Electrical Thumb Rules-(Part-15).


Luminous Efficacy, Lumen Maintenance and Color Rendition (Table-8) NBC

Light Source  Wattage Efficacy (lm/W ) Average Life Maintenance Color Rendition
Incandescent lamps  15 to 200  12 to 20  500 to 1000  Fair to good  Very good
Tungsten halogen     300 to 1500  20 to 27  200 to 2000  Good to very good  Very good
Standard fluorescent lamps       20 to 80 55 to 65 5000 Fair to good  Good
Compact fluorescent lamps (CFL)       5 to 40  60 to 70 7500 Good Good to very good
Slim line fluorescent      18 to 58 57 to 67 5000  Fair to good Good
High pressure mercury vapor lamps      60 to 1000  50 to 65 5000  Very low to fair  Federate
Blended – light lamps    160 to 250  20 to 30 5000 Low to fair  Federate
High pressure sodium vapor lamps  50 to 1000  90 to 125  10000 to 15000  Fair to good  Low to good
Metal halide lamps       35 to 2000  80 to 95 4000 to 10000 Very low  Very good
Low pressure sodium       10 to 180 100 to 200 10000 to 20000 Good to very good  Poor
LED  0.5 to 2.0  60 to 100  10000 Very good  Good for white LED


Approximate Cable Current Capacity

Cable Size Current Capacity MCB Size
1.5 Sq.mm 7.5 To 16 A 8A
2.5 Sq.mm 16 To 22 A 15A
4 Sq.mm 22 To 30 A 20A
6 Sq.mm 39 To 39 A 30A
10 Sq.mm 39 To 54A 40A
16 Sq.mm 54 To 72A 60A
25 Sq.mm 72 To 93A 80A
50 Sq.mm 117 To 147A 125A
70 Sq.mm 147 To 180A 150A
95 Sq.mm 180 To 216A 200A
120 Sq.mm 216 To 250A 225A
150 Sq.mm 250 To 287A 275A
185 Sq.mm 287 To 334A 300A
240 Sq.mm 334 To 400A 350A


Requirements  for  Physical  Protection  of Underground Cables  (As per NBC)

Protective  Element Specifications
Bricks  (a) 100 mm minimum  width 
(b) 25 mm thick 
(c) sand cushioning 100  mm  and  sand  cover 100 mm 
Concrete slabs At least 50 mm thick
Plastic  slabs (polymeric cover  strips) Fiber  reinforced plastic depending on properties  and has to be matched with the protective cushioning and cover
PVC  conduit  or  PVC  pipe  or stoneware  pipe or Hume pipe The  pipe  diameter should  be  such  so  that the  cable  is  able  to easily slip down the pipe
Galvanized pipe  The  pipe  diameter should  be  such  so  that the  cable  is  able  to easily slip down the pipe
The Trench : The trench shall be back filled to cover the cable initially by 200 mm of sand fill; and then a plastic marker strip  hall be put over the full length of cable in the trench.
The Marker Signs: The marker signs shall be provided where any cable enters or leaves a building. This will identify that there is a cable located underground near the building.
 The trench shall then be completely filled. If the cables rise above ground to enter a building or other structure, a mechanical protection such as a GI pipe or PVC pipe for the cable from the trench depth to a height of 2.0 m above ground shall be provided.



Capacity  kVA Area m2 Clear Height below the Soffit of the Beam m
25 56 3.6
48 56 3.6
100 65 3.6
150 72 3.6
248 100 4.2
350 100 4.2
480 100 4.2
600 110 4.6
800 120 4.6
1010 120 6.5
1250 120 6.5
1600 150 6.5
2000 150 6.5


Low Voltage Cabling for Building (As per NBC)

Low Voltage Cable Cables/wires, such as fiber optic cable, co-axial cable, etc. These shall be laid at least at a distance of 300 mm from any power wire or cable. The distance may be reduced only by using completely closed earthed metal trucking with metal separations for various kind of cable. Special care shall be taken to ensure that the conduit runs and wiring are laid properly for low voltage signal to flow through it.
The power cable and the signal or data cable may run together under floor and near the equipment. However, separation may be required from the insulation aspect, if the signal cable is running close to an un-insulated conductor carrying power at high voltage. All types of signal cables are required to have insulation level for withstanding 2 kV impulse voltages even if they are meant for service at low voltage.
Conduit Color Scheme Power conduit=Black
Security conduit=Blue
Fire alarm conduit=Red
Low voltage conduit=Brown
UPS conduit Green


Sub Station Guideline (As per NBC)

