Good Afternoon Everyone,

First post on the forum. Just seeking abit of guidance with an LDR based project, I have the schematic to hand which i have attached. I'm struggling to work out the voltage drop across the LDR when it changes according to light sensitivity. It has a resistance of 1 Megaohm in the dark and 500 ohm during daylight. Theres load of information on circuits with series resistors but I'm finding limited information on formulas that are parallel based as i'm unsure how you treat the LDR in the circuit?

Cheers

 

The LDR is treated as a variable Resistor.
Replace the LDR with a 1 M Ohm resistor in the "Dark" state and 500 Ohm in the "Light" state.

Read this. https://www.electronics-tutorials.ws/dccircuits/voltage-divider.html

 

Didn't actually think i

The LDR is treated as a variable Resistor.
Replace the LDR with a 1 M Ohm resistor in the "Dark" state and 500 Ohm in the "Light" state.

Read this. https://www.electronics-tutorials.ws/dccircuits/voltage-divider.html

Didn't actually think it was that simple, So just to clarify, R1+R2 treat as parallel and the 'variable resistor in series?

Thanks

 

, R1+R2 treat as parallel and the 'variable resistor in series?

No.
R1 is a load resistor that limits the current to the LED.
It is independent of R2.

The variable resistor is in series with R2.

 

No.
R1 is a load resistor that limits the current to the LED.
It is independent of R2.

The variable resistor is in series with R2.

Thanks for your help, that was my confusion with it, as you can see, I'm fairly new to this.