This document provides a cost estimation for the electrical wiring of a hospital considering safety issues. It includes calculations for the lighting requirements of various rooms in the hospital based on their size and recommended illumination levels. Additional electrical loads such as fans, appliances and medical equipment for each room are also listed. Assumptions are provided for factors like utilization rate, depreciation, and wire sizing. Calculations are shown for determining the number and wattage of lights needed in each room to meet the illumination requirements.
2. SUBMITTED BY SUBMITTED TO
STUDENT ID NAME
2001001 MD. JOBAYER
PARVEZ RATUL
2001002 SUSMITA
MOZUMDAR
2001003 MD. SADMAN RAFID
2001004 NIRVEEK SARKER
2001005 TOUSIF ARAFAT
• MD. KAMRUL HASAN
ASSISTANT PROFESSOR
DEPARTMENT OF ELECTRICAL &
ELECTRONIC
ENGINEERING (EEE)
KHULNA UNIVERSITY OF ENGINEERING &
TECHNOLOGY (KUET), KHULNA
• AREFIN AHMED SHUVO
LECTURER
DEPARTMENT OF ELECTRICAL &
ELECTRONIC
ENGINEERING (EEE)
KHULNA UNIVERSITY OF ENGINEERING &
TECHNOLOGY (KUET), KHULNA
3. Project Requirements:
A hospital has a reception and waiting area, two male and female wards, one
operating room, one emergency room, and three restrooms. A telephone is
available at the Reception desk (PSTN). The operating room must contain a
spotlight outlet. Install an exhaust fan in the restrooms. The hospital should
have smoke detectors and fire alarms installed. Except for the restrooms, each
of the rooms above has a sub-distribution board, while the main distribution
board for the department is located at the entrance.
7. Assumptions:
1. Utilization Factor (U.F) = 0.6
2. Depreciation Factor (D.F) = 0.8
3. Luminous Efficacy = 120 lm/W
4. Switch Boards are 5ft above from the ground.
5. Floor Height is 10ft
6. Blubs are positioned 2ft beneath from the surface.
7. Conduit and wire are calculated with 5% extra quantity.
8. Loads are calculated with 15% additional load.
9. Length of the operating table = 1900 mm = 6.2336ft & width = 520 mm = 1.706ft
10.Estimation for the wire size of the hospital is assumed to be perfect up to 45°C
11.Distance between the Electric Pole and Main Switch Board is assumed to be 50ft.
12.Bulbs used: Philips
Ceiling Fans used: Walton
Exhaust Fans used: Click
Wire used: BRB Cables
Conduit used: Vigo uPVC Conduit Pipe
8.
9.
10.
11. Number of Light Calculation
1. Reception & Waiting Area:
Total lumens required = 21.5×15×150×0.093/0.8×0.6 = 9372.656 lumen
Required wattage = 9372.656/120 = 78.1 watt
Here, 4 LED lights of 15 watt and 2 LED lights of 10 watt to be used to comply with the requirements.
So, power to be supplied = (15×4)+(10×2) watt = 70 watt
2. Male Ward:
Total lumens required = 37.5×20.5×150×0.093/0.8×0.6 = 22341.79688 lumen
Required wattage = 22341.79688/120 = 186.18 watt
Here, 4 LED lights of 32 watt and 2 LED lights of 30 watt to be used to comply with the requirements.
So, power to be supplied = (32×4)+(30×2) watt = 188 watt
12. 3. Restroom 1:
Total lumens required = 5×10×100×0.093/0.8×0.6 = 968.75 lumen
Required wattage = 968.75/120 = 8.0 watt
Here, 1 LED light of 8 watt to be used to comply with the requirements.
So, power to be supplied = 8 watt
4. Female Ward:
Total lumens required = 24×38×150×0.093/0.8×0.6 = 26505 lumen
Required wattage = 26505/120 = 220.875 watt
Here, 9 LED lights of 25 watt to be used to comply with the requirements.
So, power to be supplied = 25×9 watt = 225 watt
5. Restroom 2:
Total lumens required = 5×8.5×100×0.093/0.8×0.6 = 823.4375 lumen
Required wattage = 823.4375/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
13. 6. Operating Room:
Total lumens required = 300×14×22.5×0.093/0.8×0.6 = 18309.375 lumen
Required wattage = 18309.375/120 = 152.578 watt
Additionally, for spotlight outlet of operating table, total lumens required,
= 2000×6.2336×1.706×0.093/0.6×0.8 = 4120.8772 lumen
Required wattage = 4120.8772/120 = 34.34 watt
Here, 2 LED lights of 38 watt, 2 LED lights of 40 watt and 1 LED spotlight of 35 watt to be used to comply with the
requirements.
So, power to be supplied = (38×2)+(40×2)+35 watt = 191 watt
7. Emergency Room:
Total lumens required = 33×15×150×0.093/0.8×0.6 = 14385.9375 lumen
Required wattage = 14385.9375/120 = 119.8828 watt
Here, 4 LED lights of 30 watt to be used to comply with the requirements.
So, power to be supplied = 30×4 watt = 120 watt
8. Laboratory:
Total lumens required = 10×17.5×350×0.093/0.8×0.6 = 11867.1875 lumen
Required wattage = 11867.1875/120 = 98.89 watt
Here, 4 LED lights of 25 watt to be used to comply with the requirements.
So, power to be supplied = 25×4 watt = 100 watt
14. 9. Radiology Room:
Total lumens required = 10×24×100×0.093/0.8×0.6 = 4650 lumen
Required wattage = 4650/120 = 38.75 watt
Here, 4 LED lights of 10 watt to be used to comply with the requirements.
So, power to be supplied = 10×4 watt = 40 watt
10. Doctor's Room 1:
Total lumens required = 150×7.25×12×0.093/0.8×0.6 = 2528.4375 lumen
Required wattage = 2528.4375/120 = 21.07 watt
Here, 1 LED light of 12 watt and 1 LED light of 10 watt to be used to comply with the requirements.
So, power to be supplied = (12×1)+(10×1) watt = 22 watt
11. Doctor's Room 2:
Total lumens required = 150×7.25×12×0.093/0.8×0.6 = 2528.4375 lumen
Required wattage = 2528.4375/120 = 21.07 watt
Here, 1 LED light of 12 watt and 1 LED light of 10 watt to be used to comply with the requirements.
So, power to be supplied = (12×1)+(10×1) watt = 22 watt
15. 12. Nurse's Station:
Total lumens required = 150×9×15×0.093/0.8×0.6 = 3923.4375 lumen
Required wattage = 3923.4375/120 = 32.695 watt
Here, 1 LED light of 15 watt and 1 LED light of 18 watt to be used to comply with the requirements.
So, power to be supplied = (15×1)+(18×1) watt = 33 watt
13. VIP Ward 1:
Total lumens required = 10×15×150×0.093/0.8×0.6 = 4359.375 lumen
Required wattage = 4359.375/120 = 36.328 watt
Here, 1 LED tube light of 20 watt and 1 LED tube light of 18 watt to be used to comply with the
requirements.
So, power to be supplied = (20×1)+(18×1) watt = 38 watt
14. Attached Restroom 1:
Total lumens required = 4.5×10×100×0.093/0.8×0.6 = 871.875 lumen
Required wattage = 871.875/120 = 7.2656 watt
Here, 1 LED light of 8 watt to be used to comply with the requirements.
So, power to be supplied = 8 watt
16. 15. Balcony 1:
Total lumens required = 4.5×10×70×0.093/0.8×0.6 = 610.3125 lumen
Required wattage = 610.3125/120 = 5 watt
Here, 1 LED light of 5 watt to be used to comply with the requirements.
So, power to be supplied = 5 watt
16. VIP Ward 2:
Total lumens required = 10×15×150×0.093/0.8×0.6 = 4359.375 lumen
Required wattage = 4359.375/120 = 36.328 watt
Here, 1 LED tube light of 20 watt and 1 LED tube light of 18 watt to be used to comply with the requirements.
So, power to be supplied = (20×1)+(18×1) watt = 38 watt
17. Attached Restroom 2:
Total lumens required = 5×8.5×100×0.093/0.8×0.6 = 823.4375 lumen
Required wattage = 823.4375/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
17. 18. Balcony 2:
Total lumens required = 4.5×8.5×70×0.093/0.8×0.6 = 518.7656 lumen
Required wattage = 518.7656/120 = 5 watt
Here, 1 LED light of 5 watt to be used to comply with the requirements.
So, power to be supplied = 5 watt
19. VIP Ward 3:
Total lumens required = 10×15×150×0.093/0.8×0.6 = 4359.375 lumen
Required wattage = 4359.375/120 = 36.328 watt
Here, 1 LED tube light of 20 watt and 1 LED tube light of 18 watt to be used to comply with the
requirements.
