On Fri, 7 Mar 2003 13:07:30 -0500 "Tom Rauch" <w8ji@contesting.com>
writes:
> > immediately. The CT was probably about 2ft in diameter and 10"
> thick, with a ratio of 1200:5. I was really lucky (blessed) that my
knuckle on
> the same hand was touching ground, as the voltage was tremendous and I
> don't think would have survived that potential across my chest.
<snip>
> In your case the current transformer acts like a normal transformer
rather
> than an energy storage system, unless someone broke the line abruptly
at a
> sinewave peak. The problem is you can have considerable voltage drop
across
> the primary impedance, because the primary impedance can be high if the
> secondary is unterminated. With a sinewave signal you could have had
> hundreds of volts across the primary, which would be stepped up to many
> thousands of secondary volts.
Good afternoon Tom & the group,
The same formula works for unterminated secondaries in both transient and
continuous (AC) applications...
V(Secondary) = -M * dI(Primary)/dt where M is the mutual inductance.
Note: The derivative of a sine(wave) is... the cosine
73 & Have a great weekend,
Marv WC6W
> As you learned, an open termination on a current transformer can
produce
> some large voltages!! But no more that the turns ratio of the
transformer
> and operating voltage would indicate. A system with an abrupt drop in
> primary excitation, especially if flux density was high, would be a
> different matter.
>
> 73 Tom
>
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