Resistance and Voltage Question A Level

hiya, me and physics again,

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-question-paper-unit-g482-01-electrons-waves-and-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

thank you!!!

hiya, me and physics again,

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-question-paper-unit-g482-01-electrons-waves-and-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

thank you!!!

It's a potential divider circuit as well so think about ratios.

ah true, so with potential dividers is that involving 1/Rtotal = 1/R1 + 1/R2 etc, and if so then where from there?

((thank you btw!!))

ah true, so with potential dividers is that involving 1/Rtotal = 1/R1 + 1/R2 etc, and if so then where from there?

((thank you btw!!))

I would use Voltage out = (Resistance2)/(Total resistance) x Voltage In

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-question-paper-unit-g482-01-electrons-waves-and-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

ok cool, so if you used that then R2 = 400 and total would = 6000/23, or is this for the series part of the circuit??

for the second part of the reply, that means more current can flow if there are more free electrons, right?

As the light intensity on the LDR increases, its resistance falls. Therefore, the resistance of the 750 Ohm + LDR (in parallel) section decreases.

Consider the LDR+750 Ohm as a sinlge (decreasing resistance) in series with $R_1$. We see that less of the 45V drop will occur over the 750 Ohm + LDR section, as its resistance is decreasing.

More free electrons = lower resistance.
As the LDR has lower resistance, the ratio changes. The 750 ohm resistor now takes a LARGER chunk of the voltage than before.

ok cool, so if you used that then R2 = 400 and total would = 6000/23, or is this for the series part of the circuit??

for the second part of the reply, that means more current can flow if there are more free electrons, right?

ahh ok thank youu!! so the voltage $R_1$ resistor stays the same (because it's resistance remains constant) meaning the parallel loop voltage falls, so both the 750 ohm resistor and LDR fall in voltage also?

ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?

ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?

ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?

ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?

You have $R_1$ in series with the combination of 750 Ohm and the LDR (which are in parallel) - let's call that $R_2$.

As $R_2$ reduces in resistance (as the LDR does), more current flows in the circuit. As $R_1$ doesn't change, it accounts for more of the voltage drop than $R_2$.

Another way of thinking about it is that $R_1$ and $R_2$ form a voltage divider. As $R_2$ reduces, it takes less of the voltage drop.

This statement makes me think that electricity hasn't been explained very well to you, so I'll try.

Think of the circuit as plumbing. The battery is a pump - it pushes the water around the circuit. The resistors are valves that are more or less closed, resisting the flow of the water.

The LDR is a valve that is effectively being opened by the light - it's resistance decreases, so the flow of water through it will increase.

The voltage difference between any two points can be thought of as the pressure difference. It's always a difference - not an absolute value. By convention (unless otherwise stated) it's measured between the negative terminal of the battery and the point of interest.

The current is the rate at which water is flowing.

By opening the LDR valve, more water (current) flows through the pump (battery), as the total resistance has decreased. Less pressure (voltage) is needed to push water (current) through the parallel combination of the LDR and 750 Ohm resistor, so more of the pump's pressure (voltage) is dropped over $R_1$.