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Company >> Cable >> Cable Questions
 
 Grasim technical test questions  Grasim Technical Test Questions (1)  Siemens interview questions  Siemens Interview Questions (13)
 
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Question   how to calculate the current carrying capacity of cable Rank Answer Posted By  
 Interview Question Submitted By :: Krishna.g1986
I also faced this Question!!   © ALL Interview .com
Answer
fistly calculate the 
total load current,then 
multiply it by 1.2 for 
the tollerance in 
current.
 
0 Amol Folane
 
 
Answer
by using wire gauge calculate the size of the conductor and 
number of the conductors and finally see the data sheets of 
conductor size sheets
 
0 Sandeshg
 
 
Question   WHAT IS THE FUNCTION OF SEMICON IN 33KV ht CABLE ? & Is it neccessory to remove the semicon screen from the xlpe insulation at the time of termination & what happen we do not do it? Rank Answer Posted By  
 Interview Question Submitted By :: Amit Shrivas
I also faced this Question!!   © ALL Interview .com
Answer
It is used to influence the electric field within the
insulation layer.The semicon layer has to removed at the
termination at least 6-9 inches below  the lugs ,as
otherwise there will be short circuit along this layer.As
this layer is not doped, you have to treat it as a conductor.
 
0 K.prakashchandra
 
 
 
Answer
The purpose of Semi conducting tape is only to transmit the
heat from insulation to cu. tape over it. this heat is
produced in Core due to flow of current. Further, to
TERMINATE THE CABLES, ALL CONDUCTING AND SEMICONDUCTING TAPE
IS REMOVED FROM INSULATION AS current in cable can also come
into insulation and it may be damage the cable, dangerous to
persons working.
 
0 Anil
 
 
Answer
Usually HV cable has a semi conducting, one purpose it to protect your connected load in severe damage and the system as well. In the case of unexpected faults along with this cable,any small leakage current detected will trip its protection.
 
0 Marlo
 
 
Question   Concrete Mix designs for M15,M20,M25,M30 and M35 Rank Answer Posted By  
 Interview Question Submitted By :: Madhava Vk
I also faced this Question!!   © ALL Interview .com
Answer
M-15 = 3.5 N/MM2, M-25 and M-20 = 4 N/MM2, M-30 and M-35 
=5NMM2 THANKS.
 
3 Navin Morwal
 
 
Answer
m-15 1:2:4,m-20 1:1.5:3
 
5 Sadam
 
 
Answer
for M15  1:2:4
    M20  1:1.5:3
    M25  1:1:2
    M30  depend on design mix
    M35  depend on design mix
 
2 Bhai
 
 
Answer
in design mix we cannot say any propertions are fixed it is 
depends upon the trails made But what ever mr.Bhai is 
suitable for volumetric but not for design mix, Mr. bhai 
please get updated
 
3 S.prabhu
 
 
Answer
M15=1:2:4
M20=1:1.5:3
M25=1:1:2
M30=1:1:2
M35=1:5::10
 
4 Charan
 
 
Answer
M15  1:2:4
    M20  1:1.5:3
    M25  1:1:2
    M30  depend on design mix
    M35  depend on design mix
 
1 Shafique
 
 
Answer
Mix design ratio is depending on the trail mixes conducted 
to get the  particular design strength of concrete in 28 
days.It may varry depending on material, weather condition, 
batching etc.
 
2 Ibrahim S
 
 
Answer
1:2:4 for m15, 1:1.5:3 for m20, 1:1:2 for m25, for m30,m35 depends on the design mix
 
0 Balaji
 
 
Answer
yes
 
0 Sanjay
 
 
Answer
m15-1;2;4,m20-1;1.5;3,m25-1;1;2,m-30,m35- depends on the mix
 
0 Prabu
 
 
Answer
AS PER IS CODE 456 PARA 9.3.1 CLEARLY STATED THE RATIUO OF
M15 & M20, FOR M20 8 Bag Cement Is required for one cum CC
 
0 Satyavir Singh
 
 
Answer
as what are there in this content are all true n i agree with the above answer
 
0 Karma
 
 
Answer
m15 1:2:4 m20 1:1.5:3 m25 1:1:2 m30 m35 its propotionality is not fix it is depend on trial basis
 
0 Mohommud Raffique
 
 
Answer
M 15 =1:2:4(ONE CEMENT,TWO SAND AND FOUR 20mm HBM JELLY)
M 20 =1:1.5:3 ratio
M 25 = 1:1:2
M 30 = 1:1:1
OTHERS DEPENDING UPON STEELS
 
0 Jebus Dura. M.e.,civil
 
 
Answer
M20=50-138-185 (kg) 22.5ltr water
M25=50-105-147 (kg)   "
M30=50-  81-120 (kg)  21ltr  water
M35=50-  70-140 (kg)  19ltr  water
M40=50-   66-116 (kg) 17.5ltr water
 
0 Sabir
 
 
Answer
what is the practical concrete mix design for M30 AND M35
 
0 Zakir Harifal
 
 
Answer
Mix design depends on the trail mixes in the water content 
and water cement ratio is determined
to get the particular design strength of concrete in 28 
days. Water Content controls the Strength of Concrete, In 
general By Reducing The Quantity of Water we Get More 
Strength of the concrete. The Desidn Mix Also Depends on 
other factors such as Moisture Content of Materials, 
Acidity of Water, Salt content In the water etc.
 
0 Izhar Ahmed
 
 
Answer
M15-1:2:4(nominal mix)
M20-1:1.5:3 (nominal mix)
M25-1:1:2 (nominal mix)
M30&M35-Depend upon mix design (compresive strength of concrete in 28 days it is 28N/mm2 & 4N standard deveation)
 
0 Manoj Kumar Purohit
 
 
Answer
for M15 1:2:4
M20 1:1.5:3
M25 1:1:2
M30 depend on design mix
M35 depend on design mix .

Its damp sure this answer is very correct & Please contact for 
any doubts on prabhu154@gmail.com
 
0 Prabhu
 
 
Answer
m15 = 1:3:6
m20 = 1:2:4
m25 = 1:1.5:3
m30 = 1:1:2
m35 = 1:0.5:1
 
0 Qs. Patrick
 
 
Question   what is the different types of cable lugs? Rank Answer Posted By  
 Interview Question Submitted By :: Zalabhupatsinh
I also faced this Question!!   © ALL Interview .com
Answer
Aluminium or Copper Lug
Pin Type, Ring Type Fork Type
 
0 Prakash Gokhale
 
 
Answer
Lugs are categorised in following ways

1)type of conductor material
    a)Al 
    b)Cu
2)depends upon insulation
    a)Insulated (upto 6 sq.mm only) 
    b)non-insulated
3)Depends upon shape
    a)Ring type
         1)Internal Dia (ID)
             a)M4 b)M6 c)M8.(Nut size)
    b)pin type
    c)Fork or 'U'-Type
    d)tubular type
              a)M4 b)M6 c)M8.(Nut size)
    

also Bimetalic,inline connectors,reducers,T-types etc......
 