So, power to be supplied = (20×1)+(18×1) watt = 38 watt
20. Attached Restroom 3:
Total lumens required = 4.5×8.5×100×0.093/0.8×0.6 = 741.09375 lumen
Required wattage = 741.09375/120 = 6.1758 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
18. 21. Male Restroom 1:
Total lumens required = 10×4×100×0.093/0.8×0.6 = 775 lumen
Required wattage = 775/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
22. Male Restroom 2:
Total lumens required = 10×4×100×0.093/0.8×0.6 = 775 lumen
Required wattage = 775/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
23. Male Restroom 3:
Total lumens required = 10×5×100×0.093/0.8×0.6 = 968.75 lumen
Required wattage = 968.75/120 = 8.0 watt
Here, 1 LED light of 8 watt to be used to comply with the requirements.
So, power to be supplied = 8 watt
19. 24. Female Restroom 1:
Total lumens required = 10×4×100×0.093/0.8×0.6 = 775 lumen
Required wattage = 775/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
25. Female Restroom 2:
Total lumens required = 10×4×100×0.093/0.8×0.6 = 775 lumen
Required wattage = 775/120 = 6.86 watt
Here, 1 LED light of 7 watt to be used to comply with the requirements.
So, power to be supplied = 7 watt
26. Female Restroom 3:
Total lumens required = 10×5×100×0.093/0.8×0.6 = 968.75 lumen
Required wattage = 968.75/120 = 8.0 watt
Here, 1 LED light of 8 watt to be used to comply with the requirements.
So, power to be supplied = 8 watt
20. 27. Corridor:
Total lumens required =
[(15×30.5)+{(33.5+22+15)×7}+(11×5)+(10.5×5)+(18.5×4.5)+(28.5×4.5)]×100×0.093/0.8×0.6 = 24751.5625 lumen
Required wattage = 24751.5625/120 = 206.263 watt
Here, 2 LED lights of 13 watt, 4 LED lights of 12 watt, 10 LED lights of 10 watt, 1 LED light of 8 watt, 3 LED lights of
7 watt and 1 LED light of 6 watt to be used to comply with the requirements.
So, power to be supplied = (2×13)+(4×12)+(10×10)+8+(7×3)+6 = 209 watt
28. Staircase:
Total lumens required = 10×18×100×0.093/0.8×0.6 = 3487.5 lumen
Required wattage = 3487.5/120 = 29.0625 watt
Here, 2 LED light of 15 watt to be used to comply with the requirements.
So, power to be supplied = 30 watt
21. Wattage Calculation
Table: Reception and Waiting Area
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 4 15 60
2 10 20
2 Ceiling Fan Point 4 75 300
3 2 Pin Plug Point 1 42 42
4 3 Pin Plug Point 1 200 200
5 Telephone 1 5 5
6 Fire Alarm 1 1 1
Total 568 watt
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 4 32 128
2 30 60
2 Ceiling Fan Point 8 75 600
3 2 Pin Plug Point 6 2000 12000
4 Smoke Detector 1 2 2
5 Fire Alarm 1 1 1
Total 12791 watt
Table: Male Ward
22. Table: Restroom 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 8 8
2 Exhaust Fan Point 1 45 45
Total 53 watt
Table: Female Ward
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 9 25 225
2 Ceiling Fan Point 12 75 900
3 3 Pin Plug Point 6 2000 12000
4 Smoke Detector 1 2 2
5 Fire Alarm 1 1 1
Total 13128 watt
Table: Restroom 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Emergency Room
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 4 30 120
2 Ceiling Fan Point 8 75 600
3 2 Pin Plug Point 6 2000 12000
4 Smoke Detector 1 2 2
5 Fire Alarm 1 1 1
Total 12723 watt
23. Table: Operating Room
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 2 40 80
2 38 76
2 Spotlight Outlet Point 1 35 35
3 3 Pin Plug Point 1 2000 2000
4 2 Pin Plug Point 1 200 200
1 750 750
1 350 350
Total 3491 watt
Table: Laboratory
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 4 25 100
2 3 Pin Plug Point 1 20 20
1 100 100
1 110 110
1 1500 1500
1 400 400
2 1000 2000
3 Exhaust Fan Point 1 45 45
Total 4275 watt
24. Table: Radiology Room
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 4 10 40
2 3 Pin Plug Point 1 4000 4000
1 40 40
1 100 100
1 2000 2000
3 Exhaust Fan Point 1 45 45
Total 6225 watt
Table: Doctor's Room 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 10 10
1 12 12
2 Ceiling Fan Point 1 75 75
3 3 Pin Plug Point 1 1000 1000
Total 1097 watt
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 10 10
1 12 12
2 Ceiling Fan Point 1 75 75
3 3 Pin Plug Point 1 1000 1000
Total 1097 watt
Table: Doctor's Room 2
25. Table: Nurse's Station
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 15 15
1 18 18
2 Ceiling Fan Point 2 75 150
3 2 Pin Plug Point 1 2000 2000
Total 2183 watt
Table: VIP Cabin 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 Tube Light Point 1 18 18
1 20 20
2 Ceiling Fan Point 1 75 75
3 3 Pin Plug Point 1 1500 1500
4 2 Pin Plug Point 1 2000 2000
Total 3613 watt
Table: Attached Restroom 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 8 8
2 Exhaust Fan Point 1 45 45
Total 53 watt
Table: Balcony 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 5 5
Total 5 watt
26. Table: VIP Cabin 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 Tube Light Point 1 18 18
1 20 20
2 Ceiling Fan Point 1 75 75
3 3 Pin Plug Point 1 1500 1500
4 2 Pin Plug Point 1 2000 2000
Total 3613 watt
Table: Attached Restroom 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Balcony 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 6 5
Total 5 watt
Table: VIP Cabin 3
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 Tube Light Point 1 18 18
1 20 20
2 Ceiling Fan Point 1 75 75
3 3 Pin Plug Point 1 1500 1500
4 2 Pin Plug Point 1 2000 2000
Total 3613 watt
27. Table: Attached Restroom 3
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Male Restroom 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Male Restroom 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Male Restroom 3
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 8 8
2 Exhaust Fan Point 1 45 45
Total 53 watt
Table: Female Restroom 1
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
28. Table: Female Restroom 2
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 7 7
2 Exhaust Fan Point 1 45 45
Total 52 watt
Table: Female Restroom 3
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 1 8 8
2 Exhaust Fan Point 1 45 45
Total 53 watt
Table: Corridor
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 2 13 26
4 12 48
10 10 100
1 8 8
3 7 21
1 6 6
2 Smoke Detector 5 2 10
3 Fire Alarm 4 1 4
Total 168 watt
Table: Staircase
SI NO SPECIFICATION NUMBER OF POINT WATTAGE PER UNIT WATTAGE
1 LED Light Point 2 15 30
Total 30 watt
29.
30.
31. Conduit & Wire Calculation
Conduit Length Calculation:
For Main Circuit:
From Main D.B to Staircase 15.375’
From Main D.B to Corridor 8’
From Main D.B to Emergency 9.416’
From Main D.B to Restroom 14.58’
From Main D.B to Doctor’s Room (1) 5’
From Main D.B to Doctor’s Room (2) 5.167’
From Main D.B to Nurse Room 32.5’
From Main D.B to Male Ward 12.916’
From Main D.B to Female Ward 2.833’
From Main D.B to VIP Cabin (1) 11.25’
From Main D.B to VIP Cabin (2) 15.67’
From Main D.B to VIP Cabin (3) 10.4167’
From Main D.B to Operation Theater 43.83’
From Main D.B to Laboratory 3.75’
From Main D.B to Radiology 7.75’
32. For Sub Circuit:
From Sub D.B to Staircase 36’
From Sub D.B to Rest Room 130.7291’
From Sub D.B to Waiting Room 110.4167’
From Sub D.B to Operation Theatre 102.17’
From Sub D.B to Doctor’s Room (1) 46.875’
From Sub D.B to Doctor’s Room (2) 46.875’
From Sub D.B to Nurse Room 53’
From Sub D.B to Male Ward (Main room) 238.9167’
From Sub D.B to Male Ward (Rest room) 11.75’
From Sub D.B to VIP Cabin (1) (Main room) 51.75’
From Sub D.B to VIP Cabin (1) (Rest room) 15’
From Sub D.B to VIP Cabin (1) (Balcony) 9.5’
From Sub D.B to VIP Cabin (2) (Main room) 41.4167’
From Sub D.B to VIP Cabin (2) (Rest room) 13.5’
From Sub D.B to VIP Cabin (2) (Balcony) 8.75’
From Sub D.B to Laboratory 76.33’
From Sub D.B to Radiology 85’
33. Wire Length Calculation:
For Main Circuit:
From Main D.B to Staircase 20.375’x2
From Main D.B to Corridor 37.25’ x2
From Main D.B to Emergency 29.2083’ x2
From Main D.B to Waiting Room 18.0208’ x2
From Main D.B to Doctor’s Room (1) 17.145’ x2
From Main D.B to Doctor’s Room (2) 22.3125’ x2
From Main D.B to Nurse Room 46.54’ x2
From Main D.B to Male Ward 42.5625’ x2
From Main D.B to Female Ward 51.541’ x2
From Main D.B to VIP Cabin (1) 44.208’ x2
From Main D.B to VIP Cabin (2) 61.6875’ x2
From Main D.B to VIP Cabin (3) 71.04166’ x2
From Main D.B to Operation Theater 85.895’ x2
From Main D.B to Laboratory 83.4166’ x2
From Main D.B to Radiology 90.6875’ x2
34. For Sub Circuit:
From Sub D.B to Staircase 33.6667’
From Sub D.B to Rest Room 396.6664’
From Sub D.B to Reception & Waiting Area 280.3333’
From Sub D.B to Operation Theatre 309.4169’
From Sub D.B to Doctor’s Room (1) 83’
From Sub D.B to Doctor’s Room (2) 83’
From Sub D.B to Nurse Room 70.9999’
From Sub D.B to Male Ward (Main room) 558.0832’
From Sub D.B to Male Ward (Rest room) 65’
From Sub D.B to Female Ward (Main room) 912.2498’
From Sub D.B to Female Ward (Rest room) 73’
From Sub D.B to VIP Cabin (1) (Main room) 68.3334’
From Sub D.B to VIP Cabin (1) (Rest room) 69.5’
From Sub D.B to VIP Cabin (1) (Balcony) 24.75’
From Sub D.B to VIP Cabin (2) (Main room) 84.5’
From Sub D.B to VIP Cabin (2) (Rest room) 51.25’
From Sub D.B to VIP Cabin (2) (Balcony) 13.25’
From Sub D.B to Laboratory 263.4996’
From Sub D.B to Radiology 175.5’
From Sub D.B to Corridor 658.4999’
From Sub D.B to Emergency Room 456.667’
36. Size of Conductor 2 Cable d.c. or Single-phase
a.c.
