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.
Short answer is nothing will happen.
The more detailed answer is that the new capacitor is rated for a higher voltage and should, as a result, last longer. The capacitance is the same (22uf) so the circuit will function as it should and the rated max voltage is higher so all should be good. The only downside to increasing the voltage of the cap is that it will be physically larger so make sure it will fit.
Depending on the circuit, 63% of the available voltage.
A: A voltage source Will charge a capacitor to 63% of its input value, The value to get there is stated a Resistance time capacitor as time. Mathematically it will never get there but engineering consider 5 times RC time constant as close enough,
basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
a capacitor that keeps time Capacitors do not keep time they do however charge at a specific rate of 63% of the applied voltage from a source that can be used to relate to timing since the source voltage can be calculated after a time lapse
Charge the capacitor. Potential difference is a scientific term for what is more commonly called voltage. ANSWER: If big enough the battery will see a short initially and then proceed to charge the capacitor at a rate of 63% of the voltage in one time constant defined as RC For engineering purposes after 5 time the time constant the battery will and the capacitor zero potential different. The proper term should be virtual no difference.
35/63 = 5/9
Depending on the circuit, 63% of the available voltage.
35/63 = 5/9
63 3,21 3,3,7 35 5,7
It is: 63/35 = 9/5
The common factors of 35 and 63 are: 1 and 7
35/63 = 5/9 in its simplest form
35% as a decimal is 0.35 and 0.35*63 = 22.05
have you checked your computer? If your pump stays on when you first turn the ignition; without turning on the engine, the pump usually turns off after a few seconds. If your pump stays on then you can suspect the computer. Remove the computer. Remove the two Torx screws and open up the box. Then look for three capacitors. Two will be 47 micro farad 16 volt and the other will be a 10 micro farad 63 volt. If they look bloated or like they are leaking some black stuff or if the leads are detached from the PC board; replace them. You can get the capacitors at the radio shack. The trick will be that you have to buy two 4.7 micro farad capacitors in the place of the 10 micro farad capacitor and wire them in a parallel configuration. The reason for this is that they do not sell a 10 micro farad with the VOLTAGE rating that is required (63 volts). This is the only capacitor that requires a 63 volt rating. MAKE SURE THAT YOU BUY "POLARIZED" CAPACITORS. Hope this helps
The GCF is 7.
The Greatest Common Divisor of 35, 63 is 7.
The GCF is 7.