Substation Location Location of substation in the basement should be avoided, as far as possible.
If there is only one basement in a building, the substation/switch room shall not be provided in the basement and the floor level of the substation shall not be lowest point of the basement.
Substation shall not be located immediately above or below plumbing water tanks or sewage treatment plant (STP) water tanks at the same location
Substation Door/Shutter All door openings from substation, electrical rooms, etc, should open outwards
Vertical shutters (like rolling shutters) may also be acceptable provided they are combined with a single leaf door opening outwards for exit in case of emergency
For large substation room/electrical  room  having  multiple equipment,  two  or more  doors  shall  be provided which shall be remotely located from each other
No services or ventilation shafts shall open into substation or switch room unless specific to substation or switch room
Transformer Location In case of HV panel and transformers located at different floors or at a distance more than 20 m, HV isolator shall be  provided  at transformer end
In case transformer and main MV/LV panel room are located at different floors or are at a distance more than 20 m, MV/LV isolator shall be provided at  transformer  end
In  case  of  two  transformers  (dry  type  or transformers with oil quantity less than 2 000 liter)  located  next  to  each  other without intermittent wall, the distance between the two shall  be minimum  1 500 mm  for  11  kV, minimum 2 000 mm for 22 kV and minimum 2 500 mm for 33 kV. Beyond 33 kV, two transformers shall be separated by baffle wall of 4 h fire rating.
If dry type transformer is used, it may be located adjacent to medium voltage switch gear in the form of unit type substation. In such a case, no separate room or fire barrier for the transformer is required either between transformers or between transformer and the switch gear, thereby decreasing the room space requirement; however, minimum distances as specified.
Oil Filled Equipment (Transformer / C.B) Substations with oil-filled equipment/apparatus transformers and high voltage panels shall be either located in open or in a utility building
They shall not be located in any floor other than the ground floor or the first basement of a utility building  not be located below first basement slab of utility building.
They shall have direct access from outside the building for operation and maintenance of the equipment.
It shall be separated from the adjoining buildings including the main building by at least 6 m clear distance to allow passage of fire tender between the substation/utility building and adjoining building/main building.
Substation equipment having more than 2 000 liter of oil whether located indoors in the utility building or outdoors shall have  baffle walls  of  4  h  fire  rating between apparatus.
Provision of  suitable oil soak-pit, and where use of more than 9 000 liter of oil in any one oil tank, receptacle or chamber is involved, provision shall be made for the draining away or removal of any oil which may leak or escape from the tank, receptacle or chamber containing the same
Power Supply Voltage supply  is  at  240  V  single  phase  up  to  5  kVA, 415/240 V 3-phase from 5 kVA to 100 kVA, 11 kV (or 22 kV) for loads up to 5 MVA and 33 kV or 66 kV for consumers of connected load or contract demand more than 5 MVA.
In case of connected load of 100 kVA and above, the relative advantage of high voltage three-phase supply should be considered.
In case of single point high voltage metering, energy meters shall  be  installed  in  building  premise,such a place which is readily accessible to the owner/operator of the building and the Authority. The supplier or owner of the installation shall provide at the point of commencement of supply a suitable isolating device fixed in a conspicuous position at not more than 1.7 m above the ground so as to completely isolate the supply to the building in case of emergency
Trench Drain In case of cable trench in substation/HV switch room/MV switch room, the same shall be adequately drained to ensure no water is stagnated at any time with live cables.
Fence for Substation Enclose any part of the substation which is open to the air, with a fence (earthed efficiently at both ends) or wall not less than 1800 mm (preferably not less than 2400 mm) in height
HV Distribution in Building The power supply HV cables voltage shall not be more than 12 kV and a separate dedicated and  fire  compartmented  shaft  should  be provided for carrying such high voltage cables to upper floors in a building. These shall not be mixed with any other shaft and suitable fire detection and suppression measures shall be provided throughout the length of the cable on each floor.
Switch Room / MV switch room Switch room / MV switch room shall be arrived at considering 1200 mm clearance requirement from top of the equipment to the below of the soffit of the beam .In case cable entry/exit is from above the  equipment  (transformer,  HV switchgear, MV  switchgear),  height  of substation room/HV switch room/MV switch room shall also take into account requirement of space for turning radius of cable above the equipment height.



What is Correct Method of MCB Connections


  • MCB is a mechanical switching device which can carry and break currents under normal circuit conditions and also under specified abnormal conditions, such as overload and short circuit.
  • The MCB can provide protection until and unless we have install input power (LINE) connection and Output (LOAD) connections in proper Terminals of MCB.
  • Electrical engineers seem to be confused to indentify where is the Line and Load terminal of an MCB (on the top or on the bottom).

Terminal Marking of MCB:

  • There are two type of MCB available in market.
  • MCB having terminal marking (LINE / LOAD Marking) (Polarized MCB)
  • MCB having No terminal marking (No any Marking) (Non Polarized MCB)
  • Some manufacture clearly indicates where to apply Input Power and where to connect Load on MCB while some manufacture does not indicate such Terminal Marking.
  • The constructions of both MCB are almost same even though we need to understand difference between them.

(1) LINE / LOAD Terminal Marking on MCB (Polarized MCB)

  • For AC Circuit:
  • If manufactures indicate Input (LINE) making on MCB then we have to give Supply at “LINE” Terminal and Load at “LOAD” Terminal for perfect operation of MCB.
  • If we do wrong connection than MCB may or may not give proper protection in fault Condition.
  • As Per UL 489 Paragraph It is clearly indicate that “Circuit breakers shall be marked “Line” and “Load” unless the construction and test results are acceptable with the line and load connections reversed. This marking requirement specifies that UL MCB shall be marked with the word “Line” on one end of the circuit breaker and the word “Load” on the other end”, as shown in Figure


  • If MCB is not live (ON) from long time (in Cold state) than there is possibility of MCB to not operate in fault conditions.
  • In MCB ,The fixed contact is encompassed by the arc chute, and the arc products are de ionized, cooled and ejected uneventfully when the incoming power is on “Line” Terminal (when the fixed contact is ‘live’ or ‘hot’).There is less chance to re strike arc again.
  • If the power is applied to moving contact ,”Load” Terminal, the flexible connector, the trip system, everything is live/hot after the arc is quenched. Chances of restrike/flashover are much higher.
  • For DC Circuit:
  • The polarized DC MCB have a marking of ‘+’ and ‘–‘ symbol
  • If Polarized DC MCB are wired incorrectly, they are a possibility of hazard and When we turned off under load, the MCB might not be able to extinguish the arc and the circuit breaker will burn out.
  • Polarized DC MCB use a small magnet to direct the arc away from the contacts and up into the arc shoot and arc disrupter cage. If the direction of current flow through the unit is reversed, then the magnet directs the arc away from the arc shoot and into the mechanism of the unit thus destroying it.


