The charging current I’m talking about would be the one between un-shorted phases and ground when there is a short to ground in one of the phases in a distribution network or facility. I'm not talking about the charging current in a transmission network where transmission lines at extremely high voltage accumulate charge between them and may possible increase the voltage (specially if the system is not designed properly).

Please see this web page section “Estimating”. The current they are trying to estimate there is the one that I would like to define and understand. Also see this paper

Every article I have seen about estimating charging current takes for granted that the reader knows what charging current is and how it originates, and doesn't properly define it before explaining how to measure or estimate it.

I have been reading and the best definition I have found so far is :

"Each phase of a three-phase system exhibits a certain amount of distributed capacitance to ground, shown here as three capacitors. A capacitive current flows through these (1). If one of the phases shorts to ground causing a ground fault, the charging current for the other two phases will flow through the ground fault (2)."


Apparently the way to directly measure charging current in small ungrounded systems (in which the charging current value is under 10 A) is to short one of the phases to ground and measure the current. As seen:

how to measure it

What is hard for me to understand intuitively is what I have marked as (1) and (2) on the definition:

(1) There is current flowing(leaking) out the circuit to ground from each of the phases even when there are no faults and the value of this current is several Amps! Mind-blowing!!!

(2) If I short a phase to ground. What I will see is the charging current from the other 2 phases coming from ground to my shorted phase through ammeter A2. And what about the current supplied by the utility and that was normally flowing through the phase I have just shorted? What happened to it?

(3) Will ammeter A1 continue to measure 0 after I short one of the phases? Why?

  • What have you tried so far to solve this issue on your own? The essence of your question is not clear. Is this about what it takes to establish a safe high-impedance ground for a given 3-phase delta power feed? – Sparky256 Nov 6 at 0:30
  • I'm sorry I was in a hurry yesterday and couldn't elaborate a proper question. I'm not trying to estimate charging current. I'm trying to understand where it comes from and how it is measured – VMMF Nov 6 at 13:27
  • Hadn't heard the term before seeing this, so not sure if this is (a) valid, (b) helpful, but... 1. Imagine a section of the network somewhere in the middle of the line. 2. Instead of a single capacitor between each phase and ground, think of two: one to the left; one to the right. 3. Think of ground as a "virtual" wire alongside the three phases. 4. If you short one phase to the (virtual) ground, then current flows from the other two phases, through the left/right capacitors "into" the section of (virtual) ground under study, and then through the short into the final phase. – TripeHound Nov 6 at 15:58

"For a more precise estimate, use manufacturer’s data to sum up the different circuit elements that contribute to charging currents, such as cable capacitance per 100 feet, surge arrestors, motors, etc"

Charging current happens in any electrical condition, its the total capacitance of the whole system in question. You see it profoundly in transmission lines due to the insane capacity of long wire lengths. But it basically exists everywhere just to lesser extent. Hope that helped, its basically spec info that you would have to gather based on the install in question. as for the formula that will provide that, that would be another thing.

  • Thank you very much for your answer! I know that the part that you are quoting is on one of the references I provided. I'm not trying to estimate charging current, I would like to understand it intuitively. Please see my updated question – VMMF Nov 6 at 13:29

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.