How to Size a Photocell for a Lighting Installation

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Now that you know how to install and wire a photocell in a lighting installation. The next step would be to know how to determine the current rating of the photocell for a given application so that when installed, it does not burn off easily and create problems for the lighting installation.

Consider the simple lighting installation below where a photocell has been used.

Let LP 1 =LP2 = LP3 = LP4 = 250W

Power supply Voltage, V = 240V

Power Factor = 0.5 (discharge lamps see Typical power factor for common electrical loads)

Power in a single phase circuit is given by:

$P = VICosՓ$       ((see Voltage Drop and Power Formulas for Electrical Engineers)

Where I is the rated current of the photocell.

Now from the above formula for power, we get : 

$I = \frac{250}{(240 * 0.5)} = 2.0833 Amps$ 

Now the photocell should be able to withstand the inrush current of a discharge lamp which is about 1.6 times nominal current.

Hence actual current rating of photocell = 1.6 x 2.0833 = 3.33 Amps

A photocell rated 5 Amps should just do for the above application with four (4) discharge lamps. However as the number of lamps to be controlled increases, it becomes impractical to use a photocell switch to carry the lighting loads directly.

What is normally done is to use a power contactor with a higher current rating to carry the load while the photocell switch will be used to power the contactor coil. 

See How to Wire a Photocell Switch to Lighting Loads with a Contactor

How to Install and Wire a Photocell Switch in a Lighting Installation

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A photocell switch is essentially a light dependent resistor, LDR. Its resistance decreases with increasing incident light intensity. They are used in many applications for on-off control especially in lighting installations.

In lighting applications, Photocells are placed in streetlights to control when the lights are ON or OFF. During daylight, light falling on the photocell causes the streetlights to turn off and during night hours or darkness to turn on. Thus energy is saved by ensuring the lights are only on during hours of darkness.

How to Wire a Photocell 

A photocell used in lighting application has three terminals labelled as:

1. Load line (Lo)

2. Neutral line (N)

3. Supply or live line (LI)

In most photocells, the load line wire is RED, the neutral wire is WHITE and the Supply line is black. 

This colour code is not universal. It may change depending on the manufacturer of the photocell switch.

The picture of the terminals of a brand of photocell is shown below:

Photocell Terminal Markings

Wiring and installing a photocell is pretty straight forward as shown below:

How to Wire a Photocell Switch in a Lighting Installation

As shown above, the load wire (Lo) goes to the lighting installations connected in series while the neutral (N) wire through a breaker is looped to all the lights. The supply line through a breaker supplies the photocell electrical power.

See also How to wire a photocell switch to lighting loads with a contactor

How to size a photocell for a lighting installation

Voltage Drop and Power Formulas for Electrical Engineers

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Working with single phase, three-phase and DC (direct current circuits) and you quickly need to reference formulas for voltage drops and power calculations for a given conductor? The table below provides a quick reference for these calculations.

Voltage Drop and Power Calculation Formulas For Single Phase Circuits

Electrical Parameters	Formulas
Voltage Drop	$∆V = 2*I*L*(rCosՓ + xSinՓ)$
% Voltage Drop	% $∆V  =  \frac{∆V}{V_r}*100$
Active Power	$P = V*I*CosՓ$
Reactive Power	$Q = V*I*SinՓ$
Apparent Power	$S =  V*I = \sqrt{{P^2} +{Q^2}}$
Power Factor	$CosՓ = \frac{P}{S}$
Power Loss	$P_L = 2*L*r*I^2$

Voltage Drop and Power Calculation Formulas For Three-Phase Circuits

Electrical Parameters	Formulas
Voltage Drop	$∆V = \sqrt{3}*I*L*(rCosՓ + xSinՓ)$
% Voltage Drop	% $∆V = \frac{∆V}{V_r}*100$
Active Power	$P = \sqrt{3}*V*I*CosՓ$
Reactive Power	$Q = \sqrt{3}*V*I*SinՓ$
Apparent Power	$S = \sqrt{3}*V*I = \sqrt{{P^2} +{Q^2}}$
Power Factor	$CosՓ = \frac{P}{S}$
Power Loss	$P_L = 3*L*r*I^2$

