If I have a plot of current ($y$ axis) against 1/Resistance ($x$ axis).

The circuit it is measured from is a simply 2 resistors connected in parallel to battery, where the potential across the resistors is $1V$. One resistor remains fixed, and the second varies, the resistance plotted on the graph is only the resistance of the second resistor.

Does this mean that the gradient will be voltage, but will it be $1V$ or less, as the resistance on the graph is only one from one of the resistors?

Also what will the $y$ intercept signify? $I = V/R +c$, what does the $c$ represent?


1 Answer 1


Hint: Use Ohm's law $I=\frac{V}{R_p}$ and the formula $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R_0}$ for parallel resistors to derive a straight line in a $I$-$\frac{1}{R}$ diagram

$$ I~=~V \left(\frac{1}{R}+\frac{1}{R_0}\right). $$

Here $R$ and $R_0$ are the variable and the fixed resistor, respectively. To answer OP's two questions:

  1. The slope is $V=1{\rm Volt}$.

  2. The intercept of the straight line on the vertical $I$-axis is $c=\frac{V}{R_0}$. The intercept of the straight line on the horizontal $\frac{1}{R}$-axis is $\frac{-1}{R_0}$.

  • $\begingroup$ Thanks, so it will be 1 volt even though the graph does not include the resistance of the fixed resistor? Also how did you get to the y intercept? (I know using I = V/R, but why is the resistance that of the fixed resistor?) $\endgroup$
    – Jonathan.
    Commented Dec 12, 2011 at 7:44

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