0 Amolkarne
 
 
Question   How to calculate cable sizes in sq.mm. with using motor KW? Any formula there? Rank Answer Posted By  
 Interview Question Submitted By :: Suresh
I also faced this Question!!   © ALL Interview .com
Answer
normally for copper cable 2.5amps 1sq mm
for alluminium cable 1.5 amps for 1sqmm
 
2 Khalander Mulla
 
 
Answer
Normally copper has more conductivity than aluminium. So it 
is actually reverse. 1.5 mmsqr for copper and 2.5 mm sqr 
for alu
 
1 Hemant
 
 
Answer
THE ABOVE SAID IS WRONG , ITS A THUMB RULE I USE TO FALLOW 
1 sqmm = 5A &  endangering amps withstand is 8A for alu
(sqmm * 1.87A)for  cu sqmm *2
 
0 Raja
 
 
Answer
Hameent your statment is applicable for only small cable 
size ( Up to 16 sq mm ) Please reff any one catalouge
 
1 A.bala
 
 
Answer
3pahse power is P(in KiloWatt)=I.V.PF.(root 3).....
           I=1.74 x P(KW)
EXAMPLE:
so, assume 3ph 400V 15KW motor, In=1.74x15=26.1 A,
& 15KW=20HP.....so 2xIn=Breaker rating= 52.2A
...choose 63A MCB

Cable size normally 20-25% higher than breaker to avoid 
cable heat/burn accident in overcurrent fault. so, 1.25xIn= 
62.64A so choose 80A Cable rating of 25mm2.
 
1 Abdool
 
 
Answer
First of all you need to verified full load current of motor,
How can you calculate:
Ex: 1 Motor - 20HP 3ph. 415V 50Hz, 80% efficiency 

Cal: A= HPx746/(sqrt3xVoltxPFxefficiency)

      =20x746/1.732x415x0.8x0.8

      =32.43Amp.

We all know that a motor has 1.5 times starting current,

Then   32.43x1.5 = 48.65Amp

Now you can see in the current currying capacity of cable from manufacturer cable's list.

3.5 x 35 sqmm xlpe cable is sufficient for full load.
 
0 Sagir Siddiqui
 
 
Answer
normally for copper cable 2.5amps 1sq mm
for aluminium cable 1.5 amps for 1sq mm
That is the answer.......
 
0 Bethi Arun Reddy
 
 
Answer
for cooper 2.5 mm for 1 amp
for aluminum 1.5 mm for 1 amp
 
2 Ibrahim
 
 
Answer
what's wrong with you guys. You dont even know how to spell 
correctly. Please improve on that. Its not only about idea 
but also on how you convey it. Damn!
 
0 Pedro
 
 
Answer
We must consider numerous factors to calculate the exact 
safe size of the cables for a particular load.
Please let us know whether the cable is to be laid in 
ground or it will  be overhead or any other way.
Secondly, you want an aluminium cable or copper cable?
Thirdly, the load is single phase or 3-phase?
Moreover, the motor is induction motor or any other type? 
because we have to consider harmonics also.

For e.g for a 3-phase load of 80kw, the aluminium cable to 
be laid inside the ground must be 150 sq.mm for perfectly 
safe operation.
 
0 Parveen
 
 
Answer
ex  10kw 3 phase ac motor,
    10kw x 1.74 = 17.4 amp
    MCB using 25 amp/30 amp 3 pole(C)
    Motor using star-delta/overload setting 0.58 = 10.1 amp
    cable size = 6x2.5mm2/1c pvc copper cable

if using soft starter
    MCB using 25amp/30amp 3 pole MCB(C)
    cable size = 3x4mm2/1c pvc copper cable

    cable size :calculate include starter overload setting
                (right or wrong,thank's)L
 
0 K.l.lim
 
 
Answer
First of all we will chk rated amp mentioned on motor then will check from chart abt cable sq mom size
 
0 Rana Farrukh Anwsr
 
 
Answer
For Cable Current Capacity = 4X Size of Cable in Sq.mm, Ex. For 2.5 Sq.mm = 4×2.5 = 9 Amp.
 
0 Ramkashyap
 
 
Question   who is a gazetted officer who can give character certificate?? can any body give a list of gazetted officers.. Rank Answer Posted By  
 Interview Question Submitted By :: Nilu
I also faced this Question!!   © ALL Interview .com
Answer
MEO,MDO,govt. Doctors,High school HM.....
 
3 Friend
 
 
Answer
who,s name published in gazette list of state and central 
government employes and registered doctors, public 
servervents MLA,MP,Police Inspectors.
 
3 Raghuram
 
 
Answer
A gazetted officer is that person, who is authenticated or 
recognised of their service by state govt or central govt 
like: Doctor, Police officer, IAS, OAS, Collector, Head 
Master of school, Principal of collecge, Lecture or 
professor of College or University, MP, MLA, DSWO, 
Scientist service for govt., Research person of Govt. Else 
than no one can not issue this or issue is not valid in any 
purpose.
 
0 Bijaya Kumar Mishra
 
 
Answer
Gazette means news paper that owned and published by the Govt. Of.India. Gazetted officer means the name of the Post that are published in the news paper is known as Gazetted officer.
 
0 Karthik.b
 
 
Answer
is a special executive officer (S.E.O) a gazetted officer???
 
0 Husain.modi
 
 
Answer
Hi,

All officers belonging to Govt.(state or central) whose name are published in the Gazette of india are called "Gazetted Officers".

Gazetted officer(rajpatrit adhikari) have special previllages and powers.The term Gazetted is a status symbol and makes them recognizable all over India.

All Group A regular cadre officers and also some Imp post at Group B of State and central Govt(like ACP's Of Danics,Assitant Administrative officers,Asstt. Directors comes in Group B). are Gazetted.

Note: Officers of Autonomous organizations,Contractual Staff employed in Govt.,PSU,Bank did not enjoy Gazetted status hence are not Gazetted

All Gazetted officers have rites to attest certificates (ie.Photos,Marksheets etc)

Some of the Gazetted posts are as below

Defence Officers
Assistant commissionar of police(ACP)
All central services officers
Scientists of DRDO,CSIR,NPL etc
Doctors of Govt hospitals 
Assistant Programmer,IT CADRE Delhi Govt.
Assistant Director
Assistant Engineer
Librarians of Govt 

Note:
Junior Engineers,Assistant Librarian are not Gazetted
 
0 Venkat Reddy
 
 
Question   What does the term "attenuation" mean in data communication? 1 loss of signal strength as distance increases 2 time for a signal to reach its destination 3 leakage of signals from one cable pair to another 4 strengthening of a signal by a networking device Rank Answer Posted By  
 Interview Question Submitted By :: Guext
I also faced this Question!!   © ALL Interview .com
Answer
attenuation is conserned with loss of signal strength as 
distance increase
so coorect answer is 1
 
0 Jitendera Kumar Sinha
 
 
Answer
The ratio of a signal strength at the transmitting and
receiving ends of a cable.
Correct answer is 1.
 
0 A Gim
 
 
Answer
Attenuation:

->is decrease in signal amplitude over the length of a link.
->is the gradual loss in intensity of any kind of flux
through a medium.
->Or it is simply known as transmission loss.
 
5 Abebe Alambo
 
 
Question   How to calculate the current carrying capacity of electrical cables Rank Answer Posted By  
 Interview Question Submitted By :: Rakhi Kawatra
I also faced this Question!!   © ALL Interview .com
Answer
from the size of the cable we can find particular sq mm we can 
send some current
 
3 Karthik Raman.b
 
 
Answer
There is no need to calculate ampacity of the cable it would
be provided by the manufacturer its self or else if you want
to calculate it by your self you should also know its
specific resistivity,working temp,core dia & insulation
thickness.
 
