Question

The maximum power that a lossless line can deliver, in terms of the voltage magnitudes${\mathbf{V}}_{\mathbf{s}}$and${\mathbf{V}}_{\mathbf{R}}$(in volts) at the ends of the line held constant, and the series reactanceof the corresponding equivalent$\mathrm{\pi }$circuit, is given by __________, in watts.

Short answer

Therefore, The maximum power that a lossless line can deliver, in terms of the voltage magnitudes ${\mathrm{V}}_{\mathrm{s}}$and ${\mathrm{V}}_{\mathrm{R}}$(in volts) at the ends of the line held constant, and the series reactance ${\mathrm{X}}^{\text{'}}$ of the corresponding equivalent $\mathrm{\pi }$circuit, is given by , $\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{X}}^{\text{'}}}$in watts.

Step-by-step solution

Determine the formula maximum power delivered by line.

Write the formula of maximum power.

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}$ ……. (1)

Here, ${\mathrm{V}}_{\mathrm{S}}$is sending end voltage, ${\mathrm{V}}_{\mathrm{R}}$is receiving end voltage, and ${\mathrm{X}}^{\text{'}}$is the line reactance.

Determine the answer.

Write the formula of real power delivered by source through lossless line.

$\mathrm{P}=\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{X}}^{\text{'}}}\mathrm{sin\delta }$ …… (2)

Here, $\delta$is a load angle.

The maximum real power is obtained for $\delta ={90}^{°}$.

Hence, equation (2) becomes,

$$\begin{gathered} {\mathrm{P}}_{\max }=\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{X}}^{\text{'}}}\sin {90}^{°} \\ =\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{X}}^{\text{'}}} \end{gathered}$$

Therefore, The maximum power that a lossless line can deliver, in terms of the voltage magnitudes ${\mathrm{V}}_{\mathrm{S}}$and $V_R$(in volts) at the ends of the line held constant, and the series reactance ${\mathrm{X}}^{\text{'}}$ of the corresponding equivalent $\mathrm{\pi }$circuit, is given by $\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{{\mathrm{X}}^{\text{'}}}$in watts.