Impedance is the combined resistance in the flow of current because of resistive and reactive elements in the circuit. It is measured is ohm and calculated as follows;

$Z=R+jX$

**Inductive Reactance**

Consider a sinusoidal current to be flowing in the pure inductance as shown in the following Fig. that is

$i={{I}_{m}}\sin \left( wt \right)$

Since the sine wave of current is constantly changing, the coil produces a counter emf given by

${{V}_{L}}=L\frac{di}{dt}$

The voltage across the inductor is proportional to the slope of the sine wave of current and is the, therefore, a cosine wave, as shown in Fig.

Thus, in the purely inductive circuit, the current lags the voltage by a phase angle of 90°. Now substituting the equation for current into the equation for inductor voltage, we find that

${{V}_{L}}=L\frac{di}{dt}=L\frac{d\left( {{I}_{m}}sinwt \right)}{dt}={{I}_{m}}wL\cos \left( wt \right)$

The quantity I_{m}ωL is the maximum value of the voltage across the inductor (it occurs at t=0)

${{V}_{m}}={{I}_{m}}wL~~~~~\text{ }\cdots \text{ }~~~~\left( 1 \right)$

The quantity ωL is called an inductive reactance and is a measure of opposition to alternating current. Inductive reactance is measured in Ohm. The symbol X_{L} is used to denote inductive reactance.

${{X}_{L}}=wL=2\pi fL$

Since the maximum values of equation (1) are related to effective values, so we can write

${{V}_{L}}={{I}_{L}}{{X}_{L}}$

The pure inductance cannot dissipate any power. Rather, the inductance stores and releases energy in the form of magnetic field. The inductive reactive power equals the product V_{L}I_{L}.

${{Q}_{L}}={{I}_{L}}{{V}_{L}}=I_{L}^{2}{{X}_{L}}$

**Example of Inductive Reactance**

The voltage across a 1 H inductor is e=10sin200t. what is the expression for an instantaneous current?

**Solution**

${{X}_{L}}=\omega L=200*1=200~\Omega $

${{I}_{m}}=\frac{{{V}_{m}}}{{{X}_{L}}}=\frac{10}{200}=0.05~A$

In the inductance, I lags e by 90 degree, so we can write an expression for an instantaneous current as:

$i={{I}_{m}}\sin \left( wt-{{90}^{{}^\circ }} \right)=0.05~\text{sin}\left( 200t-{{90}^{{}^\circ }} \right)$

**Capacitive Reactance**

Consider a pure capacitor connected to a sinusoidal AC voltage as in the following figure.

The voltage across the capacitor is

$v={{V}_{m}}\sin \left( wt \right)$

While instantaneous current flowing to a capacitor is

$i=C\frac{dV}{dt}$

This above expression indicates that current is proportional to the slope of the voltage curve. In this case, the current is proportional to the slope of a sine wave. In fact, the slope of a sine wave is a cosine wave as shown in Fig. Hence in the pure capacitive circuit, the current leads the voltage by an angle of 90°.

Now substituting the sine wave expression for voltage into the current equation. We obtain

$i=C\frac{d\left( {{V}_{m}}\sin \left( wt \right) \right)}{dt}=wC{{V}_{m}}\cos \left( wt \right)$

$i=wC{{V}_{m}}\cos \left( wt \right)~~~~~\text{ }\cdots \text{ }~~\left( 2 \right)$

From equation (2) one finds maximum current to be;

${{I}_{m}}=wC{{V}_{m}}\text{ at t=0}$

Or

$\frac{{{V}_{m}}}{{{I}_{m}}}=\frac{1}{wC}$

The quantity 1/ωC is called the capacitive reactance, measured in ohm and represented by Xc.

${{X}_{C}}=\frac{1}{wC}=\frac{1}{2\pi fC}$

Remembering that the effective values are related to the maximum values by the same ratio, so we can write

${{V}_{C}}={{I}_{C}}{{X}_{C}}$

The pure capacitance cannot dissipate any power. Rather, capacitance stores or releases energy in the form of the electric field. The capacitive reactive power equals the product V_{C}I_{C}.

${{Q}_{C}}={{I}_{C}}{{V}_{C}}=I_{C}^{2}{{X}_{C}}$

**Example of capacitive reactance**

In above capacitor circuit, C=2μF and the source supply 1 kHz at an effective or RMS value of 10 V. (a) what current flows?

(b) What is the reactive power?

Solution:

(a):

${{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2*\pi *f*2*{{10}^{-6}}}=79.5~\Omega $

${{I}_{C}}=\frac{{{V}_{C}}}{{{X}_{C}}}=\frac{10}{79.5}=0.126~A$

(b):

${{Q}_{C}}={{V}_{C}}{{I}_{C}}=10*0.126=1.26~vars$

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