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 $M L^{2} I^{- 2} T^{- 3}$ and $M L^{2} T^{- 3} I^{- 1}$ respectively.

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Solution

Dimensional formulae of R $= [M L^{2} I^{- 2} T^{- 3}]$

Dimensional formulae os V $= [M L^{2} I^{- 1} T^{- 3}]$

Dimensional formulae of I = [I]

$∴ [M L^{2} I^{- 1} T^{- 3}] = [M L^{2} T^{- 3} I^{- 2}] [I]$

$\Rightarrow V = I R$

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Dimensional formulae of R =[ML2I−2T−3] Dimensional formulae os V =[ML2I−1T−3] Dimensional formulae of I = [I] ∴[ML2I−1T−3]=[ML2T−3I−2][I] ⇒V=IR

Dimensional formulae of R $= [M L^{2} I^{- 2} T^{- 3}]$

Dimensional formulae os V $= [M L^{2} I^{- 1} T^{- 3}]$

Dimensional formulae of I = [I]

$∴ [M L^{2} I^{- 1} T^{- 3}] = [M L^{2} T^{- 3} I^{- 2}] [I]$

$\Rightarrow V = I R$