3 or 4 cables of balanced 3-
phase
4 Cables d.c.
Normal
Area sq.
mm.
Number
and
Diameter
of Wire in
mm.
Current
Rating in
amperes
Approx. Length
of Run for 1 volt-
drop in metres
Current
Rating in
amperes
Approx. Length
of Run for 1
volt-drop in
metres
Current
Rating in
amperes
Approx.
Length
of Run for 1
volt-drop in
metres
1.5
2.5
4.0
6.0
10.0
16.0
25.0
35.0
50.0
1/1.40
1/1.80
1/2.24
1/2.80
1/3.55
7/1.70
7/2.24
7/2.50
7/3.0
10
15
20
27
34
43
59
69
91
2.3
2.5
2.9
3.4
4.3
5.4
6.8
7.2
7.9
9
12
17
24
31
38
54
62
82
2.9
3.6
3.9
4.3
5.4
7.0
8.5
9.3
10.1
9
11
15
21
27
35
48
55
69
2.5
3.4
4.1
4.3
5.4
6.8
8.5
9.0
10.0
Table: Current Rating of Copper Cables
37. Inside Trunking & Conduit Conductor
Single phase one cable AC &
DC
Three phase Three or four Core
Cable
Number &
Diameter of wires
(inch)
Cross section
Area(inch2)
Current rating Volt drop/100
ft
Current
Rating
Volt. Drop/100
feet
Amp volt Amp volt
11 14 9 9.8 1/.044 0.0015
13 12 11 9.1 3/.029 0.002
16 11 14 7.7 3/.036 0.003
21 8.4 18 6.4 7/.029 0.0045
28 7 23 5.3 7/.036 0.007
34 5.5 28 4.1 7/.044 0.01
43 5.4 36 4 7/.052 0.0145
56 4.8 48 3.5 7/.064 0.0225
66 4.3 56 3.2 19/.044 0.03
77 3.6 65 2.7 19/.052 0.04
105 3.4 88 2.5 19/.064 0.06
Table: Cable Size, Current Rating with Voltage Drop (English/British System)
38. Wire Size Calculation
Reception and Waiting Area
Total Load Of Room=568 W
Distance from MDB to SDB=18.0208’
After adding 15% Additional Load=653.2
Current=2.969 A
Cable Size=(1/1.40)10 A
Current Capacity at 45°C=0.91×10=9.1
Voltage drop for 100 ft= 14 V
Length of cable= 18.0208
So voltage drop=0.749 V
Therefore cable size should be=(1/1.40)10 A
39. LED Light-1
Total load of light=15 W
Distance from sub distribution board=9.25’
Total load after adding 15% additional load=17.25 W
Total Current=0.078 A
Cable size=(1/1.40)10 A
Current capacity at 45°c=0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=9.25’
So voltage drop=(14×9.25×0.078)/(100×9.1)=0.0111 V×
Therefore cable size should be=(1/1.4) 10 A
LED Light-2
Total load of light=15 W
Distance from sub distribution board=13’
Total load after adding 15% additional load=17.25 W
Total Current=0.078 A
Cable size=(1/1.40)10 A
Current capacity at 45°c=0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=13
So voltage drop=(14×13×0.078)/(100×9.1)=0.0156 V
Therefore cable size should be=(1/1.4) 10 A
LED Light-3
Total load of light=15 W
Distance from sub distribution board=16.75’
Total load after adding 15% additional load=17.25 W
Total Current=0.078 A
Cable size=(1/1.40)10 A
Current capacity at 45°c=0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=16.75
So voltage drop=(14×16.75×0.078)/(100×9.1)=0.0201 V
Therefore cable size should be=(1/1.4) 10 A
LED Light-4
Total load of light=15 W
Distance from sub distribution board=19.5833’
Total load after adding 15% additional load=17.25 W
Total Current=0.078 A
Cable size=(1/1.40)10 A
Current capacity at 45°c=0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=19.5833’
So voltage drop=(14×19.5833×0.078)/(100×9.1)=0.02349 V
Therefore cable size should be=(1/1.4) 10 A
40. LED Light-5
Total load of light=10 W
Distance from sub distribution board=13’
Total load after adding 15% additional load=11.5 W
Total Current=0.0522 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=13’
So voltage drop=(14×13×0.0522)/(100×9.1)=0.01044 V
Therefore cable size should be (1/1.40) 10 A
LED Light-6
Total load of light=10 W
Distance from sub distribution board=30.33’
Total load after adding 15% additional load=11.5 W
Total Current=0.0522 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Here,length of the cable=30.33’
So voltage drop=(14×30.33×0.0522)/(100×9.1)=0.024 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-1
Total load of fan=75 W
Distance from sub distribution board=35.25’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =35.35’
So voltage drop=(14×35.25×0.392)/(100×9.1)=0.213 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-2
Total load of fan=75 W
Distance from sub distribution board=38.83’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is = 38.83’
So voltage drop=(14×38.83×0.392)/(100×9.1)=0.23 V
Therefore cable size should be (1/1.40) 10 A
41. Ceiling Fan-3
Total load of fan=75 W
Distance from sub distribution board=30.33’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =30.33’
So voltage drop=(14×30.33×0.0522)/(100×9.1)= 0.18V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-4
Total load of fan=75 W
Distance from sub distribution board=28.9167’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =28.9167’
So voltage drop=(14×28.9167×0.0522)/(100×9.1)=0.171 V
Therefore cable size should be (1/1.40) 10 A
3 Pin Plug Point
Total load of 3 pin plug=200 W
Distance from sub distribution board=35’
Total load after adding 15% additional load=230 W
Total Current=1.045 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =35’
So voltage drop=(14×35×01.045)/(100×9.1)=0.5626 V
Therefore cable size should be (1/1.40) 10 A
2 Pin Plug Point
Total load of 2 pin plug=42 W
Distance from sub distribution board=20.5’
Total load after adding 15% additional load=48.3 W
Total Current=0.2195 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =20.5’
So voltage drop=(14×20.5×0.2195)/(100×9.1)=0.069 V
Therefore cable size should be (1/1.40) 10 A
42. Telephone
Total load of telephone=5 W
Distance from sub distribution board=45’
Total load after adding 15% additional load=6.25 W
Total Current=0.028 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =45’
So voltage drop=(14×45×0.028)/(100×9.1)=0.01938 V
Therefore cable size should be (1/1.40) 10 A
Fire Alarm
Total load of fire alarm=1 W
Distance from sub distribution board=13’
Total load after adding 15% additional load=1.15 W
Total Current=0.0.00522727 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =20.5’
So voltage drop=(14×13×0.00522727)/(100×9.1)=0.001045 V
Therefore cable size should be (1/1.40) 10 A
Male Ward
Total Load Of Room=12791
Distance from MDB to SDB=42.5625’
After adding 15% Additional Load=14709.65
Current=66.86
Cable Size=(7/2.5)69 A
Current Capacity at 45°C=0.91×69=62.69
Voltage drop for 100 ft= 4.1 V
Length of cable= 42.5625
So voltage drop=1.6909
Therefore cable size should be=(7/2.5)69 A
43. LED Light-1
Total load of light=32 W
Distance from sub distribution board=12’
Total load after adding 15% additional load=36.8 W
Total Current=0.1672 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =12’
So voltage drop=0.0308 V
Therefore cable size should be (1/1.40) 10 A
LED Light-2
Total load of light=32 W
Distance from sub distribution board=22.5’
Total load after adding 15% additional load=36.8 W
Total Current=0.1672 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =22.5’
So voltage drop=0.0578 V
Therefore cable size should be (1/1.40) 10 A
LED Light-3
Total load of light=32 W
Distance from sub distribution board=20.33’
Total load after adding 15% additional load=36.8 W
Total Current=0.1672 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =20.33’
So voltage drop=0.0522 V
Therefore cable size should be (1/1.40) 10 A
LED Light-4
Total load of light=32 W
Distance from sub distribution board=30.83’
Total load after adding 15% additional load=36.8 W
Total Current=0.1672 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =30.83’
So voltage drop=0.0793 V
Therefore cable size should be (1/1.40) 10 A
44. LED Light-5
Total load of light=30 W
Distance from sub distribution board=30’
Total load after adding 15% additional load=34.5 W
Total Current=0.1568 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =30’
So voltage drop=0.072 V
Therefore cable size should be (1/1.40) 10 A
LED Light-6
Total load of light=30 W
Distance from sub distribution board=40.5’
Total load after adding 15% additional load=34.5 W
Total Current=0.1568 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =40.5’
So voltage drop=0.0976 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-1
Total load of fan=75 W
Distance from sub distribution board=12’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =12’
So voltage drop=0.072 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-2
Total load of fan=75 W
Distance from sub distribution board=18.5’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =18.5’
So voltage drop=0.111 V
Therefore cable size should be (1/1.40) 10
45. Ceiling Fan-3
Total load of fan=75 W
Distance from sub distribution board=18’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =18’
So voltage drop=0.11 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-4
Total load of fan=75 W
Distance from sub distribution board=24.5’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =24.5’
So voltage drop= 0.1477 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-5
Total load of fan=75 W
Distance from sub distribution board=27.5’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =27.5’
So voltage drop=0.1658 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-6
Total load of fan=75 W
Distance from sub distribution board=34’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =34’
So voltage drop=0.205 V
Therefore cable size should be (1/1.40) 10 A
46. Ceiling Fan-7
Total load of fan=75 W
Distance from sub distribution board=36.33’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =36.33’
