Question

Show that $\left(\dfrac{\text{Henry}}{\text{Farad}}\right)^{1/2}$ represents resistance.

Hint:

Always use dimensional analysis to check consistency of physical equations.

Answer 1

Step 1: Dimensions of Henry (H).
1 Henry = $\dfrac{\text{Weber}}{\text{Ampere}}$ = $\dfrac{\text{Volt} \cdot \text{second}}{\text{Ampere}}$ \[ = \dfrac{(ML^2T^{-3}A^{-1}) \cdot T}{A} = M L^2 T^{-2} A^{-2} \]
Step 2: Dimensions of Farad (F).
1 Farad = $\dfrac{\text{Coulomb}}{\text{Volt}} = \dfrac{A \cdot T}{ML^2T^{-3}A^{-1}}$ \[ = M^{-1} L^{-2} T^4 A^2 \]
Step 3: Ratio Henry/Farad.
\[ \frac{\text{Henry}}{\text{Farad}} = \frac{M L^2 T^{-2} A^{-2}}{M^{-1} L^{-2} T^4 A^2} = M^2 L^4 T^{-6} A^{-4} \]
Step 4: Square root.
\[ \left(\frac{\text{Henry}}{\text{Farad}}\right)^{1/2} = M L^2 T^{-3} A^{-2} \]
Step 5: Dimensions of resistance.
Resistance (Ohm) = $\dfrac{\text{Volt}}{\text{Ampere}} = \dfrac{ML^2T^{-3}A^{-1}}{A} = M L^2 T^{-3} A^{-2}$
Step 6: Conclusion.
Thus, $\left(\dfrac{\text{Henry}}{\text{Farad}}\right)^{1/2}$ indeed represents resistance.