4 Lokesh.thota
 
 
Answer
current carrying capacity of the conductor is depend on 
size of cable and current density of used material of cable 
conductor

current carring capacity= size of conductor/current density 
of materils
 
3 Ghanshyam Kannaujiya
 
 
Answer
Current carrying capacity    = FLC / DF				
To refer  cable catelouge choose the cable sizes				
-Derating factor (DF)	=0.72
-Full load current (I)=  P / (1.732 x V x P.F)
 
4 Syed Rafiudeen
 
 
Answer
For calculate the current carrying capacity.
First we should calculate the cross section area of the 
conductor (sq.mm) and conductor material used in cable.

Each and every matals are having its own current carring 
capacity. This table is avail at IS.

Current carring capacity may reduce by some other factors 
like the cable length and insulation's heat withstandable 
capacity.
 
1 Venkatesh A
 
 
Answer
Refer IEC 287.If you want me to calculate the same. i will 
do it for you
 
0 Amit Jain
 
 
Answer
p=1.732vicos8
 
1 Jitendra Chauhan
 
 
Answer
It depends on many factors. 

1. Material - Copper/ Aluminum
2. Voltage Class - Some volts to 765 KV 
3. AC/ DC 
4. Type of Insulation
5. Most Importantly - Cable Cross-section Area

Refer IEC-60183 for detailed guide
 
0 Arun Reddy
 
 
Answer
The current carrying capacity of a cable is called 
Ampacity. For the porpose of the above calculation there 
are some concepts that must be taken in consideration . 
like Dia,Insulation material and thickness,type of conductor
(Copper/Aluminium),Resistivity and the working conditions. 
For BETTER CONCEPT I advice the readors to please refer 
the "Recommended Cable Sizes and their load capacity as per 
IS:13947 & IEC 60947". Below given table can help you.
Rating           Cu       Al
10A             1.5MM2    1.5MM2
15A             2.5MM2    2.5MM2
20A             2.5MM2    4.0MM2
25A             4MM2      6MM2
 
4 Dipesh Kumar
 
 
Answer
Actually current carrying capacity of the cable we can from
the manufacturer, But for the referance purpose I can give
some idea about current capacity of the cable.
So generally we are using two types of cable i.e,. copper &
Aluminium.
copper is the best conductor than the aluminium
At 1Sqmm copper conductor will Approx 2 amp in ground
At 1Sqmm copper conductor will Approx 2.4 amp in air
At 1Sqmm Aluminium conductor will Approx 1.2 amp in ground
At 1Sqmm Aluminium conductor will Approx 1.6 amp in air.

For example:
p=50kw
load current=87amps
So 
for aluminum cable size(air)= load current/1.6
                            =87/1.2
                            =72.5
But In the market there is no availability for 72.5 Sqmm so
we can propose 70Sqmm or 95Sqmm.
This is the one of the best procedure for for cable
selection with using catelogues
 
2 T.raju Goud
 
 
Answer
Well, First try to calculate the total load of the system. 
For example total load is 1000KW. The current can be 
calculated as P=1.732VICOS(PHI). Hence total current will 
be 1732A by considering V=415volt, power factor=0.8.

Step2: To calculate the cable size 

Vd = mV/L/A

where Vd= Allowable voltage drop which is normally 2.5 
percentage to 4 percent. 
L = Lenght of the cable from the source to feeder
A = Ampere rating 

Let us consider L = 1000Meters
                A = 1732A
                Vd = 2.5 percentage of 415V as in U.A.E.
                   = 2.5*415/100 = 10.375

Use these value in the above formula

mV= Vd/L*A
  = 10.375/1000*1732
  = 0.005 volt

Refer to the cable catalogue in your region and find out 
the current carrying capacity of the cable which will be 
totally depend upon the atmospheric condition, in cold 
place a smaller diameter of cable can carry larger current 
and in hot areas the same size will carry lesser current.

In U.A.E. we are normally using 11x1c 630sqmm cable with 
2500A incomer breaker. But cross check the voltage drop as 
calculated above is within the limit of voltage drop of 
these cables else you have to go to higher size of cable 
untill the voltage drop met the design milli volt drop of 
the cable.
 
0 Abdul Haseeb Khattal
 
 
Answer
BY SIMPLE 
CURRENT DENSITY = LOAD CURRENT/AREA

FOR COPPER OR ALUMINIUM CURRENT DENSITY CAN BE VARIED ACCORDING TO OUR DESIGN.FOR GOOD DESIGN CURRENT DENSITY SHOULD BE REDUCED WHICH HELPS IN EASY TEMPERATURE DISSIPATION.

CONSIDER CURRENT DENSITY OF 0.6A/SQ.MM FOR cu OR AL
LOAD CURRENT-100A
THEN
0.6=100/AREA
AREA=166 SQ.MM

MEAN THAT 166 SQ.MM REQUIRED TO CONDUCT 100A.
 
0 Manojkumar
 
 
Answer
hi frnd.. what will u say for the given ques. if interviewer asked, How a manufacturer determines the current carrying capacity of electrical cables??
 
0 Himanshu
 
 
Answer
for 50KW motor
p=50
I = 50x 1.2 (thumb rule)
  =60 A

Size of cable : 60x0.2(thumb rule)
              : 12

so min size of cable req. 12 sqmm.
 
0 At
 
 
Question   What is the CDR format generated by ZTE NEs? Rank Answer Posted By  
 Interview Question Submitted By :: Liju
I also faced this Question!!   © ALL Interview .com
Answer
Comma Seperated
 
0 Amandeep Arora
 
 
Answer
CPT format
 
0 Lodhi
 
 
Question   (1)what is last date of monthly TDS challan deposit (2) what is last date of TDS quarterly Return filling Rank Answer Posted By  
 Interview Question Submitted By :: Pradeepsinghthalwal
I also faced this Question!!   © ALL Interview .com
Answer
last date of monthly tds challan deposit is next month of 
7th
 
4 Mitul Jain
 
 
Answer
NON SALARY
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-
ANS NO.2
PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                        DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
3 Mitul Jain
[Rajasthan Stores]
 
 
Answer
7th date of every month 

April to June- 15 July
July to Sep.   15 oct
Oct to Dec     15 Jan 
jan to March   15 april
 
3 Amit Trivedi
[Rajasthan Stores]
 
 
Answer
what is last date of monthly TDS challan deposit
(2) what is last date of TDS quarterly Return filling
 
3 Bijender
[Rajasthan Stores]
 
 
Answer
1. 7th of every month
2. april to june to be filled before 15thjuly
 
0 Rajan.m
[Rajasthan Stores]
 
 
Answer
ANS.NO.1
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-       31ST MAY
ANS NO.2
PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                          DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
2 Sanjay Roy
[Rajasthan Stores]
 
 
Answer
1.7th of following month.
2. 15 july
   15 Oct
   15 Jan
   30 june
 
15 Deepu Parsad Saklani
[Rajasthan Stores]
 
 
Answer
the last date of monthly tds challan deposit is 7th of next
month 
last date of tds quartely returm filling is 15 th of every
next quarter
 
2 Sunita Mishra
[Rajasthan Stores]
 
 
Answer
answer1) 
7th of next month

for the month of april- may7th
for the month of may- June7th
for the month of July- August7th
for the month of August- September7th
for the month of September- October7th
for the month of October- November7th
for the month of November- December7th
for the month of December- Jan7th
for the month of January- Feb7th
for the month of Feb- MArch7th
for the month of March- March 31st

Answer 2) 

TDS quarterly return is to be filled by  15th of quarter 
end
 
1 Himanshu Kapoor
[Rajasthan Stores]
 
 
Answer
for the month of march, the last date is 31 may.
 
