16
Io=2πfCU=2×3.14×0.1CU(mA)…
.....................................
(formula 1)
C is the electric capacity of the cable to the ground, unit: UF; u is the effective value of the test
voltage, unit: kV.
Example 2: a 10KV (UN = 10kV, uo = 8.7kv) cable is 4km long, with a single phase to ground
capacitance of 0.21uf/km and 0.1Hz super
If the low frequency test voltage is 26kv (peak value), the leakage current is
approximately:
Io=2
π
fCU=2
×
3.14
×
0.1CU=0.628
×
0.21
×
4
×
26/
2
=9.69
（
mA
）
Setting current value of overcurrent protection: I=kIo
...........................................
(formula 2)
