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  • Power Calc

    From Dan Green@21:1/5 to All on Wed Jul 26 00:53:52 2023
    Okay, here's anohter dumb quesiton.

    Say I have a resistor across the mains power supply. Say its 10k for
    argeuments sake. Mains here is about 115VAC. So the resistor will
    disspate 1.3 watts give or take by my rekoning. Now say I put a diode
    in series with the resistor. Does the power dispataion now halve?

    just curious.

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  • From Phil Hobbs@21:1/5 to Dan Green on Wed Jul 26 00:44:47 2023
    Dan Green <[email protected]> wrote:
    Okay, here's anohter dumb quesiton.

    Say I have a resistor across the mains power supply. Say its 10k for argeuments sake. Mains here is about 115VAC. So the resistor will
    disspate 1.3 watts give or take by my rekoning. Now say I put a diode
    in series with the resistor. Does the power dispataion now halve?

    just curious.


    Yes. The dissipation is the same in the positive and negative half-cycles,
    so blocking the current in one direction halves the average dissipation.

    The diode drops some voltage when conducting, so the dissipation is
    slightly less than half in real life.

    Cheers

    Phil Hobbs

    --
    Dr Philip C D Hobbs Principal Consultant
    ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics,
    Electro-optics, Photonics, Analog Electronics

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  • From Phil Allison@21:1/5 to Dan Green on Tue Aug 15 21:14:41 2023
    Dan Green wrote:
    -----------------------------
    Okay, here's anohter dumb quesiton.

    Say I have a resistor across the mains power supply. Say its 10k for argeuments sake. Mains here is about 115VAC. So the resistor will
    disspate 1.3 watts give or take by my rekoning. Now say I put a diode
    in series with the resistor. Does the power dispataion now halve?


    ** Conducting for half the time, or any other fraction, reduces the heat in a resistor by that same fraction.
    The *average value* of the current through or voltage across the resistor is also halved but this is not so for the *RMS* value. RMS values follow the heating effect on a resistance so will be reduced by 0.7071 ( 1/sq rt 2) or in your example from 115
    down to 80.6 volts.



    .... Phil

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