4 Vijay Patil
[Rajasthan Stores]
 
 
Answer
1. monthly tds challan deposit date next 7th
2. 15th july
   15th oct.
   15th jan.
   15th jun
 
1 Keshar Chand Sharma
[Rajasthan Stores]
 
 
Answer
1.7th of the next month but in the case of march last date 
is 31st may.
2.Aprilto June             July,15
  July to September        October,15
  October to December      January,15
  January to March         June,15
 
0 Bharat Verma
[Rajasthan Stores]
 
 
Answer
for every month 7th of next month but for in march only for 
slary it 31 of apari
 
0 Dinesh
[Rajasthan Stores]
 
 
Answer
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-       31ST MAY
ANS NO.2
PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                          DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
0 V.s. Chouhan
[Rajasthan Stores]
 
 
Answer
next month 7th
apirl to june  july15
july to september  october15
octorber to december  january15
january to march june15
 
0 Srinivas
[Rajasthan Stores]
 
 
Answer
(Ans.1)
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-       31ST MAY
(Ans.2)
apirl to june  july15
july to september  october15
octorber to december  january15
january to march june15
 
1 Krishna Kant Rathore
[Rajasthan Stores]
 
 
Answer
next month 7th except 31st March which is 31st May i.e. 
April to Feb- 7the of next month and 1st March to 30 March- 
7th of next month and for 31st March-31st May

Qtrly Returns-both salary and non salary
apirl to june  july15
july to september  october15
octorber to december  january15
january to march june15
 
0 Pawan Agrawal
[Rajasthan Stores]
 
 
Answer
1. 7th of next month/31st may for transactions booked on 
31st march.
2. A- for first qtr. 15th july
   B- for 2nd qtr. 15th october
   C- for 3rd qtr. 15th january
   D- for 4th qtr. 15th june
 
0 Surender Bhardwaj 9999510435
[Rajasthan Stores]
 
 
Answer
ANS.NO.1
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-       31ST MAY
ANS NO.2
PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                          DUE DATE
FOR THE 1st QTR ENDING JUNE-30           JULY 15
FOR THE 2nd QTR ENDING SEP-30            OCT-15
FOR THE 3rd QTR ENDING DEC-31            JANUARY 15
FOR THE 4th QTR ENDING MAR-31            JUNE-15
 
0 Jagat Babu Sharma
[Rajasthan Stores]
 
 
Answer
Ans No.1:
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH(up to 30th march)- 07th 
April      
TDS which deduct on 31 March           31st May
ANS NO.2
Last Date of Quarterly Return:
PARTICULARS                          DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
0 Amit Sharma
[Rajasthan Stores]
 
 
Answer
i am satisfy with answer no-19
 
0 Ali
[Rajasthan Stores]
 
 
Answer
every month 7 th
and quarterly retrun 15 th on salaries
 
0 Vijay
[Rajasthan Stores]
 
 
Answer
ANSWER NO 17 IS CORRECT
 
0 Sanjeev Gupta
[Rajasthan Stores]
 
 
Answer
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-       31ST MAY
ANS NO.2
PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                          DUE DATE
FOR THE 1st QTR ENDING JUNE-30           JULY 15
FOR THE 2nd QTR ENDING SEP-30            OCT-15
FOR THE 3rd QTR ENDING DEC-31            JANUARY 15
FOR THE 4th QTR ENDING MAR-31            JUNE-15
 
5 Manoj Kumar [advocate]
[Rajasthan Stores]
 
 
Answer
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH(up to 30th march)- 07th 
April      
TDS which deduct on 31 March           31st May
ANS NO.2
Last Date of Quarterly Return:
PARTICULARS                          DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
0 Manoj Kumar [advocate]
[Rajasthan Stores]
 
 
Answer
NON SALARY
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-

PAYMENT SALARY & OTHER THAN SALARY TO A RESIDENT
PARTICULARS                        DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            JUNE-15
 
0 Gaurav Jain
[Rajasthan Stores]
 
 
Answer
every months challan deposited to the 7th of the following 
month and for the month of march you can deposit the amount 
on 30th april.
and for return submission 15 th day  of succedding month of 
quartern ending and for the last quarter it is 15th of may.
 
0 Prashant Kumar Rai
[Rajasthan Stores]
 
 
Answer
from which date tds amount deposit date tds return 
submission date changed for march
 
0 Satish Kumar
[Rajasthan Stores]
 
 
Answer
ANS NO.1
PARTICULARS                              LAST DATE  
FOR THE  ENDING MONTH OF APRIL-        7TH MAY
FOR THE  ENDING MONTH OF MAY-          7TH JUNE             
FOR THE  ENDING MONTH OF JUNE          7TH JULY
FOR THE  ENDING MONTH OF JULY-         7TH AUG
FOR THE  ENDING MONTH OF AUG-          7TH SEP
FOR THE  ENDING MONTH OF SEP-          7TH OCT
FOR THE  ENDING MONTH OF OCT-          7TH NOV
FOR THE  ENDING MONTH OF NOV-          7TH DEC
FOR THE  ENDING MONTH OF DEC-          7TH JAN
FOR THE  ENDING MONTH OF JAN-          7TH FEB
FOR THE  ENDING MONTH OF FEB-          7TH MAR
FOR THE  ENDING MONTH OF MARCH-
ANS NO.2
PARTICULARS                        DUE DATE
FOR THE QTR ENDING JUNE-30           JULY 15
FOR THE QTR ENDING SEP-30            OCT-15
FOR THE QTR ENDING DEC-31            JANUARY 15
FOR THE QTR ENDING MAR-31            APRIL-15
 
0 Deepak Jain
[Rajasthan Stores]
 
 
Answer
Next month of Before 7th day

Before 15th day Quarterly TDS Return Filing
 
0 Jayesh Suresh Gurav
[Rajasthan Stores]
 
 
Answer
1. 7th of every next month... but in the case of march the 
date is 31st march.
2. 15th of next month of quarter ended..
 
0 Deepak Bansal
[Rajasthan Stores]
 
 
Answer
1.last date of monthly tds challan deposit is next month of 
  7th  

2. TDS quarterly return is to be filled by 15th of quarter 
end
 
0 Auxilia
[Rajasthan Stores]
 
 
Question   I want formula to calculate cable size as per load given in kw & amp.I searched many sites but didn't right answer.Plz reply me asap. Rank Answer Posted By  
 Interview Question Submitted By :: Ksawasthi
I also faced this Question!!   © ALL Interview .com
Answer
sorry i dont have the xat formula for that but 
 for ALUMINIUM CABLE:  1sq.mm = 1.2 A of current
      CU:              1sq.mm = 5.0 A of current
  and the relation b/w A & KW IS 
  1KW= 1000*A*240*COS#
   A- CURRENT
   240= VOLTAGE
   COS# = POWER FACTOR (.8 TO .99)
NOW IF U KNOW THE EXACT CURRENT LOAD OR LOAD IN THE FORM OF 
KW U CAN EASILY FIND THE AMPERE RATING.
AFTER FINDING THE AMPERE RATING U CAN CALCULATE THE THE 
SIZE OF THE CABLE.  ANKUSH MIGLANI 
                     9718151073
                  ankushmiglani2000@gmail.com
 
3 Er. Ankush Miglani 0971815107
 
 
Answer
You want the appropriate cable size then just apply the 
frml P=V*I*Cos#
Where voltage = 0.240
Load is given and cos# lie between 0.8 to 0.9
you can easily get the ampere rating and choosecable size.
 
