We have a circuit where there is a variable resistor, and we increase this resistance at a steady rate, while increasing current. Thus we have increasing voltage. The gradient is defined by $dy/dx$. Yet if we state that $R$ is the gradient, then $R$ can also be calculated by $V/I$, which does not involve limits! Thus, my question boils down to this: if $R$ is the gradient function of such a graph (as described above) then how can it be calculated by such mundane means as simply $V/I$, while in other situations this does not work and we have to differentiate?
[Physics] Is resistance the gradient in a $V/I$ graph
differentiationelectric-circuitselectric-currentelectrical-resistancevoltage
Best Answer
$R(V,I) = \frac{V}{I}$ by definition, it is not a gradient. $r = \frac{dV}{dI}$ is called the fractional, differential, dynamical or small-signal resistance. It just happens that for resistors $R(V,I) = R_0$ is a constant, thus the two quantities are the same: $r = R_0$.