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 converient for use in the factory. The transmission line resistance is 0.63 Q/cable, and the power of the generator is 256 kW. If V_t = 83 kV, what are the voltage decrease Delta V along the transmission line and the rate P_4 at which energy is dissipated in the line as thermal energy? If V_e = 8.8 kV, what are Delta V and P_d ? If V_t = 1.1 kV, what are Delta V and P_d?

Explanation / Answer

a) current = P/V = 256 kW/83kV = 3.08 A

voltage drop is V = IR = 2 x 3.08 x 0.63 = 3.88 volts total.

b) P = VxI = 3.88 x 3.08 = 11.95 watts

c) current = P/V = 256 kW/8.8kV = 29.1 A

voltage drop is V = IR = 2 x 29.1 x 0.63 = 36.67 volts total.

d) P = VxI = 36.67 x 29.1 = 1067.097 watts

e) current = P/V = 256 kW/1.1kV = 232.73 A

voltage drop is V = IR = 2 x 232.73 x 0.63 = 293.24 volts total.

f) P = VxI = 293.24 x 232.73 = 68245.7452 watts

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