Question

Dimensions of electrical resistance are:
$\text{A}\text{. }\left[M{L}^2{T}^{-3}{A}^{-1}\right]$
$\text{B}\text{. }\left[M{L}^2{T}^{-3}{A}^{-2}\right]$
$\text{C}\text{. }\left[M{L}^3{T}^{-3}{A}^{-2}\right]$
$\text{D}\text{. }\left[M{L}^2{T}^3{A}^2\right]$

Answer 1

Hint: According to Ohm’s law, V = iR, where V is potential difference, i is current and R is resistance. Hence, $R={\displaystyle \frac{V}{i}}$. Use this formula to find the dimensional formula of R. V is work done per unit charge.

Formula used:
V = iR
$\left[W\right]=\left[M{L}^2{T}^{-2}\right]$
[q] = [AT]

Complete step by step answer:
Electrical resistance of a given substance is the ability of the substance to resist the flow of electrons or charges when a potential difference is created across the substance.
The flow of charges per unit time is called current in the circuit.

Suppose a conductor with a resistance of R is connected across a cell of emf V. This cell will create a potential difference V across the conductor. Due to the potential difference there will be current in the circuit. Let the current in the circuit be i.
According to Ohm’s law, V = iR.
Hence, we get that
$R={\displaystyle \frac{V}{i}}$.
Hence, the dimensional formula of resistance will be equal to the ratio of dimensional formulas of potential difference to current. $\left[R\right]={\displaystyle \frac{\left[V\right]}{\left[i\right]}}$ …. (i).
Therefore, let us find the dimensional formula of V and i.

Potential difference is equal to the work done per unit charge.
The dimensional formulas of work done is $\left[W\right]=\left[M{L}^2{T}^{-2}\right]$.
The dimensional formula of charge is [q] = [AT].
Hence, the dimensional formula of potential difference is $\left[V\right]={\displaystyle \frac{\left[W\right]}{\left[q\right]}}={\displaystyle \frac{\left[M{L}^2{T}^{-2}\right]}{\left[AT\right]}}=\left[M{L}^2{T}^{-3}{A}^{-1}\right]$.
The dimensional formula of current is [A].

Substitute the dimensional formulas of potential difference and current in equation (i).
$\left[R\right]={\displaystyle \frac{\left[M{L}^2{T}^{-3}{A}^{-1}\right]}{\left[A\right]}}=\left[M{L}^2{T}^{-3}{A}^{-2}\right]$.
Hence, the correct answer is option B.

Note:
We can also use the units of the quantities for finding the dimensional formula of resistances.
The unit of work is $kg{m}^2{s}^{-2}$.
The unit of charge is As.
Therefore, the unit of potential difference is $kg{m}^2{s}^{-3}{A}^{-1}$.
The unit of current is A.
Hence,the unit of resistance is ${\displaystyle \frac{kg{m}^2{s}^{-3}{A}^{-1}}{A}}=kg{m}^2{s}^{-3}{A}^{-2}$.
The dimensions of units kg, m, s and A are M, L, T and A.
Hence, the dimensions of resistance are $\left[M{L}^2{T}^{-3}{A}^{-2}\right]$.