Xi = 2πfL = 2 × 3.1416 × 150 × .041 = 38.64 ohms
(note that 41 milli-henrys are reduced to .041 henry before substituting in the above equation).
The capacity reactance, or
| 1 | 1 | |||||
| Xc | = | = | = | 20.8 ohms | ||
| 2πfC | 2 × 3.1416 × 150 × .000051 |
(note that 51 microfarads are reduced to .000051 farad before substituting in the above equation).
Substituting the values as calculated for 2πfL and 1 / (2πfC) in equation (2)
Z = √(232 + (38.64 - 20.8)2) = 29.1 ohms.
To solve the problem graphically, lay off in fig. 1,340, the line AB equal to 23 ohms resistance, using any convenient scale. Draw upward and at right angles to AB the line BC = 38.64 ohms inductance reactance, and from C lay off downward CC' = 20.8 ohms capacity reactance. The resultant reactance is BC' and being above the horizontal line AB shows that inductance reactance is in excess of capacity reactance by the amount BC'. Join AC' which gives the impedance sought, and which by measurement is 29.1 ohms.
In order to obtain the impressed pressure in circuits containing resistance, inductance and reactance, an equation similar to (2) on page [1,095] is used which is made up from the following:
| Eo | = | RI | (3) |
| Ei | = | 2πfLI | (4) |
| I | |||
| Ec | = | (5) | |
| 2πfC |