# Is resistance the gradient in a $V/I$ graph?

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?

$R(V,I) = \frac{V}{I}$ 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: $r = R_0$.

A resistor is defined as the circuit element for which the voltage across is proportional to the current through and the constant of proportionality is the resistance $R$:

$$V_R = R\cdot I_R$$

Clearly, for this linear relationship, it is also true that

$$\frac{dV_R}{dI_R} = R$$

However, for general circuit elements, the derivative of $V(I)$ is not a constant. Thus, it is not generally meaningful to take the ratio $\frac{V}{I}$.

As always, we can Taylor expand $V(I)$ around some operating point:

$$V(I_{OP} + i) = V(I_{OP}) + \frac{dV(I_{OP})}{dI} \cdot i+ \frac{1}{2}\frac{d^2V(I_{OP})}{dI^2}\cdot i^2 + ...$$

Note that the second term in the expansion 'looks' like a resistance multiplying the change in current $i$ from the operating point.

Thus, for $i$ small enough such that we can ignore the higher order terms in the expansion, we can meaningfully speak of a small-signal resistance $r$ and write:

$$V(I) = V_{OP} + v(i) = V(I_{OP}) + r\cdot i$$

where

$$r = \frac{dV(I_{OP})}{dI}$$

• You said "for general circuit elements, the derivative of V(I) isn't constant." Yes i agree completely but you also said that we can't use the ratio to find R. But we can use it if we have both V and I, thus we find the gradient R without differentiating. This is the anomaly which I don't understand. Commented Nov 19, 2015 at 23:59
• @S.Mo, R equals the gradient (slope) of the V-I curve only in the case V is proportional to I. So your statement is false; the ratio of the voltage to current, for a non-linear circuit element gives the slope of the secant line through the origin and the point on the curve, not the slope of the tangent line through the point. Commented Nov 20, 2015 at 0:45