Simple circuit question involving an LDR and a fixed resistor - I have a question.

EDIT: Ohh, do you mean because for both resistors (LDR and fixed), the value of I will be the same. So in V=IR for both resistors, the highest resistor size will have the most voltage across it, and the smaller resistor would have a smaller voltage through it? Is that right? If so, I love you <3

EDIT: Ohh, do you mean because for both resistors (LDR and fixed), the value of I will be the same. So in V=IR for both resistors, the highest resistor size will have the most voltage across it, and the smaller resistor would have a smaller voltage through it? Is that the right reason of explaining it?

Yaaayyyy!!!! That's exactly right.

And just to clarify, .......the value of I) is the same for each resistor because they are in series

so if you were to eliminate/ignore the current value, the smaller the resistor, the smaller the voltage and the larger the resistor, the larger its voltage would be (as V=IR)?

Correct.



Yes.

Ohms law still applies.

V = IR

Remember that in our example, V is the supply voltage, I is the total current and R is the summed series resistance (R_LDR + R_series)

Rearranging to find I_total

I_total = V_supply / R_total

And we know that because this is a series circuit, I_total is the same in both resistances.


Now that we have I_total, we can find the pd developed across either resistance by simply replacing the subscript:

V_LDR = I_total x R_LDR

V_series = I_total x R_series



As a final check, sum the two pd's and they should always equal the supply voltage.

V_series + V_LDR = V_supply

Thanks so much for your help, I really really appreciate it!

I have a question that isn't really related to this exam question: Does a resistor reduce both voltage and current?

I know it reduces the total circuit current (as I=V/R, bigger R means smaller I), but does it reduce the voltage too? I guess they do reduce voltage too, as voltage across a resistor is energy being dissipated, so other components get less voltage (and current). Is that correct (that it reduces both voltage and current)?

In essence, yes. But be careful, this applies to series resistances only.

Placing a resistor in series with another will increase the circuit resistance, but also reduces the current through both. The supply voltage will be divided across both resistors in the same ratio as the resistances to each other.

i.e. if the resistors were in the ratio 2:1, then the larger resistor will have 2/3 of the supply voltage dropped across it and the smaller resistor will drop 1/3 of the supply across it. 2/3V + 1/3V = V = supply voltage.

The situation is different when resistors are placed in parallel. But that is for a different question.

Well done, that sounds right to me

Just occurred to me that you've recently been doing transformers and power supplies... might be worth mentioning that potential dividers (also know as voltage dividers) like this aren't useful as power supplies. usually you'll see them drawn with a volt meter across the output or with the output just left unconnected.
They are really useful for producing a signal voltage. e.g. in the last diagram Vout would be a signal voltage indicating the amount of light falling on the LDR

...But if you try take a significant amount of power out through at Vout by putting a lightbulb or similar load there, you mess up the potential divider effect... some current is going out into your load, so there is now a difference between the current through the top and bottom resistors.

it's OK to put Vout across a Voltmeter or an oscilloscope (or another instrument) which has very high resistances and don't cause the current in the top resistor to change significantly.

They're for making voltage measurements with, not supplying power.

hope this hasn't added confusion - I remember I got taught potential dividers and power supplies at the same time and tbh I don't think it was explained sufficiently that they're not supposed to do the same job as each other... and IIRC it took some struggle to figure that out myself.

Thanks. For a normal wire at a constant temperature, I know that the graph would have a positive correlation, as when V increases, I increases too because V=IR.

However, I see that for internal resistance experiments, they vary the value of R with an external variable resistor, and plot a graph of V against I, where as V increases, I decreases.

Why is that? Can you explain that to me please? I have a feeling it is very simple but i cannot understand why when you vary the resistance, the current would increase when the voltage decreases.

Thanks

Hi
the setup here is similar to the potential divider circuits you were looking at yesterday.
R and r are in series and at any instant the current through both R and r is the same.
the Pd across R plus the Pd r across must add up to the value of the internal EMF of the cell.
The cell's internal EMF is constant and the internal resistor r is constant.

what the experimenter is doing is varying I and measuring the V across the external resistor. V is the dependent variable and I is the independent variable... Changing I is the 'cause' and changes in V are the 'effect'

as with the potential dividers yesterday the current is the same at every point in the circuit but the PD across each component can vary in proportion to each other... but thePD's can't just be anything, the PD across r and R has to add up to the EMF.

Because the EMF is constant and the r is constant and we know the rules about the current being equal through every series component and the PD's adding up to the EMF we're able to find the value of r as shown in the book.

Ok thanks. When you slide the variable resistor to vary the current, does that not also vary the voltage (as V=IR)?

And why does the resistor vary the current and not the voltage (if it doesnt vary the voltage)?

If the resistor also varied the voltage, then would that change in voltage not affect the result? I suppose the resistor would cause a voltage drop across the circuit (as well as reducing the total current), am I right?

If the variable resistance decreases, the ratio between the variable resistor and the battery internal resistance changes. More volts are dropped across the internal resistance because the current has increased. .

less volts are dropped across the variable resistance because its' resistance has decreased .

The current has increased, as I=V/Rtotal, and Rtotal is less because the variable resistor is less. Have I understood that right?

And then V=IR, R being constant for the internal resistance, so as I increases, V must increase? (V being the pd across the internal resistance, so when that increases, there are more lost volts, so the terminal pd is less)

So because the resistance of the variable resistor has decreased, V=IR means the voltage dropped across it decreases (even though the current has increased a little)?

How come that even though the current has increased, the voltage always decreases across the variable resistor? Is it because both resistors (internal and variable) are in series, so the current across them both is the same, so if you were to use V=IR and divide by I (as it would be the same value in both resistor calculations of voltage), then the internal resistor would have a bigger voltage across it because the current has increased? Is that the right concept?