Why does potential difference increase when resistance increases?

More context needed please.

https://imgur.com/65xTDkt
not sure if this works but this is the question

The fixed resistor and LDR in series form a potential divider, if you've got some notes on potential dividers you could investigate how the PD across each of the resistors changes when the resistance of one of them is varied.

I'm mainly wondering because of this question in a GCSE Combined Science Trilogy Physics Paper:


It's referring to the relationship between potential difference, current and resistance when the light intensity changes (could be referring to temperature as well, however, the mark scheme mentions light intensity only).
I've grasped my head around current increasing when resistance decreases as well as resistance decreasing when light intensity increases.

However, the mark scheme mentions the potential difference increasing across the fixed resistor while it decreases across the LDR, and I'm unsure why. In fact, I'm confused as to the relationship between resistance and potential difference in general.

https://imgur.com/65xTDkt

not sure if this works but this is the question

Sometimes, it is good to do some numerical computation before moving to the “general relationship”.
Assume that the emf of the ideal cell to be 12 V, the resistance of the fixed resistor and the LDR to be 3 ohms and 9 ohms, respectively.
The total current in the series circuit is 1 A, so
p.d. across the fixed resistor = 3 V
p.d. across the LDR = 9 V
Next, says, as the light intensity increases, the resistance of LDR decreases to 3 ohms.
The total current increases to 2 A (12 ÷ 6), while
the p.d. across the fixed resistor increases from 3 V to 6 V,
and the p.d. across LDR decreases from 9 V to 6 V.
Can you summarise the computation in words without the numerical value now?

Potential divider = voltage divider
https://www.bbc.co.uk/bitesize/guides/z22mmp3/revision/3
https://www.bbc.co.uk/bitesize/guides/zk37hyc/revision/8

oooh right, so when the light intensity increases, the current throughout the circuit increases as the resistance of the LDR decreases

and then because resistance of the LDR decreases, the p.d across it decreases as well, leading to an increase in the p.d of the resistor since there's more to share

this also means that the reading on the voltmeter decreases as well (as it measures it across the LDR) and then the reading on the ammeter increases

and then because resistance of the LDR decreases, the p.d across it decreases as well, leading to an increase in the p.d of the resistor since there's more to share

Most of the writing is good except
Note that the resistance of LDR decreases but the current in the LDR increases, this does not explain why the p.d. across LDR decreases.

what would i have to say for that?
all i can think of is how there's less "push" needed for the current to flow through the LDR - as its resistance has decreased - therefore less potential difference is needed for it and more of the p.d. share can go to the resistor

there's probably a much more accurate explanation for it since i don't really know how to explain the link between current and p.d or the link between resistance and p.d

Be careful in the exam, we cannot conclude the dependent variable to increase or decrease when one variable increases while the other variable decreases. A common error.
Why does the p.d. across LDR decrease when the current increases?
Note that the sum of the p.d. across the fixed resistor and LDR is fixed, that is the emf of the cell.
When the current increases, the p.d. across the fixed resistor increases and this would cause the p.d. across the LDR to decrease because the total p.d. across the fixed resistor and LDR is the emf of the cell which does not change.