APSPDCL Electrical Technical Question Papers Answers Free download pdf 2016.
Free download pdf Andhra Pradesh Southern power Distribution Company Limited for Engineers APSPDCL Electrical Technical Last 10 Years Papers with solution, Last Year APSPDCL APSPDCL Electrical Technical Question Papers Answers 2011 pdf Model Papers APSPDCL Electrical Technical Solved
For V/F control, when frequency is increased in transformer
a) Core loss component current increases, Magnetizing component current decreases
b) Core loss component current increases, Magnetizing component current increases
c) Core loss component current decreases, Magnetizing component current decreases
d) Core loss component current decreases, Magnetizing component current increases
Ans: Core loss component current decreases, Magnetizing component current decreases
In ceiling fan the angle between auxiliary winding a main winding is
a) 0 deg
b) 90 deg
c) 180 deg
d) 360 deg
Ans: 90 deg
In a shaded pole motor, shaded rings are used to
a) Field flux production
b)
c)
d)
Ans: Field flux production
Practical method of improving string efficiency
a) Increasing crass arms length
b) Using different size of insulators
c) Using different insulator materials
d) Using of guard rings
Ans: Increasing crass arms length*
In which type of fault all 3-phase components are equal
a) L-G
b) L-L
c) L-L-G
d) 3-Phase fault
Ans: 3-Phase fault
11/220 KV 100 MVA transformers, the primary base voltage rating is 10 KV then secondary base KV is
a) 10 KV
b) 220 KV
c) 220/sqrt3
d)
Ans: 220/sqrt3*
APSPDCL Electrical Technical Question Papers Answers Free download pdf 2016
Water hamming effect is occurs in
a) Surge tank
b) Penstock
c) Turbine
d) Reservoir
Ans: Penstock
Transient stability can be improved by
a) By putting series capacitor
b) By using dynamic resister
c) Auto re-closers
d) All of the above
Ans: All of the above
If sending end voltage is Vs at no-load in a transmission line then receiving end voltage is if ABCD parameters of line is given
a) Vs
b) Vs/A
c) 0
d) Infinity
Ans: Vs/A
Harmonics are eliminated by using
a) Skewing of rotor
b) Distribution winding
c) Short pitch winding
d) All of the above
Ans: All of the above
For parallel operation of transformer at no load, then load shared are equal when
a) Impedance is proportional with respect to own KVA rating
b)
c)
d)
Ans: Impedance is proportional with respect to own KVA rating
In Induction motor Slip frequency of rotor current, when rotor speed is Nr. Then rotor producing slip field rotates with respect to stator is
a) Slip frequency
b) Nr
c) Ns
d) None
Ans: Ns
Good regulation means
a) Less fluctuations from no-load to full-load
b)
c)
d)
Ans: Less fluctuation from no-load to full-load
At dead short circuit at terminals of Alternator then the current I is
a) ZPF Lag
b) ZPF lead
c) Unity power factor
d) 0.8 Power factor lag
Ans: ZPF Lag
Full scale Ammeter reading is 10 mA, Voltage across Ammeter is 100 mV. Then for 100 A measurements the power loss in the circuit is
a) 1 W
b) 10 W
c) 100 W
d) 1000 W
Ans: 10 W*
