placed across the same supply, find the
current taken. If now the frequency is reduced to 40 Hz, the voltage being maintained
constant, find the current taken.
3. A current is represented by the equation i = 40 sin 50t where t is in seconds. A second
current of the same frequency but twice the amplitude lags behind the first by 60 O.
Calculate (a)The value of the first current when the second is at a positive peak and
(b)The values of both current 0.02 sec. later.
4. Moisture damage, as a result of condensation occurring inside of the cargo winch master
switches, can be reduced by _________________.
(a)installing a light bulb in the pedestal stand
(b)coating the switch box internals with epoxy sealer
(c)venting the switch box regularly
(d)using strip heaters inside the switch box
Briefly Justify your Answer.
5. 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 over speed 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.
6. 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.
7. With reference to electrical switch gear explain the purpose of each of the following:
(a)Preferential tripping,
(b)Dash pots.
(c)Reverse power tripping;
(d)Under voltage tripping.
8. With reference to electrical equipment in areas aboard ship having potentially flammable
atmospheres:
(a) Explain the hazards involved.
(b) State the design features that render the equipment safe.
(c) Explain the precautions necessary when maintenance work is being carried out.
9. (a) Sketch a simplified circuit of a reverse power relay.
(b)Explain briefly how the reverse power relay operates.
(c)Explain why there is a time delay incorporated before the reverse power relay operates.
10. When troubleshooting most electronic circuits, "loading effect" can be minimized by
using a voltmeter with a/an ___________________.
(a) input impedance much greater than the impedance across which the voltage is
being measured
(b) input impedance much less than the impedance across which the voltage is being
measured
(c) sensitivity of less than 1000 ohms/volt
(d) sensitivity of more than 1000 volts/ohm
--------------------X----------------------
95MET-4
Sr. No. EMR
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Electrical, Electronics and Control Engineering
at Operational Level
MARINE ELECTRO TECHNOLOGY
CLASS IV
(Time allowed - 2 hours)
Morning Paper
INDIA (2002)
N.B. -
ANSWERS
B
0.2  110 V
Ans 1.
I2
0.25 100 V
5
I1I2
D
200 V
+
A
C
Current ABC
I1  5 - 0.2 I2 = 90
(i)
Circuit ADC
5I1 + (I1 + I2)  0.25 = 100 (ii)
(i)  0.25 I1  1.25 - 0.05 I2 = 22.5
(ii)  0.2 I1  1.05 + 0.05 I2 = 20.0
I1  2.3
I1 =
= 42.5
42.5
2.3
Total Marks 100
(1) All Questions are compulsory
(2) All Questions carry Equal Marks.
(3) Neatness in handwriting and clarity in expression carries weightage
= 18.478 amps
From (i)
5  18.478 - 90 = I2 = 11.95 amps
0.2
I1 + I2 = 11.95 + 18.478 = 30.43 amps
110 V battery discharges at 11.95 amps
100 V battery charged at 30.43 amps
Current taken from mains = 18.478 amps
ANS : 2
10 A
12 A
220 V
220 V
220 V
XL = (Z Coil) = 220/10 = 22  (as the resistance is negligible)
R of resistor = 220/12= 18.33 
Trial impedance = 222 + 18.332 = 820 = 28.64 
I=
220 = 7.685 amp
28.64
2fl
=22 L1
=
22
=
22
2f
100 
XL (=Z of the choke) at 40 Hz
= 80   22= 7.6 
100 
Total impedance = 17.62 + 18.332
= 645.7 = 25.41 
Current =
220
= 8.658 amp
25.41
ANS : 3
iZ
+
L1 = 40 sin 50t
L2 = 80 sin (50 t - /2)
80 = 80 sin (50t - /3)
sin (50t - /3) = 1 = sin /2
50t =  +  = 5 
3
2
6
T = /60
L1 = 40 sin 50   = 40 sin 5 = 40 sin 5 /6= 40 sin 150
= 40  0.5 = 20 amp.
Vectors would have travelled = 0.2  50 = 1 rad = 57.3O
L1
= 40 sin (150 + 57.3) = 40 sin (180 + 27.3)
= - 40 sin 27.3 = -40  0.4591 = - 18.37 amp
L2
= 80 sin (90 + 57.3)
= 80 sin (180 -32.7) = 80 sin 32.7
= 80  0.5402 = 43.216 amp.
ANS: 4
Correct Answer : d
ANS: 5
Correct Answer : a
ANS: 6
Correct Answer :c
ANS: 10
Correct Answer: a