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Please explain this fairly simple thing to me about voltages in a circuit Watch

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    Hey

    So I am doing this past paper question, and I see the mark scheme says that the lower the resistance of the LDR, the lower the p.d. across the LDR OR the higher the p.d. across the fixed resistor.

    Why would that be? Can you explain why 1) a lower LDR resistance would mean a lower p.d across the LDR, and can you also explain why 2) a lower LDR resistance would mean a higher p.d. across the fixed resistor?

    I really don't get it. I know that V=IR, but when the resistance of the LDR decreases, the voltage would decrease, but the current wouldn't stay the same in the circuit would it because the total resistance would have changed? I am getting really confused.

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    (Original post by blobbybill)
    Hey

    So I am doing this past paper question, and I see the mark scheme says that the lower the resistance of the LDR, the lower the p.d. across the LDR OR the higher the p.d. across the fixed resistor.

    Why would that be? Can you explain why 1) a lower LDR resistance would mean a lower p.d across the LDR, and can you also explain why 2) a lower LDR resistance would mean a higher p.d. across the fixed resistor?

    I really don't get it. I know that V=IR, but when the resistance of the LDR decreases, the voltage would decrease, but the current wouldn't stay the same in the circuit would it because the total resistance would have changed? I am getting really confused.

    Attachment 612480
    It doesn't matter if the current changes; the same current is going through both resistors.

    If they have resistances R_1, R_2, then the total resistance is R_1+R_2 and so the current is \dfrac{V}{R_1+R_2}. So then the voltage across the first resistor is V \dfrac{R_1}{R_1+R_2} and across the second V \dfrac{R_2}{R_1+R_2},

    In other words, the voltage is "shared" between the resistors in proportion to how the resistance is shared. So if R_1 decreases, it's now a smaller share of the total resistance and so you end up with a lower voltage across R1. Similarly if R_1 increases.
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    Okay so basically since they are both in series they share voltage. The components with a higher resistance have a greater ratio of potential difference acting through them. If the resistance of the LDR decreases there will be less collisions in the LDR so therefore there is less energy transferred through the LDR as more of it will be transferred through the fixed resistor instead. Since the overall voltage stays the same the fixed resistor would get the remaining voltage which will now be a higher value than before.
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    The LDR and fixed resistor are making a potential divider (voltage divider).
    it's quite a common thing to do with LDRs and other variable resistance components you want to use as sensors.

    at any instant the current has to be the same in both the LDR and fixed resistor because they are in series

    the PD across the components in series has to equal the PD between the battery terminals
    because of this if the resistance of one component goes down the potential across that component goes down (and at the same time the potential across the other component goes up because the potential across both still has to add up to 6V )
 
 
 
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