### 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 discharge 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:

$P = VICosՓ$       (

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.

### 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 for some other brand of photocell.
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.

### 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