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Asked in GUJCET 2017 · Units and dimensions
Given: R (resistance) and L (inductance).
Idea: in the growth of current in an L-R circuit, i=i₀(1-e^-Rt/L). An exponent must be a pure number, so L/R has to have the dimension of time.
Check it directly: [L]=M¹L²T⁻²A⁻² and [R]=M¹L²T⁻³A⁻².
([L])/([R])=T¹, a time.
A square root would leave T^1/2 and the inverted ratio T⁻¹, so only L/R works.
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