R2 and the LDR will ALWAYS have the same potential across them if you are talking about them as a set.
The potential across one or the other will change.
So R2 (10K) and the LDR at 500Ω will let more current flow through them because the total resistance through them both is 10.5k. We can do the math and determine that:
9V / 10.5k = 0.8mA through both the LDR and R2
0.85mA * 10k = 8.5V potential across R2 only.
0.85mA * 500 = 0.5V potential across the LDR. (This happens to be the voltage at point B I was talking about)
Now if the LDR is in the dark, lets say they are both at 10k. The total resistance is 20k so we can determine:
9V / 20k = 0.45mA through both the LDR and R2.
0.45 * 10k = 4.5V across R2
0.45 * 10K = 4.5V across the LDR.
The resistance of the LDR changes which will change the current flowing through the LDR and R2... but using Ohms Law we can expect what voltage potential will exist across the LDR once we can figure out current and resistance.
You'll notice the two math examples, that even though more current flows through the LDR when it is bright, that is has a lower voltage drop... This may be hard to imagine or explain because voltage, current, and resistance changes at the same time.
Try to picture this from R2. The higher the current, the more potential will be across R2, which means that less will be across the LDR.
No tips for your presentation unless you want to experiment and get comfortable with what happens when you change R2. Maybe you can change it to stay off in normal light and turn on when you put your hand over it
The potential across one or the other will change.
So R2 (10K) and the LDR at 500Ω will let more current flow through them because the total resistance through them both is 10.5k. We can do the math and determine that:
9V / 10.5k = 0.8mA through both the LDR and R2
0.85mA * 10k = 8.5V potential across R2 only.
0.85mA * 500 = 0.5V potential across the LDR. (This happens to be the voltage at point B I was talking about)
Now if the LDR is in the dark, lets say they are both at 10k. The total resistance is 20k so we can determine:
9V / 20k = 0.45mA through both the LDR and R2.
0.45 * 10k = 4.5V across R2
0.45 * 10K = 4.5V across the LDR.
The resistance of the LDR changes which will change the current flowing through the LDR and R2... but using Ohms Law we can expect what voltage potential will exist across the LDR once we can figure out current and resistance.
You'll notice the two math examples, that even though more current flows through the LDR when it is bright, that is has a lower voltage drop... This may be hard to imagine or explain because voltage, current, and resistance changes at the same time.
Try to picture this from R2. The higher the current, the more potential will be across R2, which means that less will be across the LDR.
No tips for your presentation unless you want to experiment and get comfortable with what happens when you change R2. Maybe you can change it to stay off in normal light and turn on when you put your hand over it