When Alternator excitation increases and machine is operating at lagging power factor then
a) I increase
b) I decreases
c) No effect on current
d) None
Ans: I increase
Flue gases coming out from furnace is first going through
a) Turbine
b) Economizer
c) Air pre-heater
d) Chimney
Ans: Economizer
Two alternators rated are 200 KW at 4% regulation, 400 KW 5% regulation operating in parallel at 50HZ , when supply 600 KW the new frequency is
a) 50
b) 49
c) 51.3
d) 47.7
Ans: 47.7
Va and Vb are negative sequence component voltages the difference angle between Va and Vb with respect to Va is
a) 240
b) 120
c) 180
d) 360
Ans: 120
4-Quadrent operation of chopper I is positive but V is may be positive are negative operates in which quadrant
a) 1 and 2
b) 2 and 3
c) 1 and 3
d) 1 and 4
Ans: 1 and 4
Heat convection, radiation, and conduction all are takes place in
a) Ice
b) Boiler
c) Refrigerator
d) Flue gases in pipe
Ans:
When moving iron meter is used to measure d.c current what is the disadvantage
a) It calibration in rms value
b)
c)
d)
Ans:
maximum power transfer to load is
a) 20 W
b) 50 W
c) 100W
d) 0W
Ans: 0W
Find Zbus Parameters
a) Z11=Z12=Z21= j0.1 ohm, Z22=j0.3 ohm
b)
c)
d)
Ans: Z11=Z12=Z21= j0.1 ohm, Z22=j0.3 ohm
Find Z11 and Z12
a) Z11=4 ohm, Z12= 2 ohm
b) Z11=4 ohm, Z12= 5 ohm
c) Z11=5 ohm, Z12= 2 ohm
d) Z11=4 ohm, Z12= 3 ohm
Ans:
Current in 5 ohm resister is
a) 10 A
b) -10 A
c) 5 A
d) -5 A
Ans: -10 A
I in the circuit at t=0+
a) 2 A
b) 5 A
c) 7 A
d) 10 A
Ans:
voltage across inductor at t=0+
a) 0 V
b) 2.5 V
c) 5 V
d) 10 V
Ans: 5 V
This wave is applied to the PMMC meter, meter reads
a) 0 V
b) 3 V
c) 2 V
d) 1 V
Ans: 1 V
Find Vab
a) Zero
b) 50 at an angle of 90 deg
c) 50 at an angle of 90 deg
d) None
Ans:
71. Find E in the circuit
a) 5 V
b) 10 V
c) 20 V
d) None
Ans: None
Vab is reference then angle between Vab and Ib is
a) -15 deg
b) 240 deg
c) 180 deg
d) -45 deg
Ans: -15 deg*
When maximum power transfer to load is
Click Here to Show Diagram
a) 20 W
b) 50 W
c) 100W
d) 0W
Ans: 0W
Find Zbus Parameters
Click Here to Show Diagram
a) Z11=Z12=Z21= j0.1 ohm, Z22=j0.3 ohm
b)
c)
d)
Ans: Z11=Z12=Z21= j0.1 ohm, Z22=j0.3 ohm
Find Z11 and Z12
Click Here to Show Diagram
a) Z11=4 ohm, Z12= 2 ohm
b) Z11=4 ohm, Z12= 5 ohm
c) Z11=5 ohm, Z12= 2 ohm
d) Z11=4 ohm, Z12= 3 ohm
Ans:
Current in 5 ohm resister is
Click Here to Show Diagram
a) 10 A
b) -10 A
c) 5 A
d) -5 A
Ans: -10 A
Find I in the circuit at t=0+
Click Here to Show Diagram
a) 2 A
b) 5 A
c) 7 A
d) 10 A
Ans: 2 A
Find voltage across inductor at t=0+
Click Here to Show Diagram
a) 0 V
b) 2.5 V
c) 5 V
d) 10 V
Ans: 5 V
This wave is applied to the PMMC meter, meter reads
Click Here to Show Diagram
a) 0 V
b) 3 V
c) 2 V
d) 1 V
Ans: 1 V
Find Vab
Click Here to Show Diagram
a) Zero
b) 50 at an angle of 90 deg
c) 50 at an angle of 90 deg
d) None
Ans:
Find E in the circuit
Click Here to Show Diagram
a) 5 V
b) 10 V
c) 20 V
d) None
Ans: None
Vab is reference then angle between Vab and Ib is
Click Here to Show Diagram
a) -15 deg
b) 240 deg
c) 180 deg
d) -45 deg
Ans: -15 deg*
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