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 ML² I⁻²T⁻³ and
ML²T³ I⁻¹ respectively.
PhysicsHelper:
Concept of Physics - 1 , HC VERMA , Chapter "Introduction to Physics".
Answers
Answered by
65
Hello Dear.
According to the Ohm's law, Current flowing in the conductor is directly proportional to the potential difference across it conductor providing that the physical conditions such as temperature remains constant.
Numerically Ohm's law can be stated that ⇒
V = I × R
I = V/R
We know, Dimension of Current is I.If we can prove that its dimension is I through the dimension of Potential V and the Resistance R, then the Ohm's law can be proved.
Given ⇒
Dimension of R (Resistance) = M L² I⁻² T⁻³
Dimension of V (Potential) = M L² T⁻³ I⁻¹
Now, From the Numerically Stated 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
Thus, it is proved that the Dimension of Current is I.
Hence, it is proved that I = V/R. Since, the dimension of the Current is correct.
Hope it helps.
According to the Ohm's law, Current flowing in the conductor is directly proportional to the potential difference across it conductor providing that the physical conditions such as temperature remains constant.
Numerically Ohm's law can be stated that ⇒
V = I × R
I = V/R
We know, Dimension of Current is I.If we can prove that its dimension is I through the dimension of Potential V and the Resistance R, then the Ohm's law can be proved.
Given ⇒
Dimension of R (Resistance) = M L² I⁻² T⁻³
Dimension of V (Potential) = M L² T⁻³ I⁻¹
Now, From the Numerically Stated 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
Thus, it is proved that the Dimension of Current is I.
Hence, it is proved that I = V/R. Since, the dimension of the Current is correct.
Hope it helps.
Answered by
25
This might help u....
Attachments:
Similar questions