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 V_t to a much lower value that is safe and convenient for use in the factory. The transmission line resistance is 0.43 Ohm/cable, and the power of the generator is 250 kW. If V_t = 100 kV, what are (a) the voltage decrease Al/along the transmission line and (b) the rate P_d at which energy is dissipated in the line as thermal energy? If V_t = 8.2 kV, what are (c) Ay and (d) P_d? If V_t = 0.87 kV, what are (e) Ay and (f) P_d? Number Units Number Units Number Units Number Units Number Units Number Units

Explanation / Answer

Power=250KW,   Resistance R=0.43 ohms

If Vt=100KV,

              I=P/V=250KW/100KV=2.5A

(a).......Voltage Drop V=IR=2.5*2*0.43=2.15V

(b).......Power P=VI=2.15*2.5=5.375W

If Vt=8.2KV

      I=P/V=250KW/8.2KV=30.48A

(c)......Voltage Drop V=I*R=30.48*2*0.43=26.2V

(d)......Power P=V*I=26.2*30.48=798.576W

If Vt=0.87KV

       I=P/V=250KW/0.87KV=287.35A

(e)......Voltage Drop V=I*R=287.35*2*0.43=247.17V

(f)......Power P=V*I=247.17*287.35=71KW

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