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

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    (Original post by blobbybill)
    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
    Well done, that sounds right to me :top:

    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.
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    (Original post by blobbybill)

    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.
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    (Original post by uberteknik)
    Yaaayyyy!!!! That's exactly right.
    And just to clarify, that is because the "current multiplier ratio" so to speak (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)?
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    (Original post by blobbybill)
    And just to clarify, .......the value of I) is the same for each resistor because they are in series
    Correct.

    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)?
    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 (RLDR + Rseries)

    Rearranging to find Itotal

    Itotal = Vsupply / Rtotal

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


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

    VLDR = Itotal x RLDR

    Vseries = Itotal x Rseries



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

    Vseries + VLDR = Vsupply
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    (Original post by uberteknik)
    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 (RLDR + Rseries)

    Rearranging to find Itotal

    Itotal = Vsupply / Rtotal

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


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

    VLDR = Itotal x RLDR

    Vseries = Itotal x Rseries



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

    Vseries + VLDR = Vsupply
    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)?
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    (Original post by blobbybill)
    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.
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    (Original post by uberteknik)
    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.
    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

    Attachment 614872
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    (Original post by Joinedup)
    Well done, that sounds right to me :top:

    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

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    (Original post by blobbybill)
    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

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    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.
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    (Original post by Joinedup)
    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?
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    (Original post by blobbybill)
    Ok thanks. When you slide the variable resistor to vary the current, does that not also vary the voltage (as V=IR)?
    Yes. As the resistance is varied, the p.d. developed across the variable resistor will also vary.

    And why does the resistor vary the current and not the voltage (if it doesnt vary the voltage)?
    See above. It varies both the p.d. voltage across both the variable resistor and the p.d. voltage across the internal resistance.

    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?
    Be very clear which voltage you are talking about. This is critical.

    The variable resistor changes the load resistance which will alter both the total current in the circuit and also the p.d. developed across itself.

    As the current also flows through the battery internal resistance, changes in the current will also change the p.d. voltage developed across the internal resistance.

    But changing the variable resistor WILL NOT change the battery emf because we assume that the battery is capable of sourcing unlimited current.

    It's the internal resistance of the battery which causes the voltage at the battery terminals to vary because volts are lost across the internal resistance.

    Remember, this is the same as the potential divider action discussed in this and other threads.

    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.

    If the variable resistance increases, the ratio between the variable resistor and the battery internal resistance changes. Less volts are dropped across the internal resistance because the current has decreased. more volts are dropped across the variable resistance because its' resistance has increased.
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    (Original post by uberteknik)
    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. .
    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)

    less volts are dropped across the variable resistance because its' resistance has decreased .
    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?
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    (Original post by blobbybill)
    The current has increased, as I=V/Rtotal, and Rtotal is less because the variable resistor is less. Have I understood that right?
    Correct.

    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)
    Correct

    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)?
    Correct

    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?
    Exactly right.

    As the variable resistor decreases and the total current increases, the battery internal resistance hogs more of the available battery emf.

    The battery emf must be shared between the variable resistor and the internal resistance. More current equals more volts lost to that pesky internal resistance and the variable resistance p.d. must go down as a result.

    In practical use, it's exactly the reason why the battery terminal voltage is demonstrated to fall as heavier loads demand more current.

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    By Jove Caruthers, I think he's got it!

 
 
 
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