Voltage Drop and Power Calculation Formulas For Direct Current (DC) Circuits

Electrical Parameters	Formulas
Voltage Drop	$∆V = 2*I*L*r$
% Voltage Drop	% $∆V = \frac{∆V}{V_r}*100$
Active Power	$P = V*I$
Reactive Power	$ - $
Apparent Power	$ - $
Power Factor	$ - $
Power Loss	$P_L = 2*L*r*I^2$

Meaning of symbols used in the formulas above:
$L$       =   Total length of conductor
$r $       =    Resistance of conductor per unit length
$x$       =    Reactance of conductor per unit length
$∆V$    =   Voltage drop 
$P$       =   Active power
$Q$      =    Reactive power
$I$        =   Current

Formulas For Star - Delta Transformations in Three - Phase Electrical Circuits

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In three - phase electrical circuits, there is often the need to transform from a Delta configuration to a Star configuration and vice versa. The formulas for making these conversions are detailed below:

For a Star  to Delta Transformation, as shown in the formula above:



$Z_{12} = Z_1 + Z_2  + \frac{Z_1 Z_2}{Z_3}$

$Z_{23} = Z_2 + Z_3  + \frac{Z_2 Z_3}{Z_1}$

$Z_{13} = Z_3 + Z_1  + \frac{Z_3 Z_1}{Z_2}$




For a Delta to Star Transformation, we have:
$Z_1 = \frac{Z_{12} Z_{13}}{Z_{12}+Z_{13}+ Z_{23}}$

$Z_2 = \frac{Z_{12} Z_{23}}{Z_{12}+Z_{13}+ Z_{23}}$

$Z_3 = \frac{Z_{23} Z_{13}}{Z_{12}+Z_{13}+ Z_{23}}$

With the above formulas, you can easily transform from a Delta configuration to a Star configuration or vice versa.

How to Size the Neutral Conductor in an Electrical Installation

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In a balanced three phase systems, the current in the neutral conductor is theoretically zero. However in a practical electrical installation, this is not the case. In fact there is always some current flow in the neutral although small if the loads in the three phase are sufficiently balanced. However increasing current will flow through the neutral in an installation with high harmonics necessitating the need to appropriately determine the minimum cross sectional area of the neutral that will be safe for the installation.

Given the implications of under sizing the neutral conductor, the neutral conductor, shall have the same cross section as the line conductor:

1. in single-phase, two-wire circuits whatever the section;

2. in poly-phase and single-phase three-wire circuits, when the size of the line conductors is less             than or equal to 16mm2 in copper, or 25mm2 in Aluminium.

The cross section of the neutral conductor can be less than the cross section of the phase conductor when the cross section of the phase conductor is greater than 16mm2 with a copper cable, or 25mm2 with an aluminium cable, if both the following conditions are met:

1. The cross section of the neutral conductor is at least 16mm2 for copper conductors and 25mm2          for aluminium conductors;

2. There is no high harmonic distortion of the load current. If there is high harmonic distortion (the          harmonic content, THD, is greater than 10%), as for example in equipment with discharge                  lamps, the cross section of the neutral conductor cannot be less than the cross section of the                phase conductors.

The table below shows the minimum cross sectional area of the neutral conductor in a given electrical installation under different types of circuits:

Type of Circuit	Phase Conductor Cross Section, S, (mm2)	Minimum Neutral Conductor Cross Section, SN (mm2)
Single Phase/Two Phase Circuits - Copper/Aluminium	Any	S
Three-Phase Circuits - Copper	S ≤ 16	S
	S > 16	16
Three - Phase Circuits - Aluminium	S ≤ 25	S
	S > 25	25

Recommended Wire and Protective Device SIzes for Domestic Electrical Installations

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Safe wiring in domestic installations requires the use of the correct sizes of wire and protective device to avoid over loading switch gears and creating the potential for a fire hazard. The wiring regulations recommended the sizes for wire and protective devices enumerated in the table below as the minimum requirement for any domestic wiring  and installation