(2) No Terminal Marking on MCB (Non Polarized MCB)

  • For AC Circuit:
  • If manufacture has not indicated any Terminal Marking than we are free to connect line or load at any side as we wish.
  • If construction / Operating principle of both MCB are same then what are the different between them.
  • Without Terminal Marking MCB has following additional features.
  • (1) By Design improvement (Manufacture has provided some more provision for quenching of arc (So it cannot reproduce it again).
  • (2) By doing some more extra test as per IEC 60947-2 and UL 489


  • The performance of single-break circuit breakers is slightly different when the “LINE” and “LOAD” feed either from the bottom or Top hence IEC 60947-2 specifies that one additional SC test be carried out with connections required when the terminals are not specifically marked ‘Line’ and/or ‘Load’

Table 10- Number of samples for test (IS / IEC 60947-2)

Test Sequences

Terminal Marking (Line / Load) No of Sample for Testing


Sample For *





1 1
Ics (Rated service short-circuit breaking capacity)  (Ics=25%Icu)

2 1

3 1

3 1

4 1
Icu  (Rated ultimate short-circuit breaking capacity)

2 1

3 1

3 1

4 1
* Sample For Indications
1 In of a given frame size.
2 This sample is omitted in the following cases:
A circuit-breaker having a single non-adjustable current setting for a given frame size;
A circuit-breaker provided only with a shunt release (i.e. without an integral over current release);
A circuit-breaker with electronic over current protection, of a given frame size, having an adjustable current rating by electronic means only (i.e. without change of current sensors).
3 Connections reversed.
4 Connections reversed, if terminals unmarked.
  • As Per UL 489, Paragraph “if a circuit breaker is not marked “Line” and “Load,” one sample of each set tested, or one additional sample, shall be connected with the line and load connections reversed during the overload, endurance and interrupting tests”.
  • This UL test requirement specifies that for MCC to be UL Listed for reverse-feed applications, samples shall be tested with the line and load terminals reverse-fed, as shown in Figure, and that the test results shall be the same as those of “normally” fed circuit breakers. Depending on the design configuration and construction, the circuit breaker may or may not be affected by the application of power in a reverse-feed connection during these tests.


  • If Line / Load are not marked, we can connect Line or Load either on Top or bottom of MCB. However, it is a good practice to keep the fixed contact side connected to the bus bar.
  • For DC Circuit:
  • The Non polarized DC MCB have a No marking as ‘+’ and ‘–‘ symbol
  • Non polarized DC MCB operate safely as load breaking isolators and for fault current protection regardless of the direction of current flow through them.



  • When a MCB are marked “Line” and “Load,” the power supply conductors must be connected to the marked “Line.” These MCB cannot be reverse-fed.
  • If “Line” and “Load” are not marked on MCB, the power supply conductors may be connected to either end. These devices are suitable for reverse-feed applications.

Pirating of Technical Works-2

It has been observed that some website totally copy paste of this blog and parallel republished all posts of this blog again on their commercial website  .

Lots of time has been spent to read Books,Manuals,Handbooks and combined it with  practical experience to serve Handy Electrical tools,Notes to serve the Electrical Community.This Blog is a fusion of Theoretical and Practically knowledge to make all technical things easier to understand.

Please look at following totally copy paste material of  this  Blog. 

Originally published

(1) https://electricalnotes.wordpress.com/2016/10/04/how-to-select-mcb-mccb-part1/

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Totally Copy & Paste word by word


Calculate Size of Contactor / Fuse / CB / OL Relay of Star-Delta Starter

  • Calculate Size of each Part of Star-Delta starter for 10HP, 415 Volt Three Phase Induction Motor having Non Inductive Type Load, Code A, Motor efficiency 80%, Motor RPM 600, Power Factor 0.8. Also Calculate Size of Overload Relay if O/L Relay Put in the wingdings (overload is placed after the Winding Split into main and delta Contactor) or in the line (Putting the overload before the motor same as in DOL).


Basic Calculation of Motor Torque & Current:

  • Motor Rated Torque (Full Load Torque) =5252xHPxRPM
  • Motor Rated Torque (Full Load Torque)=5252x10x600=88 lb-ft.
  • Motor Rated Torque (Full Load Torque) =9500xKWxRPM
  • Motor Rated Torque (Full Load Torque)=9500x(10×0.746)x600 =119 Nm
  • If Motor Capacity is less than 30 KW than Motor Starting Torque is 3xMotor Full Load Current or 2X Motor Full Load Current.
  • Motor Starting Torque=3x Motor Rated Torque (Full Load Torque).
  • Motor Starting Torque==3×119=356 Nm.
  • Motor Lock Rotor Current =1000xHPx figure from below Chart/1.732×415
Locked Rotor Current
Code Min Max
A 1 3.14
B 3.15 3.54
C 3.55 3.99
D 4 4.49
E 4.5 4.99
F 5 2.59
G 2.6 6.29
H 6.3 7.09
I 7.1 7.99
K 8 8.99
L 9 9.99
M 10 11.19
N 11.2 12.49
P 12.5 13.99
R 14 15.99
S 16 17.99
T 18 19.99
U 20 22.39
V 22.4
  • As per above chart Minimum Locked Rotor Current =1000x10x1/1.732×415=14 Amp
  • Maximum Locked Rotor Current =1000x10x3.14/1.732×415=44 Amp.
  • Motor Full Load Current (Line) =KWx1000/1.732×415
  • Motor Full Load Current (Line) = (10×0.746)x1000/1.732×415=13 Amp.
  • Motor Full Load Current (Phase) =Motor Full Load Current (Line)/1.732.
  • Motor Full Load Current (Phase) ==13/1.732=7 Amp.
  • Motor Starting Current (Star-Delta Starter) =3xFull Load Current.
  • Motor Starting Current (Line)=3×13=39 Amp

(1) Size of Fuse:

Fuse  as per NEC 430-52
Type of Motor Time Delay Fuse Non-Time Delay Fuse
Single Phase 300% 175%
3 Phase 300% 175%
Synchronous 300% 175%
Wound Rotor 150% 150%
Direct Current 150% 150%
  • Maximum Size of Time Delay Fuse =300% x Full Load Line Current.
  • Maximum Size of Time Delay Fuse =300%x13= 39 Amp.
  • Maximum Size of Non Time Delay Fuse =1.75% x Full Load Line Current.
  • Maximum Size of Non Time Delay Fuse=1.75%13=23 Amp.

