question_answer

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 \[\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{3}}}{{\text{A}}^{-\text{2}}}\]and \[\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{3}}}{{\text{A}}^{-\text{1}}}\]respectively.

Answer:

                Let                                                            \[\text{V}={{\text{I}}^{\text{a}}}{{\text{R}}^{\text{b}}}\] ...(i) \[\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{3}}}{{\text{A}}^{-\text{1}}} \right]\text{ }={{\text{A}}^{\text{a}}}{{\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{3}}}{{\text{A}}^{-\text{2}}} \right]}^{\text{6}}}=\text{ }{{\text{M}}^{\text{b}}}{{\text{L}}^{\text{2b}}}{{\text{T}}^{-\text{3b}}}{{\text{A}}^{\text{a}-\text{2b}}}\] Applying the principle of homogeneity of dimensions, we gel \[\text{b}=\text{1},\text{ 2b}=\text{2},-\text{3b}=-\text{3a}-\text{2b}=-\text{1},\text{ a}=-\text{1}+\text{2b}=-\text{1}+\text{2}=\text{1}\] . Putting these values in (i), we get                                                    \[\text{V }=\text{ }{{\text{I}}^{\text{1}}}{{\text{R}}^{\text{1}}}\] . This is Ohm's law.