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  1. Since only electron's flow in electricity and electrons have negative charge, then why we don't say —1amps (—1C/s)?

  2. Secondly, as conventional way we write down independent variable in $x$ axis and dependent on $y$ axis then $I/V$ ($V$ on $x$ and $I$ on $y$ axis) should not indicate conductance?

  3. If conductance is constant then would the resistance also be constant?

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  • $\begingroup$ Related: physics.stackexchange.com/q/17109/2451 and links therein. $\endgroup$ – Qmechanic Feb 26 '16 at 6:55
  • $\begingroup$ I edited this post to make it clear that you're asking three questions. However, we really prefer to have one question per post. Can you please select one of these questions to ask here and ask the other two in separate posts? $\endgroup$ – DanielSank Feb 26 '16 at 7:08
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1) The convention of electron charge sign does not make any real difference. If you speak of current magnitude you use positive numbers. If you have some oriented device, e.g. battery, you may surely write "+1 A" for discharging and "-1 A" for charging etc.

2) In fact in nonlinear systems (e.g. tunnel diodes) the IV curve can become quite complicated. It is better to look at it as a set of all points of voltage and current that may be ever measured at the device. And yes, I/V is conductance, but its use makes sense to me only in linear systems.

3) Yes, conductance is 1/resistance. If one is constant, the second is as well.

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  • $\begingroup$ then,can we conclude this by agreeing that THOUGH I/V graph indicates conductance but because of convention we use this graph for defining resistance from its gradient.(which is only for linear cases)? $\endgroup$ – ffahim Mar 3 '16 at 11:36
  • $\begingroup$ I am a little confused by your question, but this is for sure - if I is proportional to V, then we can easily deduce the conductance from its slope (as I/V) and resistance as well (as V/I). $\endgroup$ – dominecf Mar 4 '16 at 9:01
  • $\begingroup$ I meant that by one graph we can describe both conductance & resistance. hope its clear. Now i think i am right. Ain't I? $\endgroup$ – ffahim Mar 5 '16 at 17:11

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