An ac generator provides emf to a resistive load in a remote factory over a two-

Question

An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a step-down transformer reduces the voltage from its (rms) transmission value Vt to a much lower value that is safe and convenient for use in the factory. The transmission line resistance is 0.34 /cable, and the power of the generator is 284 kW. If Vt = 110 kV, what are (a) the voltage decrease V along the transmission line and (b) the rate Pd at which energy is dissipated in the line as thermal energy? If Vt = 8.1 kV, what are (c) V and (d) Pd? If Vt = 0.91 kV, what are (e) V and (f) Pd?

Explanation / Answer

a. current = P0wer/Voltage = 284 kW/110kV = 2.582 amps

voltage drop is V = I*R = 2x 2.582x 0.34 = 1.75576volts.

b. P dissipated = V*I = 1.7557*2.582 = 4.5334 watts

(c) V

current = Power/Voltage = 284 kW/8.1kV = 35.06 amps

voltage drop is V = I*R = 2x 35.06x 0.34 = 23.8408volts

and

(d) Pd= V*I =23.8408*35.06 =835.858 Watts

e. current = Power/Voltage = 284 kW/0.91kV = 312.09 amps

voltage drop is V = I*R = 2x 312.09x 0.34 = 212.2212volts

f. Pd = V*I =212.2212*312.09 =66232.11431 watts =66.232 kW

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