# Light dependent resistor

I am trying to understand how this works. I've read that in dark conditions, the resistance of the LDR is high so it takes in the larger voltage. As the LDR and filament bulb are in parallel, they will have the same voltage.

What I can't understand is why the voltage increases when resistance increases.. can someone please explain?
(edited 1 year ago)
You are correct, the PD across the bulb and the LDR have to be the same because they are connected in parallel.

I don't think the person who drew that diagram ever tried to test the circuit

That circuit would have to work by changing the resistance of the bulb and LDR in parallel so much that the PD across the bulb varied a lot

typical maximum (dark) resistance of a LDR ~1M Ohm
typical minimum (light) resistance of a LDR ~ 10K Ohm

lets assume a 12V 5W tungsten filament bulb
at rated power that's passing 0.42 A which gives it a resistance of 28 Ohm - we'll assume that's constant even though in reality the resistance goes up as the filament gets glowing hot.

calculating the effective resistances of the bulb in parallel with the LDR

Dark
1/R=1/1000000 + 1/28
R=27.99

Light
1/R=1/10000 + 1/28
R=27.92
It's a consequence of the rule for combining resistors in parallel that when the resistances are widely different the lower resistance overpowers the effect of the high resistance in parallel with it.

it should be easy to satisfy yourself that these small changes in resistance are insufficient to make any significant difference in the voltage across the bulb in the circuit provided.

The real way to control a light with an LDR is to put the LDR in series with a fixed resistor to make a potential divider and use the midpoint voltage to drive a voltage controlled electronic switch. the electronic switch controls the lamp.