TWO EXAMPLES
A - Currents of long duration and low intensity
In order that level V reaches threshold D, a certain volume of water is necessary (likened to a certain
quantity of electrical charges). If this water is supplied slowly by the pump (current of long duration and
low intensity), some of the water has time to go through L and raise piston E therefore increasing the
threshold level (accommodation). The quantity of liquid (the current) will therefore have to be greater
because level V has to reach point D higher up. Moreover, a large amount of liquid returns from B to A
through tap K. It is easy to understand that all these extra quantities that P has to transport indicate that
we have an unfavourable stimulation current.
B - Currents of short duration and higher intensity
The durations intended here are close to the excitation constant value k.
In this case, as the flow is high, the pump action is short. As almost no liquid has gone through L, the float
does not rise and accommodation is therefore negligible. Nevertheless, a certain quantity of water returns
through K and has to be compensated for by P.
The Weiss law applies to these kinds of current (please refer to the fundamental
law of electrostimulation).
Q q + it
or
I t q + it
Q
is the total quantity of liquid provided by P with
I
= intensity of the stimulation current
t
= pulse duration
q
is the volume of liquid separating
Vo
from
So
i.e. the quantity of charges that
would have to be provided if there were no leak
K
. In other words, if the membrane
potential varied instantaneously and not exponentially in
accordance with a time constant
K
.
it the quantity of liquid that returns from
B
to
A
through tap
K
.
Fig. 4
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