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 0.30 /cable, and the power of the generator is 250 kW. If Vt 80 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

Answer 1

Answer:

a) 1.875 volts

b) 5.86 W

Explanation:

Given data:

transmission line resistance = 0.30/cable

power of the generator = 250 kW

if Vt = 80 kV

a) Determine the decrease V along the transmission line

Rms current in cable = P / Vt = 250 / 80 = 3.125 amps

hence the rms voltage drop ( Δrmsvoltage )

Δrmsvoltage = Irms* R  = 3.125 * 2 * 0.30 = 1.875 volts

b) Determine the rate Pd at which energy is dissipated in line as thermal energy

Pd = $I^2rms*R$

    =  3.125^2 * 2* 0.30  = 5.86 W