Part-A(Objective Test Items )2005
Q:1 .You have taken four readings from a earth resistance testers,placing the spikes at different position
around the electrode under test . Distance between electrodes is the same for all the test. The
measured values are 2.8 ohm 2.1 ohm 2.5 ohm and 1.6 ohm.
measured values are 2.8 ohm 2.1 ohm 2.5 ohm and 1.6 ohm.
Determine the value of earth resistance.
Ans:1. Total earth resistance
=2.8+2.1+2.5+1.6 ohm
=9 ohm
Average value=9 ohm/4=2.25 ohm
Q:2. The statically induced emf has a polarity which opposes the voltage that created it.
What is the other name by which this induced emf is know as?
Ans:2. Counter emf (Back emf).
Q:3. To identical 3 phase induction motors in good condition of same HP,voltage, frequency and speed
are connected to the rated supply voltage. On no load you notice one motor is drawing 1.6 A and
the other 4.8 A.
are connected to the rated supply voltage. On no load you notice one motor is drawing 1.6 A and
the other 4.8 A.
What is the reason for the above current different?
Ans:3. Winding of motors are connected in different ways. Motor drawing 1.6 A is connected in star.Motor
drawing 4.8 A is connected in delta.
Q:4. What is the method of heat transfer accomplished in a
a) Hot water boiler.
b) Electric iron.
Ans:4. a) Convection.
b) Conduction.
Q:5. After rectifying the short circuit in a electric toaster you have to replace the blown out fuse. The old
fuse was rated for a current of 1 A.
a)What is the function of the fuse in normal operation?
b)What will happen if you use a fuse with 0.3 A?
Ans:5. a) A fuse most not open the circuit under normal operating conditions.
b) A 0.3 A fuse will open the circuit immediately after you switch ON the toaster as the operating
current in the circuit is higher than 0.3 A.
Q:6. State the device used for coarse excess current protection.
Ans:6. Re-wirable type fuse.
fuse was rated for a current of 1 A.
a)What is the function of the fuse in normal operation?
b)What will happen if you use a fuse with 0.3 A?
Ans:5. a) A fuse most not open the circuit under normal operating conditions.
b) A 0.3 A fuse will open the circuit immediately after you switch ON the toaster as the operating
current in the circuit is higher than 0.3 A.
Q:6. State the device used for coarse excess current protection.
Ans:6. Re-wirable type fuse.
Q:7. Apart from high mechanical strength, the
metal conduit system of Wiring has one definite
advantage over the PVC conduit system.
What is the advantage with respect to electrical protection?
Ans:7. Metal conduit also functions as earth
continuity conductor.
Q:8. Sheet metal edges of thin gauges are made
by hem.
State any two purposes of the hem in sheet metal work?
Ans:8. a)Hem stiffens the sheet of metal and
improves mechanical strength.
b)Hem prevents sharp edges.
c)prevents damages to the edges.
Q:9. From making light punch marking a particular type of punch is used.
What is the name of punch and its angle of point?
Ans:9. Prick punch. Angle of point 30 degree
or 60 degree
Q:10.
The amount of current flow in pure resistive circuit on AC is influenced by
five factors. Among
the five factors three factors are DC Resistance,
Eddy current and Hysterisis.
Name
two other factors.
Ans:10. a)Skin effect.
b)Dielectric stress
Q:11.
There are different methods of connections in domestic installation.Which
method is shown
in the figure?
Ans:11.
Loop-in method.
Q:12.
You are connecting a 3 phase induction motor. If you connect starting ends of
all the three
winding together and give three phase supply to other
three ends, which type of connection
you have made.
Ans:12.
Star connection.
Part B
Q:1. a) Name the active chemical substance used in lead acid battery.
b) State common defects of a lead acid battery.
Ans:1. a)
i) Lead peroxide (Pbo2)
ii) Spongy lead (pb)
iii) Diluted sulphric acid (H2SO4)
b) Common defects on the lead acid battery.
i) Sulphation on active plates
ii) Buckling of plates
iii) Sedimentation/Shedding at the bottom of container
iv) Internal short circuit.
Q:2. a) A.D.C shunt generator when started up does not build up its voltage. State the possible
causes for its failure.
b) Explain why the P.D at the terminal of shunt generator falls as the load increased. What
are the factors responsible for this fall in P.D?
Ans:2. a) the shunt generator fails to build up voltage due to
i) Absence of residual magnetism
ii) The field circuit may be reversed.
iii) The shunt field resistance above the critical value (too high) and do not produce appreciable
flux to initiate the build up process.
iv) Imperfect brush contact on commutator.
v) The prime mover rotates opposite to the marked direction and speed may be too low.
b) The voltage developed by a shunt generator is dropping on load due to
i) Increased IR drop
ii) Increased Armature reaction
iii) Decrease in PD across field terminal due to the above two reasons.
