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 0.52 ?/cable, and the power of the generator is 268 kW. If Vt = 90 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 = 8.9 kV, what are (c) ?V and (d) Pd? If Vt = 0.93 kV, what are (e) ?V (in kV) and (f) Pd?

Explanation / Answer

a)the rms current in the cable

Irms =P/Vt =268*103/90*103 =2.98A

the rms voltage drop is

V =IrmsR =2.98*(2*0.52) =3.1V

b)the rate of energy dissipation is

Pd =I2rmsR =2.782*(2*0.52)=8.04 W9.2W

c)Irms =268*103/8.9*103 =30.1A

V=30.1*(2*0.52) =31.3 V

d)Pd =30.12*(2*0.52) =942.3W or 0.94KW

e)Irms=268*103/0.93*103 =288.2 A

V =288.2*(2*0.52) =299.73V 300V

f)Pd =288.22*(2*0.52) =86,381.6 W or 86.28KW

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