(2) Size of Circuit Breaker:

Circuit Breaker as per NEC 430-52
Type of Motor Instantaneous Trip Inverse Time
Single Phase 800% 250%
3 Phase 800% 250%
Synchronous 800% 250%
Wound Rotor 800% 150%
Direct Current 200% 150%
  • Maximum Size of Instantaneous Trip Circuit Breaker =800% x Full Load Line Current.
  • Maximum Size of Instantaneous Trip Circuit Breaker =800%x13= 104 Amp.
  • Maximum Size of Inverse Trip Circuit Breaker =250% x Full Load Line Current.
  • Maximum Size of Inverse Trip Circuit Breaker =250%x13= 32 Amp.

(3) Thermal over Load Relay:

Thermal over Load Relay (Phase):

  • Min Thermal Over Load Relay setting =70%xFull Load Current(Phase)
  • Min Thermal Over Load Relay setting =70%x7= 5 Amp
  • Max Thermal Over Load Relay setting =120%xFull Load Current(Phase)
  • Max Thermal Over Load Relay setting =120%x7= 9 Amp

Thermal over Load Relay (Line):

  • For a star-delta starter we have the possibility to place the overload protection in two positions, in the line or in the windings.
  • If O/L Relay Placed in Line: (Putting the O/L before the motor same as in DOL).Supply>Over Load Relay>Main Contactor
  • If Over Load Relay supply the entire motor circuit and are located ahead of where the power splits to the Delta and Star contactors, so O/L Relay size must be based upon the entire motor Full Load Current.
  • Thermal over Load Relay setting =100%xFull Load Current (Line).
  • Thermal over Load Relay setting =100%x13= 13 Amp
  • Disadvantage: O/L Relay will not give Protection while Motor runs in Delta (Relay Setting is too High for Delta Winding)
  • If O/L Relay Placed In the windings: (overload is placed after the Winding Split into main and delta Contactor).Supply>Main Contactor-Delta Contactor>O/L Relay
  • If overload is placed after the Point where the wiring Split into main and delta Contactor, Size of over load relay at 58% (1/1.732) of the motor Full Load Current because we use 6 leads going to the motor, and only 58% of the current goes through the main set of conductors (connected to the main contactor).
  • The overload then always measures the current inside the windings, and is thus always correct. The setting must be x0.58 FLC (line current).
  • Thermal over Load Relay setting =58%xFull Load Current (Line).
  • Thermal over Load Relay setting =58%x13= 8 Amp.
  • Disadvantage: We must use separate short-circuit and overload protections

(4) Size and Type of Contactor:

  • Main and Delta Contactor:

  • The Main and Delta contactors are smaller compared to single contactor used in a Direct on Line starter because they Main and Delta contactors in star delta starter are controlling winding currents only. The currents through the winding are 1/√3 (58%) of the current in the line. These two contactors (Main contactor and Delta Contactor) are close during run. These rated at 58% of the current rating of the motor.
  • Star Contactor:

  • The third contactor is the star contactor and that only carries star current while the motor is connected in star in starting. The current in star winding is 1/√3= (58%) of the current in delta, so this contactor can be rated at 1/3 (33%) of the motor rating. Star contactor can be selected smaller than the others, providing the star contactor pulls first before the main contactor. Then no current flows when third contactor pulls.
  • In star connection at start, the motor draws and delivers 1/3 of its full rated power.
  • When the starter switches over to Delta, the motor draws full power, but since the contactors and the overload relay are usually wired within the Delta, you need to use contcators and relay which are only rated 1/√3 =58% of the full rated power of the motor.
Application Contactor Making Cap
Non-Inductive or Slightly Inductive ,Resistive Load AC1 1.5
Slip Ring Motor AC2 4
Squirrel Cage Motor AC3 10
Rapid Start / Stop AC4 12
Switching of Electrical Discharge Lamp AC5a 3
Switching of Electrical Incandescent Lamp AC5b 1.5
Switching of Transformer AC6a 12
Switching of Capacitor Bank AC6b 12
Slightly Inductive Load in Household or same type load AC7a 1.5
Motor Load in Household Application AC7b 8
Hermetic refrigerant Compressor Motor with Manual O/L Reset AC8a 6
Hermetic refrigerant Compressor Motor with Auto O/L Reset AC8b 6
Control of Restive & Solid State Load with opto coupler Isolation AC12 6
Control of Restive Load and Solid State with T/C Isolation AC13 10
Control of Small Electro Magnetic Load ( <72VA) AC14 6
Control of Small Electro Magnetic Load ( >72VA) AC15 10
  • As per above Chart
  • Type of Contactor= AC1
  • Making/Breaking Capacity of Contactor= Value above Chart x Full Load Current (Line).
  • Making/Breaking Capacity of Contactor=1.5×13= 19 Amp.
  • Size of Star Contactor (Starting Condition) = 33%X Full Load Current (Line).
  • Size of Star Contactor =33%x13 = 4 Amp.
  • Size of Main Contactor (Starting-Transition-Running) = 58%X Full Load Current (Line).
  • Size of Main Contactor =58%x13 = 8 Amp.
  • Size of Delta Contactor (Running Condition) = 58%X Full Load Current (Line).
  • Size of Delta Contactor =58%x13 = 8 Amp.


