Dimensions of `ohm` are same as that of (where h is Planck's constant and e is charge)

Options

A. `(h)/(e)`
B. `(h^(2))/(e)`
C. `(h)/(e^(2))`
D. `(h^(2))/(e^(2))`

Answer

C

Explanation

Dimension of (ohm) `R`
`=(h)/(e^(2))`, (e = charge = current `xx` time)
`=([Et])/([it]^(2))=(P)/(i^(2))=(R)" as "P=((E)/(t))`