4 Er.Nitesh
 
 
Answer
i will give you the basic eqaution,

PL = V*I*Pf=S*Pf ; i need to calculate the load current ,

I= PL/(V*Pf); as per ADDC(abu dhabi electricity), Pf=0.9 
V=230 volt,

From the load CURRENT i can choose the rated current of 
cable, take into account the voltage drop, so refer to 
cataloge of any cables and wires company like oman cables

regarding with three phase just put cu(3) in the equations 
 best regards
 
3 Mohammed Motawe
 
 
Answer
good

3phase cable size calculation formula
 
2 Tharik
 
 
Answer
See dear the basic soultion to cable size calculatiion in 
sismple form can be done like this

Vd= ( mv* I * lenghth)*100/(400*1000)

I  u can calualte from the Kw value for the particular pf
simple P/554

L required length

mV is the volatage drop/amp/km that can be obtained for the 
particluar size of cable from supplier 

eg. for 4cx16 its around 2.5 for most suppliers

100/400 is just amultiplication factor for converting it in 
to 3 phase


Now do calulation and check if  voltage drop limit is below 
2.5 or 4 as per your standards


Simple.

But be cutious this is the simplest method and suitabel 
method,many factors like grouping factor,derating 
factor,spacing factor,type of laying,etc  has to be taken 
for apt analysis

Software like E Tap can also be utilised
 
0 Premgin
 
 
Answer
mv= (VD*1000)/L*I

VD= VOLTAGE DROP(AS PER COUNTRY STANDARDS) 
L = LENGTH OF CABLE
I = CURRENT RATING
I = P/(1.73*V*PF)
PF= POWER FACTOR
MV= MILLI VOLTAGE DROP/AMP/METER

SELECT THE CABLE DEPENDING UPON THE MV RATING IN PRODUCT CATALOGS
 
3 Robert Nithin
 
 
Answer
Voltage drop works on Ohms Law.Multiply the route length in
metres by the the number of ampers of current that the
appliance draws,and then multiply by the number of milivolts
per ampere metre (mV/A.m)
Formula: Vd = L x I x Vc /1000
Where: Vd is the voltage drop measured in volts
       L is the length of the cable in metre
       I is the circuit current in Ampere
       Vc is the millivolts per amper per metre droped
           along the cable
       1000 is a correction factor applied because Vc is in 
            millivolts and Vd is in volts.
 
0 Nour Travis
 
 
Answer
Cable size depends on 1)Voltage Drop, 2)Distance, and 3)Load
So u have to calculate first the permissible voltage drop.
For lighting circuit permissible voltage drop:
2% of the rated voltage + 1 volt
2 X 230 /100 +1 = 460/100 + 1 volt = 5.6V

For motoring load
5% of the rated voltage

5 X 415/100 = 20.75 V

Now the formula for calculating voltage drop:

Voltage Drop = I X D (r X 0.85 + Xc X 0.53)/No. of runs X 1000
I : Current
D : Distance
r : Resistance
Xc : Capacitive reactance

If the voltage drop comes within the limits then ur cable
selection is ok.
 
0 Anupam Mathur
 
 
Answer
Hai 
First u will convert the KW OR kvA loads to  AMph
After u can select the suitable cable from cable manufuture 
catalouge . that u will consider derating factor = 0.72
For Example 
50 A
 16 SQ XLPE Al cable may be suitable 
The 16 sqmm cable suitable for 70A ( DATA TAKEN FROM        
HAVEELS CATALOUGE)
70 * 0.72 = 50.4 A
So the cable suitable for the above load. 
 after u will cross check voltage drop for the cable
 
0 Bala
 
 
Answer
Hai
 The formula is  delda v =
               rt3 x I X Z X L/ 1000 x no OF runs           
% OF VOLTAGE DROPS       =  Deta v 
                            ------          X 100 
                             APPLIED VOLTAGE
 The % of voltage voltage drops should be below 3 % 
 I = Requird current 
 Z = Cable impedance 
 L = cABLE LENGTH 
  
First  how many Amp you required 
Example - 80
then you go to cable catalouge find out  suitable cable 
size 
then calculate original current carring capacity of that 
cable  and multifly with 0.72 ( derating factor )
 example in catalouge may be 35 sq mm cable  and current 
carring capacity may be 125 A 
 125 X 0.72 =  90
The carring carring capacity point of view  this cable will 
be ok 
next you will check voltage drop  with help of above said 
fomula
 
5 Bala
 
 
Answer
it is easy,just refer some catalogue from a cable retail shop u idiot.
 
0 Shine
 
 
Answer
dear you can calculate the size of the cable as follow:

first you need to measure the current capacity and it's low as follow:
Kw/(P.F * Voltage that we wanna use)= current

so when you get the current capacity for the load you can get the size from catalogs there is tables gives you the rating of current and how much the size for every rate for this current.  

and this current will be ok if we dont have a voltage drop in line of distance between the source and the load.

actually all the distance between the load and the source have a voltage drop so we have to use this law in order to measure the percentage Voltage drop and it has be 2.5% from the load to panels:

V.D = {((x sin@ + Y cos @) x I x L)/ V } x 100%

I is the current that you measured.
L is the distance between the load and the panel.
V is the voltage that we use.
X is the AC resistance.
y is the AC reactance but sometime we ignore it.
X,Y it is depends on the size of cable so when you chose size this factors will control the voltage drop and you can change it sometimes to get the voltage drop rate as 2.5%
 
0 Engr. Abed Alraheem
 
 
Answer
hey hi he needs formula for cable calulation without refer 
any catloge
 
0 Saran
 
 
Answer
What is the correct size of cable[4core armoured cable]for
aload  of 15kva.
 