  •  Type of Contactor= AC1
  • Making/Breaking Capacity of Contactor=19 Amp.
  • Size of Star Contactor =4 Amp.
  • Size of Main Contactor = 8 Amp.
  • Size of Delta Contactor =8 Amp.

Effects of unbalanced Electrical Load (Part:2)

  • Harmonics in system by UPS:

  • UPS or inverter supplies also perform with poor efficiency and inject more harmonic currents in case of unbalances in the system
  • Decrease Life cycle of Equipment:

  • Unbalanced Voltage increase I2R Losses which increase Temperature. High temperatures, exceeding the rated value of a device, will directly decrease the life cycle of the device and speed up the replacement cycle for the device, and significantly increase the costs of operation and maintenance.
  • Relay malfunction

  • Unbalanced Voltage flows Negative and Unbalanced Voltage of Voltage or Current.
  • The high zero-sequence current in consequence of voltage imbalance may bring about malfunctions of relay operation or make the ground relay less sensitive. That may result in serious safety problems in the system.
  • Inaccurate Measurement

  • Negative and zero-sequence components of voltages or currents will give rise to inaccurate measurements in many kinds of meters.
  • The imprecise measured values might affect the suitability of settings and coordination of relay protection systems and the correctness of decisions by some automated functions of the system.
  • Decrease Capacity of transformers, cables and lines

  • The capacity of transformers, cables and lines is reduced due to negative sequence components. The operational limit is determined by the RMS rating of the total current, due to ‘useless’ non-direct sequence currents the capacity of equipment is decrease.
  • Increase Distribution Losses

  • Distribution network losses can vary significantly depending on the load unbalance.
  • Unbalance load increase I2R Losses of distribution Lines.
  • Increase Energy Bill by increasing Maximum Demand

  • Unbalanced Load increase maximum Demand of Electrical supply which is significantly effects on energy bill. By load balancing we can reduce energy bill.
  • For Energy Consumption Energy Supply Company does not charge on kVA but on kW for Residential customers. This means that they are charged for the “actual” energy used and not charged for the “total” energy supplied. Thus the power factor and Maximum Demand do not impact residential customers.
  • But Commercial, Industrial and H.T Connection charged by its maximum demand . We have to specify the maximum “demand“(in kVA) at the time of connection. During the month if you exceed your maximum “demand” you have to pay penalty (or extra price) for the same. That is the MDI penalty that appears on electricity bills.
  • Let’s assume That Two Company has same approved load of 40 KW and runs 30KW for 100 hours.
  • Electricity charge = 65 Rs per kWh
  • Demand charge = 210Rs per kW
  • Example 1: Company A runs a 30 KW loads continuously for 100 hours but It’s Maximum Demand is 50KW
  • 30 KW x 100 hours = 3,000 KWh
  • Energy Consumption Charge =3000×65=195000Rs
  • Demand difference = 50 KW-40KW=10KW
  • Demand Charges = 10X210=2100Rs
  • Total Bill:  195000+2100=197100Rs
  • Example 2: Company A runs a 30 KW loads continuously for 100 hours but It’s Maximum Demand is 40W
  • 30 KW x 100 hours = 3,000 KWh
  • Energy Consumption Charge =3000×65=195000Rs
  • Demand difference = 40 KW-40KW=0KW
  • Demand Charges = 0X210=00Rs
  • Total Bill:  195000+0=195000Rs
  • Failure of Transformer

  • Three-phase voltage with high unbalanced may cause the flux inside the transformer core to be asymmetrical.
  • This asymmetrical flux will cause extra core loss, raise the winding temperature and may even cause transformer failure in a severe case.
  • Ideally any distribution transformer gives best performance at 50% loading and every electrical distribution system is designed for it. But in case of unbalance the loading goes over 50% as the equipments draw more current.
  • The efficiency of transformer under different loading conditions
  • Full Load- 98.1%
  • Half Load- 98.64%
  • Unbalanced loads- 96.5%
  • For a distribution transformer of 200KVA rating, the eddy currents accounts for 200W but in case of 5% voltage unbalance they can rise up to 720W.
  • Bad / Loose connection of neutral wire

  • In balance Load condition Bad connection of Neutral wire does not make more impact on distribution System but in unbalance load condition such type of Bad neutral connection make worse impact on distribution.
  • The Three Phase power supplies a small a three-floor building. Each floor of this three-floor building is serviced by a single-phase feeder with a different phase. That is the first, second and third floor are serviced by phase R, Y and B. The external lighting load is connected only on R Phase.
  • The supply transformer is rated at 150 kVA and connected delta-grounded wye to provide for 430/220 V three-phase four-wire service.
  • This Transformer has a loose or Bad Neutral connection with the earth.
  • The transformer delivers a load of 35 kVA at 220 V with 0.9 power factor lagging to each floor.
  • During the daytime on, most of the Load of the Building are distributed equally over the three floors which is R Phase=30A, Y Phase =32A, B Phase=38A.
  • In Daytime The Bad connection of Neutral does not effected the Distribution system due to equal load distribution of the System
  • However it is not case in Nighttime. the Load on Y Phase and B Phase are negligible but R Phase Load is high compare to Y and B Phase.
  • In R Phase due to High Electrical Load and The fluorescent lamps flash frequently during the Nighttime of external Lighting Load
  • In Night time a bad electrical contact of the neutral wire of the supply makes the high contact resistance between the neutral wire and connector .which was about 15 kΩ.
  • This extra high impedance caused an unusually high voltage drop in the phase a circuit. In this case, the voltage of phase a dropped from the normal 220V to 182.5V, about 17% based on the nominal voltage. If the contact impedance goes higher than 20 kΩ, it may result in more serious conditions such as extinguishing all lamps.
  • This problem can be removed by fixing the bad connection and keeping the contact impedance near to zero.
Neutral Wire Contact Resistance Voltage across  bad Connection Point Voltage across  Transformer Secondary Side
Day Time Night Time Day Time Night Time
Proper Connection (0Ω) 0v 0v 0v 0v 0v 0v 220v 220v 220v 220v 220v 220v
Bad Connection (15Ω) 0v 0v 0v 40v 0v 0v 220v 220v 220v 182v 220v 220v
  • Neutral wire broken

