Question

Suppose a power plant produces 811 kW of power and is to send that power for many miles over a copper wire with a total resistance of 12.0 Ω.

If the power is sent across the copper wires at 48.0 kV rms, how much current flows through the wires?

If the power is sent across the copper wires at 48.0 kV rms, what is the power dissipated due to the resistance of the wires at this current?

What percent of the total power output of the plant is the power dissipated due to the resistance of the wires?

Although a series of transformers step the voltage down to the 120 V used for household voltage, assume you are using a single transformer to do the job. If the single transformer has 10500 primary turns, how many secondary turns should it have?

Answer #1

Power of the plant, P = 811 KW = 811 x 10^3 W

(a) Voltage level, V = 48.0 KV = 48 x 10^3 V

Current flows through the wires, I = P / V = (811 x 10^3) / (48 x 10^3) = 16.9 A (Answer)

(b) Power dissipated due to resistance of the wire, P(L) = I^2 x R

= 16.9^2 x 12.0 = 3426 W (Answer)

(c) Required percentage = [P(L) / P] x 100 = [3426 / 811000] x 100 = 0.42 % (Answer)

(d) N1 / N2 = V1 / V2

=> 10500 / N2 = 48000 / 120

=> N2 = (10500 x 120) / 48000 = 26 (Answer)

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