Question 2.5: Good Versus Poor Power Factor. A utility supplies 12,000 V (......
Good Versus Poor Power Factor. A utility supplies 12,000 V (12 kV) to a customer who needs 600 kW of real power. Compare the line losses for the utility when the customer’s load has a power factor of 0.5 versus a power factor of 1.0.
To find the current drawn when the power factor is 0.5, we can start with (2.43):
P_{avg} = VI \cos \theta = VI \times PF (2.43)
\begin{matrix} \quad \quad P & = & VI \cdot PF \quad \quad \quad \quad \ \\ 600 kW & = & 12 kV \cdot I \left(A\right) \cdot 0.5 \end{matrix}so
I = \frac{600}{12 \times 0.5} = 100 AWhen the power factor is improved to 1.0, (2.43) now looks like
600 kW = 12 kV \cdot I \left(A\right) \cdot 1.0so the current needed will be I = \frac{600}{12} = 50 A
When the power factor in the plant is improved from 0.5 to 1.0, the amount of current needed to do the same work in the factory is cut in half. The utility line losses are proportional to current squared, so line losses for this customer have been cut to one-fourth of their original value.