Andy Wade wrote:
wrote:
if its 2 core cable the L shuold be vanishingly small, as i is flowing
in opposite directions in each core.
There's a current loop of finite size, so the inductance is finite, not
zero. Whether or not it's vanishingly small depends on the application.
In 50 Hz work we're used to being able to neglect wiring inductance,
but it ceases to be negligible at higher frequencies, or even at 50 Hz
for large cables (50mm^2 upwards, say). In this case the frequency is
over three orders of magnitude above mains frequency.
How do you get your 0,9uH figure?
Any text book on E-M theory /transmission line theory will give you the
following expression for the inductance per unit length of a
parallel-wire line
L = (mu / pi) * ln (s/r) [for s r]
where
mu is the permeability, in this case = mu_0 = 4*pi*10^-7 H/m
(so mu / pi = 0.4 uH/m),
s is the spacing (between centres) of the conductors, and
r is the radius of each conductor.
For 1.5 mm^2 T&E cable the wire diameter is 1.38 mm, so r =~ 0.7 mm.
I guessed s = 6 mm.
Substituting these values, the ln() term evaluates to 2.15, so
L = 0.4 * 2.15 = 0.86 uH/m. QEF.
I see, he said looking surprised, thanks for the details. But one
critical detail isnt mentioned the is this figure the L of both
wires carrying i in same direction, or for counter flow? The two Ls
will be very different.
NT