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Question about regulating amperage

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    I normally approach these calculations from a standpoint of power (Watts). Your charger can output 6V * 10A = 60 W of power. At 5.3A required current, your backing plate will consume 5.3A * 6V = 32W of power. To achieve proper current drop you will need to burn off the excess 28Watts in bulbs connected in series with your plating setup.
    An easier solution would be to plate both parts at once, if the tank is big enough. 10A will plate these really well in about 25 minutes.

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  • Matt Harwood
    Nobody knows?

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  • Matt Harwood
    started a topic Question about regulating amperage

    Question about regulating amperage

    OK, let me see if I've got this straight: I want to cadmium plate a large brake backing plate 13" in diameter. Surface area is approximately 270 square inches. At 0.02 A per square inch, I need a little more than 5.3 amps to make it work. So far, so good.

    Now, I have a battery charger capable of 6 or 12 volts, and up to 10 amps. From what I've read here, I need to regulate this down using light bulbs. For instance, I need to string together enough bulbs so that their combined amperage is 5.3 A, correct? So if I have 5.3 A worth of illumination, I should have 5.3 A worth of plating power, correct?

    This seems counter-intuitive to me, since a bulb is nothing more than a resistor, and it seems that I am adding resistance to the circuit. It seems that I should have 10A - 5.3A = 4.7A worth of bulbs instead.

    Can anybody clarify this for me? Thanks very much, it is appreciated!
    Matt Harwood
    Cleveland, OH