0 Antony Ngigi
 
 
Answer
FIRST YOU CALCULATE VOLTAGRDROP
 VD=1.732*V*L*I*Z/1000
 %VD=VD/RATED VOLTAGE*100
LV CABLE MINIMUM=5%
MV CABLE MINIMUM=3%
V=VOLTAGE
L=LENGTH FOR  FEED
I=AMPS
Z=IMPEDANCE
 
0 Inudeen
 
 
Answer
hai am also no exact answer
may be
eg 5HP motor select cable size 2.5sqmm cu cable, r 4 sqmm 
Al cable.
work out
5Hp motor taken 90% full load current 7 amps.
that current * 5 times = 35 Amps (motor not run in full load
its my experiance)
2.5sqmm cu cable full load withstand current 36A
4sqmm Al cable full load with stand current 34A
so we choice Cu r Al cable
 
0 S.venkatraman
 
 
Answer
We can use fromula for Voltage Drop Calculation
 VD = mVxIxL/1000
VD = Voltage Drop
mV = Milli Volt(Given Rating Cable Manufacture)
Example         16sqmm        mV =2.5 mv/a/m
I = Load Current
L = Lenght of Cable
Exapmle #
Voltage = 415vAC, Ampere = 25, Length of Cable = 100Meter

VD = 2.5x25x100/1000
   = 6250/1000
   = 6.25 (in volt)
   = 1.50% (in percent)
 
0 Tauqeer Hussain
 
 
Answer
Kv*load/volt


for single phase use 230 volt& for three phase use 440 volt
Kv =1000
 
0 Sukhvinder Singh
 
 
Answer
Voltage drop Calculation for Exaple 2Run cable
=mV/A/M(Full load current/2)length
for single Run
length x Full load current x mV/A/M/1000
mV/A/M find from cable Catalogue
this is only for Voltage drop
 
0 Anubaby
 
 
Answer
PERCENTAGE OF VOLTAGE DROP IS EQUAL TO
% V DROP = 1.732*Ir*L/1000 ( R COSØ + X SIN Ø ) / V * N
WHERE
Ir = FULL LOAD CURRENT
L = LENGTH OF THE CABLE
R = RESISTENCE OF THE CABLE
X = REACTENCE OF THE CABLE
COSØ = 0.8
SINØ = 0.53
V = VOLTAGE
N = NO. OF RUN CABLE
MAXIMUM % VOLTAGE DROP CONSIDERED 6% As per IE Rule & it cable size designs is Economical.

If Power rating & required cable length is increase , so your cable size is also increase.
 
0 Amit Kumar Tripathi
 
 
Answer
The Power Cable size shall be selected on the basis of 
(a)  Full load of Current Carrying Capacity 
(b)  Fault current Rating
(c)  Steady state voltage drop

(a)  Full load of Current Carrying Capacity 					

Transformer TRF-1 Rating (P) 		=	1600 KVA	
Rated System Voltage (V)		=	0.4 KV
Minimum size is required for full load current 	=P/_/3 x V		
		                                =2309.47Amps
Overall derating factor of multicore LV cables 	=0.660	

From Cable catalogue, 1C x 400 Sqmm cable is selected and					
Current rating of the cable in air = 680Amps

Therefore 6 cables Run is selected= 4080	Amps
Hence selected cable size is adequate

(b)    Fault Current Rating criterion 

Since feeder is protected by circuit breakekr, cable size is decided based on the bus of Fault current the transformer feeder .					
  					
The required size of cable for fault current is given by 					
formula =Isc. _/t		
	   K		
where,					
Isc - Short circuit current in KA = 50KA
t - Fault Clearing time in seconds =1Sec
K - XLPE insulation factor = 0.094		
Therefore required cable size 	= 531.91Sq mm 

Hence, Actual cable size choose 6 Run - 1C x 400 Sq mm is adequate

(C) Steady state voltage drop

Ir - Full load current of motor = 2309.47Amp
Rated System Voltage (V)= 400V
Cos Ø =	0.85		
Sin Ø =	0.53		
L - length of the cable =5		
N - No of cable runs per feeder =6 Run
R - Resistance of 1C x 400 Sq mm cable 	=0.104	Ohm / km
X - Reactance of 1C x 400 Sq mm cable 	=0.07	ohm / km
The voltage drop is given by formula 	=
	√3 x Ir x L/1000 (R cosø1 + X sinø1)/N X V 						
Therefore voltage drop 	=0.1044	%
Voltage drop is less than permissible value of 6%
 
0 Amit Kumar Tripathi
 
 
Answer
Dear See 1 eg. of Cable Designing

From - 5 MVA power T/f LV side

To      - 11 kV LA’s

•	RMS Symmetrical Short Ckt current = 17.39 kA (As per Tender spec.)

•	Fault Clearing Time = 1 sec

•	Min. Cable Size = (17.39x1000xsq root 1)/94= 185 sqmm.

•	Min. Cable Size = 1-3C 185 sqmm.

•	Cable Length = 0.06 Km. (Assumed).

•	Ifl = (5000x1000)/(1.734x11000x0.9x0.95) = 307 A.

•	Derating Factor = Ca X Cg = .88x1 = 0.88

•	Cable sized considered = 3C X 185 sqmm. (XLPE+Al).

•	Cable Resistance (Ohm/Km) = 0.21

•	Cable Reactance (Ohm/Km) = 0.087

•	Cable Ampacity = 330 A.

•	Cable Derated Current = 330x0.88 = 290.4 A.

•	Min. No. of cable required = 307/290 = 1.0586 = 1 No.

•	No. of Cable Runs Selected = 1 No.

•	% Voltage Drop = (sqrt 3x307x0.06(0.21x0.9+0.087x0.43)x100)/(1x11000) 
                                 = 0.066 % (Acceptable)
 
0 Er. Naushad Ashraf
 
 
Answer
I dont understand why people are complicating the issue.... 
the answer is very simple.

Consider Cable selection for the 3ph motor of 1kW

Formula>> P= root(3) x V x I x Cos(Phi) x Eff
p= power... Say 1kW
V= voltage (415Volt)
I= Unknown
Cos(phi)= 0.85.. say
Eff= Efficiency..= 0.85... say

Now,calculating this, we get>> I= 1.9Amp.. say approx 2A

Now derate this current (bcoz, cable will be laid in duct 
or undergroung,etc) consider derating factor as 0.7 (say)
Then current is 2A/0.7=2.85A... say 3A..

now see the cable catlogue of any company.. say polycab..
See which size of cable can carry 3A current and use it... 
thats it..

Approx.. for 1kW motor, 3Cx 2.5 Sq.mm cable is sufficient 
upto 275m length.

NOTE: This is rough method of calculation. If cable length 
is more or motor size is more... then voltage drop 
calculation is also to be done.

Contact- 9665065050
 
0 Tushar Jana
 
 
Answer
you can calculate cable size by this link
http://www.solar-wind.co.uk/cable-sizing-DC-cables.html
 
0 Muhammad Usama Qadir
 
 
Answer
i have a load 230 ampere and the length to the MDP is 285
meter the is going to mdp in a pvc duct under ground i need
the formula
 
0 Berialay
 
 
Answer
if u want to find the cable size for a specific load. so u 
find the load current. and just divid by 4, similsrly u 
find the accurate size of cable for specific load. for 
example. a load of 50kw, ist u find the load current, and 
then divide by 4.
                 P = 50kw or 50,000watt
                 V = 440 volt
                 I = ?
         p = 1.7320*440*I*0.8
         I = p/1.7320*440*.8
         I = 50,000/598.4
         I = 83 Ampere

now u divid the load current by 4

       83/4 =  20 sq.mm

its simple if u find the cbale size...
 
0 Tariqusman
 
 
Answer
sorry i remind you that if u determind the above value of 
cable size, for example i find 20 sq. mm cble for 83 ampere 
load . so u use the 25 sq. mm cable.

sorry for 20 sq mm. there no have 20 sq.
 
