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Inductive Reactance Topics Covered in Chapter 20 20-1: How X L Reduces the Amount of I 20-2: X L = 2πfL 20-3: Series or Parallel Inductive Reactances 20-4:

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Presentation on theme: "Inductive Reactance Topics Covered in Chapter 20 20-1: How X L Reduces the Amount of I 20-2: X L = 2πfL 20-3: Series or Parallel Inductive Reactances 20-4:"— Presentation transcript:

1 Inductive Reactance Topics Covered in Chapter 20 20-1: How X L Reduces the Amount of I 20-2: X L = 2πfL 20-3: Series or Parallel Inductive Reactances 20-4: Ohm's Law Applied to X L 20-5: Applications of X L for Different Frequencies 20-6: Waveshape of v L Induced by Sine-Wave Current Chapter 20 © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

2 20-1: How X L Reduces the Amount of I  An inductance can have appreciable X L in ac circuits to reduce the amount of current.  The higher the frequency of ac, and the greater the L, the higher the X L.  There is no X L for steady direct current. In this case, the coil is a resistance equal to the resistance of the wire. McGraw-Hill© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 20-1: How X L Reduces the Amount of I Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 20-1:  In Fig. 20-1 (a), there is no inductance, and the ac voltage source causes the bulb to light with full brilliance.  In Fig. 20-1 (b), a coil is connected in series with the bulb.  The coil has a negligible dc resistance of 1 Ω, but a reactance of 1000 Ω.  Now, I is 120 V / 1000 Ω, approximately 0.12 A. This is not enough to light the bulb.  In Fig. 20-1 (c), the coil is also in series with the bulb, but the battery voltage produces a steady dc.  Without any current variations, there is no X L and the bulb lights with full brilliance.

4 20-2: X L = 2πfL  The formula X L = 2πfL includes the effects of frequency and inductance for calculating the inductive reactance.  The frequency is in hertz, and L is in henrys for an X L in ohms.  The constant factor 2π is always 2 x 3.14 = 6.28.  The frequency f is a time element.  The inductance L indicates the physical factors of the coil.  Inductive reactance X L is in ohms, corresponding to a V L /I L ratio for sine-wave ac circuits.

5 20-3: Series or Parallel Inductive Reactances Fig. 20-5  Since reactance is an opposition in ohms, the values X L in series or in parallel are combined the same way as ohms of resistance.  With series reactances, the total is the sum of the individual values as shown in Fig. 20-5 (a).  The combined reactance of parallel reactances is calculated by the reciprocal formula.

6 20-4: Ohm's Law Applied to X L Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 20-6: The amount of current in an ac circuit with only inductive reactance is equal to the applied voltage divided by X L. I = V/X L = 1 A I = V/X LT = 0.5 AI 1 = V/X L1 = 1 A I 2 = V/X L2 = 1 A I T = I 1 + I 2 = 2 A

7 20-5: Applications of X L for Different Frequencies  The general use of inductance is to provide minimum reactance for relatively low frequencies but more for higher frequencies.  If 1000 Ω is taken as a suitable value of X L for many applications, typical inductances can be calculated for different frequencies. Some are as follows:  2.65 H60 HzPower-line frequency  160 mH10,000 HzMedium audio frequency  16 mH10,000 HzHigh audio frequency  1.6 μH100 MHzIn FM broadcast band

8 20-6: Waveshape of v L Induced by Sine-Wave Current  Induced voltage depends on rate of change of current rather than on the absolute value if i.  A v L curve that is 90° out of phase is a cosine wave of voltage for the sine wave of current i L.  The frequency of V L is 1/T.  The ratio of v L /i L specifies the inductive reactance in ohms.

9 20-6: Waveshape of v L Induced by Sine-Wave Current Current 0  di/dt I inst. = I max × cos  Sinusoidal Current dt di Lv L = di/dt for Sinusoidal Current is a Cosine Wave Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10 20-6: Waveshape of vL Induced by Sine-Wave Current Inductor Voltage and Current 0 I V I V Time Θ = -90  Amplitude Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

11 20-6: Waveshape of vL Induced by Sine-Wave Current Application of the 90° phase angle in a circuit  The phase angle of 90° between V L and I will always apply for any L with sine wave current.  The specific comparison is only between the induced voltage across any one coil and the current flowing in its turns.

12 20-6: Waveshape of vL Induced by Sine-Wave Current Fig. 20-8  Current I 1 lags V L1 by 90°.  Current I 2 lags V L2 by 90°.  Current I 3 lags V L3 by 90°. NOTE: I 3 is also I T for the series- parallel circuit.


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