0
$\begingroup$

This is the standard derivation for Ohm's Law :

$$I \propto V$$ $$V \propto I$$ $$\therefore V = k.I$$

Where $k$ is the constant of proportionality.

We define this $k$ as resistance of $R$.

Why can't we derive it like this: $$I \propto V$$ $$\therefore I = k_2V$$

Where $k_2$ is the constant of proportionality

So, in this case : $$k_2=\dfrac{1}{R}$$

Why was resistance not defined as how $k_2$ is defined and why was it defined as it is?

I actually think that this might be the reason :

Most of the time, current(($I$) is less than the voltage($V$)

Let's say that resistance was defined as $k_2$, in that case, $$k_2=\dfrac{I}{V}$$ if $I$ is less than $V$ (like it generally is), then the value of $k_2$ will come out to be a fraction which is not too convenient. So, if we define the constant of proportionality as $R$, then $$R=\dfrac{V}{I}$$ and since $I$ is less than $V$ in most cases, we would get $R > 0$ unlike the case with $k_2$ in which we would get $0 <k_2 < 1$

Is this reasonable?

$\endgroup$
8
  • 1
    $\begingroup$ How can current be less or more than voltage? They are different things. That's like saying I can run faster than the volume of my water bottle. $\endgroup$ Commented Apr 29, 2020 at 13:53
  • $\begingroup$ @AaronStevens What I was comparing was magnitude of current and voltage $\endgroup$ Commented Apr 29, 2020 at 13:59
  • 1
    $\begingroup$ Yeah, that still doesn't work. Magnitudes have units also. $\endgroup$ Commented Apr 29, 2020 at 14:01
  • $\begingroup$ I thought magnitude is a pure number. Like if we say $1$ $ampere$, the magnitude is $1$ and the unit is $amperes$. $\endgroup$ Commented Apr 29, 2020 at 14:04
  • 1
    $\begingroup$ But our choice of units is arbitrary. 1 amp is 1000 milliamps. So is 1000 mA more or less than 5 V? It would also be just as valid to make up a new current unit called the Rajdeep, such that $1\ {\rm A}=5.2\times 10^6\ {\rm Rd}$. Now is 5.2 million Rd more or less than 5 V? $\endgroup$
    – The Photon
    Commented Apr 29, 2020 at 14:59

1 Answer 1

5
$\begingroup$

What you are talking about is conductance. You can use that instead if you want to.

I actually think that this might be the reason : Most of the time, current(I) is less than the voltage(V)

This isn't correct. You can't compare values that are physically different. Currents can't be more or less than voltages.

...then the value of $k_2$ will come out to be a fraction which is not too convenient

I'm not sure where this is coming from. Fractions are totally fine. Beginners in algebra might hate fractions, but they aren't inconvenient in their actual application. Plus, many numbers that are fractions have reciprocals that are also fractions. e.g. $3/2$ and $2/3$.


Is this reasonable?

To summarize, it's totally fine to use $1/R$ which is the conductance. However, your arguments as to why one would want to do this either aren't valid, or are nonissues.

$\endgroup$
7
  • $\begingroup$ Let's say that you are George Ohm, and you notice a relationship between current and voltage. How would you decide which constant to call resistance and which one to call conductance? This is probably a silly question $\endgroup$ Commented Apr 29, 2020 at 14:02
  • 4
    $\begingroup$ @Rajdeep_Sindhu The larger the Resistance for a given voltage, the smaller the current. Hence "resistance" resists current. A larger conductance (smaller resistance) means you can conduct more current. I think that makes a lot of sense. $\endgroup$ Commented Apr 29, 2020 at 14:04
  • $\begingroup$ Thanks! I just realized how stupid my arguments were. And by the way, when I said fractions, I was actually referring to proper fractions ($0 < x < 1$ or $\dfrac{p}{q}$, where $p < q$) $\endgroup$ Commented Apr 29, 2020 at 14:08
  • $\begingroup$ @Rajdeep_Sindhu Proper fractions aren't inconvenient either :) $\endgroup$ Commented Apr 29, 2020 at 14:11
  • $\begingroup$ And as I said, I know my arguments were stupid and childish. Also, I'm no beginner to algebra too. :) $\endgroup$ Commented Apr 29, 2020 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.