Question

In $\mathbf{\pi }$ equivalent circuits of lossless lines,$\mathbf{Z}\mathbf{\text{'}}$ is pure____ and $\mathbf{Y}\mathbf{\text{'}}$pure______.

Short answer

The$\mathrm{\pi }$ equivalent circuit of lossless line,$\mathrm{Z}\text{'}$ is purely inductive and$\mathrm{Y}\text{'}$ is purely capacitive.

Step-by-step solution

Determine the equation to calculate the parameter of π .

The equation for the calculation of the$\mathrm{Z}\text{'}$ is given below,

$\mathbf{Z}\mathbf{\text{'}}\mathbf{=}\mathbf{j\omega L}\mathbf{\left(}\frac{\mathbf{sin\beta l}}{\mathbf{\beta l}}\mathbf{\right)}$ …… (1)

Here, $\mathbf{\beta }$ is the phase constant.

The equation for the calculation of$\mathrm{Y}\text{'}$ is given below,

$\frac{\mathbf{Y}\mathbf{\text{'}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{j\omega C}}{\mathbf{2}}\left[\frac{\tan \left(\frac{\mathrm{\beta l}}{2}\right)}{\frac{\mathrm{\beta l}}{2}}\right]$ …… (2)

Determine the nature of the Z' and Y' .

Since the inductance is present in the equation of the $\mathrm{Z}\text{'}$parameter of the $\mathrm{\pi }$equivalent circuit, therefore the $\mathrm{Z}\text{'}$parameter purely inductive.

Similarly, the capacitance is present in the$\mathrm{Y}\text{'}$ parameter of the$\mathrm{\pi }$ equivalent circuit, therefore,$Y\text{'}$ parameter is purely is purely capacitive.

Therefore,role="math" localid="1656492160063" $\mathrm{\pi }$ equivalent circuit of lossless line,$\mathrm{Z}\text{'}$ is purely inductive and $Y\text{'}$is purely capacitive.