  • The effect of a broken neutral makes voltage imbalance in a Three Phase Four Wire System.
  • For a Three Phase Four Wire System, high neutral wire impedance might enlarge a voltage imbalance (Some Phase Voltage increase while some Phase Voltage decreases).
  • High Voltage damage the equipment connected and even destroy on other hand low voltage effect operation of equipments.
  • The Three Phase star connected lighting loads are fed by a 430 V balanced three-phase voltage source. The fluorescent lamps are all rated at 220 V, 100 W each. The lamps are not equally in R Phase 5 No’s of Bulbs are connected, in Y Phase 3 No’s of Bulbs are connected and on B Phase 3 No’s of Bulbs are Connected. And, the normal impedance of the neutral wire is 1Ω
  • In unbalanced three phase load arrangement, high neutral wire impedance will enlarge the voltage across the neutral wire. The voltages of phases B and C at the load terminal raised to 255 V and 235 V, respectively, and gaining 16.15% and 5.77% based on rated voltage. These abnormally high phase voltages might damage the lamps in phase B and C.
  • On the other hand, the voltage in phase A was reduced from 220V to 185V. That might cause the lamps to flash.
  • If the broken neutral line problem is fixed, then the three phase voltages will go back to normal in near balanced status .however, if the loads are distributed equally to the three phases this problem can also be removed or minimized.
Conditions Voltage across the neutral wire Voltage at  the load terminal
Normal Condition 1v 1v 1v 220v 220v 220v
Neutral Broken 0v 0v 0v 182v 255v 235v
  • Unsuitable capacitor bank installation

  • For reducing energy loss, utilities always force their customers to maintain the power factor within a limit. Penalty will be applied to the customers if their loads’ power factors run outside the limits.
  • Installation of shunt capacitor banks is the most common and cheapest manner to improve the power factor. However, unsuitable installation (single Phase Capacitor instead of Three Phase Capacitor ) may make it worse.
  • The supply transformer is rated at 150 kVA, 11kV/430 V, and supplies a three-phase load of 105 kVA with power factor 0.7 lagging.
  • A single-phase 20KVAR capacitor bank is connected to B phase to improve system power. The impedance of the shunt capacitor bank is 1.805Ωper phase.
  • This kind of single phase Capacitor installation should make the system unbalanced. This unsuitable installation consumes extra real power of 44355 W.
  • The extra real power consumption = 1.732x2XV(RB) / 4xXc =(1.732x2x430) / (4×1.805) =44355W
  • This case shows that the system balance should be considered when installing a capacitor bank to correct the system power factor for a three-phase power distribution system.

 Remedial Action to prevent unbalances Load:

  • All the single phase loads should be distributed on the three phase system such that they put equal load on three phases.
  • Replacing the disturbing equipments i.e. with unbalanced three phase reactance.
  • Reducing the harmonics also reduces the unbalance, which can be done by installing reactive or active filters. These filters reduce the negative phase sequence currents by injecting a compensating current wave.
  • In case the disturbing loads cannot be replaced or repaired, connect them with high voltage side this reduces the effects in terms of percentage and even controlled disturbance in low voltage side.
  • Motors with unbalanced phase reactance should be replaced and re-winded.
  • Distribution of single-phase loads equally to all phases.
  • Single-phase regulators have been installed that can be used to correct the unbalance but care must be exercised to ensure that they are controlled carefully not to introduce further unbalance.
  • Passive network systems and active power electronic systems such as static var compensators and line conditioners also have been suggested for unbalance correction.
  • Load balancing.
  • Use of passive networks and static VAR compensators.
  • Equipment that is sensitive to voltage unbalance should not be connected to systems which supply single-phase loads.
  • Effect of voltage unbalance on ac variable speed drives can be reduced by properly sizing ac side and dc link reactors
  • Tight all Neutral Connections of the System.
  • Install Proper size of Capacitor Bank to the System.
  • Load Scheduling, where the loads in an electrical network are scheduled in a way to turn on and off at precise times to prevent the overloading of any one phase.
  • Manual Load Shifting, where an electrician opens a breaker panel and physically removes the loads from one phase and inserts them onto another phase.
  • Load Shedding, where the loads in an electrical network are immediately turned off in order to instantly “rebalance” the phases. This is usually done by ranking the loads in a network by how long they can be turned off before it affects operations


Effects of unbalanced Electrical Load (Part:1)


  • Generally, three phase balance is the ideal situation for a power system and quality of delivered Electrical Power. However Voltage unbalance may makes worse effect on Power quality of Electrical Power at distribution level.
  • The voltages are quite well balanced at the generator and transmission levels. but the voltages at the utilization level can become unbalanced due to the unequal system impedances, the unequal distribution of single phase loads, asymmetrical three-phase equipment and devices (such as three-phase transformers with open star-open delta connections), unbalanced faults, bad connections to electrical connectors.
  • An excessive level of voltage unbalance can have serious impacts on power quality. In the system the level of current unbalance is several times the level of voltage unbalance. Such an unbalance in the line currents can lead to excessive line losses, losses in the stator and rotor of Motor Malfunctioning of Relay, unsymmetrical measuring of Meters. Voltage unbalance also has an impact on ac variable speed drive systems where the front end converter consists of three-phase rectifier systems
  • Phase balancing is very important and usable to reduce distribution feeder losses and Improve system stability and security

