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, safe and convenient for use in the factory. The transmission line resistance is 0.26 /cable, and the power of the generator is 250 kW. Calculate the voltage drop V along the transmission line and the rate Pd at which energy is dissipated in the line as thermal energy for the following values of Vt.

Explanation / Answer

For Vt = 79 kV

current = P/V = 250 kW/79kV = 3.165 A

Voltage drop is V = IR = 2 x 3.165 x 0.26 = 1.65 V(a)

P = VxI = 1.65 v x 3.165 = 5.22 W (b)

We have to repeat the above for the other values of V

For Vt = 7.9 kV

current = P/V = 250 kW/7.9kV = 31.65 A

Voltage drop is V = IR = 2 x 31.65 x 0.26 = 16.5 V(a)

P = VxI = 16.5 v x 31.65 = 520.8 W (b)

For Vt = 0.79 kV

current = P/V = 250 kW/0.79kV = 316.5 A

Voltage drop is V = IR = 2 x 316.5 x 0.26 = 165 V(a)

P = VxI = 165 v x 316.5 = 52082.3 W (b)

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