So voltage drop=0.219 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-8
Total load of fan=75 W
Distance from sub distribution board=42.83’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =42.83’
So voltage drop=0.2583 V
Therefore cable size should be (1/1.40) 10 A
2 Pin Plug Point-1
Total load of 2 pin plug=2000 W
Distance from sub distribution board=4.5’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =4.5’
So voltage drop=0.7235 V
Therefore cable size should be (1/1.80) 15 A
2 Pin Plug Point-2
Total load of 2 pin plug=2000 W
Distance from sub distribution board=14.5’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =14.5’
So voltage drop=2.33 V
Therefore cable size should be (1/1.40) 10 A
47. 2 pin plug point-3
Total load of 2 pin plug=2000 W
Distance from sub distribution board=24.5’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =24.5’
So voltage drop= 2.63V
Therefore cable size should be (1/1.80) 15 A
2 pin plug point-4
Total load of 2 pin plug=2000 W
Distance from sub distribution board=33.75’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =33.75’
So voltage drop=3.617 V
Therefore cable size should be (1/1.80) 15 A
2 pin plug point-5
Total load of 2 pin plug=2000 W
Distance from sub distribution board=48.25’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =48.25’
So voltage drop=5.17 V
Therefore cable size should be (1/1.80) 15 A
2 pin plug point-6
Total load of 2 pin plug=2000 W
Distance from sub distribution board=62.75’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =62.75’
So voltage drop=6.73 V
Therefore cable size should be (1/1.80) 15 A
48. Restroom 1
Total Load Of Room=53
Total Load of Room = 53 W
After adding 15% Additional Load=60.95
Total Current=0.277
Cable Size= (1/1.40)10 A
Current Capacity at 45°C=0.91×10=9.1
Voltage drop for 100 ft= 14 V
Length of cable= 86’
So voltage drop=0.333508 V
Therefore cable size should be=(1/1.40)10 A
LED Light
Total load of light=8 W
Distance from sub distribution board=40.5’
Total load after adding 15% additional load=9.2 W
Total Current=0.0418 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =40.5’
So voltage drop=0.026 V
Therefore cable size should be (1/1.40) 10 A
Exhaust Fan
Total load of fan=45 W
Distance from sub distribution board=45.5’
Total load after adding 15% additional load=51.75 W
Total Current=0.235 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =45.5’
So voltage drop=0.1645 V
Therefore cable size should be (1/1.40) 10 A
49. Female Ward
Total Load Of Room = 13128 W
After adding 15% Additional Load = 15097.2W
Total Current = 68.6236 A
Cable size = (7/3.0) 91 A
Current capacity at 45⁰ C = 0.91×91 = 82.81 A
Voltage drop for 100 ft for this cable = 3.5 V
Here, the length of cable = 51.541’
So, voltage drop= [(3.5×51.541×68.6236)÷(100×91)] = 1.36 V < 5.5 V
Therefore, size of the cable should be (7/3.30) 91 A
LED Light 1
Total load of light = 25 W
Distance from sub distribution board = 11.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 11.75’
So, voltage drop= [(14×11.75×0.1306)÷(100×10)]= 0.021 V < 5.5 V
Therefore, cable size should be= (1/1.40) 10 A
LED Light 2
Total load of light = 25 W
Distance from sub distribution board = 17.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 17.75’
So,Voltage drop = {14×17.75× 0.1306/100×10}=0.032 V< 5.5 V
Therefore, cable size should be (1/1.40) 10 A
50. LED Light 3
Total load of light = 25 W
Distance from sub distribution board = 23.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 23.75’
So,Voltage drop = {14×23.75×0.1306}/{100×10} = 0.0434 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 4
Total load of light = 25 W
Distance from sub distribution board = 21.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10=9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 21.75’
So,voltage drop = {14×21.75×0.1306}/{100×9.1} =.0447 V < 5.5 V
Therefore, cable size should be = (1/1.40) 10 A
LED Light 5
Total load of light = 25 W
Distance from sub distribution board = 27.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10=9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable =27.75’
So,voltage drop = {14×27.75×0.1306}/{100×9.1} = 0.0557 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 6
Total load of light = 25 W
Distance from sub distribution board = 33.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 33.75’
So,voltage drop = {14×33.75×0.1306}/{100×9.1} = 0.0678 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
51. LED Light 7
Total load of light = 25 W
Distance from sub distribution board = 31.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31.75’
So,voltage drop = {14×31.75×0.1306}/{100×9.1} = 0.0638 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 8
Total load of light = 25 W
Distance from sub distribution board = 37.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 37.75’
So,voltage drop = {14×37.75×0.1306}/{100×9.1} = 0.0758 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 9
Total load of light = 25 W
Distance from sub distribution board = 43.75’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.1306 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 43.75’
So,voltage drop = {14×37.75×0.1306}/{100×9.1} = 0.0879 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 1
Total load of fan = 75 W
Distance from sub distribution board = 12.5‘
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 12.5’
So,voltage drop = {14×12.5×0.392}/{100×9.1} = 0.0753 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
52. Ceiling Fan 2
Total load of fan = 75 W
Distance from sub distribution board = 18.5’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 18.5’
So,voltage drop = {14×18.5×0.392}/{100×9.1} = 0.1116 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 3
Total load of fan = 75 W
Distance from sub distribution board = 24.5’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 24.5’
So,voltage drop = {14×24.5×0.392}/{100×9.1} = 0.1477 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 4
Total load of fan = 75 W
Distance from sub distribution board = 15.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 15.8333’
So,voltage drop = {14×15.8333×0.392}/{100×9.1} = 0.0954 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 5
Total load of fan = 75 W
Distance from sub distribution board = 21.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 21.8333’
So,voltage drop = {14×21.8333×0.392}/{100×9.1} = 0.1316 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
53. Ceiling Fan 6
Total load of fan = 75 W
Distance from sub distribution board = 27.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 27.8333’
So,voltage drop = {14×27.8333×0.392}/{100×9.1} = 0.1678 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 7
Total load of fan = 75 W
Distance from sub distribution board = 25.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 25.8333’
So,voltage drop = {14×25.8333×0.392}/{100×9.1} = 0.1558 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 8
Total load of fan = 75 W
Distance from sub distribution board = 31.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31.8333’
So,voltage drop = {14×31.8333×0.392}/{100×9.1} = 0.1919 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 9
Total load of fan = 75 W
Distance from sub distribution board = 37.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 37.8333’
So,voltage drop = {14×37.8333×0.392}/{100×9.1} = 0.2282 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
54. Ceiling Fan 10
Total load of fan = 75 W
Distance from sub distribution board = 35.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 35.8333’
So,voltage drop = {14×35.8333×0.392}/{100×9.1} = 0.2161 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 11
Total load of fan = 75 W
Distance from sub distribution board = 41.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 41.8333’
So,voltage drop = {14×18.5×0.392}/{100×9.1} = 0.2522 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan 12
Total load of fan = 75 W
Distance from sub distribution board = 47.8333’
Total load of fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 47.8333’
So,voltage drop = {14×47.8333’×0.392}/{100×9.1} = 0.2884 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point 1
Total load of plug point = 2000 W
Distance from sub distribution board = 9.5’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 9.5’
So,voltage drop = {11×9.5×10.4545}/{100×9.1} = 1.2 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
55. 3 Pin Plug Point 2
Total load of plug point = 2000 W
Distance from sub distribution board = 18.5’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 18.5’
So,voltage drop = {11×18.5×10.4545}/{100×9.1} = 2.33 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
3 Pin Plug Point 3
Total load of plug point = 2000 W
Distance from sub distribution board = 27.5’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 27.5’
So,voltage drop = {11×27.5×10.4545}/{100×9.1} = 3.47 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
3 Pin Plug Point 4
Total load of plug point = 2000 W
Distance from sub distribution board = 36.6667’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 36.6667’
So,voltage drop = {11×36.6667×10.4545}/{100×9.1} = 4.64 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
