Yes you can - in principle, and in many cases. As long as the form factor of the new capacitor is the same as the form factor of the former capacitor, and as long as the maximum voltage of the new capacitor is higher than the max voltage of the former capacitor, you can often replace a lower-rated with a higher rated capacitor.
However...
Apart from the nominal capacity and maximum voltage rating, you should also consider the capacitor's design. There are many ways to make a capacitor, from tiny ceramics over various plastics and metal foil combinations to electrolytic capacitors. Apart from the capacity and voltage rating, the different types of capacitors also vary in a lot of other aspects: impedance, precision, drift with time and drift with temperature, are just some.
Electronics designers typically choose the capacitors carefully, and considering all those aspects. Only when those aspects don't matter much, designers tend to chose the cheapest technology. In such a case, you're safe to replace.
In most cases of capacitors rated at 1 micro Farad and above, electrolytic capacitors are used (typically small cylinders). Those are typically selected because they offer excellent value (capacitance) for money, and you're typically safe to replace them, subject to above rules.
CAUTION: some capacitors have a distinct polarity. You must make sure to observe the polatiry when replacing polarized capazitors. Failure to do so can result in serious harm (basically, the part might explode).
You risk having an unexpected explosion or a circuit fire if you use the wrong type of capacitor just as much as if you use one of too low a voltage rating for the particular end-application. You can replace a low-voltage capacitor for one of the same microfarads size only if the "type of construction" of the replacement capacitor is the same as the original AND the rated voltage is equal to or higher than that of the the original. In other words if the original is a "non-electrolytic" type - meaning that the insulation between the plates is plain paper or plastic and not a chemical compound - the replacement capacitor must also be a "non-electrolytic" type, even if the rated voltage is higher.
If the insulation between the plates is "electrolytic" the replacement capacitor must also be "electrolytic", even if the rated voltage is higher.
The technical reason for having to use exactly the same type when a replacement has to be made is to make sure that only a non-electrolytic will be used if the circuit carries high-voltage ac (alternating current), because an electrolytic type will not be suitable, or safe, to use in that situation.
Electrolytic types are much smaller than equivalent-sized non-electrolytics - which is very handy for miniature products. But electrolytics can only be used on dc (direct current) circuits which carry ac "signal" currents - and hence voltages - which are tiny in comparison with the standing dc voltage.
This is likely to produce problems, but the exact nature of the problem and if it might be acceptable anyway will depend on the purpose of the capacitor in the circuit. Value of a component is NOT the only criterion for successful substitution when the original is not available.
If it's use is an input, no you can not, if it is an output yes you can.
about 500 uF
It takes 1 farad for every 1000 watts so u need 2 farads.
12kv is equal to 12000 volt.(12*1000).
usually yes. what is it being used for?
The formula you are looking for is Watts = Amps x Volts. killo = 1000 1 kw = 1000 watts.
9200 volts my 1000 uF capacitor only holds 10 volts
The capacitive reactance is approximately 4 kΩ .
Zero watts can be installed in 1000 micro farads. Watts are the product of amperage times volts. Micro farads is a value used in talking about capacitance.
1000
1000 microfarads is its rated capacitance, while 35 volts is its rated voltage.
Farad = Coloumb / Volt; solving for Coloumb, you get Coloumb = Farad x Volt. Just plug in the numbers - 1 microfarad is a millionth farad; 0.001 microfarad - if that is what you mean - is 0.000000001 Farad; wherease 1 KV = 1000 Volts.
1000
Kilovolt or 1000 volts. 1KV is 1000 volts, 2KV is 2000 volts. Kilo stands for 1000.
It will take slightly less than one second (0.92 seconds) to charge a 1000uF capacitor to 12 volts through a 1000 ohm resistor if your power source is 20 volts.The time constant of a 1000uF capacitor in series with a 1000 Ohm resistor is 1 second. (1x10-3 Farads times 1x10+3 Ohms = 1 second) It takes 1 time constant to reach 63% of a step change, 2 time constants to reach 86% of step change, and so forth using the equation VT = V0 (1 - e (-T/RC)). See notes below12 is 60% of 20, so it will take about 0.92 seconds for the capacitor to reach 12 volts. In two seconds the capacitor will reach about 17 volts. In five seconds, five time constants, the capacitor will be considered to be fully charged, 99.3%, to 20 volts.Notes:VT is voltage after a given number of seconds. V0 is the total initial voltage, in this case, 20 volts. -T/RC is the negative number of time constants for the exponential equation e-TC, which a charging capacitor exhibits. (More specifically, e-TC is the proportion of voltage across the resistor, while 1 - e-TC is the proportion of voltage across the capacitor.)This equation is based on the fundamental equation of a capacitor...dv/dt = i/c... which states that the slope of the voltage is proportional to current and inversely proportional to capacitance. Plug this into an initial state differential equation, for the case of charging an initially discharged capacitor through a resistor, and you getVT = V0 (1 - e (-T/RC))(Derivation requires calculus, and that seems a little bit out of scope for this question.)
The prefix, 'milli', literally means 1/1000. Just like a millimeter is 1/1000th of a meter. So replace milli with 1/1000 and you have the answer in volts. For example, 583 millivolts = 583 x 1/1000 volts =583/1000 volts = 0.583 volts.
Ohms and volts are different things -- it's not possible to equate ohms to volts.
about 500 uF