# What happens to the resistance in the LDR if there is dark

In darkness the intensity of light will be low then resistance increases in LDR,due to increases in resistance the current will flow less according to ohm’s law. But in actual practice (automatic streetlight) the current will flow high in dark which helps to glow the street light in dark…what is the reason behind that.
Can anyone explain what is the reason behind that

Hello Mrunali !The voltage is high in darkness so according to the ohm’s law(V=IR)in the darkness the voltage is high & the current is also going high so to limit the current resistor is used but resistance is limiting the amount of current,but the required current is flowing to glow the street,It depends on the resistor which we use to limit the current,because the limitation on current is depends on the resistance

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If you want to see an application where output voltage should increase as light intensity decreases(as it gets darker), try the following configuration 3.3v-R1-A0-LDR-Gnd.
As resistance of LDR increases in dark, more of the voltage can be observed at A0 since it is proportional to Rldr/(R1+Rldr).
This is strictly not the case with streetlights, but I guess you could make such a model using the above configuration.

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The LDR is a sensor unit whose internal resistance changes based on the light falling on it.

This change in resistance can be detected as a change in voltage by using the LDR circuit as a resistive component in a voltage divider circuit.

For automatic lighting, this signal is given to some form of a switch, which turn on or turns off the streetlight depending on the voltage generated by the LDR voltage divider circuit.

This circuit can be made in such a way that when the voltage output of the voltage divider circuit is high, the streetlight turns on, and turns off when the voltage is low. The circuit can also be made to act completely opposite. It completely depends on the components you have used for the voltage detection and the switching circuit.

The current flowing through the LDR never directly affects the streetlight.

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Bro wat is the use of using 10ohms resistor,caz ldr is itself a resistor

Its 10k ohms resistor

Tq for ur info. Now my doubt is clarified…

@ritugurav99 how the current increases when resistance in ldr increases in dark

usually the change in resistor value of an ldr is measured using analog pin.
so in automatic streetlight we set a threshold value such that in dark light ldr must crossed that threshold value and accordingly light should turned on.

When light falls on the LDR, current starts to increase & the resistance decreases & opposite action will be in the darkness

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In darkness, Resistance decreases and LED glows and vice-versa.
Real-life example : Street Lights
Hope it helps. Thankyou

LDR depends on the light.When light falls on the LDR then the resistance decreases (because when light falls on the LDR,currentb starts to increase and according to ohms law,increase in current leads to decrease in the resistance) and increases in the dark(vice versa as explained for when LDR is in light).Hope your doubt is cleared.Feel free ro ask if you have any doubts…Thank you…!!