Question { BDL, 3500 }

what do you mean by power factor ? in new view

Answer

It is the value of cosine angle between voltage and current due to inductive load (legging power factor) i.e the the value of cos 36 degree is 0.8 ( 0.8 cos fi ),cos 26 degree is 0.9 approx.(0.9 cos fi) and so on. It is also calculate by
True power/Apparent power,Resistance/Impedance or conductance/Admittance.here current leg behind voltage And in capacitive load that angle is minimized i.e improving the power factor.

Question { BSL, 4756 }

Actually why the transformers are rated in KVA.?

Answer

Transformer gives apparent power,it is the supply power to load and the power actually used by the load is the
real power/true power/active power. Apparent power of 100 KVA has capacity to give 100 KW resistive load and for inductive load of o.8 pf it give 80 KW,and at 0.7 pf the same has capacity to give 70 KW . but the manufacturing company does not know which type of load to be connected across the transformer,therefore in simple we can say " DUE TO KNOWN POWER FACTOR THE TRANSFORMER ARE RATED IN KVA" and it is meaning less discuss about Iron loss & copper loss because the Inverted and UPS are rated in KVA but there is no iron loss & copper loss And in other hand in DC Generator there is Iron loss &copper loss but it is rated in KW not in KVA, because there is no angle between current and voltage(pf) in DC supply.

PRADYUMNA CHANDRA PANDA
Mob No- 9439721208

Question { BSL, 4756 }

Actually why the transformers are rated in KVA.?

Answer

Edit- due to unknown power factor transformers are rated in kva

Question { BSL, 4756 }

Actually why the transformers are rated in KVA.?

Answer

Transformer gives apparent power,it is the supply power to load and the power actually used by the load is the real power/true power/active power. Apparent power of 100 KVA has capacity to give 100 KW resistive load and for inductive load of o.8 pf it give 80 KW,and at 0.7 pf the same has capacity to give 70 KW . but the manufacturing company does not know which type of load to be connected across the transformer,therefore in simple we can say " DUE TO UNKNOWN POWER FACTOR THE TRANSFORMER ARE RATED IN KVA" and it is meaning less discuss about Iron loss & copper loss because the Inverted and UPS are rated in KVA but there is no iron loss & copper loss And in other hand in DC Generator there is Iron loss &copper loss but it is rated in KW not in KVA, because there is no angle between current and voltage(pf) in DC supply N:B . Transformer capacity -transformer iron loss & copper loss = transformer efficiency in KVA
PRADYUMNA PANDA BRG
Mob No. 9437917849

Question { 2384 }

back emf of motor at the time of starting ?

Answer

Zero

Question { 2067 }

In which iron losses is higher induction motor /transformer

Answer

Induction motor.

Question { XYZ, 3799 }

Actually why the transformers ate rated in KVA.?

Answer

Transformer gives apparent power,it is the supply power to load and the power actually used by the load is the real power/true power/active power. Apparent power of 100 KVA has capacity to give 100 KW resistive load and for inductive load of o.8 pf it give 80 KW,and at 0.7 pf the same has capacity to give 70 KW . but the manufacturing company does not know which type of load to be connected across the transformer,therefore in simple we can say " DUE TO UNKNOWN POWER FACTOR THE TRANSFORMER ARE RATED IN KVA" and it is meaning less discuss about Iron loss & copper loss because the Inverted and UPS are rated in KVA but there is no iron loss & copper loss And in other hand in DC Generator there is Iron loss &copper loss but it is rated in KW not in KVA, because there is no angle between current and voltage(pf) in DC supply N.B transformer capacity -transformer iron loss & copper loss = transformer efficiency in KVA
PRADYUMNA CHANDRS PANDA
Mob No. 9437917849

Question { 31085 }

What are formula we can apply to determine the HT/LT current of 11KV/433V transformer except KVA/LV×√3. ?

Answer

Fourmela No 1 power in one phase/Voltage in one phase(I.e phase voltage)
example there is a 300 KVA 11KV/433 V transfomula. Power in one phase 300KVA/3 =100KVA both in HT & LT side, Phase voltase in HT side 11/√3 = 6351Volt .Phase voltage in LT side 433/√3 = 250 Volt.
According to formula HT side current 100,000/6351 = 15. 7 amps. and LT side current 100,000/250 = 400 amps
Formula No 2 for HT side transformer in kVA/20 and for LT side KVA×4/3. field lineman are using this formula.
According to this formula 300 KVA transformer's HT side current 300/20 =15 amps apporximately +5℅ may added for big size to accurate value. And LT side current 300×4/3 = 400 amps apprx.
In general formula for 300 KVA 11KV/433V transformer HT side current KVA/LV×√3 i.e 300000/11000×1.732 = 15.7 amps .LT side current 300000/433×√3 = 400 amps

PRADYUMNA CHANDRA PANDA
mob No 943
7917849

Question { 1906 }

There is a carem board where 4 No of player can play at a time but 5 No of player have to play with every for gain personal score(not double).In this case how many total minimum No of game /meach will be play..

Answer

N=N1×N2 whole by N3
N=total minimum No of game /meach N1=No of player/team to play N2=to how many player/team to pkay.
N3=No of team/Player at once
i.e 5×4 whole by 4/=5 Ans

Pradyumna Panda
Berhampur
Mob No 9861357949

Question { 2281 }

There is a carem board where 4 No of player can play at a time but 5 No of player have to play each to every. In this case how many minimum game will be play.?

Answer

N= N1× N2 whole by N3
Where N=total minimum No of game. N1= No of player to play. N2=to how many player to play. N3= No of player in each game
5×4/4=4 And
P C Panda Berhampur
Mob No 9861357949

Question { 2281 }

There is a carem board where 4 No of player can play at a time but 5 No of player have to play each to every. In this case how many minimum game will be play.?

Answer

N= N1× N2 whole by N3
Where N=total minimum No of game. N1= No of player to play. N2=to how many player to play. N3= No of player in each game
5×4/4=5 Ans
P C Panda Berhampur
Mob No 9437917849