0 Tariqusman
 
 
Answer
For 400V 3ph AC system 

# For 1% permissible voltage drop 
     
   size of cable in mmsq (aluminium cable)=( I*L) /80 

# For 2% permissible voltage drop 

   size of cable in mmsq (aluminium cable)=( I*L) /160

# For 3% permissible voltage drop 
 
   size of cable in mmsq (aluminium cable)=( I*L) /240

   I = Full load current
   L = Length in meter from load center to motor
 
0 Abhishek Gupta
 
 
Answer
as one of our friend said above 
for Al its 1.2 amps and for Cu its 5.0 amps per 1 Sq.mm 
but actually its
 1 .0 amp for Al and 8-12 amps for Cu [per 1 Sq.mm].

the basic thing is that suppose a cable for load of 100 amps is to be laid then 
for Al 100 amps/3 phases=33.33 amps/phases,

as we don't have 33 amps cable take 36 Sq.mm cable (near and above the value) i.e,3 C*36 Sq.mm.
for Cu consider 8 amps for Cu
then 8 Amps*5 Sq.mm =40 Amps per core or phase

i.e,for 100 amps load u need to lay 
a cable Al= 3 Core *36 Sq.mm 

and for Cu =3 core *5 Sq.mm
 
0 Santhosh
 
 
Answer
as u know power=v*i*power factor
the load is given in KW
SUPPOSE THE LOAD IS 30KW
FOR THREE PHASE VOLTAGE IS 440V
THEN,   30*1000=440*I*O.9(SUPPOSING PF)
       30*1000=396*I
THEN,   I=75AMP
FOR ALUMINIUM CABLE 1Sq.mm=1.5 amp of current
for copper cable    1sq.mm=5 amp of current
then,  your cable size is 15sq.mm for copper cable
 
0 Tanu Saxena
 
 
Answer
sir  plz send me proper formula for cable size calculation
 
0 Qasim Riaz
 
 
Answer
Oh Dear..Its so easy...Follow these simple steps (below
link)... 

http://www.electricaltechnology.org/2013/10/How-to-determine-the-suitable-size-of-cable-for-Electrical-Wiring-Installation-with-Solved-Examples-in-both-British-and-SI-System.html

Titled as "How to determine the suitable size of cable for
Electrical Wiring Installation with Solved Examples (in both
British and Si System) "
 
0 Engr Wasim Khan
 
 
Answer
CABLE SIZING FORMULAS:-
1ST STEP CALCULATE THE VOLTAGE DROP (Vd).
(1 ph.) Vd= 2 x I x L x P.F/(K x Area mm2)
(3 ph.) Vd=1.732 x I x L/(KxArea mm2)
where:-
I:- Load current
L:- Length of the conductor.(in meters)
K:- 58 for copper conductor and 36 for aluminum conductor.
2 for single phase and sqrt 3 for three phase.
____________________________________________________________
2nd step.
AREA mm2 = 2 X I X L /(K x Vd) (FOR SINGLE PHASE)
AREA mm2 = 1.732 x I x L/(K x Vd)(for three phase)
____________________________________________________________
FORMULAS DESIGNED BY
ENGR.
MUHAMMAD YOUNAS KHAN
M.Sc.Engg.,MIEE(UK)
NEAR KEEKER WALI MASJID.
MOHELLAH LAKER MANDI
WAZIRABAD CITY,PUNJAB
PAKISTAN
 
0 Engr.muhammad Younas Khan
 
 
Question   how to choose the cable rating for load (KW)? EXAMPLE.50KW load Rank Answer Posted By  
 Interview Question Submitted By :: Radhakrishnan
I also faced this Question!!   © ALL Interview .com
Answer
Copper Cable: Sqmm*1.5, Exam: If we have 50 Sqmm copper 
cabe, The it will take load is 50*1.5=75Amp



Al Cable is multiplying factor of 1.
Exam: 50 Sqmm cable will take 50A load.
 
1 Suresh Kumar
 
 
Answer
In a three phase system p=1.732x3xVxIxCOS@
When load is given 50KW
Then I=50000/1.732x3x415x.8
=86A near about
taking some future expectancy you can take a cable of 
having 100A capacity.SO by this u can take a cable of 70mm 
square in aluminium and 50 mm square in copper
 
3 Vinod Sharma
 
 
Answer
Cable carring capasity is there.. we can get form cable 
manufacturer...

for 50kW load,
current(I) = 50/(1.732*415*.8(p.f))
        I = 28.75A
For 1.5Sqmm cable curent carring capasity = 20A in ground
    2.5 = 27,
    4   = 34,
    6   = 43,
    10  = 57,
    16  = 73A like that.
now our load current is 28.75A
     No.of cable with size =28.75/43A (6Sqmm)
                           = .6686/.7 (Diff factor)
                           =.955
there fore 1run 6 Sqmm aluminium cable we selected for 50 
kW load.
 
0 Parasuram
 
 
Answer
Right cable size Sellection depends on 3 factors:
1-   Load Current.current(I) = 50kW /(1.732*415*0.85   
(p.f))   =139A

2-  Distance(From feeder to your load)
3-  Allawable Voltage droop in your system (Should be less 
than 5% ) 

As experience you can use 10sqrmm for 40A,for 120meter.
 
0 Engineer
 
 
Answer
Fisrt calcultate Load in Ampere as basic formula below

I (Amp ) =  P ( KW) / 1.732*Voltage (400 or 415 or 380 )* 
p.f

Voltage depend on frequency rating (50 Hz or 60 Hz)
then check where you have to laid in AIR or Ground 
you have to consider so many factor.
suppose you laid in Ground then factor to be consider as 
below
(1) Soil Thermal Rseistivity
(2) Group factor
(3)Rating factor of ground temprature 
(4)Rating Factor depth of laying

then calculte Voltage drop  on behalf of amper rating by 
load.
once you calculate then check length

V.D = mv x I X L/1000 OR mv =V.D X 1000/IXL

VD= Maximum acceptable volt drop (in volt)
mv =appropriate volt drop (in mv /amp /metre
I = Current per phase (in ampere)
L= Length ( in meter)
 
0 Mohammad Farman Khan
 
 
Answer
Calclate the load Current 
I = p/((root3)*V*P.F )
where I line current, v line voltage
then go to cable tabe 
and check the Drateing factors.such as
if this cable underground,free,soil,depth of laying, 
ambient temperture, armoed or unarmoed etc.

then calcute the voltage drop
if more than 4%
you should to select the next bigger cross section at table.

by experince for copper PVC cables 
1mm take 2.5 AMP
 
0 M.saad
 
 
Answer
1 H.P = 1.25 Amps
IN CASE ALUMINIUM CONDUCTOR USE CONSTANT VALUE IS =0.75
IN CASE COPPER  CONDUCTOR USE CONSTANT VALUE IS =1.25

SO CONDUCTOR SIZE= 1.25*1.25(CU)=1.25SQMM
 
0 Er.p.somasundaram
 
 
Answer
10 H.P = 10.25 Amps
IN CASE ALUMINIUM CONDUCTOR USE CONSTANT VALUE IS =0.75
IN CASE COPPER  CONDUCTOR USE CONSTANT VALUE IS =1.25

SO CONDUCTOR SIZE+cable sqmm  *075  =10.25amps, now a days increase current so cable size is small size increases. 
10.25*0.75=10 sq mm
 
0 Er.p.somasundaram
 
 
Answer
50kw load will take 87 amps (approx.). There is table of 
different types of cables available in the manufactures 
site. More over there is so many factors are there to 
select a cable. Such as the type of material (copper / 
aluminium), single core or multicore, how and where the 
cable is being laid, the length of cable(for voltage drop 
calculation), type of load and its starting current (for 
motors), future load etc. Normally if the load is 50KW, 
70sq.mm Aluminium cable or 50sq.mm copper cable can be used.
 
