V = I x R -- Voltage (V) as a function of Resistance (R) is Linear; however, Current (I) as a function of Resistance is not Linear -- why?

Question

I am totally blanking on the following simple equation (Ohm's law):

V = I x R -- Voltage is Current x the Resistance.

Clearly, One can prove f(R) = I x R is linear: f(aR) = af(R) f(R1 + R1) = f(R1) + f(R2) holds.

However,

g(R) = V / R -- Current as a function of Resistance is not linear -- its an inverse function, as so f(x) = 1/x.

But, how is an algebraic equation that describes the relationship between variables in one form linear, whilst in another form not linear.

Answer 1

The relationship between Voltage and Resistance is linear (if you consider current to be constant), whether you are expressing V in terms of R, or R in terms of V.

The relationship between Current and Resistance in inverse (if you consider voltage to be constant), whether you are expressing I in term of R, or R in terms of I.

You got confused because you changed which pair of variables you were considering the relationship between: from V and R to I and R.

Answer 2

Just as JonathanZ supports MonicaC said,

$$V \propto R \therefore R \propto V$$ however, suppose $V$ is equal to a constant $k$

$ I\cdot R = k$, which implies that the function of either $I(R)$ or $R(I)$ are hyperbolic, and therefore non-linear.

A method of proof would be taking the derivative of either function because the derivative of linear functions is always a constant.

We can show $\frac{\partial V}{\partial R} =\frac{\partial}{\partial R} [IR]$

Assuming I is a constant $\epsilon$ yields $\frac{\partial V}{\partial R} =\frac{\partial}{\partial R} [\epsilon R]=\epsilon$

And for Current, $\frac{\partial V}{\partial R} =\frac{\partial}{\partial R} [IR]$

Assuming R is a constant $\epsilon$ yields $\frac{\partial V}{\partial I} =\frac{\partial}{\partial I} [\epsilon I]=\epsilon$

Since the result is a constant, one can infer $V$ is a linear function of $I$ and $R$ on the other hand, $$I \propto {1\over R} \ and \ R \propto {1\over I}$$ Differentiating using the same idea that a linear function's first derivative will be constant proves $I$ and $R$ do not have a linear relationship.

$$\frac{\partial I}{\partial R} = \frac{\partial}{\partial R}R^{-1}$$ $$ = -R^{-2}$$

A non-constant result proves $I(R)$ is non-linear, and for $R(I)$, $$\frac{\partial R}{\partial I} = \frac{\partial}{\partial I}I^{-1}$$ $$ = -I^{-2}$$

Also non-linear; therefore, current and resistance have a non-linear relationship.