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 transmission-line value Vt to a much lower value that is safe and convenient for use in the factory. The transmission line resistance is 0.3 Ω per cable, and the power of the generator is Pt = 250,000 watts (rms).1 If 80,000 volts Vt = (rms), what are:

(a) the voltage drop ∆V across the transmission line?

(b) the rate at which energy is dissipated in the line as thermal energy?

(c) If instead Vt were 8000 V (rms), what would be the voltage drop and power dissipated in the transmission line?

(d) Repeat part c for 800 volts Vt = (rms)

a) 1.875 V b) 5.86 W c) 18.75 V 586 W d) 187.5 V 5.86 kW.

Explanation:

a) As the load is purely resistive, we can get the Irms, applying directly the definition of electrical power, as follows:

I = P / V = 250,000 W / 80,000 V = 3.125 A

Applying Ohm's Law to the resistance of the conductors in the transmission line, we have:

∆V = I. R = 3.125 A. 0.6 Ω = 1.875 V

b) The rate at which energy is dissipated in the line as thermal energy, can be obtained applying Joule's law, using the RMS value of the current that we have already got:

Pd = I2. R = (3.125) 2 A. 0.6Ω = 5.86 W

c) If Vt were 8,000 V instead of 80,000 V, we would have a different value for I, as follows:

I = 250,000 W / 8,000 V = 31.25 A

So, we would have a different ∆V:

∆V = 31.25 A. 0.6 Ω = 18.75 V

As the RMS current is different, we will have a different value por the dissipated power in the line:

P = (31.25) 2. 0.6 Ω = 586 W

d) As above, we will have a new value for I, as follows:

I = 250,000 W / 800 V = 312. 5 A

The new voltage loss in the transmission line will be much larger:

∆V = 312.5 A. 0.6 Ω = 187.5 V

Consequently, we will have a higher dissipated power:

P = (312.5) 2. 0.6 Ω = 58.6 kW