A) \[[{{A}^{-1}}]\]
B) \[[{{A}^{-2}}]\]
C) \[[A]\]
D) \[[{{A}^{2}}]\]
Correct Answer: A
Solution :
\[\frac{L}{R}\] has the dimensions of time. Because\[\frac{L}{R}\]is the time constant of\[L-R\]circuit. \[\therefore \] \[\left[ \frac{L}{R} \right]=[T]\] \[Charge=Capacitance\times Potential\text{ }difference\] \[Q=CV\] and \[[Q]=[AT]\] \[\therefore \] \[\left[ \frac{L}{RCV} \right]=\left[ \frac{L}{Q} \right]\] \[=\left[ \frac{T}{AT} \right]\] \[\left[ \frac{L}{RCV} \right]=[{{A}^{-1}}]\]You need to login to perform this action.
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