3 Pin Plug Point 5
Total load of plug point = 2000 W
Distance from sub distribution board = 20.5’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 20.5’
So,voltage drop = {11×20.5×10.4545}/{100×9.1} = 2.59 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
56. 3 Pin Plug Point 6
Total load of plug point = 2000 W
Distance from sub distribution board = 36.75’
Total load of plug point after adding 15% additional load = 2300 W
Total Current = 10.4545 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 36.75’
So,voltage drop = {11×36.75×10.4545}/{100×9.1} = 4.64 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
Smoke Detector
Total load of Smoke Detector = 2 W
Distance from sub distribution board = 14.67’
Total load of Smoke Detector after adding 15% additional load = 2.3 W
Total Current = 0.01 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 14.67’
So,voltage drop = {14×14.67×0.01}/{100×9.1} = 0.0023 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Fire Alarm
Total load of Fire Alarm = 1 W
Distance from sub distribution board = 10’
Total load of Fire Alarm after adding 15% additional load = 1.15 W
Total Current = 0.005 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 10’
So,voltage drop = {14×10×0.01}/{100×9.1} = 0.0015 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
57. Restroom 2
Total Load Of Room = 52 W
After adding 15% Additional Load = 59.8 W
Total Current = 0.2718 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 73’
So, voltage drop= [(14×73×0.2718)÷(100×9.1)] = 0.31 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 7 W
Distance from sub distribution board = 37’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 37’
So, voltage drop= [(14×37×0.2718)÷(100×9.1)] = 0.1547 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Exhaust Fan
Total load of Exhaust Fan = 45 W
Distance from sub distribution board = 36’
Total load of LED Light after adding 15% additional load = 51.75 W
Total Current = 0.2352 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 36’
So, voltage drop= [(14×36×0.2718)÷(100×9.1)] = 0.151 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
58. Emergency Room
Total Load Of Room=12723 W
After adding 15% Additional Load=14631.45 W
Current=66.506 A
Cable size = (7/2.5)69 A
At 45⁰ c Current capacity = 0.91×69 = 62.79
Voltage Drop for 100 ft = 4.1 V
The length of cable = 29.2083’
So,voltage drop = (4.1×29.2083×66.506)/(100×69) = 1.15 V
Therefore,the cable should be (7/2.5)69 A
LED Light-1
Total load of light = 30 W
Distance from sub distribution board = 12’
Total load of light after adding 15% additional load = 34.5 W
Total Current = (34.5/220) = 0.1568 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 12’
So, voltage drop = [(14×12×0.1568)/(100×10)] = 0.0263424 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-2
Total load of light = 30 W
Distance from sub distribution board = 17’
Total load of light after adding 15% additional load = 34.5 W
Total Current = (34.5/220) = 0.1568 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 17’
So, voltage drop = [(14×17×0.1568)/(100×10)] = 0.0373184 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
59. LED Light-3
Total load of light = 30 W
Distance from sub distribution board = 26’
Total load of light after adding 15% additional load = 34.5 W
Total Current = (34.5/220) = 0.1568 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 26’
So, voltage drop = [(14×26×0.1568)/(100×10)] = 0.057 A < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-4
Total load of light = 30 W
Distance from sub distribution board = 31’
Total load of light after adding 15% additional load = 34.5 W
Total Current = (34.5/220) = 0.1568 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31’
So, voltage drop = [(14×31×0.1568)/(100×10)] = 0.068 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-1
Total load of fan = 75 W
Distance from sub distribution board = 10’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 10’
So, voltage drop = [(14×10×0.39)/(100×10)] = 0.0546 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-2
Total load of fan = 75 W
Distance from sub distribution board = 15’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 15’
So, voltage drop = [(14×15×0.39)/(100×10)] = 0.0819 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
60. Ceiling Fan-3
Total load of fan = 75 W
Distance from sub distribution board = 17.25’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 17.25’
So, voltage drop = [(14×17.25×0.39)/(100×10)] = 0.094 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-4
Total load of fan = 75 W
Distance from sub distribution board = 22.25’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 22.25’
So, voltage drop = [(14×22.25×0.39)/(100×10)] = 0.1214 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-5
Total load of fan = 75 W
Distance from sub distribution board = 24’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 24’
So, voltage drop = [(14×24×0.39)/(100×10)] = 0.13 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-6
Total load of fan = 75 W
Distance from sub distribution board = 31’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31’
So, voltage drop = [(14×31×0.39)/(100×10)] = 0.16926 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
61. Ceiling Fan-7
Total load of fan = 75 W
Distance from sub distribution board = 31.5’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31.5’
So, voltage drop = [(14×31.5×0.39)/(100×10)] = 0.17199 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Ceiling Fan-8
Total load of fan = 75 W
Distance from sub distribution board = 36.5’
Total load of fan after adding 15% additional load = 86.35 W
Total Current = (86.35/220) = 0.39 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 36.5’
So, voltage drop = [(14×36.5×0.39)/(100×10)] = 0.19926 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
2 Pin Plug Point-1
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 15.25’
Total load of 2 Pin Plug Point after adding 15% additional load =
(2000+2000×15%) = 2300 W
Total Current = (2300/220) = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 15.25’
So, voltage drop = [(11×15.25×10.45)/(100×15)] = 1.168 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point-2
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 4.25’
Total load of 2 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 4.25’
So, voltage drop = [(11×4.25×10.45)/(100×15)] = 0.325 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
62. 2 Pin Plug Point-3
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 33.75’
Total load of 2 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 33.75’
So, voltage drop = [(11×33.75×10.45)/(100×15)] = 2.586 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point-4
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 52.25’
Total load of 2 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 52.25’
So, voltage drop = [(11×52.25×10.45)/(100×15)] = 4.004 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point-5
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 63.25’
Total load of 2 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 63.25’
So, voltage drop = [(11×63.25×10.45)/(100×15)] = 4.847 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point-6
Total load of 2 Pin Plug Point = 2000 W
Distance from sub distribution board = 14.4167’
Total load of 2 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 14.4167’
So, voltage drop =[(11×14.4167×10.45)/(100×15)] = 1.1 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
63. Smoke Detector
Total load of smoke ditector = 2 W
Distance from sub distribution board = 10’
Total load of smoke ditector after adding 15% additional load = (2+2×15%) =
2.3 W
Total Current = (2.3/220) = 0.1045 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 10’
So, voltage drop = [(14×10×0.1045)/(100×10)] = 0.01463 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Fire Alarm
Total load of fire alarm = 1 W
Distance from sub distribution board = 9.4167’
Total load of fire alarm after adding 15% additional load = (1+1×15%) = 1.15
W
Total Current = (1.15/220) = .005227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 9.4167’
So, voltage drop = [(14×9.4167×0.005227)/(100×10)] = .000689 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
64. Operating Room
Total Load Of Room=5491
15% Additional Load=6314.65
Current=28.7
Cable size = (1/3.55) 34 A
At 45⁰ c Current capacity = 0.91×34 = 30.94
Voltage Drop for 100 ft = 5.7 V
The length of cable = 85.895’
So,voltage drop = (5.7×85.895×28.7)/(100×34) = 4.13 V
Therefore,the cable should be (1/3.55) 34 A
LED Light-1
Total load of light = 40 W
Distance from sub distribution board = 8’
Total load of light after adding 15% additional load = 46 W
Total Current = (46/220) = 0.209 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 8’
So, voltage drop = [(14×8’×0.209)/(100×10)] = 0.023 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-2
Total load of light = 40 W
Distance from sub distribution board = 12.5’
Total load of light after adding 15% additional load = 46 W
Total Current = 0.209 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable =
