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.