I'm not an expert, but do know something.
If it is a LED, there already is a series resistor. Doubling the resistor value halves the current.
If a light bulb, you can use the following formula to calculate the required series resistor for a bulb voltage you want.
R = ((Vb
- U) * (Vb
/ P)) / U
R is the required resistance.
if the battery voltage (nominal 12V).
U is the voltage you wish to have over the lamp.
P is the nominal lamp power in watts.
= 12V, U is 8V, P is 1W.
R = ((12 - 8) * (12 * 12 / 1)) / 8 = 72 ohm
Once you have the resistance, you get the power loss in the
- U) * (Vb
- U) / R
In the example above:
= (12 - 8) * (12 - 8) / 72 = 0.22W
The problem with the above calculation of course is that it does not account for the decreasing lamp resistance with decreasing lamp voltage. Therefore you need to select somewhat lower resistance than the formula gives.
To switch the illumination on when it gets dark you need more components than just the LDR. There could be ready schematics for it in the internet, but I couldn't come up with the right keywords to find any right away.