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Inductive Circuits Topics Covered in Chapter 21 21-1: Sine-Wave i L Lags v L by 90° 21-2: X L and R in Series 21-3: Impedance Z Triangle 21-4: X L and R in Parallel Chapter 21 © 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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Topics Covered in Chapter 21 21-5: Q of a Coil 21-6: AF and RF Chokes 21-7: The General Case of Inductive Voltage McGraw-Hill© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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21-1: Sine-Wave i L Lags v L by 90° Fig. 21-1 When sine-wave variations of current produce an induced voltage, the current lags its induced voltage by exactly 90°, as shown in Fig. 21-1. The phasors in Fig. 21-1 (c) show the 90° phase angle between i L and v L. The 90° phase relationship between i L and v L is true in any sine- wave ac circuit, whether L is in series or parallel.
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21-1: Sine-Wave i L Lags v L by 90° The phase angle of an inductive circuit is 90° because v L depends on the rate of change of i L. The i L wave does not have its positive peak until 90° after the v L wave. Therefore, i L lags v L by 90°. Although i L lags v L by 90°, both waves have the same frequency.
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21-2: X L and R in Series 21-2: X L and R in Series When a coil has series resistance, the current is limited by both X L and R. This current I is the same in X L and R, since they are in series. Each has its own series voltage drop, equal to IR for the resistance and IX l for the reactance.
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21-2: X L and R in Series 21-2: X L and R in Series Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-2:
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21-2: X L and R in Series 21-2: X L and R in Series Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-3: Instead of combining waveforms that are out of phase, they can be added more quickly by using their equivalent phasors, as shown in Fig. 21-3 (a). These phasors show only the 90° angle without addition. The method in Fig. 21-3 (b) is to add the tail of one phasor to the arrowhead of the other, using the angle required to show their relative phase.
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21-3: Impedance Z Triangle Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-4: A triangle of R and X L in series, as shown in Fig. 21-4, corresponds to a voltage triangle. The resultant of the phasor addition of R and X L is their total opposition in ohms, called impedance, with the symbol Z T. The Z takes into account the 90° phase relation between R and X L.
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21-3: Impedance Z Triangle Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. = 53° 30 40 = Tan -1 R XLXL Θ= Tan -1 V A leads I by 53° 30 Ω 40 Ω 50 Ω I = 2 A V A = 100 R = 30 Ω X L = 40 Ω I VLVL VAVA 53° Phase Angle of a Series R L Circuit
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21-4: X L and R in Parallel Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. IRIR ILIL ITIT A R VAVA IRIR 4 30 120 === A XLXL VAVA ILIL 3 40 120 === AI R 2 + I L 2 I T =54 2 + 3 2 == V A = 120 R = 30 Ω X L = 40 Ω I T = 5 A Currents in a Parallel R L Circuit
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21-4: X L and R in Parallel Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The total current lags the source voltage by 37°. 4 A 3 A5 A I T = 5 A V A = 120 R = 30 WX L = 40 Ω = −37° 4 3 IRIR ILIL Θ = Tan −1 −= Tan −1 − Phase Angle in a Parallel R L Circuit
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21-4: X L and R in Parallel Phasor Current Triangle Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-6: Fig. 21-6 illustrates a phasor triangle of inductive and resistive branch currents 90° out of phase in a parallel circuit. This phasor triangle is used to find the resultant I T.
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21-4: X L and R in Parallel Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. = = 24Ω 5 120 ITIT VAVA Z EQ = 4 A 3 A5 A I T = 5 A V A = 120 R = 30 WX L = 40 W Impedance of X L and R in Parallel
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21-4: X L and R in Parallel In a parallel circuit with L and R: The parallel branch currents I R and I L have individual values that are 90° out of phase. I R and I L are added by phasors to equal I T, which is the main-line current. The negative phase angle −Θ is between the line current I T and the common parallel voltage V A. Less parallel X L allows more I L to make the circuit more inductive, with a larger negative phase angle for I T with respect to V A.
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21-5: Q of a Coil The ability of a coil to produce self-induced voltage is indicated by X L, since it includes the factors of frequency and inductance. A coil, however, has internal resistance equal to the resistance of the wire in the coil. This internal resistance r i of the coil reduces the current, which means less ability to produce induced voltage. Combining these two factors of X L and r i, the quality or merit of a coil is, Q = X L /r i.
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21-5: Q of a Coil Figure Fig. 21-7 shows a coil’s inductive reactance X L and its internal resistance r i. The quality or merit of a coil as shown in Fig. 21-7 is determined as follows: Q = X L /r i Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-7:
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21-6: AF and RF Chokes Fig. 21-9 In Fig. 21-9, X L is much greater than R for the frequency of the ac source V T. L has practically all the voltage drop with very little of V T across R. The inductance here is used as a choke to prevent the ac signal from developing any appreciable output across R at the frequency of the source.
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21-7: The General Case of Inductive Voltage The voltage across any inductance in any circuit is always equal to L(di/dt). This formula gives the instantaneous values of v L based on the self-induced voltage produced by a change in magnetic flux from a change in current. A sine waveform of current I produces a cosine waveform for the induced voltage v L, equal to L(di/dt). This means v L has the same waveform as I, but v L and I are 90° out of phase for sine-wave variations. The inductive voltage can be calculated as IX L in sine- wave circuits.
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