Question

Let I = current through a conductor, R = its resistance and V = potential difference across the 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 ML²I^-2T^-3and ML²T^-3I^-1 respectively.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{assume} \\ \mathrm{I}={\mathrm{V}}^{\mathrm{a}}{\mathrm{R}}^{\mathrm{b}} \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{I}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0{\mathrm{I}}^1\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{V}=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-1}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{R}=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-2}\right] \\ \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0{\mathrm{I}}^1\right]={\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-1}\right]}^{\mathrm{a}}{\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-2}\right]}^{\mathrm{b}} \\ \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0{\mathrm{I}}^1\right]=\left[{\mathrm{M}}^{\mathrm{a}+\mathrm{b}}{\mathrm{L}}^{2\mathrm{a}+2\mathrm{b}}{\mathrm{T}}^{-3\mathrm{a}-3\mathrm{b}}{\mathrm{I}}^{-\mathrm{a}-2\mathrm{b}}\right] \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}, \\ \mathrm{a}+\mathrm{b}=0............\left(1\right) \\ 2\mathrm{a}+2\mathrm{b}=0..........\left(2\right) \\ -3\mathrm{a}-3\mathrm{b}=0........\left(3\right) \\ -\mathrm{a}-2\mathrm{b}=1........\left(4\right) \\ \mathrm{from}\mathrm{equation}\left(1\right), \\ \mathrm{a}=-\mathrm{b} \\ \mathrm{Substitute}\mathrm{in}\mathrm{equation}\left(4\right), \\ \mathrm{b}-2\mathrm{b}=1 \\ -\mathrm{b}=1 \\ \mathrm{or},\mathrm{b}=-1 \\ \mathrm{a}=1 \\ \mathrm{Hence}, \\ \mathrm{I}={\mathrm{VR}}^{-1} \\ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \end{gathered}$$