To compensate for the reduced wattage of your replacement lamp, you have installed additional resistance. But a quick calculation of the original arrangement (assuming that the given 30-watt is correct), shows that nearly all the volt-drop between the supply and the lamp appears across the series inductance: 1.5 H in total. (Series impedance = 475 Ω @ 84°; reactance = 471 Ω). On that basis, when you fit a 22-watt lamp, I would have thought that increasing the series inductance rather than the resistance would have been appropriate. My calcs. are as follows.
Old wattage ÷ new wattage = 30 ÷ 22 = 1.36.
So, inductance needs increasing by 1.36.
1.36 x 1.5 = 2 H.
That is an increase or 0.5 H.
So an additional choke of 0.5 H is required, with a resistance not exceeding approx. 50 Ω.
In effect, this is all about power factor. However, I am not at all familiar with the behaviour - incl. PF - of fluorescent lamps, so maybe someone can provide a better analysis and resultant calculation or explanation.
HTH,
Al.
Old wattage ÷ new wattage = 30 ÷ 22 = 1.36.
So, inductance needs increasing by 1.36.
1.36 x 1.5 = 2 H.
That is an increase or 0.5 H.
So an additional choke of 0.5 H is required, with a resistance not exceeding approx. 50 Ω.
In effect, this is all about power factor. However, I am not at all familiar with the behaviour - incl. PF - of fluorescent lamps, so maybe someone can provide a better analysis and resultant calculation or explanation.
HTH,
Al.
