Schedule of load preparation is essential and a basic calculation for electrical engineers. In this process the proper sizing of conductors, overload protection and conduits are determined.

While there are different methods doing an electrical design but there is only one thing that cannot be altered --- code requirements must be followed.

• This example emphasized the procedure rather than mimicking the actual loads of a residential unit.
• In this example the voltage drop and short circuit calculation is not included.
• The system voltage of this example is 220 VAC, 60 Hz.

 Ckt Load Ph Rating Per outlet No. of Outlets VA Volts Amps Wire CB Cond. 1 L.O. 1 100 VA 12 1,200 220 5.45 #14 TW 15 AT, 1P ½” dia. 2 L.O. 1 100 VA 9 900 220 4.09 #14 TW 15 AT, 1P ½” dia. 3 L.O. 1 100 VA 6 600 220 2.73 #14 TW 15 AT, 1P ½” dia. 4 C.O 1 180 VA 10 1,800 220 8.18 #12 TW 20 AT, 1P ¾” dia. 5 C.O 1 180 VA 12 2,160 220 9.82 #12 TW 30 AT, 1P ¾” dia. 6 ACU 1 2.5 HP 1 2331 220 10.60 #10 TW 30 AT, 1P ¾” dia. 7 ACU 1 2.5 HP 1 2331 220 10.60 #10 TW 30 AT, 1P ¾” dia. 8 ACU 1 2.5 HP 1 2331 220 10.60 #10 TW 30 AT, 1P ¾” dia. 9 Range Load 1 5000 W 1 5000 220 22.71 # 8 TW 80 AT, 1P 1.0” dia.

Schedule of loads are just a summary of data to easily identify and facilitate the necessary values and equipment rating to be used in any electrical installation. Any data given in the schedule of loads were backed by calculation based on a well settled electrical principles and code requirements.

The general rule: The cable and circuit breaker must always be coordinated.

Computations

Circuit 1:

I = 1,200 VA/ 220 V = 5.45 Ampere
Wire = 5.45 / 80% = 6.82 Amperes , Use 2.0 sqmm TW wire or #14 AWG 
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.

Circuit 2:

I = 900 VA/ 220 V = 4.09 Ampere
Wire = 4.09/ 80%= 5.11 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.

Circuit 3:

I = 600 VA/ 220 V = 2.72 Ampere
Wire = 2.72/ 80%= 3.41 Amperes , Use 2.0 sqmm TW wire or #14 AWG
Circuit Breaker = 6.82 Amperes, Use 15 A Circuit Breaker
Conduit = Use 1/2" diameter PVC conduit.

Circuit 4:

I = 1,800 VA/ 220 V = 8.18 Ampere
Wire = 8.18/ 80%= 10.23 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.

Circuit 5:

I = 2,160 VA/ 220 V = 9.82 Ampere
Wire = 9.82/ 80% = 12.27 Amperes , Use 3.5 sqmm TW wire or #12 AWG
Circuit Breaker = Use 20 A CB
Conduit = Use 3/4" diameter PVC conduit.

Circuit 6-8:

VA = [ 2.5 HP  x ( 746 Watts/ HP ) ] / 0.8 pf (assume 0.8 pf)
VA = 2331 VA

I = 2,331 VA/ 220 V = 10.60 Ampere
Wire = 10.60  x 125% = 13.24 Amperes , Use 3.5 sqmm TW wire or #12 AWG 
Circuit Breaker = 5.45 x 250% = 26.5 Amperes, Use 30 A Circuit Breaker 
Conduit = Use 3/4" diameter PVC conduit.

Note: since the breaker is 30 Ampere, we need to increase the cable size to 5.5 sqmm (rated 30 amps by NEC ) to maintain the coordination of cable and the circuit breaker.

Circuit 9:

VA = 5000 W / 1.0 pf (heating load is a resistive load w/ 100% pf)
VA = 5,000 VA

I = 5, 000 VA/ 220 V = 22.72 Ampere
Wire = 22.71 / 80% = 28.41 Amperes , Use 8.0 sqmm TW wire or #8 AWG
Circuit Breaker = Use 40 A Circuit Breaker
Conduit = Use 1.0" diameter PVC conduit.

#### Read: How to Select Proper Type of Miniature Circuit Breakers MCBs

Main Feeder

By inspection:

Continuous loads = 9,963 VA  or  45.29 A @ 220V (lighting loads and ACU)
Non- Continuous = 8, 960 VA  or 40.72 @ 220V (conv. outlet & range load)
Total Loads         = 19, 923 VA

Main Feeder Current =  (45.29 x 100% ) + (40.72 x 125%) = 96.19 Amperes 

Use 50 sqmm TW cable as main feeder or service entrance wire
Use 100 Ampere MCCB, 1 pole - 10 kAIC*

note: 10 kAIC is just an assumed value, we need short circuit calculation to determine the right specs of the OCPD to be used in this example

Rules Applied:

1. NEC 210-9a - Maximum to be served by branch circuit must not be less than 80% of the ampacity of the condutor

2. NEC 430 -22 = The size of the wire supplying motorized load shall not be less than 125% of the rated full load current of the motor.

3. NEC 430- 52 = The size of the branch circuit protection for motor loads shall not be greater than 250% of motor full load current for CB and 300% for non-time delay fuses on full voltage starting.

4. NEC 210-22(C) = Over-Current Protection Device shall be calculated as 100% of non-continuous load + 125% of the continuous load.

References:

1. National Electrical Code 2011 (Handbook)
2. General Electric Circuit Breaker Catalogue

1. 2. This comment has been removed by a blog administrator.

3. Nice blog… Thanks for sharing very useful information about electrical circuits.
Learn Electronic Circuits

4. help:
question 1: circuit 4; how did you compute circuit breaker: 20.45 amps?
question 2: circuit 6-8; how did you compute circuit breaker: 5.45 amps?

5. Oops!? you multiplied the continous load by 100% and the noncontinous load by 125%.

6. in what chapter in the nec stated that 1 LO is rated 100VA?

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8. This comment has been removed by the author.

9. ..Hi, I'm just Wondering. Why did you always divide the AMPERAGE RESULT to 80% pf except Circuit 6-8 that you Multiply into 125%? thus this 80% pf different from 80% Demand Factor (DF)?

10. calculating voltage drop in the system. This code provides data of standard conductor properties that can be used in voltage drop calculation. CCTV Installers Melbourne

11. National Electrical Code 2011 (Handbook)
2. CESEEPS Red Book, Low Voltage Systems and Applications in the Industries
3. General Electric Circuit Breaker Catalogue http://www.electricaltest.cn/

12. 13. 