So, voltage drop = [(14×12.5×0.209)/(100×10)] = 0.036 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
65. LED Light-3
Total load of light = 38 W
Distance from sub distribution board = 42.9167’
Total load of light after adding 15% additional load = 43.7 W
Total Current = 0.1986 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 42.9167’
So, voltage drop = [(14×42.9167×0.1986)/(100×10)] = 0.119 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-4
Total load of light = 38 W
Distance from sub distribution board = 50.9167’
Total load of light after adding 15% additional load = 43.7 W
Total Current = 0.1986 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 50.9167’
So, voltage drop = [(14×50.9167×0.1986)/(100×10)] = 0.14 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Spotlight Outlet Point
Total load of spotlight outlet point = 35 W
Distance from sub distribution board = 32.9167’
Total load of spotlight outlet point after adding 15% additional load =
40.25 W
Total Current = (40.25/220) = 0.18 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 32.9167’
So, voltage drop = [(14×32.9167×0.18)/(100×10)] = 0.08 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point-1
Total load of 3 Pin Plug Point = 2000 W
Distance from sub distribution board = 25.9167’
Total load of 3 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80 )×15 A
Current Capacity at 45⁰ C = .91×15 =13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 25.9167’
So, voltage drop = [(11×25.9167×10.45)/(100×15)] = 1.98 V < 5.5 V
Therefore,cable size should be (1/1.80 )×15 A
66. 3 Pin Plug Point-2
Total load of 3 Pin Plug Point = 200 W
Distance from sub distribution board = 36.9167’
Total load of 3 Pin Plug Point after adding 15% additional load = 230 W
Total Current = 1.045 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 36.9167’
So, voltage drop = [(14×36.9167×1.045)/(100×10)] = 0.54 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
3 Pin Plug Point-3
Total load of 3 Pin Plug Point = 750 W
Distance from sub distribution board = 40.9167’
Total load of 3 Pin Plug Point after adding 15% additional load = 862.5 W
Total Current = 3.92 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 40.9167’
So, voltage drop = [(14×40.9167×3.92)/(100×10)] = 2.24 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
3 Pin Plug Point-4
Total load of 3 Pin Plug Point = 350 W
Distance from sub distribution board = 36.9167’
Total load of 3 Pin Plug Point after adding 15% additional load = 402.5 W
Total Current = 1.8295 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 36.9167’
So, voltage drop = [(14×36.9167×1.8295)/(100×10)] = 0.945 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
67. Laboratory
Total Load Of Room=4230
15% Additional Load=4864.5
Current=22.11
Cable size = (1/2.80) 27 A
At 45⁰ c Current capacity = 0.91×27 = 24.57
Voltage Drop for 100 ft = 7 V
The length of cable = 83.4166’
So,voltage drop = (7×83.4166×22.11)/(100×27) = 4.78 V
Therefore,the cable should be (1/2.80) 27 A
LED Light-1
Total load of light = 25 W
Distance from sub distribution board = 14.8333’
Total load of light after adding 15% additional load=(25+25*15%)=28.75 W
Total Current = (28.75/220) = 0.13068 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 14.8333’
So, voltage drop = [(14×14.8333’×0.13068)/(100×10)] = 0.027 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-2
Total load of light = 25 W
Distance from sub distribution board = 22.3333’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.13068 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 22.3333’
So, voltage drop = [(14×22.3333’×0.13068)/(100×10)] = 0.0408 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
68. LED Light-3
Total load of light = 25 W
Distance from sub distribution board = 22.3333’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.13068 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 22.3333’
So, voltage drop = [(14×22.8333’×0.13068)/(100×10)] = 0.0417 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-4
Total load of light = 25 W
Distance from sub distribution board = 30.3333’
Total load of light after adding 15% additional load = 28.75 W
Total Current = 0.13068 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 30.3333’
So, voltage drop = [(14×30.3333’×0.13068)/(100×10)] = 0.055 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point-1
Total load of 3 Pin Plug Point = 20 W
Distance from sub distribution board = 10.3333’
Total load of 3 Pin Plug Point after adding 15% additional load = 23 W
Total Current = 0.1045 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 10.3333’
So, voltage drop = [(14×10.3333’×0.1045)/(100×10)] = 0.015 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
3 Pin Plug Point-2
Total load of 3 Pin Plug Point = 100 W
Distance from sub distribution board = 14.0833’
Total load of 3 Pin Plug Point after adding 15% additional load = 115 W
Total Current = 0.5227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 14.0833’
So, voltage drop = [(14×14.0833’×0.5227)/(100×10)] = 0.1 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
69. 3 Pin Plug Point-3
Total load of 3 Pin Plug Point = 110 W
Distance from sub distribution board = 17.8333’
Total load of 3 Pin Plug Point after adding 15% additional load= 126.5 W
Total Current = 0.575 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 17.8333’
So, voltage drop = [(14×17.8333’×0.575)/(100×10)] = 0.14 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
3 Pin Plug Point-4
Total load of 3 Pin Plug Point = 1500 W
Distance from sub distribution board = 31.8333’
Total load of 3 Pin Plug Point after adding 15% additional load = 1725 W
Total Current = 7.84 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31.8333’
So, voltage drop = [(14×31.8333’×7.84)/(100×10)] = 3.49 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point-5
Total load of 3 Pin Plug Point = 400 W
Distance from sub distribution board = 20.3333’
Total load of 3 Pin Plug Point after adding 15% additional load = 460 W
Total Current = 2.09 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 20.3333’
So, voltage drop = [(14×20.3333’×2.09)/(100×10)] = 0.59 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point-6
Total load of 3 Pin Plug Point = 2000 W
Distance from sub distribution board = 24.0833’
Total load of 3 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15 = 13.65
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 24.0833’
So, voltage drop = [(11×24.0833×10.45)/(100×10)] = 2.768 V < 5.5 V
Therefore,cable size should be (1/1.80) 15 A
70. Radiology Room
Total Load Of Room=6225 W
15% Additional Load=7158.75 W
Current=32.539 A
Cable size = (1/3.55) 34 A
At 45⁰ c Current capacity = 0.91×34 = 30.94
Voltage Drop for 100 ft = 5.7 V
The length of cable = 90.6875’
So,voltage drop = (5.7×90.6875×32.539)/(100×34) = 4.94 V
Therefore,the cable should be (1/3.55) 34 A
LED Light-1
Total load of light = 10 W
Distance from sub distribution board = 8.5’
Total load of light after adding 15% additional load = 11.5 W
Total Current = 0.05227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 8.5’
So, voltage drop = [(14×8.5×0.05227)/(100×10)] = 0.006 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light-2
Total load of light = 10 W
Distance from sub distribution board = 14.5’
Total load of light after adding 15% additional load = 11.5 W
Total Current = 0.05227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable =14 V
Here, the length of cable = 14.5’
So, voltage drop = [(14×14.5×0.05227)/(100×10)] = 0.01 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
71. LED Light-3
Total load of light = 10 W
Distance from sub distribution board = 16.5’
Total load of light after adding 15% additional load = 11.5 W
Total Current = 0.05227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 16.5’
So, voltage drop = [(14×16.5×0.05227)/(100×10)] = 0.01 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
LED Light-4
Total load of light = 10 W
Distance from sub distribution board = 22.5’
Total load of light after adding 15% additional load = 11.5 W
Total Current = 0.05227 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 22.5’
So, voltage drop = [(14×22.5×0.05227)/(100×10)] = 0.016 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point-1
Total load of 3 Pin Plug Point = 4000 W
Distance from sub distribution board = 7.5’
Total load of 3 Pin Plug Point after adding 15% additional load = 4600 W
Total Current = 20.9 A
Cable size = (1/2.80) 27 A
Current Capacity at 45⁰ C = 0.91×27 = 24.57
Voltage drop for 100 ft for this cable = 7 V
Here, the length of cable = 7.5’
So, voltage drop = [(7×7.5×20.9)/(100×27)] = 0.406 V < 5.5 V
Therefore, cable size should be (1/2.80) 27 A
3 Pin Plug Point-2
Total load of 3 Pin Plug Point = 40 W
Distance from sub distribution board = 14’
Total load of 3 Pin Plug Point after adding 15% additional load = 46 W
Total Current = 0.209 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 14’
So, voltage drop = [(14×14×.209)/(100×10)] = 0.04 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
72. 3 Pin Plug Point-3
Total load of 3 Pin Plug Point = 100 W
Distance from sub distribution board = 28’
Total load of 3 Pin Plug Point after adding 15% additional load = 115 W
Total Current = 0.068 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 28’
