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1000 microfarads is its rated capacitance, while 35 volts is its rated voltage.
This should work if the form factor fits your application and other ratings are not exceeded. You want the same capacitance and make sure there are not any embedded components like a bleed-off resistor in the replacement that aren't in the original.
NO! There are 50 bulbs in a 2.5 volt string or 125 volts. There are 35 bulbs in a 3.5 volt string or 122.5 volts. If you put 50 3.5 volt bulbs in you will have 175 volts and most american house have 110 volt service so you will have dim bulbs.
1000 * 35 = 35,000
(35 / 1000) * 100 = 3.5%
$1000 subtracting 35%
35/1000 = 7/200
35 1000/35
Expressed as a decimal, 35/1000 is equal to 0.035.
C=Q/V12 where Q/ V12 is the Charge per Potential Difference between the plates of the capacitor. If you solve for Q, you see that the charge is proportional to this potential difference. You are likely to surpass your load requirements by increasing the charge/discharge amplitude with the 35 volt cap. In other words, your cap will charge up to 35 volts and then discharge that 35 volts onto your load that was set at resonance to operate with 16 volts discharging. Any separation of circuits using this cap would probably fry something on one side or the other over time. Hope this helps.
Divide by 1000. 35 / 1000 = 0.035 grams.
There are 1000 milligrams in one gram. Therefore, 35 milligrams is equal to 35/1000 = 0.035 grams.