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A high-voltage powerline operates at 500000 V-rms and carries an rms current of 500 A. If the resistance of the cable is 0.050Ω/km, what is the resistive power loss in 200 km of the powerline?

Respuesta :

pstnonsonjoku

Answer:

2,500,000W or 2.5MW

Explanation:

The power lost due to resistance is given by I^2R. We must first obtain R as follows;

Resistance per kilometer= 0.050Ω/km

Distance covered= 200km

R = 200km x 0.050Ω/km = 10Ω

The lost power as a 500A current passes through the powerline is:

P = I²R

P= 500² x 10

P= 2,500,000 W or 2.5MW

snehashish65

The resistive power loss in 200 km of the powerline is of 2.5 MW.

Given data:

The root mean square voltage is, V' = 500000 V.

The magnitude of current through the power line is, I =500 A.

The magnitude of resistance of cable is, R = 0.050 Ω/km.

The length of powerline is, L = 200 km.

Whenever there is a flow of current through the wire, then there are various losses out of which the power loss is a major factor. The mathematical expression for the power loss is given as

P = I²R

Solving as,

P= 500² x 10

P= 2,500,000 W or 2.5MW

Thus, we can conclude that the resistive power loss in 200 km of the powerline is of 2.5 MW.

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