Q:3. a) Briefly explain the working principle of an induction motor.
b) What is meant by slip in an induction motor?
c) A 5 KW, 50 Hz, 6 pole slip ring induction motor runs at 960 r.pm calculate.
i) sync. speed ii) % slip
Ans:3. a) Induction motor operates on the principle of faraday's laws of electromagnetic induction. The
stator winding is connected to the AC supply which produces a rotating magnetic field . Due
to this field an emf is induced in the rotor conductors by electro magnetic induction and hence
the name induction motor. The flux produced by the rotor currents makes the rotor rotate. The
direction of rotation is same as the of rotating magnetic field of the stator. The speed of the rotor
is slightly less than the speed of the rotating magnetic field.
b) The difference between synchronous speed (Ns) and actual rotor speed (N) is known as
slip. It is expressed as percentage of synchronous speed Ns,
% slip=(Ns-N)/Ns=100
c) i) f = PN/120 = N = f.120/P
= 50 x 120/6 =1000
synchronous speed= 1000 r.p.m
ii) % slip = Ns-N/N x 100
= 1000-960/1000 x 100
= 4%
Q:4. a) What is power factor?
b) Why current of 10 amps flows in a circuit whit 30 degree lag, when the applied voltage is 100
volts. Find the resistance and reactance of the circuit.
Ans:4. a) Cosine angle subtended between voltage and current
P.F=Resistance/impedence (or) P.F=True power/Apparent power
b) I=10 Amps
Thita=30 degree lag
v=100 volts
impendence=(z)=V/I=100/10=10 ohm
P.F=cos thita= T/Z; Cos30 degree R/10; 0.866=R/10 R=8.66 ohm
Xl = Root zsq-Rsq= Root 10sq - 8.66sq =5 ohm
Inductive Reactance = (Xl)=5 ohm
Q: 5. a) Why are the transformers used in distribution system?
b) Why are thir neutral earthed on the secondary side?
Ans:5 a) To step doen the 11 kv into 440 volt which is suitable to conect 3 phase power loads and single
phase lighting loads distribution transformer is used.
b) i) Transformer neutral point is earthed to restrict the voltage of live conductors whit repect to
the general mass of earth to a value consistent whit the insulation level.
ii) To limit the supply voltage fluctuation.
Q:6. a) What are the different method of transistor amplifier configuration?
b) Draw a typical circuit whit voltage divider biased PNP transistor amplifier which has phase
invertion.
Ans: 6.a)
i) Common emitter amplifier
ii) Common base amplifier
iii) Common collecttor amplifier
b)
Common emitter amplifier.
Q:3. a) Briefly explain the working principle of an induction motor.
b) What is meant by slip in an induction motor?
c) A 5 KW, 50 Hz, 6 pole slip ring induction motor runs at 960 r.pm calculate.
i) sync. speed ii) % slip
Ans:3. a) Induction motor operates on the principle of faraday's laws of electromagnetic induction. The
stator winding is connected to the AC supply which produces a rotating magnetic field . Due
to this field an emf is induced in the rotor conductors by electro magnetic induction and hence
the name induction motor. The flux produced by the rotor currents makes the rotor rotate. The
direction of rotation is same as the of rotating magnetic field of the stator. The speed of the rotor
is slightly less than the speed of the rotating magnetic field.
b) The difference between synchronous speed (Ns) and actual rotor speed (N) is known as
slip. It is expressed as percentage of synchronous speed Ns,
% slip=(Ns-N)/Ns=100
c) i) f = PN/120 = N = f.120/P
= 50 x 120/6 =1000
synchronous speed= 1000 r.p.m
ii) % slip = Ns-N/N x 100
= 1000-960/1000 x 100
= 4%
Q:4. a) What is power factor?
b) Why current of 10 amps flows in a circuit whit 30 degree lag, when the applied voltage is 100
volts. Find the resistance and reactance of the circuit.
Ans:4. a) Cosine angle subtended between voltage and current
P.F=Resistance/impedence (or) P.F=True power/Apparent power
b) I=10 Amps
Thita=30 degree lag
v=100 volts
impendence=(z)=V/I=100/10=10 ohm
P.F=cos thita= T/Z; Cos30 degree R/10; 0.866=R/10 R=8.66 ohm
Xl = Root zsq-Rsq= Root 10sq - 8.66sq =5 ohm
Inductive Reactance = (Xl)=5 ohm
Q: 5. a) Why are the transformers used in distribution system?
b) Why are thir neutral earthed on the secondary side?
Ans:5 a) To step doen the 11 kv into 440 volt which is suitable to conect 3 phase power loads and single
phase lighting loads distribution transformer is used.
b) i) Transformer neutral point is earthed to restrict the voltage of live conductors whit repect to
the general mass of earth to a value consistent whit the insulation level.
ii) To limit the supply voltage fluctuation.
Q:6. a) What are the different method of transistor amplifier configuration?
b) Draw a typical circuit whit voltage divider biased PNP transistor amplifier which has phase
invertion.
Ans: 6.a)
i) Common emitter amplifier
ii) Common base amplifier
iii) Common collecttor amplifier
b)
Common emitter amplifier.
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