What is unbalance Voltage

  • Any deviation in voltage and current waveform from perfect sinusoidal, in terms of magnitude or phase shift is termed as unbalance
  • In ideal conditions the phases of power supply are 120 degree apart in terms of phase angle and magnitude of their peaks should be same. On distribution level, the load imperfections cause current unbalance which travel to transformer and cause unbalance in the three phase voltage. Even minor unbalance in the voltage at transformer level disturbs the current waveform significantly on all the loads connected to it
  • If three phase voltages have the same magnitude and are in exactly 120deg phase displacement, then the Three-phase voltage is called balanced, otherwise, it is unbalanced.
  • There are no negative- and zero-sequence voltages in a balanced system, only positive-sequence components of balanced three-phase voltage exist. On the contrary, if the system is unbalanced, negative-sequence components or zero-sequence components or both may exist in the system.


Causes of unbalance Voltage

  • Switching of three phase heavy loads results in current and voltage surges which cause unbalance in the system.
  • Unequal impedances in the power transmission or distribution system cause differentiating current in three phases.
  • Any large single phase load, or a number of small loads connected to only one phase cause more current to flow from that particular phase causing voltage drop on line
  • With continuous operation of motor’s in various environment cause degradation of rotor and stator windings. This degradation is usually different in different phases, affecting both the magnitude and phase angel of current waveform
  • A three phase equipment such as induction motor and Transformer with unbalance in its windings. If the reactance of three phases is not same, it will result in varying current flowing in three phases and give out system unbalance.
  • A current leakage from any phase through bearings or motor body provides floating earth at times, causing fluctuating current.
  • Unbalanced incoming utility supply
  • Unequal transformer taps settings
  • Large single phase distribution transformer on the system
  • Open phase on the primary of a 3 phase transformer on the distribution system
  • Faults or grounds in the power transformer
  • Open delta connected transformer banks
  • A blown fuse on a 3 phase bank of power factor improvement capacitors
  • Unequal impedance in conductors of power supply wiring
  • Unbalanced distribution of single phase loads such as lighting
  • Heavy reactive single phase loads such as welders

 How to calculate unbalance

  • %voltage unbalance= 100x (maximum deviation from average voltage) / (average voltage)
  • Example: With phase-to-phase voltages of The System is 430V, 435V, and 400V.
  • The average Voltage=(430+435+400)/3=421V.
  • The maximum Voltage deviation from Average Voltage=435-421=14V
  • %voltage unbalance=14×100/421=3.32%
  • The permissible limit in terms of percentage of negative phase sequence current over positive sequence current is 1.3% ideally but acceptable up to 2%.

 Effects of unbalance Voltage on System and Equipment:

  • The factors for voltage unbalances can be classified into two categories: normal factors and abnormal factors.
  • Voltage imbalances due to normal factors, such as single-phase loads and three-phase transformer banks with open star-open delta connections, can generally be reduced by properly designing the system and installing suitable equipment and devices.
  • Abnormal factors include series and shunt faults of circuits, bad electrical contacts of connectors or switches, asymmetrical breakdown of equipment or components, asynchronous burnout of three phase power fuses, single-phase operation of motors, etc. The abnormal factors just mentioned above might result in critical damage of systems and equipment.
  • Increase Neutral Return Current

  • The unequal distribution of loads between the three phases of the system cause the flow of unbalanced currents in the system, that produce unbalanced voltage drops on the electric lines. This increase in neutral current which cause line losses.
  • If the system has balanced phase then Neutral current flow will be less on a system. We can save thousands to millions of rupees money by reduce losses be the reducing the neutral current flow in the system
  • Thus unbalance in LV distribution network resulting in increase of neutral current.
  • Voltage or Current Shift

  • If the system is unbalanced, negative-sequence components or zero-sequence components or both may exist in the system.
  • The resistance for negative sequence current is 1/6th of the positive sequence current, which means a small unbalance in voltage waveform will give more current and thus losses.
  • Excessive power loss

  • The unbalance Voltage always causes extra power loss in the system. The higher the voltage unbalance is the more power is dissipated means higher power bills.
  • The imbalance of current will increase the I2R Losses
  • Let’s look at a simple exercise, In balance System The Load current in R Phase=200A, Y Phase=200A,B Phase=200A and in Unbalance System The Load current in R Phase=300A, Y Phase=200A,B Phase=100A,Consider Resistance of line are same in both case and all phases.
  • In Balanced System:
  • Total Load current =R+Y+B = 200+200+200=
  • Total Losses =R(I2R)+Y(I2R)+B(I2R)=40000+40000+40000=120,000Watt.
  • In Un Balanced System:
  • Total Load current =R+Y+B = 300+200+100=
  • Total Losses =R(I2R)+Y(I2R)+B(I2R)=90000+40000+10000=140,000Watt.
  • Here Total Load current is same in both case but Losses in unbalance system is more than balance system.
  • An unbalance of 1% is acceptable as it does not affect the cable. But above 1% it increases linearly and at 4% the de-rating is 20%. This means – 20% of the current flowing in the cable will be unproductive and thus the copper losses in the cable will increase by 25% at 4% unbalance.
  • Motor failure

