Obtain the dimensions of (1) resistance, (2) inductance, (3) capacitance.

Question

Accepted Answer

1 Resistance is electric potential difference divided by electric current. From Equation (1.8) the dimensions of electric potential difference are [latex][M L^{2} T ^{-3} A^{-1}][/latex]. Thus [r] = [latex][M L^{2} T^{-3} A^{-1}] / [ A ] [/latex], so

[latex][pd] = [M L^{2} T^{-3} A^{-1}][/latex] (1..8)

[latex][r] = [M L^{2} T^{-3} A^{-2}][/latex] (1.9)

2 The magnitude of the emf induced in a coil of inductance L when the current through it changes at the rate of I ampere in t seconds is given by e = L I / t , where e is measured in volts and is a potential difference. Thus the dimensions of L are given by [l] = [pd][T]/[A]. From Equation (1.8),

[latex][pa] = [M L^{2} T^{-3} A^{-1}],[/latex] so
[latex][l] = [M L^{2} T^{-2} A^{-2}][/latex] (1.10)

3 Capacitance (C) is electric charge (Q) divided by electric potential difference (V). From Equation (1.7), [q] = [A T]. From Equation (1.8), [latex][pd] = [M L^{2} T^{-3} A^{-1}][/latex]. Thus [latex][c] - [A T]/[M L^{2} T^{-3} A^{-1}][/latex], so

[latex][c] = [M^{-1} L^{-2} T^{4} A^{2}] [/latex] (1.11)

Question 1.4: Obtain the dimensions of (1) resistance, (2) inductance, (3)...

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