Agenerator produces 2 mw of electric power at

Ответ:

SavgeKid7578

10.01.2022

To solve this problem it is necessary to apply the concepts related to Power in function of the current and the resistance.

By definition there are two ways to express power

P = I^2R

P =VI

Where,

P = Power

I = Current

R = Resistance

V = Voltage

In our data we have the value for resistivity and not the Resistance, then

$R = \rho \frac{1}{A}$

$R = 2.7*10^{-8}\frac{1}{\pi r^2}$

$R = 2.7*10^{8}\frac{1}{\pi (6.8*10^-3)^2}$

$R = 1.85*10^{-4}\Omega$

The loss of the potential can mainly be given by the resistance of the cables, that is,

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

$I = \frac{2*10^6W}{15*10^3V}$

I = 133.3A

Therefore the expression for power loss due to resistance is,

P = I^2 R

$P = 133.3^2 * 1.85*10^{-4}$

P_l = 3.2872W

The total produced is 2 * 10 ^ 6 MW , that is to say 100%, therefore 3.2872W is equivalent to,

$x = \frac{3.2872*100}{2*10^6}$

$x = 1.6436*10^{-4}\%$

Therefore the percentage of lost Power is equivalent to $1.6436 * 10 ^ 4 \%$ of the total

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