Capacitors and Ideal Current Consumption

Hello. Capacitors are rated for working voltages - not for ampers they consume. And, by the way, if your capacitor , say, consume 45A in a 230V ac device, its capacitance is about 630 microfarads.

Capacitor current rating is the maximum RMS current that can flow through the capacitor without overheating the capacitor.

For a 45 amp rated capacitor, the current can be anywhere between 0 and 45 amp.

Voltage ratings also has to be observed.

<...45 amp rated capacitor...>

Does that criterion relate to the end connection leads, rather than the capacitor itself?

Hi!!, What capacitor are we talking about?.. If we are talking about power capacitors, what I can give you is an example.

Ex:

A capacitor is rated at 400kVar at 14.4kV (this particular value is the actual nameplate of a power capacitor, and you should use this value), what you want to know is the ideal or effective and the rated current.

For rated current: (e.g. single capacitor)

Current = Kvar / KV = 400kVar / 14.4kV = 28.57 Amps

For the effective or ideal current consumption: (first, you have to know at what particular voltage level your capacitor is going to be used. it should less than the rated kV probably. in this example, we use = 13.8kV)

At this point, your capacitor, doesn't provide you 400kVar since this kVar is only released at 14.4kV, you have to calculate the effective kVar rating of the capacitor.

kVar_new = kVar_old * (kV_new/kV_old)^2

so

kVar_new = 400*(13.8/14.4)^2 = 367.3611 kVar

with this kVar, you can calculate the ideal current

Current = kvar_new / kv_new = 367.3611 / 13.8 = 26.62 amps

that's how it is done.

PS. please take note that capacitor are rated according to kVar, Mvar, and then its corresponding kV. you have to convert or referred them to its voltage level being used.

If you have any questions, or maybe i made some mistakes, please notify me.

Thanks,

hkian

Thanks Mr HKIAN,

the example you quoted made it ample clear and easy in understanding.

I again thank you for your help.

Regards,

Rooney

I fully agree with your calculations.

As a matter of information, I have calculated de capacity of the capacitor as follows:

C= 400 A x 1 second x 1 k V / 14,5^2 kV. kV;

C= 400 Q/ s x 1 s x 1 kV / 14,5 ^2 kV. kV;

C= (400 Q/V ) / 14,5^2 x 10^3=

C= 0,4 / 14,5 x 14,5 F= 0,0019024 F=0,019024 micro F

A question:

Which is the definition of rated amperage?

I think it is the number of A that could "pass" through the capacitor under an alternating given voltage, at given frequency.

Best regards,

Arturo