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 Company >> Cable >> Cable Questions Grasim Technical Test Questions (1) 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
```fistly calculate the
multiply it by 1.2 for
the tollerance in
current.```

0 Amol Folane

```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
```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

```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

`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
```M-15 = 3.5 N/MM2, M-25 and M-20 = 4 N/MM2, M-30 and M-35
=5NMM2 THANKS.```

3 Navin Morwal

`m-15 1:2:4,m-20 1:1.5:3`

```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

```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

3 S.prabhu

```M15=1:2:4
M20=1:1.5:3
M25=1:1:2
M30=1:1:2
M35=1:5::10```

4 Charan

```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

```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

`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

`yes`

0 Sanjay

`m15-1;2;4,m20-1;1.5;3,m25-1;1;2,m-30,m35- depends on the mix`

0 Prabu

```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

`as what are there in this content are all true n i agree with the above answer`

0 Karma

`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

```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

```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

`what is the practical concrete mix design for M30 AND M35`

0 Zakir Harifal

```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

```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

```for M15 1:2:4
M20 1:1.5:3
M25 1:1:2
M30 depend on design mix
M35 depend on design mix .

any doubts on prabhu154@gmail.com```

0 Prabhu

```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

```in m30 how to mix concrete exampal cement sand jall water

0 Kumar

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
```Aluminium or Copper Lug
Pin Type, Ring Type Fork Type```

0 Prakash Gokhale

```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
```normally for copper cable 2.5amps 1sq mm
for alluminium cable 1.5 amps for 1sqmm```

2 Khalander Mulla

```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

```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

```Hameent your statment is applicable for only small cable
size ( Up to 16 sq mm ) Please reff any one catalouge```

1 A.bala

```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

```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

```normally for copper cable 2.5amps 1sq mm
for aluminium cable 1.5 amps for 1sq mm

0 Bethi Arun Reddy

```for cooper 2.5 mm for 1 amp
for aluminum 1.5 mm for 1 amp```

2 Ibrahim

```what's wrong with you guys. You dont even know how to spell
but also on how you convey it. Damn!```

0 Pedro

```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

```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

`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

`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
`MEO,MDO,govt. Doctors,High school HM.....`

3 Friend

```who,s name published in gazette list of state and central
government employes and registered doctors, public
servervents MLA,MP,Police Inspectors.```

3 Raghuram

```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

`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

`is a special executive officer (S.E.O) a gazetted officer???`

0 Husain.modi

```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 Director
Assistant Engineer
Librarians of Govt

Note:
Junior Engineers,Assistant Librarian are not Gazetted```

0 Venkat Reddy

`Kantabanji revenue officer not given cherecter certificate plese hi light`

0 Kumar Sahoo

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
```attenuation is conserned with loss of signal strength as
distance increase

0 Jitendera Kumar Sinha

```The ratio of a signal strength at the transmitting and
receiving ends of a cable.

0 A Gim

```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
```from the size of the cable we can find particular sq mm we can
send some current```

3 Karthik Raman.b

```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

```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

```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

```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

```Refer IEC 287.If you want me to calculate the same. i will
do it for you```

0 Amit Jain

`p=1.732vicos8`

1 Jitendra Chauhan

```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

```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.
the "Recommended Cable Sizes and their load capacity as per
Rating           Cu       Al
10A             1.5MM2    1.5MM2
15A             2.5MM2    2.5MM2
20A             2.5MM2    4.0MM2
25A             4MM2      6MM2```

4 Dipesh Kumar

```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
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

```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

```BY SIMPLE

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
THEN
0.6=100/AREA
AREA=166 SQ.MM

MEAN THAT 166 SQ.MM REQUIRED TO CONDUCT 100A.```

0 Manojkumar

`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

```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
`Comma Seperated`

0 Amandeep Arora

`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
```last date of monthly tds challan deposit is next month of
7th```

4 Mitul Jain

```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]

```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]

```what is last date of monthly TDS challan deposit
(2) what is last date of TDS quarterly Return filling```

3 Bijender
[Rajasthan Stores]

```1. 7th of every month
2. april to june to be filled before 15thjuly```

0 Rajan.m
[Rajasthan Stores]

```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]

```1.7th of following month.
2. 15 july
15 Oct
15 Jan
30 june```

[Rajasthan Stores]

```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]

```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

TDS quarterly return is to be filled by  15th of quarter
end```

1 Himanshu Kapoor
[Rajasthan Stores]

`for the month of march, the last date is 31 may.`

4 Vijay Patil
[Rajasthan Stores]

```1. monthly tds challan deposit date next 7th
2. 15th july
15th oct.
15th jan.
15th jun```

1 Keshar Chand Sharma
[Rajasthan Stores]

```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]

```for every month 7th of next month but for in march only for
slary it 31 of apari```

0 Dinesh
[Rajasthan Stores]

```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]

```next month 7th
apirl to june  july15
july to september  october15
octorber to december  january15
january to march june15```

0 Srinivas
[Rajasthan Stores]

```(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]

```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]

```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]

```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]

```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]

