[Physics] What are the dependent and independent variables in Ohm’s law

electric-currentelectrical-resistanceintuitionvoltage

I can't quite wrap my head around Ohm's law. The relationship itself is quite intuitive to me. What I don't understand is when a system has dynamic voltages, currents, and resistances. I don't quite understand which variables are dependent and which are independent. For example, one could take multiple 9V batteries, connect them, and get a larger potential difference between the positive and negative terminal. If we assume the resistance didn't change between the terminals, that would imply there must be a higher current between the terminals. This example leads me to believe that voltage and resistances and independent variables, and current is dependent on those two variables. But, if we had a circuit with multiple resisters in parallel, then there would be voltage drops between them, making voltage dependent on resistance. I appreciate all help.

Best Answer

First off, Ohm's law is not the equation $V = IR$ alone. Instead, $V = IR$ is significant in at least two different ways, only one of which is properly called as "Ohm's law":

  • One of these is that it is a definition of "resistance" as a physical quantity. In that case, it would perhaps better be written as $$R := \frac{V}{I}$$ . In this sense, the equation is analogous to the definition of capacitance: $$C := \frac{Q}{V}$$ The reason this is not a "law" is because a "law" in scientific parlance means a rule that describes an observed relationship between certain quantities or effects - basically, it's a . A definition, on the other hand, synthesizes a new quantity, so that the relationship is effectively trivial because it's created by fiat.
  • The other, however, is what is properly called "Ohm's law", and it refers to a property of materials, the "law" being that they generally follow it: a material that behaves in accordance with Ohm's law (often only approximately) is called an "ohmic" material, and Ohm's law here says that the voltage-current relationship looks like $$V = IR$$ for a constant value of $R$. Note that in the definition sense, there is no reason at all that $R$ needs to be a constant. In this sense, though, Ohm's law should be understood perhaps as analogous to the idea of modelling friction in elementary mechanics by $$F_\mathrm{fric} = \mu F_N$$ giving a linear dependence between the friction and the normal force $F_N$ through the coefficient of friction $\mu$. (Once more, though, you can also take this as a definition of a CoF - the "law" part is in that $\mu$ is constant so the linear relationship holds.)

And so I presume that your question is asking about the first sense: if we consider $V = IR$ just a defining relation between three quantities, which one is the "dependent" and which is the "independent" quantity? The answer is that this is not a really good question given the parameters. The terms "dependent" and "independent" quantities are kind of an old-fashioned terminology from the less-rigorous earlier days of maths that keeps getting knocked around in not-so-great school texts, and relate to functions: if we have a function $f$ with one variable $x$, which in a fully modern understanding would be called the function's argument or input, then in the specific case where we bind (i.e. mandate it has the same value as) another variable $y$, to have the value of the function in question, so that $y = f(x)$ following the binding, then $y$ is called as the dependent variable, and $x$ the independent variable.

To see why that doesn't work so well in this case, note the logical structure of the above statement: the givens, argument, and conclusions. We are given a function $f$, then we create a binding between a variable $y$ and the value $f(x)$ of the function, then finally, we name the two. But in the case of "$V = IR$", we are simply giving this relationship; there is no "function" here of any type, much less being employed in this very specific manner.

(What do I mean by "binding"? Well, that's what the symbol $:=$ earlier means: to bind variable $y$ to some expression means that we are to declare that $y$ now can only be substituted for the expression given, and not something else, at least within a particular context. Writing $y := \mathrm{(expr)}$ means $y$ is bound to expression $\mathrm{(expr)}$.)

And this is also why I say it is "old-fashioned" from a modern point of view - in modern usage functions are far more general and flexible than they used to be, and a modern point of view is that an expression like

$$x + y > \cos(xy)$$

is in fact entirely built from functions: not only $\cos$ but the multiplication $\cdot$ (here suppressed in favor of juxtaposition) and addition $+$ but also interestingly, the symbol $>$ itself: that is a special kind of function called a "Boolean function" or a relation, which asserts that something is true or false about the arguments you put into it. When you say that an "equation holds", you mean the Boolean function $=$ evaluates to "True".

Likewise, in modern usage, the terminology of "dependent" and "independent" variables really is more at home in a scientific/empirical context: in conducting an experiment, the independent variable is the one we modify, while the dependent variable is the one we seek to analyze with regard to if and how it responds to changes in the independent variable. In the case of an experiment involving electric circuits, any of the three variables here may serve those roles (yes, even $R$ - think about swapping resistors, or using a variable resistor, and for $R$ as dependent variable, think about heating up a resistor with suitably high current, causing its resistance to change [i.e. behave non-ohmically]).

That said, if we are going to really insist on sticking to this regardless, I'd say that in most cases, we would want to say that the current is the dependent variable, the other two are independent variables. This is because we can typically control voltage and resistance much more easily, and we think of voltage as the "causative" element in the situation. Hence, in light of our previous discussion, we take $I$ to be a function of $V$ and $R$:

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

and note that $V = IR$ then holds.

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