but no instantaneous change in current.
For a normal circuit, there will be some initial charging current for the conductor.
The formula we use in distribution systems is:
Vl-l / sqrt(3) / Xc
I've never calculated how much for a small circuit, but for a 3-phase, 25kV distribution line we get:
4/0 ACSR: 0.0913 amps/phase/mile
1/0 ACSR: 0.0858 amps/phase/mile
1/0 CU: 0.0845 amps/phase/mile
#6 CU: 0.0740 amps/phase/mile
So the same could be calculated for a building cable:
The charging current is:
2*pi*f*C*Vln * 10^-12
where:
f = frequency in hertz
C = cable capacitance in picofarads per ft
Vln = Voltage line-to-neutral
I could not find a table of cable capacitance value so I picked some reasonable values. The cable testing guys probably have some better data as they have to deal with this when testing cables. To avoid high AC charging currents, they will use DC tests. When using DC tests, they have to deal with the total leakage current which is made up of leakage current, charging current, and absorption current.
Anyway,
using 120 volts, 7 pf/ft, and a 100 ft run, you get 3.17 mA of charging current.
Using 120 volts, 50 pf/ft, and a 100 ft run, you get 22.6 mA of charging current.
For DC, the charging current in uA/kft is given by:
V/R*2.718281^(−t/RC)
where:
V = Voltage line-to-ground
R=dc resistance of cable in megohms for 1000 feet
t=time in seconds
C=capacitance of circuit in microfarads per 1000 feet