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Impedance, Inductive Reactance and Capacitive Reactance

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;

$Z=R+jX$

Impedance Diagram

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.

Current Lags the Voltage

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 ImωL 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 XL 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  VLIL.

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

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°.

Current Leads the Voltage

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 VCIC.

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

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):

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

About Ahmad Faizan

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.

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