# 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?

## 2 Answers

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. – Airdish Nov 19 '15 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. – Alfred Centauri Nov 20 '15 at 0:45

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