Aviation Training Experts™

handbook

Aviation Maintenance Technician Handbook–General

FAA-H-8083-30B Version 2023

Chapter 12

Fundamentals of Electricity & Electronics

The square root of both sides of the equation gives

Formula: Z = R2 + XL 2

This formula can be used to determine the impedance when the values of inductive reactance and resistance are known. It can be modified to solve for impedance in circuits containing capacitive reactance and resistance by substituting XC in the formula in place of XL. In circuits containing resistance with both inductive and capacitive reactance, the reactances can be combined, but because their effects in the circuit are exactly opposite, they are combined by subtraction:

Formula: X = XL − XC or X = XC − XL (the smaller number is ; always subtracted from the larger)

In Figure 12-135, a series circuit consisting of resistance and inductance connected in series is connected to a source of 110 volts at 60 cps. The resistive element is a lamp with 6 ohms resistance, and the inductive element is a coil with an inductance of 0.021 henry. What is the value of the impedance and the current through the lamp and the coil?

Solution:

First, the inductive reactance of the coil is computed:

Formula: XL = 2π × f × L ; XL = 6.28 × 60 × 0.021 ; XL = 8 ohms inductive reactance

Next, the total impedance is computed:

Formula: Z = R2 + XL 2 ; Z = 62 + 82 ; Z = 36 + 64 ; Z = 100 ; Z = 10 ohms impedance

Then the current flow,

Formula: E ; I = ; Z ; 110 ; I = ; 10 ; I = 11 amperes current
Figure 12-130. AC circuit containing inductance.
Figure 12-130. AC circuit containing inductance.
Figure 12-131. Inductances in series.
Figure 12-131. Inductances in series.

The voltage drop across the resistance (ER) is:

Formula: ER = I × R ; ER = 11 × 6 = 66 volts

The voltage drop across the inductance (EXL) is:

Formula: EXL = I × XL ; EXL = 11 × 8 = 88 volts

The sum of the two voltages is greater than the impressed voltage. This results from the fact that the two voltages are out of phase and, as such, represent the maximum voltage. If the voltage in the circuit is measured by a voltmeter, it is approximately 110 volts, the impressed voltage. This can be proved by the equation:

Formula: E = (ER)2 + (EXL)2 ; E = 662 + 882 ; E = 4,356 + 7,744 ; E = 12,100 ; E = 110 volts