Question

An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the energy is transmitted through a long-distance power line which has a resistance of 100 ohms. What percentage of the power delivered by the generator is dissipated as heat in the long-distance power line?

Answer 1

Given:

$I_{rms} = 50 A$

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, $R = 100\ohm$

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = $50\times 3600$ W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = $\frac{P}{V'}$

I = $\frac{1800000}{100000}$

I = 1.8 A

Power line voltage drop is given by:

$V_{drop} = I\times R$

$V_{drop} = 1.8\times 100$

$V_{drop} = 180 V$

Power dissipated in long transmission line $P_{dissipated} = V_{drop}\times I$

Power dissipated in long transmission line $P_{dissipated} = 180\times 1.8$ = 324 W

% Heat loss in power line, $P_{loss} = \frac{P_{dissipated}}{P}\times 100$

% Heat loss in power line, $P_{loss} = \frac{324}{180000}\times 100$

$P_{loss} = 0.18%$