Question

Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm's law, product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for R and V are ML2I-2T-3 and ML2T-3I-1 respectively.

Answer 1

I = V / R

Explanation:

Assuming I = VaRb

Dimensional formula of I = [M0L0T0I1]

Dimensional formula of V = [M1L2T−3I−1]

Dimensional formula of R = [M1L2T−3I−2]

Substituting the formula, we get:

[M0L0T0I1] = [M1L2T−3I−1]^a * [M1L2T−3I−2]^b

[M0L0T0I1] = [M^a+b L^2a+2b T^−3a−3b I^−a−2b]

When we compare the co-efficients, we infer that:

a+b=0  ............ (1)

2a+2b=0  .......... (2)

−3a−3b=0  ........ (3)

−a−2b=1 ........ (4)

From equation (1), we get a = −b.

Substituting that in equation (4), we get:

b −2b = 1 −b = 1 or  b = −1a =1

So  I = VR^−1

I =V/R

Answer 2

Obtained Ohm's law from dimensional analysis of current is I.

Explanation:

According to Ohm's law,

Current flowing in the conductor is directly proportional  to the potential difference across its conductor providing that the physical conditions such as temperature remains constant.

       (∝ = directly proportional)

Ohm's law ,

              V = I × R

                I = V/RS

               V = potential

               R = resistance

We know, Dimension of Current is I. If we can prove that its dimension is I through the dimension of V and R, then the Ohm's law can be proved.

Given :  

Dimension of R  $=M L^{2} I^{-2} T^{-3}$

Dimension of $V=M L^{2} T^{-3} I^{-1}$

Ohm's law,

                I = V/R

                [I] = (M L² T⁻³ I⁻¹) ÷ (M L² I⁻² T⁻³)

                [I] = 1/I⁻¹

                 [I] = I

It is  proved that I = V/R. Since the dimension of the Current is correct.

Thus, it's proved that the Dimension of Current is I.