A) \[[M{{L}^{2}}{{T}^{2}}{{A}^{-2}}]\]
B) \[[M{{L}^{2}}{{T}^{-1}}{{A}^{-2}}]\]
C) \[[M{{L}^{2}}{{T}^{-1}}{{A}^{-1}}]\]
D) \[[M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]\]
Correct Answer: C
Solution :
Numerically, self-inductance L is given by \[L=\frac{e}{\Delta i/\Delta t}=\frac{e\Delta t}{\Delta i}\] where e is emf, Ac the time and Ai the current. \[\therefore \] \[Unit\text{ }of\text{ }L=\frac{Volt-\text{ }second}{Ampere}\] \[=\frac{(Joule/Coulomb)\text{ }second}{Ampere}\] \[=\frac{Newton-metre-second}{Coulomb-ampere}\] \[\Rightarrow \] \[Unit\text{ }of\text{ }L=\frac{Newton-metre}{Ampere}\] \[\therefore \]\[[L]=\frac{[ML{{T}^{-2}}][L]}{{{[A]}^{2}}}=[M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]\]
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