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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:

$P = VICosŐ“$ ((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.