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Possible Duplicate:
Current against the inverse of resistance graph, I = V/R +c

How would you set up a circuit with a fixed resistor in parallel with a variable one. We are told to measure the current across the combination resistors and our Voltage is $1V$ and then construct an $I$ against $1/R$ graph what sort of correlation would this be?

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marked as duplicate by Qmechanic, dmckee Feb 5 '12 at 17:44

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    $\begingroup$ I think the entire point of this exercise is for you to discover the relationship between the current I and the resistance of the variable resistor R. It'd be a spoiler to give you the relationship here, wouldn't it? $\endgroup$ – Adam Zalcman Feb 5 '12 at 11:21
  • $\begingroup$ We have already done the experiment but the relationship of the graph ws a positive correlation which is were im cnfused about $\endgroup$ – Jeff Feb 5 '12 at 11:24
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Assuming that R refers to the variable resistance, the graph should be: $Y=V(x +\frac{1}{r})$ (r is the fixed resistance), which is a positive correlation. Derivation:

Net resistance $R'$ comes from (Resistances in parallel) $$\frac{1}{R'}=\frac{1}{R}+\frac{1}{r}$$. Writing $x$ (coordinate) as $\frac{1}{R}$,$$\frac{1}{R'}=x+\frac{1}{r}$$. Using $I=\frac{V}{R'}$ (Ohm's law), $$I=\frac{V}{R'}=V(x+\frac{1}{r})$$. Writing I as $y$ (coordinate), $$y=V(x+\frac{1}{r})$$. This is a linear graph with positive slope $\implies$ positive correlation.

Probably your confusion here was the fact that you were graphing $I$ versus $\frac{1}{R}$ and not $I$ vs $R$ (which is a negative correlation). Taking reciprocals flips the correlation (for positive values).

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