Obtain Ohm’s law from dimensional analysis
Question:

Let $I=$ current through a conductor, $R=$ its resistance and $V=$ potential difference across its ends. According to Ohm’s law, product of two of these quantities equals the third. Obtain Ohm’s law from dimensional analysis. Dimensional formulae for $R$ and $V$ are $\left[\mathrm{ML}^{2} \mathrm{I}^{-2} \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~J}^{-1}\right]$ respectively

Solution:

Dimension of $\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{I}^{-2} \mathrm{~T}^{-3}\right]$

Dimension of $\mathrm{V}=\left[\left.\mathrm{ML}^{2} \mathrm{~T}^{-3}\right|^{-1}\right]$

Now, $R=\left[\mathrm{ML}^{2} \mathrm{I}^{-2} \mathrm{~T}^{-3}\right]=\left[\mathrm{ML}^{2} \mathrm{I}^{-1} \mathrm{~T}^{-3}\right] /[I]=\mathrm{V} /[\mathrm{I}]=\mathrm{V} / \mathrm{I}$ or, $V=I R$.

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