So, voltage drop = [(14×28×.068)/(100×10)] = 0.027 V < 5.5 V
Therefore,cable size should be (1/1.40) 10 A
3 Pin Plug Point-4
Total load of 3 Pin Plug Point = 2000 W
Distance from sub distribution board = 41’
Total load of 3 Pin Plug Point after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current Capacity at 45⁰ C = 0.91×15
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 41’
Therefore, cable size should be (1/1.80) 15 A
Exhaust Fan-1
Total load of Exhaust Fan = 45 W
Distance from sub distribution board = 23’
Total load of Exhaust Fan after adding 15% additional load = 51.75 W
Total Current = 0.235 A
Cable size = (1/1.40) 10 A
Current Capacity at 45⁰ C = 0.91×10 = 9.1
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 23’
Therefore, cable size should be (1/1.40) 10 A
73. Doctor's Room 1
Total Load Of Room=1097 W
Distance from M.D.B. to S.D.B.=17.145’
After adding 15% Additional Load=1261.55 W
Current=5.73 A
Cable Size=(1/1.40)10 A
Current Capacity at 45°C=0.91×10=9.1
Voltage drop for 100 ft= 14 V
Length of cable= 17.145
So voltage drop=1.51 V
Therefore cable size should be=(1/1.40)10 A
LED Light-1
Total load of light=10 W
Distance from sub distribution board=10.9167’
Total load after adding 15% additional load=11.5 W
Total Current=0.05227 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =10.9167’
So voltage drop=0.00839 V
Therefore cable size should be (1/1.40) 10 A
LED Light-2
Total load of light=12 W
Distance from sub distribution board=30.16’
Total load after adding 15% additional load=13.8 W
Total Current=0.0627 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =30.16’
So voltage drop=0.029 V
Therefore cable size should be (1/1.40) 10 A
74. Ceiling Fan
Total load of fan=75 W
Distance from sub distribution board=15.58’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =15.58’
So voltage drop=0.094 V
Therefore cable size should be (1/1.40) 10 A
3 pin plug point
Total load of 3 pin plug=1000 W
Distance from sub distribution board=26’
Total load after adding 15% additional load=1150 W
Total Current=5.22 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =26’
So voltage drop=2.088 V
Therefore cable size should be (1/1.40) 10 A
75. Doctor's Room 2
Total Load Of Room=1097 W
Distance from M.D.B. to S.D.B.=17.145’
After adding 15% Additional Load=1261.55 W
Current=5.73 A
Cable Size=(1/1.40)10 A
Current Capacity at 45°C=0.91×10=9.1
Voltage drop for 100 ft= 14 V
Length of cable= 17.145
So voltage drop=1.51 V
Therefore cable size should be=(1/1.40)10 A
LED Light-1
Total load of light=10 W
Distance from sub distribution board=10.9167’
Total load after adding 15% additional load=11.5 W
Total Current=0.05227 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =10.9167’
So voltage drop=0.00839 V
Therefore cable size should be (1/1.40) 10 A
LED Light-2
Total load of light=12 W
Distance from sub distribution board=30.16’
Total load after adding 15% additional load=13.8 W
Total Current=0.0627 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =30.16’
So voltage drop=0.029 V
Therefore cable size should be (1/1.40) 10 A
76. Ceiling Fan
Total load of fan=75 W
Distance from sub distribution board=15.58’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =15.58’
So voltage drop=0.094 V
Therefore cable size should be (1/1.40) 10 A
3 pin plug point
Total load of 3 pin plug=1000 W
Distance from sub distribution board=26’
Total load after adding 15% additional load=1150 W
Total Current=5.22 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =26’
So voltage drop=2.088 V
Therefore cable size should be (1/1.40) 10 A
77. Nurse's Station
Total Load Of Room=2183 W
Distribution from M.D.B. to S.D.B.=46.54’
After adding 15% Additional Load=2510.45
Current=11.41 A
Cable Size=(1/1.80)15 A
Current Capacity at 45°C=0.91×15=13.65
Voltage drop for 100 ft= 14 V
Length of cable= 46.54
So voltage drop=5.44 V< 5.5 V
Therefore cable size should be=(1/1.80)15 A
LED Light-1
Total load of light=15 W
Distance from sub distribution board=13.83’
Total load after adding 15% additional load=17.25 W
Total Current=0.078 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =13.83’
So voltage drop=0.016596 V
Therefore cable size should be (1/1.40) 10 A
LED Light-2
Total load of light=18 W
Distance from sub distribution board=15.83’
Total load after adding 15% additional load=20.7 W
Total Current=0.094 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =15.83’
So voltage drop=0.02289 V
Therefore cable size should be (1/1.40) 10 A
78. Ceiling Fan-1
Total load of fan=75 W
Distance from sub distribution board=10’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.40) 10 A
Current capacity at 45°c= 0.91×10=9.1
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =10’
So voltage drop=0.0603 V
Therefore cable size should be (1/1.40) 10 A
Ceiling Fan-2
Total load of fan=75 W
Distance from sub distribution board=15’
Total load after adding 15% additional load=86.25 W
Total Current=0.392 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =15’
So voltage drop=0.09046 V
Therefore cable size should be (1/1.80) 15 A
2 pin plug point
Total load of 2 pin plug=2000 W
Distance from sub distribution board=16.33’
Total load after adding 15% additional load=2300 W
Total Current=10.45 A
Cable size=(1/1.80) 15 A
Current capacity at 45°c= 0.91×15=13.65
Voltage drop for 100 feet for this cable=14 V
Hence cable size is =16.33’
So voltage drop=1.75 V
Therefore cable size should be (1/1.80) 15 A
79. VIP Cabin 1
Total Load Of Room = 3613 W
After adding 15% Additional Load = 4154.95 W
Total Current = 18.88 A
Cable size = (1/2.60) 27 A
Current capacity at 45⁰ C = 0.91×27 = 24.57 A
Voltage drop for 100 ft for this cable = 8.4 V
Here, the length of cable = 51.541’
So, voltage drop= [(8.4×51.541×18.88)÷(100×24.57)] = 3.32 V < 5.5 V
Therefore, size of the cable should be (1/2.60) 27 A
Tube Light 1
Total load of Tube Light = 18 W
Distance from sub distribution board = 11.92’
Total load of Tube Light after adding 15% additional load = 20.7W
Total Current = 0.0941 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 11.92’
So, voltage drop= [(14×11.92×0.0941)÷(100×9.1)] = 0.0173V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Tube Light 2
Total load of Tube Light = 20 W
Distance from sub distribution board = 16.167’
Total load of Tube Light after adding 15% additional load = 23 W
Total Current = 0.1045 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 16.167’
So, voltage drop= [(14×16.167×0.1045)÷(100×9.1)] = 0.026 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
80. Ceiling Fan
Total load of ceiling fan = 75 W
Distance from sub distribution board = 11.5’
Total load of Ceiling Fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 11.5’
So, voltage drop= [(14×11.5×0.392)÷(100×9.1)] = 0.069 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point
Total load of 3 Pin Plug = 1500 W
Distance from sub distribution board = 10’
Total load of 3 Pin Plug after adding 15% additional load = 1725 W
Total Current = 7.84 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 10’
So, voltage drop= [(11×10×7.84)÷(100×13.65)] = 0.63 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point
Total load of 2 Pin Plug = 2000 W
Distance from sub distribution board = 18.75’
Total load of 2 Pin Plug after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 18.75’
So, voltage drop= [(11×18.75×10.45)÷(100×13.65)] = 1.58 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
81. Attached Restroom 1
Total Load Of Room = 53 W
After adding 15% Additional Load = 60.95 W
Total Current = 0.277 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 69.5’
So, voltage drop= [(14×69.5×0.277)÷(100×9.1)] = 0.296 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 8 W
Distance from sub distribution board = 28.75’
Total load of LED Light after adding 15% additional load = 9.2 W
Total Current = 0.0418 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 28.75’
So, voltage drop= [(14×28.75×0.0418)÷(100×9.1)] = 0.018 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Exhaust Fan
Total load of Exhaust Fan = 45 W
Distance from sub distribution board = 40.75’
Total load of LED Light after adding 15% additional load = 51.75 W
Total Current = 0.2352 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 40.75’
So, voltage drop= [(14×40.75×0.2352)÷(100×9.1)] = 0.147 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
82. Balcony 1
Total Load Of Room = 5 W
After adding 15% Additional Load = 5.75 W
Total Current = 0.0261 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 24.75’
So, voltage drop= [(14×24.75×0.0261)÷(100×9.1)] = 0.009 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 5 W
Distance from sub distribution board = 24.75’
Total load of LED Light after adding 15% additional load = 5.75 W
Total Current = 0.0261 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 24.75’
So, voltage drop= [(14×24.75×0.0261)÷(100×9.1)] = 0.009 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
83. VIP Cabin 2
Total Load Of Room = 3613 W
After adding 15% Additional Load = 4154.95 W
Total Current = 18.88 A
Cable size = (1/2.60) 27 A
Current capacity at 45⁰ C = 0.91×27 = 24.57 A
Voltage drop for 100 ft for this cable = 8.4 V
Here, the length of cable = 61.69’
