Question

Does$\mathit{r}\mathbf{\text{'}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}}\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{788}\mathit{r}$, which comes in calculation of inductance, play a role in capacitance computations?

(a) Yes

(b) No

Short answer

The correct option is (b): No

Step-by-step solution

Determine the expression for the single-phase circuit of inductance.

The expression for the calculation of inductance for the single-phase circuit is given by,

$\mathit{L}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{\times }\mathit{l}\mathit{n}\left(\frac{D}{r\text{'}}\right)\mathbf{ }\mathit{H}\mathbf{/}\mathit{m}$

Here, is the distance between the two conductors and is of the radius of the conductor.

Hence the r^' plays a role in the calculation of the inductance.

Determine whether plays a role in the calculation of capacitance or not.

The expression for the calculation of the capacitance of the single-phase circuit is given by,

$C=\frac{2\pi \epsilon }{\ln \left(\frac{D}{r}\right)} \text{F/m}$

Here, D is the distance between the two conductors and r is the radius of the conductor.

From the above discussion, it is clear that the r' does not play a role in the calculation of the capacitance.

Therefore, capacitance does not dependon r'.