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I am working with a simple DC cirquit. It is a power supply (internal_______________________________________________
resistance 1.6 ohms) with a voltmeter at its output. An electrolytic
cell
(platinum anode and nickel cathode) is connected to the power supply,
through an ammeter. What can be more simple?
The current versus voltage relation is different from what one would
expect from a metallic resistor. First the current is ~zero (below ~10
microamps that I would notice), up to about 2.9 volts. Then the current
starts changing linearly at the rate of about 12.4 V/A, up to 650 mA at
11 volts. When the concentration of the electrolyte (Li2SO4 in water)
doubles (goes from 0.21 M to 0.42 M) the threshold remains 2.9 but the
slope of the line lbecomes 11 V/A.
I have three questions.
1) How to interpret the threshold of 2.9 volts? (I do not have a table
of electromotive potentials for different metals at home. Is 2.9 a
difference between Ni and Pt?)
2) I expected the slope to change from 14 to be 7 (because by doubling
the concentration I doubled the number of free carriers in my liquid
resistor). Why was the second slope 11 and not 7? (Note that the
internal r of my power supply is by nearly one order of magnitude
smaller that the slopes.)
3) The PD never exceeds 10 V. Large electrolytic capacitors for low
voltages are available. Suppose a capacitor of that kind is connected
in parallel with my voltmeter (at the output of the power supply).
Would this make the internal resistance of the source much smaller? My
currents never exceed 0.6 A.