**Inductive Reactance**

Opposition to the flow of an alternating current by the inductance of the circuit; equal to 2πfL and measured in ohms.

When a changing current flows through an inductor, a self-induced voltage is developed. Its polarity is such that it opposes the change. The emf varies directly with the rate of change of current. This opposition is called inductive reactance. And it is a property of all inductors. Inductive reactance is measured in ohms and is represented by XL, where X indicates reactance and the subscript L implies that it is inductive. If the current change is sinusoidal, the inductive reactance can be calculated from the following equation:

${{X}_{L}}=2\pi fL\text{ }\cdots \text{ (1)}$

Where

2π=one cycle in radian measure

F=frequency in Hertz

L=inductance in Henry

Because angular velocity, 2πf, is frequently written as ω, we can write equation 1 as follows:

${{X}_{L}}=\omega L\text{ }\cdots \text{ (2)}$

And analysis of equation 1 indicates that inductive reactance is directly proportional to frequency and inductance. The following example explores this relationship:

**Example**

Calculate the inductive reactance of an 8 H filter choke at 60 Hz and at 120 Hz?

**Solution**

At 60 Hz:

${{X}_{L}}=2\pi fL=6.28*60*8\cong 3016\Omega $

At 120 Hz:

${{X}_{L}}=2\pi fL=6.28*120*8\cong 6032\Omega $

This example illustrates that if the frequency is doubled, the reactance is doubled the same way.

If the reactance of a particular inductance is known at a given frequency, its inductance can be determined by rearranging equation 1 as follows:

$L=\frac{{{X}_{L}}}{2\pi }$

**Example**

A small inductor used in a TV receiver has a reactance of 5kΩ at 45 M Hz. What is its inductance?

**Solution**

$L=\frac{{{X}_{L}}}{2\pi }=\frac{5*{{10}^{3}}}{6.28*45*{{10}^{6}}}=1.769*{{10}^{-5}}=17.69\mu H$

**Graph of X**_{L}

_{L}

Equation 1 is linear because X_{L} varies directly as the product of the linear terms f and L. if we plot inductive reactance against frequency, we will have a graph like the one shown by the solid line in figure 1, which happens to be for 8 H filter choke of example 1. If the frequency were increased beyond 120 Hz, the reactance line would increase beyond the limits shown.

Fig.1: Graph of X_{L} versus frequency for an 8 H and a 4 H inductor

Suppose the inductance used in example 1 were changed to 4 H. how would the reactance vary with frequency? Since there is a direct relation between inductive reactance and frequency, we can expect X_{L} to be reduced to one-half of its’s value at 60 Hz and 120 Hz. Plotting the reactance of the 4 H inductor on the graph of figure 1 results in the dotted line indicated. Observe that the slope (steepness) is one-half that of the 8 H inductance line. If the value of L were increased to 16 H, we would expect the slope line of its line to be twice that of the 8 H line.

**X**_{L} in series and in parallel

_{L}in series and in parallel

Inductive reactance’s in series or in parallel are treated in the same manner as resistances, as shown in figure 2.

Fig.2: Reactances: (a) In Series (b) In Parallel

Hence, the total inductive reactance of two or more series-connected inductors is as follows:

${{X}_{LT}}={{X}_{{{L}_{I}}}}+{{X}_{{{L}_{2}}}}+\cdots +{{X}_{{{L}_{n}}}}$

For two parallel-connected inductive reactance’s, the formula is as follows:

${{X}_{LT}}=\frac{{{X}_{{{L}_{I}}}}{{X}_{{{L}_{2}}}}}{{{X}_{{{L}_{I}}}}+{{X}_{{{L}_{2}}}}}$

When three or more inductive reactances are in parallel, the reciprocal formula is used:

$\frac{1}{{{X}_{LT}}}=\frac{1}{{{X}_{{{L}_{1}}}}}+\frac{1}{{{X}_{{{L}_{2}}}}}+\frac{1}{{{X}_{{{L}_{3}}}}}+\cdots +\frac{1}{{{X}_{{{L}_{n}}}}}$

Voltage drops and current flows are calculated by using Ohm’s law, except that X_{L} is substituted for R.

Very thankful

Thank you.