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# Answer on how to measure intensity wave

### A colleague asks me:

"

How can the intensity wave be measured without having the practice box?

"

### Answer

In chapter 2 of the book: PRÁCTICAS
DE ELECTROTERAPIA EN FISIOTERAPIA, on the point: **How to
measure intensity wave easily**. An easy way to show the intensity
wave without having the practice box is explained in detail.

Simple assemblies can be seen in Figures 2-26 a), b) and c). The one in figure a) is very practical but it is not appropriate to use it with a patient, due to the fact that part of the cables are exposed and the safety regulations are broken. The one in figure c), fully complies with safety regulations and can be used with guarantees on patients.

The difference between figure a) and b) is important, since in a) there is only one resistor that absorbs a power of 10 watts (W); But this resistance is wound and has a certain induction behavior that influences the result of the wave.

The resistors in figure b) form a pack of resistors joined in parallel, each one absorbs a power of 2 W resulting in a power equal to the sum of all of them. BUT THEY ARE NO LONGER COILED and their behavior is pure resistive, to avoid alterations in the wave.

I recommend making a pack of 10 resistors of 100 Ohm each, 2 W (not wound) and soldered in parallel. This gives us a resulting resistance of 10 Ohm with 20 W of power, capable of perfectly absorbing the usual electrotherapy applications.

A resistance around 5 Ohm is usually recommended, but why do I suggest 10 Ohm?

When it comes to arithmetically, 10 is a value that allows quick and very simple mental operations.

Do not forget that the oscilloscope does not measure intensity, it only measures voltage, but using Ohm's Law, the intensity can be perfectly calculated with the oscilloscope.

Supongamos un análisis de la onda de intensidad aplicando una corriente formada por pulsos cuadrados y nos preguntamos ¿Cuántos mA están pasando en cada pulso?

We observe in the oscilloscope that it shows pulses of, for example, 45 mV (0.045 V) at the plateau of the pulses; and based on the information we have, we want to know the mA that circulate through the plateau of the pulses.

By Ohm's Law we know that: I = V / R

at that time I = 0,045 V / 10 Ohm = 0,0045 A = 4,5 mA in every pulse

We can already deduce the why of the 10 Ohm, simply because it is a matter of mentally shifting the comma to the left or right to divide or multiply respectively.

On the other hand, if we look at the operation, the numerical value of the mV coincides with the value of the mA, but with a "," in the middle. Thus, it can be concluded that: the value read in millivolts coincides with the value in milliamps divided by 10 (provided that the resistance is 10 Ohm). For other resistance values, you have to do the full calculation and find the result.

For the value of the pass-through resistance to be exactly 10 Ohm, it is important that the 10 resistors have a value as close to 100 Ohm.The resistors usually have a variability of 5% in their value (unless resistance is used to one%). Good practice is to get more than you need at 5%, measure them with a multimeter, and select the 10 that are closest to 100 Ohms.

This pack of resistors can be inserted in both the [+] and [-] poles,
and in order to correctly analyze the waveform on the oscilloscope, it
is essential to control the polarity set on the stimulator and the
sockets with the probe clamps. If you do not do it incorrectly, the
image will be rendered inverted as it is in reality. Read at the
beginning of chapter 2 the epigraphs: Capture
of the voltage wave **and** Capture of the intensity wave.

Posted on 06/03/202021