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  1. #41

    Join Date
    May 2011
    Shooter
    35mm RF
    Posts
    5
    A little bit more physics if you can bear with me. I have a Chinese made 22.5V zinc carbon battery (which I am putting nowhere near my precious Leica and Rollei flashes). I have just been playing with my multimeter and 490 Ω resistor and as far as I can work out, the internal resistance of the battery is in round numbers 20 Ω. A #26FP flashbulb has a resistance of 5 Ω. Therefore if you were to fire with battery rather than capacitor, the flash current would in theory be 0.9 amps (22.5v / 25 Ω) but due to the limiting current of these very small cells, probably around half of this number but let's be generous and say it's 0.6 amp. The capacitor has effectively no internal resistance and the current would be 4.5 amps (22.5v / 5 Ω) i.e. nearly 7 times the amount via the battery. This is the attraction of using a capacitor fired flash.

    Wilson

  2. #42

    Join Date
    Aug 2006
    Location
    Cincinnati Ohio USA
    Shooter
    Medium Format
    Posts
    674
    I have a PDF file of the GE Flashbulb data booklet up at:

    http://www.butkus.org/chinon/flashes_meters.htm
    It is under the title
    GE Flashbulb Data 1961

    I don't get any money for this. But do donate to Mike if you use his site. He as 1,000s of photo related instruction manuals and you can see how much time he has dedicated to his site.

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