  • In general, a three-phase motor fed by a balanced three-phase voltage with only positive-sequence component which produces only positive-sequence torque.
  • Reduce Motor life by heating: Extra loss due to voltage imbalance will heat the motor windings, by increasing the operating temperature of Motor leads to the breakdown of winding insulation and might finally in motor failure. This may also decompose the grease or oil in the bearing and de-rate the motor windings. The voltage unbalance of 3% increases the heating by 20% for an induction motor.
  • Winding insulation life is reduced by one-half for each 10°C increase in operating temperature
  • Vibration of Motor: The negative-sequence voltage caused by voltage imbalance produces opposite torque and leads to motor vibration and noise. Severe voltage imbalance may even result in motor collapse.
  • Reduce Motor Life: Heat generated by Unbalance Voltage may also reduce the Motor life
  • Reduce Efficiency: In induction motors connected to unbalanced supply, the negative sequence currents flow along with positive sequence current resulting in decreased percentage of productive current and poor motor efficiency. Any unbalance above 3% hampers the motor efficiency.
Motor Efficiency %
Motor Load % Full Voltage Unbalance
Nominal 1% 2.5%
100 94.4 94.4 93.0
75 95.2 95.1 93.9
50 96.1 95.5 94.1
  • Assume that the 100-HP motor tested was fully loaded and operated for 800 hours per year with an unbalanced voltage of 2.5%. With energy priced at 23Rs/KWH. the annual energy and cost savings calculation are
  • With Normal Voltage
  • Annual Energy Consumption=100HPx0.746X800X(100/94.4)x23=1454068Rs
  • With Unbalanced Voltage
  • Annual Energy Consumption=100HPx0.746X800X(100/93)x23=1475957Rs
  • Annual Cost Savings = 1475957-1454068=21889Rs
  • Overall savings may be much larger because an unbalanced supply voltage may power numerous motors and other electrical equipment.
  • Tripping of Motor: Negative phase sequence current flowing due to unbalance can cause faults in the motor, resulting in, tripping or permanent damage of the electrical equipment
  • Reduce Capacity: For motors, an unbalance of 5% will result in capacity reduction by 25%.
  • Tripping of VFD Drives: The variable frequency or speed drives connected to an unbalanced system can trip off. VFD treats high level unbalances as phase fault and can trip on earth fault or missing phase fault.

What is Fixture’s Beam Angle & Beam Diameter (Part-2)

How to Measure Beam Diameter at Floor:

  • If we install lights at a certain height then how much light will be on the surface will be calculated by following equation.
  • Diameter of light Speared on Floor = 0.018 × Beam angle × The distance
  • For example if we need to calculate the diameter of light for a spotlight of 14° at 3 meter distance.
  • Diameter of Light Spread on Floor=0.018×14×3=0.756
  • As light moves away from a light source, it spreads out and becomes less intense.
  • The beam spread chart below gives a quick reference for common light angles and distances.

Beam Spread at various Beam angle and distance

Beam Angle At 5 Feet At 10 Feet At 15 Feet At 20 Feet
10° 0.9 feet 1.8 feet 2.7 feet 3.6 feet
15° 1.35 feet 2.7 feet 4.05 feet 5.4 feet
20° 1.8 feet 3.6 feet 5.4 feet 7.2 feet
25° 2.25 feet 4.5 feet 6.75 feet 9 feet
40° 3.6 feet 7.2 feet 10.8 feet 14.4 feet
60° 5.4 feet 10.8 feet 16.2 feet 21.6 feet
90° 8.1 feet 16.2 feet 24.3 feet 32.4 feet
120° 10.8 feet 21.6 feet 32.4 feet 43.2 feet

Lamp has Same Lumen but Different Lux due to change in Beam Angle:

  • Amount of Lux at Floor is depending upon Distance between lamp and working floor and Beam Angle of Lamp.


  • If the Lumens and distance between working plan and lamp is the same for all the four lights having beam angle of 10°,28°,38° and 60°.
  • The amount of Lux at working plan is different. At narrow beam angle 10° it is more Lux at the center of Light (1390 Lux) and it will be reduce as we move from the center. While for wide angle 60° it is less Lux at the Center (39 Lux).

Narrow Beam Angle have Good Light (Lux) at Central

  • A LED light bulb with a narrower beam angle may also seem brighter but the overall total luminous flux (Lumen) will be the same as the same LED light bulb with a lens which produces a wider beam angle. The brighter light is created by focusing the light within a more localized area, much like a magnifying glass can be used to focus the light of the sun. This is sometimes referred to the angular intensity of the light. 
  • If we use a narrower beam angle, we will increase light intensity but reduce the size of the area being illuminated for the same height.
  • The 10 degree beam will be brightest in the center; however, the lux drops very fast away from the center. Thus, it totally is wrong to conclude that 10 degree beam is brighter than the 60 degree beam and hence10 degree beam is a better light.
  • The 60 degree beam has low center lux because it has more light spread over a larger area. The 10 degree beam is good to provide spot lighting. The 60 degree beam may be good for different lighting ambiance.

Illumination as per Distance (Inverse Square Law of Illumination):

  • Only natural light provides even illumination on earth even though it pass from clouds, environment and shadows.
  • But all artificial light are affected from various factor and when the distance increases from the light source then the illuminance reduces according to distance.
  • This is phenomena is called the inverse square law of illumination where the illuminance falls to a quarter of its value if the distance is doubled.


  • As the luminous flux (Lumen) travels away from the light source the area over which it spreads increases, therefore the illuminance (lux) must decrease. The relationship is called as the inverse square law.
  • Illumination (E) = Lighting Intensity (Lumen) / (Distance)2
  • The inverse square law describes how the intensity of a light is inversely proportional to the square of the distance from the light source (the illuminator).
  • As light travels away from the point source it spreads both horizontally and vertically and therefore intensity decreases. In practice this means that if an object is moved from a given point, to a point double the distance from the light source it will receive only a ¼ of the light (2 times the distance squared = 4).
  • Taking this theory further, if an object at 10m from a light source receives 100 LUX, moving the object to 40m, it will receive only 1/16th of the light (4 times the distance, squared = 16) resulting in the object receiving only 6.25 LUX.
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