When I attach a Digital Multimeter (DMM) to read the resistance between my right and left hands, the resistance starts off high and reduces over time. It appears almost like "RC-type" behavior. Can someone explain to me what it is that I am seeing on the DMM in some detail?
[Physics] Measuring human resistance with a DMM
electric-circuitselectrical-resistancehome-experiment
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I'm not sure I understood all your points. I suggest you to read this beautifull paper
- Romer, R. H. (1982). What do “voltmeters” measure?: Faraday’s law in a multiply connected region, American Journal of Physics, 50(12), 1089. http://dx.doi.org/10.1119/1.12923
if you can find it. It's seems it's exactly what Prof. Levin is doing in his lecture. The proof is clear. If you have some difficulties to obtain this paper, I have some notes about this paper that I can share on SE too.
Edit: If you can find this reference, you will see that the only thing which matters is the position of the circuitry relative to the solenoid. More explicitly, this is the topology of the circuit which matters. It will then be clear for you that
- you would record nothing if you don't enclose the solenoid, subsequently you would measure nothing until you close (i.e. make a turn) the circuit,
- putting your wires far away from the solenoid does not change the measured voltage, nor putting them close to he solenoid, the only thing which matters is the previous point,
- doubling the number of turn around the solenoid doubles the corresponding voltage,
- the wires participate for nothing in the measurement except for encircling the solenoid (as stated in the previous points). This last point is true only for idealised conditions of course (infinite solenoid, no resistive effect in the wire, ...).
Now, some of your questions:
There is a famous law which says that a potential difference is produced across a conductor when it is exposed to a varying MF. But, how do you measure it to prove? It is quite practical.
You have two ways to see induction, as clearly discuss in the Feynman lecture. To be honest, I do not know a better book to start with.
you close a loop with a moving bar, whereas a coil was previously inside the loop. Then you record a voltage drop through Faraday's law because the circuit is moving (if you prefer, the path you calculate your integral with).
you close a loop and encapsulate a time dependent magnetic field via time dependent current passing through the coil. Then you record a voltage drop because of the time variation of the magnetic field itself.
In both case you have a time varying flux. As Prof. Levin says: "Misterrr Farrraday is happy with that !"
Look at dr. Levin's voltmeter. What does it measure? I ask this question partially because he tells about non-conservative measurements, which depends on the path, but does not explain how to set up the paths to measure -0.1 v in one case and +0.9 v in the other, between the same points.
Yes he does ! It is always from top to bottom (A to D point if I remember correctly), first passing on the left, second time passing on the right. Edit: The previous statement was confusing. When discussing magnetic flux, you need to define convention for following the circuit path, i.e. rotation direction. The path for -0.1 V is the counterclockwise path from A to D, the path giving +0.9 V is the clockwise path from A to D. That explain the signs also.
How the typical voltmeter works? There should be some known high resistance and small current through it shows how large the voltage is. But here, in addition to the D-A induced voltage, the EMF may be added because the is also induced current flowing through the voltmeter. How much is the effect?
I suggest you to read the wikipedia page http://en.wikipedia.org/wiki/Voltmeter. A perfect voltmeter has no EMF. (Edit: see also http://en.wikipedia.org/wiki/Electromotive_force for the definition(s) of an EMF and its different interpretations.)
For me your last paragraph is incomprehensible. Faraday's law relates voltage drop to time varying magnetic flux, so it doesn't apply to open circuit.
Final edit: A way to avoid the flux drop: From http://dx.doi.org/10.1119/1.12923, see http://i.stack.imgur.com/X5qPb.png.
… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. …
In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit.
As a mathematical model, the behavior of your circuit is undefined; it is an inconsistent overdetermined system. There is no value of the current which satisfies the equations defining the ideal components as you have connected them.
As a physical system, when you close the circuit, the current will rise from zero (at a rate determined by the inductance of the system, which depends on its shape and size) and quickly reach a limit where either
- P < ∞: the source is no longer operating as a voltage source, as its output voltage lowers (either due to internal resistance or deliberate foldback),
- P > 0: the “resistor” is no longer operating as a superconductor, as its maximum current density is exceeded,
or some of both effects at once. You cannot escape these limitations.
Best Answer
It has nothing to do with any capacitance. It's all about skin resistance. The resistivity inside your body is much lower than that of the skin. As a result, your measurement is really showing you the sum of two resistances thru the skin.
The main reason skin has higher resistance than the body internally is because the skin is dry. However, the skin can get moister by sweating and external blocking of evaporation. When you grip the probes, evaporation is blocked in the immediate vicinity, so moisture builds up. Your body may also produce a bit more sweat when you grip hard.
Try licking your fingers before touching the probes and you will see the resistance reduced substantially. You can also reduce the resistance by gripping harder. That increases the contact area and also makes better contact.