So, voltage drop= [(8.4×61.69×18.88)÷(100×24.57)] = 3.98 V < 5.5 V
Therefore, size of the cable should be (1/2.60) 27 A
Tube Light 1
Distance from sub distribution board = 12.5’
Total load of Tube Light after adding 15% additional load = 20.7W
Total Current = 0.0941 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 12. Total load of Tube Light = 18 W
5’
So, voltage drop= [(14×12.5×0.0941)÷(100×9.1)] = 0.018 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Tube Light 2
Total load of Tube Light = 20 W
Distance from sub distribution board = 31’
Total load of Tube Light after adding 15% additional load = 23 W
Total Current = 0.1045 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31’
So, voltage drop= [(14×31×0.1045)÷(100×9.1)] = 0.045 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
84. Ceiling Fan
Total load of ceiling fan = 75 W
Distance from sub distribution board = 20.5’
Total load of Ceiling Fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 20.5’
So, voltage drop= [(14×20.5×0.392)÷(100×9.1)] = 0.0296 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point
Total load of 3 Pin Plug = 1500 W
Distance from sub distribution board = 19’
Total load of 3 Pin Plug after adding 15% additional load = 1725 W
Total Current = 7.84 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 19’
So, voltage drop= [(11×19×7.84)÷(100×13.65)] = 1.2 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point
Total load of 2 Pin Plug = 2000 W
Distance from sub distribution board = 1.5’
Total load of 2 Pin Plug after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 1.5’
So, voltage drop= [(11×1.5×10.45)÷(100×13.65)] = 0.126 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
85. Attached Restroom 2
Total Load Of Room = 52 W
After adding 15% Additional Load = 59.8 W
Total Current = 0.2718 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 51.25’
So, voltage drop= [(14×51.25×0.2718)÷(100×9.1)] = 0.214 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 7 W
Distance from sub distribution board = 22.75’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 22.75’
So, voltage drop= [(14×22.75×0.365)÷(100×9.1)] = 0.013 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Exhaust Fan
Total load of Exhaust Fan = 45 W
Distance from sub distribution board = 28.5’
Total load of LED Light after adding 15% additional load = 51.75 W
Total Current = 0.2352 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 28.5’
So, voltage drop= [(14×28.5×0.2352)÷(100×9.1)] = 0.103 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
86. Balcony 2
Total Load Of Room = 5 W
After adding 15% Additional Load = 5.75 W
Total Current = 0.0261 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 13.25’
So, voltage drop= [(14×13.25×0.0261)÷(100×9.1)] = 0.005 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 5 W
Distance from sub distribution board = 13.25’
Total load of LED Light after adding 15% additional load = 5.75 W
Total Current = 0.0261 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 13.25’
So, voltage drop= [(14×13.25×0.0261)÷(100×9.1)] = 0.005 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
87. VIP Cabin 3
Total Load Of Room = 3613 W
After adding 15% Additional Load = 4154.95 W
Total Current = 18.88 A
Cable size = (1/2.60) 27 A
Current capacity at 45⁰ C = 0.91×27 = 24.57 A
Voltage drop for 100 ft for this cable = 8.4 V
Here, the length of cable = 71.04’
So, voltage drop= [(8.4×71.04×18.88)÷(100×24.57)] = 4.41 V < 5.5 V
Therefore, size of the cable should be (1/2.60) 27 A
Tube Light 1
Total load of Tube Light = 18 W
Distance from sub distribution board = 13’
Total load of Tube Light after adding 15% additional load = 20.7W
Total Current = 0.0941 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 13’
So, voltage drop= [(14×13×0.0941)÷(100×9.1)] = 0.019 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Tube Light 2
Total load of Tube Light = 20 W
Distance from sub distribution board = 13.75’
Total load of Tube Light after adding 15% additional load = 23 W
Total Current = 0.1045 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 13.75’
So, voltage drop= [(14×13.75×0.1045)÷(100×9.1)] = 0.022 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
88. Ceiling Fan
Total load of ceiling fan = 75 W
Distance from sub distribution board = 21’
Total load of Ceiling Fan after adding 15% additional load = 86.25 W
Total Current = 0.392 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 21’
So, voltage drop= [(14×21×0.392)÷(100×9.1)] = 0.127 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
3 Pin Plug Point
Total load of 3 Pin Plug = 1500 W
Distance from sub distribution board = 19.5’
Total load of 3 Pin Plug after adding 15% additional load = 1725 W
Total Current = 7.84 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 19.5’
So, voltage drop= [(11×19.5×7.84)÷(100×13.65)] = 1.23 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
2 Pin Plug Point
Total load of 2 Pin Plug = 2000 W
Distance from sub distribution board = 1.5’
Total load of 2 Pin Plug after adding 15% additional load = 2300 W
Total Current = 10.45 A
Cable size = (1/1.80) 15 A
Current capacity at 45⁰ C = 0.91×15 = 13.65 A
Voltage drop for 100 ft for this cable = 11 V
Here, the length of cable = 1.5’
So, voltage drop= [(11×1.5×10.45)÷(100×13.65)] = 0.126 V < 5.5 V
Therefore, cable size should be (1/1.80) 15 A
89. Attached Restroom 3
Total Load Of Room = 52 W
After adding 15% Additional Load = 59.8 W
Total Current = 0.2718 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 61.75’
So, voltage drop= [(14×61.75×0.2718)÷(100×9.1)] = 0.258 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light
Total load of LED Light = 7 W
Distance from sub distribution board = 27.25’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 27.25’
So, voltage drop= [(14×27.25×0.0365)÷(100×9.1)] = 0.015 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
Exhaust Fan
Total load of Exhaust Fan = 45 W
Distance from sub distribution board = 34.5’
Total load of LED Light after adding 15% additional load = 51.75 W
Total Current = 0.2352 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 34.5’
So, voltage drop= [(14×34.5×0.2352)÷(100×9.1)] = 0.125 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
90. Corridor
Total Load Of Corridor = 168 W
After Adding 15% Additional Load = 193.2 W
Total Current = 0.878 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 37.25’
So, voltage drop= [(14×37.25×0.878)÷(100×9.1)] = 0.503 V < 5.5 V
Therefore, size of the cable should be (1/1.40) 10 A
LED Light 1
Total load of LED Light = 7 W
Distance from sub distribution board = 28.83’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 28.83’
So, voltage drop= [(14×28.83×0.0365)÷(100×9.1)] = 0.016 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 2
Total load of LED Light = 7 W
Distance from sub distribution board = 31.83’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 31.83’
So, voltage drop= [(14×31.83×0.0365)÷(100×9.1)] = 0.0178 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
91. LED Light 3
Total load of LED Light = 7 W
Distance from sub distribution board = 21.5’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 21.5’
So, voltage drop= [(14×21.5×0.0365)÷(100×9.1)] = 0.012 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 4
Total load of LED Light = 7 W
Distance from sub distribution board = 26.5’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 26.5’
So, voltage drop= [(14×26.5×0.0365)÷(100×9.1)] = 0.015 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 5
Total load of LED Light = 7 W
Distance from sub distribution board = 12.4167’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 12.4167’
So, voltage drop= [(14×12.4167×0.0365)÷(100×9.1)] = 0.007 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 6
Total load of LED Light = 7 W
Distance from sub distribution board = 17.4167’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 17.4167’
So, voltage drop= [(14×17.4167×0.0365)÷(100×9.1)] = 0.0098 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
92. LED Light 7
Total load of LED Light = 7 W
Distance from sub distribution board = 13.33’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 13.33’
So, voltage drop= [(14×13.33×0.0365)÷(100×9.1)] = 0.0075 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 8
Total load of LED Light = 7 W
Distance from sub distribution board = 18.33’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 18.33’
So, voltage drop= [(14×18.33×0.0365)÷(100×9.1)] = 0.01 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 9
Total load of LED Light = 7 W
Distance from sub distribution board = 19.25’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 19.25’
So, voltage drop= [(14×19.25×0.0365)÷(100×9.1)] = 0.01 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A
LED Light 10
Total load of LED Light = 7 W
Distance from sub distribution board = 26.25’
Total load of LED Light after adding 15% additional load = 8.05 W
Total Current = 0.0365 A
Cable size = (1/1.40) 10 A
Current capacity at 45⁰ C = 0.91×10 = 9.1 A
Voltage drop for 100 ft for this cable = 14 V
Here, the length of cable = 26.25’
So, voltage drop= [(14×26.25×0.0365)÷(100×9.1)] = 0.015 V < 5.5 V
Therefore, cable size should be (1/1.40) 10 A