Chapter 21, Problem 34P

To determine

The distance of current travel from plant to house as per the given Fig 21-25

Answer to Problem 34P

Solution:

At the distance of $\text{X=58}\text{.83 m}$, power loss will be 50%

Explanation of Solution

Given info:

As per fig 21-25

Power Plant ${\text{P}}_{\text{p}}\text{=100KW}$

${\text{ =100×10}}^{\text{3}}$

Delivered to step-up transformer at $V_p\text{=12000 V}$

Resistance per unit length ${\text{R = 5×10}}^{\text{-5}}\text{Ω}/\text{m}$

Formula to be used:

$\text{P=VI}$

Calculation:

$\Rightarrow I\text{=}\frac{P}{V}$

$\begin{array}{l}\text{=}\frac{{\text{100×10}}^{\text{3}}}{\text{12000}}\\ \text{=8}\text{.3 A}\end{array}$

As it is a step-up transformer,which stepped up the voltage by a faster of 20 so that the current is stepped down by a faster of 20.

According to the current ratio formula (equation)

$\frac{I_s}{I_p}\text{=}\frac{N_p}{N_s}$

$\begin{array}{l}\Rightarrow I_s{\text{= I}}_p\text{×}\frac{N_p}{N_s}\\ \text{ = 8}\text{.3×}\frac{\text{1}}{\text{20}}\\ \text{ = 0}\text{.415}\end{array}$

The power loss in the lines,

Power loss ${\text{=I}}^{\text{2}}\text{RX}$ [given resistance is in per unit length]

$50\times {10}^3=\left({\left(0.415\right)}^2\times {\text{5×10}}^{\text{-5}}\right)\text{X}$

$\Rightarrow \text{X=58}\text{.83}\times {\text{10}}^9\text{ m}$