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Using a multimeter to measure wood moisture

Next Iresearchedhow moisture

metersactually workon the web. Ifound that allthey do ismeasure DCelectricalresistance ofthe wood, anduse that tocalculatemoisturecontent. So Ifigured, whynot just use a

multimeter to measure the resistance and calculate the moisture content myself?

The catch with this is that a typical multimeter will read up to about 20 megohms inresistance mode, and the resistance between the two probes for any "normal"moisture content is often in the hundreds of megohms. So the multimeter wouldn't beable to measure it.

But there is atrick tomeasuringmuch higherresistancewith a cheapmultimeter.

Most of thesemultimeters,when involtagemeasurementmode, havean internal

resistance of about 10 megohms. That is to say, the meter itself is like a 10 megohmresistor that just happens to display the voltage across it. I checked, using anothermultimeter, that this meter was one of those with the 10 megohm resistance.

So now if I use a 9-volt battery, and hook the volt meter and my wood probes inseries with each other, the voltage drop across my meter and the wood will be equal

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to the 9 volts from the battery. And the voltage drops across the meter and theprobes will be in the same proportion as their resistances.

Knowing thatcurrent isequal tovoltagedivided byresistance,and that thecurrentthrough eachelement is thesame, we can

express thisas:

Cu

rre

nt

=

Vbat

=

V

met

er

=

V

woo

d

Rmeter

+Rwo

od

R

met

er

R

woo

d

For example,

if the woodbetween the probes has an 80 megohm resistance, my meter has a 10 megohmsresistance, and my battery is 9 volts, I should see an 8 volt drop across the woodand a 1 volt drop across the meter. So if my meter reads 1 volt, I know that theresistance is 80 megohms.

Vbat = Vmeter + Vwood Now, through a bunch of algebraic manipulation, we can solve forRwood as afunction ofknownvariables Vbat,

Vmeterand Rmeter:

Rwood = Rmeter (Vbat

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