`i am satisfy with answer no-19`

0 Ali
[Rajasthan Stores]

```every month 7 th
and quarterly retrun 15 th on salaries```

0 Vijay
[Rajasthan Stores]

`ANSWER NO 17 IS CORRECT`

0 Sanjeev Gupta
[Rajasthan Stores]

```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```

[Rajasthan Stores]

```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```

[Rajasthan Stores]

```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]

```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]

```from which date tds amount deposit date tds return
submission date changed for march```

0 Satish Kumar
[Rajasthan Stores]

```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]

```Next month of Before 7th day

Before 15th day Quarterly TDS Return Filing```

0 Jayesh Suresh Gurav
[Rajasthan Stores]

```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]

```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
```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

```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

```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

```good

3phase cable size calculation formula```

2 Tharik

```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

```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

```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

```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

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

```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

```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

`it is easy,just refer some catalogue from a cable retail shop u idiot.`

0 Shine

```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

```hey hi he needs formula for cable calulation without refer
any catloge```

0 Saran

```What is the correct size of cable[4core armoured cable]for

0 Antony Ngigi

```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

```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

```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
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

```Kv*load/volt

for single phase use 230 volt& for three phase use 440 volt
Kv =1000```

0 Sukhvinder Singh

```Voltage drop Calculation for Exaple 2Run cable
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

```PERCENTAGE OF VOLTAGE DROP IS EQUAL TO
% V DROP = 1.732*Ir*L/1000 ( R COSØ + X SIN Ø ) / V * N
WHERE
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

```The Power Cable size shall be selected on the basis of
(a)  Full load of Current Carrying Capacity
(b)  Fault current Rating

(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

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 	=
&#8730;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

```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)```

```I dont understand why people are complicating the issue....

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

```you can calculate cable size by this link
http://www.solar-wind.co.uk/cable-sizing-DC-cables.html```

```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

```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

```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

```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

L = Length in meter from load center to motor```

0 Abhishek Gupta

```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

```as u know power=v*i*power factor
the load is given in KW
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

`sir  plz send me proper formula for cable size calculation`

0 Qasim Riaz

```Oh Dear..Its so easy...Follow these simple steps (below

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

```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:-
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.
M.Sc.Engg.,MIEE(UK)
NEAR KEEKER WALI MASJID.
MOHELLAH LAKER MANDI
PAKISTAN```

`Ye sab galt jawab de rahe hai koi v correct answer nahi diya`

0 Md. Ekram Ali

```As we Now that,

Area= (IscXrtT)/K, Factor for PVC or XLPE,

As we know that Derating Factor = AgXAaXAsXAf

and I= P/(rt3XVXCos#)
and I=DfXI```

Interview Question Submitted By :: Radhakrishnan
I also faced this Question!!   © ALL Interview .com
```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

```In a three phase system p=1.732x3xVxIxCOS@
Then I=50000/1.732x3x415x.8
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

```Cable carring capasity is there.. we can get form cable
manufacturer...

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

0 Parasuram

```Right cable size Sellection depends on 3 factors:
1-   Load Current.current(I) = 50kW /(1.732*415*0.85
(p.f))   =139A

3-  Allawable Voltage droop in your system (Should be less
than 5% )

As experience you can use 10sqrmm for 40A,for 120meter.```

0 Engineer

```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
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)```

```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```

```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

```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

```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
70sq.mm Aluminium cable or 50sq.mm copper cable can be used.```

0 Shabeeb

```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

```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

`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
```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

```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

`Full load current and voltage drop`

0 Sahil>m

```There r so many factors for selection of cable:
2. Length of cable.
3. Derating factor.
4. Insulation level```

0 Amit3456

```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

```cable size depends on:
2)lenght between the source and the load.
3)temprture
4)isulator type
6)wiring place& type```

0 Eng.yousef

```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

```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.```

```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

```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

```My point of view, To find the cable size need to divide the
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
```AYFY Means Alluminum PVC Insulated Fltat strip armour and
PVC outer sheeth cable```

0 Vasanth

```A -ALLUMINIUM
Y - INNER SHEATH
F - FLAT ARMOURED
Y - OUTER SHEATH```

0 Har

`pvc insulatd aluminium armoured aluminum conductor cable.`

0 Vinayak

```AYFY means Aluminium Conductor, PVC Insulated, Galvanised
Flat Steel Wire (Strip) Armoured and PVC Outer Sheathed
Heavy Duty Cables```

0 K.r.sakthivelan

```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

`ALLUMIUM PVC INSULATED FLAT STRIP ARMOUR PVC OUTER CABLE`

0 M.a.abraham

`ALLUMINIUM INNER SHEATH FLAT ARMOURED OUTER SHEATH`

0 Yoge

`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
```Ok > prtenv boot-device

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

ok> boot```

0 Raviga006

```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

or

format c0t0dx

6) configure the disk```

0 Phani-a
[Xxxxxxxxxxx]

```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
verify it  ls -ld /dev/rdsk/c1t1dos2 this should return no
devices

it is safe to physically remove the disk  n replace

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

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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