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Explanation of low resistance wires in a transformer Watch

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    All I know is that P(AC main) = IV and P(loss) = I^2R, so if you decrease R, you get less power loss for the same power input.

    I don't understand why P(loss) = I^2R, or why low resistance in the wires doesn't affect the current from the mains. Could anyone provide a basic, but full explanation of why low resistance wires are used in transformers to reduce power loss? Thank you.
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    (Original post by Teenage Hype)
    All I know is that P(AC main) = IV and P(loss) = I^2R, so if you decrease R, you get less power loss for the same power input.

    I don't understand why P(loss) = I^2R, or why low resistance in the wires doesn't affect the current from the mains. Could anyone provide a basic, but full explanation of why low resistance wires are used in transformers to reduce power loss? Thank you.
    P=IV [1]
    and
    V=IR [2]

    so sub [2] into [1]

    P=I (IR)
    P=I2R

    ---
    this relationship also explains why it's more efficient to do long distance transmission of an amount of electrical power at high voltages and low currents rather than the other way around - it's easier to make the voltage 1000 times higher (and consequently the current 1000 times lower) than it is to make the cables have 1*106 times the cross sectional area.
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    (Original post by Joinedup)
    P=IV [1]
    and
    V=IR [2]

    so sub [2] into [1]

    P=I (IR)
    P=I2R

    ---
    this relationship also explains why it's more efficient to do long distance transmission of an amount of electrical power at high voltages and low currents rather than the other way around - it's easier to make the voltage 1000 times higher (and consequently the current 1000 times lower) than it is to make the cables have 1*106 times the cross sectional area.
    Thanks for the help.

    I understood why P = IV = I^2R, but why is the power loss I^2R, and not IV or V^2R? So I understood what you were saying before I posted, but why doesn't the village affect the power loss?
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    (Original post by TheKevinFang)
    Thanks for the help.

    I understood why P = IV = I^2R, but why is the power loss I^2R, and not IV or V^2R? So I understood what you were saying before I posted, but why doesn't the village affect the power loss?
    Well in normal use there's power coming out of the secondary - that power wasn't lost so just measuring the power input to the primary alone can't tell you anything about efficiency.

    ---
    you can measure the resistance of your primary using a DMM or similar which operates on DC and that'll give you a resistance value that you can use to see what the resistive power loss will be at a given AC current (using RMS)
    some of the impedance to AC in that primary isn't going to be resistive though you'll have a total impedance which is part resistive and part reactive and you won't be able to tell what the proportion is just by measuring voltage across the primary terminals.
    IIRC they don't tell you about reactance at A level anymore
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    (Original post by TheKevinFang)
    Thanks for the help.

    I understood why P = IV = I^2R, but why is the power loss I^2R, and not IV or V^2R? So I understood what you were saying before I posted, but why doesn't the village affect the power loss?
    (This is a d.c. example but the principles apply to a.c. except it's more complicated).

    P = IV = I2R = V2/R

    The answer is the same in each case so long as we remember that the voltage is that placed (or developed) across the load resistance and that the current is flowing through the load resistance.

    It's a matter of choosing the appropriate equation to solve a given problem when there are unknowns.

    For instance, if we know the voltage (mains = 230V) and we know the load resistance, we would choose P = V2/R.

    If we know the current and the load resistance, then choose I2R.

    It's rather like using SUVAT in that respect.


    If we use a transformer, then the conductors of that device together with the cables connecting it to the load, have their own resistance.

    This places a series resistance in the path of the current before it passes through the load. That means volts will be dropped across the series resistance and as a result, some of the power available from the supply will be used up by that series resistance (transformer conductor and cables).

    It's exactly the same effect as a potential divider. The greater the resistance in the series path, the more volts will be dropped across that resistance.



    Vin = Power station voltage at the generator end

    I = current output demanded by the village load and distribution cables combined.

    R2 = village load resistance

    R1 = transformer and cable resistance

    VR2 = voltage available to mains outlets in the village

    VR1 = voltage lost across the transformer and distribution cables.
 
 
 
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