Ei = 2πfLI = 2 × 3.1416 × 60 × .15 × 8 = 452 volts
The capacity drop is
| I | 8 | |||||
| Ec | = | = | = | 170 volts. | ||
| 2πfC | 2 × 3.1416 × 60 × .000125 |
Substituting the values thus found,
| impressed pressure | = | √(Eo2 + (Ei - Ec)2) |
| = | √(2002 + (452 - 170)2) | |
| = | √(2002 + 2822) | |
| = | √(119524) | |
| = | 345.7 volts. |
CHAPTER XLVIII
THE POWER FACTOR
The determination of the power in a direct current circuit is a simple matter since it is only necessary to multiply together the volts and amperes to obtain the output in watts. In the case of alternating current circuits, this holds true only when the current is in phase with the pressure—a condition rarely found in practice.
When the current is not in phase with the pressure, the product of volts and amperes as indicated by the voltmeter and ammeter must be multiplied by a coefficient called the power factor in order to obtain the true watts, or actual power available.
There are several ways of defining the power factor, any of which requires some explanation. The power factor may be defined as: The number of watts indicated by a wattmeter, divided by the apparent watts, the latter being the watts as measured by a voltmeter and ammeter.