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 0.56 ?/cable, and the power of the generator is 284 kW. If Vt = 86 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 = 9.1 kV, what are (c) ?V and (d) Pd? If Vt = 1.1 kV, what are (e) ?V (in kV) and (f) Pd?

Explanation / Answer

a)   The rms current in the cable is I(rms)=P/Vt=284000/86000=3.302A

      therefore rms voltage drop is: V=I(rms)*R=3.302*2*0.56=3.698V

b)    Rate of energy dissipation Pd=I(rms)^2*R=3.302^2*(2*0.56)=12.211W

Same calculation applies to other parts of question

i found

c) I(rms)=31.208A   V=34.953V

d) Pd=1090.8W

e) I(rms)=258.18A   V=289.163V

d) Pd=37327.87W

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