Type of Circuit	Supply Voltage	Recommended Wire Size	Recommended Size of Protective Device (Amps)
Lighting	230VAC	1.5mm2	10 Amps Fuse, 10 Amps Circuit Breaker
Sockets	230VAC	2.5mm2	16 Amps Circuit Breaker, 20 Amps Fuse
Water Heater	230VAC	2.5mm2	16 Amps Circuit Breaker, 20 Amps Fuse
Washing Machine	230VAC	2.5mm2	16 Amps Circuit Breaker, 20 Amps Fuse
Air Conditioners	230VAC	2.5mm2	16 Amps Circuit Breaker, 20 Amps Fuse
Cooker	230VAC	6.0mm2	32 Amps Circuit Breaker, 32 Amps Fuse

How to Wire Socket Outlets in a Domestic Installations

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The power points in an electrical installation are called socket outlets. These socket outlets commonly called plug points are wired in two ways. These are the ring circuit wiring and the radial circuit wiring.

Ring Circuit Wiring

A ring circuit commences from the consumer unit or distribution board through an MCB or fuse of specific rating usually 30 Amps and loops into each socket outlet and returns to the same MCB or fuse in the consumer unit or distribution board. Looping of the ring circuit wires must be done for the live conductor, neutral conductor and the protective conductor in separate rings. The ring method of connection is done only for the 13 Amps socket outlets, as the individual 13Amps plugs are separately having fuses. A ring circuit may have an unlimited number of socket outlets provided that the floor area served by the ring does not exceed 100 square meters and that the maximum demand of the circuit does not exceed the MCB or fuse rating. A kitchen should usually have a separate ring circuit.

Radial Circuit Wiring

A Radial circuit commences from the consumer unit/distribution board through an MCB/fuse of specific rating e.g. 20Amps, loops into each socket outlet but ends at a socket outlet and does not return to the original fuse/MCB at the consumer unit or distribution board.

How to Wire a Two-Way Switch

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When it comes to electrical wiring in the domestic arena specifically in a house, two-way switches are very popular. A two-way switch is used when it is necessary to operate a lamp from two positions, such as at the top and bottom of a staircase and at the ends of a long corridor. The connection and operation are shown in the diagram below:

Typical Two-Way Circuit in Domestic  Installations

Basics of Harmonics in Electrical Systems:

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Harmonics are essentially distortions in the electrical power systems. An electrical power system can be represented by a sinusoidal waveform which varies with time. The harmonic with frequency corresponding to the period of the original waveform is called fundamental and the harmonic with frequency equal to “n” times that of the fundamental is called harmonic component of order “n”. The presence of harmonics in an electrical system is an indication of the distortion of the voltage or current waveform and this implies such a distribution of the electric power could result in the malfunctioning of equipment and protective devices.

The harmonics are nothing less than the components of a distorted waveform and their use allows the analysis of any periodic non-sinusoidal waveform through different sinusoidal waveform components. 

Harmonics Distortions Waveform. Photo Credit: ABB

Causes of Harmonics in Electrical Systems

Harmonics are generated by nonlinear loads. When we apply a sinusoidal voltage to a load of this type, we shall obtain a current with non-sinusoidal waveform.

The main equipment generating harmonics are:

1. personal computer

2. fluorescent lamps

3. static converters

4. continuity groups

5. variable speed drives

6. welders.

7. Transformers (mostly third harmonics which becomes insignificant with increasing loading of the transformer).

In general, waveform distortion is due to the presence, inside of these equipment, of bridge rectifiers, whose semiconductor devices carry the current only for a fraction of the whole period, thus originating discontinuous curves with the consequent introduction of numerous harmonics.

Effects of Harmonics in Electrical Systems

The effects of harmonics can be felt in both the current and voltage.

The main problems caused by harmonic currents are:

1) overloading of neutrals

2) increase of losses in the transformers

3) increase of skin effect.

The main effects of the harmonics voltages are:

4) voltage distortion

5) disturbances in the torque of induction motors (since Torque is proportional to supply voltage)

Total Harmonic Distortion (THD) in an Electrical System

If the rms values of the harmonic components are known, the total rms value can be easily calculated by the following formula:

Total Harmonic Distortion (THD)

The total harmonic distortion is defined as:

The harmonic distortion ratio is a very important parameter, which gives information about the harmonic content of the voltage and current waveforms and about the necessary measures to be taken should these values be high.

For THDi < 10% and THDu < 5%, the harmonic content is considered negligible and such as not to require any provisions