By calculation, impressed pressure = √(2402 + 1702) = 294 volts.
EXAMPLE.—In the circuit shown in fig. 1,330, what is the angle of lead?
The tangent of the angle of lead is given by the quotient of the reactance divided by the resistance of the circuit. That is,
| reactance | reactance drop | ||||||
| tan φ | = | = | |||||
| resistance | resistance drop | ||||||
| Ec | I | ||||||
| tan φ | = | = | ÷ | Ea | (1) | ||
| Ea | 2πfC |
The tangent is given a negative sign because lead is opposed to lag and because the positive value is assigned to lag. Substituting in (1)
| 170 | 2.125" | |||||
| tan φ | = | or | = | -.71 | ||
| 240 | 3" |
the angle corresponding is approximately 35¼° (see table page 451).
Figs. 1,332 and 1,333.—Diagrams for circuits containing inductance and capacity. Since inductance and capacity act 180° apart, their reactances, or their ohmic drops may be represented by oppositely directed lines. These may be drawn above and below a reference line, as in fig. 1,332, and their algebraic sum taken, or both may be drawn on the same side of the reference line and their difference in lengths, as CD, fig. 1,333, measured. Recourse to a diagram for obtaining the resultant reactance in circuits containing inductance and capacity is unnecessary as it is simply a matter of taking the difference of two quantities.