Question

In a dc 2 wire feeder the voltage drop per wire is 2.5 the transmission efficiency of the feeder is

Answer 1

Answer:5

Explanation:

Answer 2

Answer:

η =90 %

Explanation:

Given that

Voltage drop = 2.5 V per wire

Input power

Pi = I.Vi

Vi=Input voltage

Output power

Po= I.Vo

SO the transmission efficiency of feeder

η = Po/Pi

η = Vo/Vi

Vo= Vi- Drop in both the wire

η = (Vi - drop)/Vi

Drop = IR

$\eta=1- \dfrac{IR }{V_i}$

100 x IR/Vi is the percentage drop per wire .

$\eta=1- \dfrac{IR }{V_i}$

$\eta=(1- 2\times 5)\times 100$

η =90 %

So the transmission efficiency is 90%.