The current is determined by voltage, frequency and capacitance.

A capacitor for AC will not have a problem with the current. It is only at higher frequencies that you need to be careful. And your frequency is probably 60 or 50 Hz. So, no sweat.

There are Power Factor Compensation capacitors. Use them instead. They are designed for this.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

Thank you for the help


Now I brought the calculator:

For 1 Ampere, I need 127 ohms. At 60Hz that would require a 20uF capacitor.
Too bulky

Guess I'll have to find another solution

Thanks again

You said 0.5 A in the OP. Has that changed?

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

If you are speaking about cable inductive reactance this reduces the capacitive reactance then the current will be higher.As Skogsgurra said:
X=1/Cap/(2*pi()*f); Cap=1/X/(2*pi()*f) ; X=V/I ; X=127/0.5= 254 ohm Cap=1/254/(2*pi()*50)*10^6=12.53 microF.
See [for instance]:
http://www.ebay.com/bhp/12uf-capacitor

Tell us about your circuit. Your results may not be as anticipated.
Is this a brew your own Nola scheme?
Capacitors are used with motors to improve the power factor and avoid PF penalties.
In small systems and most residential applications you are not penalized for a poor power factor.
Let's look at a motor with 5 Amps of magnetizing current. That is the current that may be reduced by a capacitor.
The circuit conductors for this motor are probably sized for a maximum of 5% drop from the meter to the motor This would be an ampacity of about 20 amps.
So, at no load, the voltage drop that may be avoided by the use of a capacitor may be about 5% of 5/20 = about 1.25%
This is a simplification, but the point is that there is not much saving to be had on small circuits.
Most of the calculations showing Nola savings are misleading.
There are interesting applications of capacitors to compensate for losses on long transmission lines. One application that comes to mind was the connection of capacitors in series with a 500,000 Volt transmission line running around 1000 Amperes.
Another application was to raise the voltage for an entire city at the end of an overloaded transmission line.
For 1/2 Amp at 127 Volts, generally not worth the effort.
But, share your circuit.
You may have a unique application where it will be worth while to use a capacitor.

Bill
--------------------
"Why not the best?"
Jimmy Carter

I cannot believe that I wrote "inductive reactance" on the opening post. Please scratch that.

It is my understanding that power correction capacitors are used in parallel with the inductive load (right ?). That is not the case I'm curious about

I have another thread where I make use of this technique (capacitor in series to reduce AC current)

I have attached two links. One for the thread, another for the schematic.

In this case it's a very small current.
But half ampere? 10 Amps... The more the better. There is no reason not to explore the limit. I'd like to know where it is.


Thread:
http://www.eng-tips.com/viewthread.cfm?qid=408055

Schematic:
http://files.engineering.com/download.aspx?folder=...

• http://www.eng-tips.com/viewthread.cfm?qid=408055

One application that comes to mind was the connection of capacitors in series with a 500,000 Volt transmission line running around 1000 Amperes
This was achieved with series/parallel connected banks of individual capacitors rated at 17,000 Volts.
I have heard of 750,000 Volt lines with capacitor stations installed.
How high a voltage do you want?
What are you powering? Many loads do not like to see the current reduced.
Note: On a long transmission line most of the voltage drop may be reactive rather than resistive. The capacitors are used to cancel the line inductive reactance and cancel the reactive voltage drop. The capacitors DO NOT reduce the current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

I have often made such "transformerless, capacitor-drop" power supplies.
In my experience, it makes no sense to build them for currents higher than ~50 mA. The capacitor gets so big, that you might as well use a transformer or a flyback.

For reliability and safety, you need to add a couple of resistors:
1: a current limiting resistor in series with the drop capacitor to keep inrush current in check so rectifiers and Zener diode don't suffer.
2: a bleed resistor in parallel with the drop cap so people don't get a shock when the circuit is unplugged and someone touches the plug prongs.

Benta.

I am experiencing what benta described: larger currents require a larger capacitor.

Indeed, it's not the right strategy for a high current

My question was "how much current could a capacitor withstand"

In a scenario where I make use of a huge capacitor that allow one ampere, I was asking if it could actually withstand this current


But the answer is now irrelevant.

Even if I make a bank of capacitors, as waross suggests, the solution is far from optimal
For this sake, the thread can be closed


I now depart thanking everyone for the patience, and helpful responses