0 Shabeeb
 
 
Answer
LOAD : 50kW , 415V , 0.8p.f 
1- I = 28.75A 
2- 28.75 * 1.2 = 34.5A
3- temp. factor * group factor = 0.75 "ex."
4- 34.5/.75 = 46A
5- choose 16mmsq
6- cal. Vd = (I * L(m) * delta(v)/ (V * 1000))*100% 
its should less than 5%
7- short circuit temp. rise
 
0 Mathhar
 
 
Answer
Given- KW-50
  Now from formula I= KW X1000/(V X I X 1.732 X .8) 
                    =86.95 A  
 Now by refering fenolex cable rating table
   For 25mm2 =80A Which is less than our o/p current so we  
can select 35mm2 having current capacity is 102A
 
1 Sunil Patil
 
 
Answer
380V/50HZ,50KW=4core50mm
 
0 Ajith
 
 
Question   How can w select the cable size for different rating load & also waht r factors affecting the during selection of cable? Rank Answer Posted By  
 Interview Question Submitted By :: Guest
I also faced this Question!!   © ALL Interview .com
Answer
Cable size depend upon the full load current rating.Suppose 
current rating is  86A then csble size is 35 SQMM.The 
factores are the cunductor size and used cunductor
 
2 Nilkanth
 
 
Answer
cable size primarily depends on its voltage rating or kW 
rating of load. From the kW rating, we can calculate the 
current that should be carried by cable and the number of 
runs required. we also have to calculate voltage drop 
during starting and running in case of a motor load.
 
3 Ranga
 
 
Answer
Full load current and voltage drop
 
0 Sahil>m
 
 
Answer
There r so many factors for selection of cable:
1. full load of equipment or load.
2. Length of cable.
3. Derating factor.
4. Insulation level
 
0 Amit3456
 
 
Answer
Cnductor size depends upon the load which in turn depends 
upon the current carrying by the cable. Generally any cable 
can carry current  2.5 times the cable size. Suppose 4 
suare mm cable can carry 10.5 amps current.
 
0 Sanjay Talwaria
 
 
Answer
cable size depends on:
1)total connected load via cable.
2)lenght between the source and the load.
3)temprture
4)isulator type
5)load type
6)wiring place& type
 
0 Eng.yousef
 
 
Answer
For sizing of any feeder following are required:-
1. The load for which we are sizing
2.Dearating factor
3.pf
4.Starting voltage drop
5.running voltage drop
6.Spacing/Touching factor
7Effeciency
 
1 Ranjan
 
 
Answer
If cable size calulate on load in Amps.basis then please 
give the formula, or advice cable size in 3.3kv 104Amps and 
52 Amps,67 Amps, 33 Amps, and 15 Amps load.

Powe engrs.
 
0 D.k.kakade
 
 
Answer
A) Firstly we have to check for wht type of system we require the cable i.e LV, MV, HV AND EHV
B) then we have to calculate first: the rated current required by load on 3phase or 1phase system i.e

1) For 3-Phase System
KVA= KW/SQRT3*PF
2) For 1-Phase System
KVA=KW/PF

C) In companies there is feeder schedule for every type of cable w.r.t the amp.

Enjoy!
 
0 Manishhctm
 
 
Answer
first we consider the type of load,voltage 
rating,location,power etc will be known after that we 
calculate the in put current and 20% extra value will be 
added to that currnet and based on that current we refer 
the cable manufacture booket and select the suitable size 
of the cable
 
0 Ch.venkateswra Rao
 
 
Answer
My point of view, To find the cable size need to divide the 
LOAD CURRENT by 3.
i.e:
Cable size= Load current (10A)/ 3
          = 3.3
Now the answer is 3.3, But we don't have the 3.3 size of 
cable. so we can choose next related size of cable, which 
is 4 sq.mm.
 
0 Rathinam
 
 
Question   what is the meaning and full form of AYFY CABLE? Rank Answer Posted By  
 Interview Question Submitted By :: Guest
I also faced this Question!!   © ALL Interview .com
Answer
AYFY Means Alluminum PVC Insulated Fltat strip armour and 
PVC outer sheeth cable
 
0 Vasanth
 
 
Answer
A -ALLUMINIUM
Y - INNER SHEATH
F - FLAT ARMOURED
Y - OUTER SHEATH
 
0 Har
 
 
Answer
pvc insulatd aluminium armoured aluminum conductor cable.
 
0 Vinayak
 
 
Answer
AYFY means Aluminium Conductor, PVC Insulated, Galvanised 
Flat Steel Wire (Strip) Armoured and PVC Outer Sheathed 
Heavy Duty Cables
 
0 K.r.sakthivelan
 
 
Answer
All the above appear to be correct.As per IS 1554 to which
the mentioned  HT cable belongs: A-Al conductor,Y-PVC
insulation,F-suitable for low temperature,Y-PVC outer
sheath.In all others F is for single strip armour
 
0 K.prakashchandra
 
 
Answer
ALLUMIUM PVC INSULATED FLAT STRIP ARMOUR PVC OUTER CABLE
 
0 M.a.abraham
 
 
Answer
ALLUMINIUM INNER SHEATH FLAT ARMOURED OUTER SHEATH
 
0 Yoge
 
 
Answer
ALLUMINIUM INNER SHEATH FLAT ARMOURED CABLE
 
0 Jitendra Kumar
 
 
Question   How to repalce a failed disk? Rank Answer Posted By  
 Interview Question Submitted By :: Ramanji.gundala
I also faced this Question!!   © ALL Interview .com
Answer
Ok > prtenv boot-device

ok> setenv boot-device <new boot disk address>

ok> boot
 
0 Raviga006
 
 
Answer
1) echo |format

check the all disks and find out the failed disk

2) check the dick is configured in veritas or meta

3) remove the disk from veritas or meta controller 

4) insert a new disk

5) run devfsadm -c disk
or

devfsadm

format c0t0dx

6) configure the disk
 
0 Phani-a
[Xxxxxxxxxxx]
 
 
Answer
Check the status of the disks using command: iostat -e
or if the disk is a metadevice then use metastat -p 

If  the disk is faulty and needs to be removed then use the 
prtvtoc /dev/rdsk/c#t#d#s2 > /diskprtvtoc to save the disk 
partition.

Use the command cfgadm -c unconfigure c1::dsk/c1t1do to 
unconfigure the device
verify using cfgadm -al if the disk is unconfigured

remove the failed disk using command cfgadm -x 
remove_device c1::/dsk/c1t1do

run devfsadm -C -c disk to invoke cleanup to remove device 
pathsand unreferenced links
verify it  ls -ld /dev/rdsk/c1t1dos2 this should return no 
devices

it is safe to physically remove the disk  n replace 
cfgadm -x insert_device c1::/dsk/c1t1do

now configure it cfgadm -c configure c1::dsk/c1t1do 

Run devfsadm and verify about the configured disk using ls -
ld command.

Run the fmthard -s /diskprtvtoc /dev/rdsk/c1t1d0s2 to copy 
back the saved partitions to the new configured and 
inserted disk
 
0 Priya
[Xxxxxxxxxxx]
 
 
 
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