Inductive Circuits Topics Covered in Chapter : 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.

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.

21-1: Sine-Wave i L Lags v L by 90° Fig  When sine-wave variations of current produce an induced voltage, the current lags its induced voltage by exactly 90°, as shown in Fig  The phasors in Fig (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.

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.

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.

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:

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 (a).  These phasors show only the 90° angle without addition.  The method in Fig (b) is to add the tail of one phasor to the arrowhead of the other, using the angle required to show their relative phase.

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.

21-3: Impedance Z Triangle Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. = 53° = 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

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 === A XLXL VAVA ILIL === AI R 2 + I L 2 I T = == V A = 120 R = 30 Ω X L = 40 Ω I T = 5 A Currents in a Parallel R L Circuit

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

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 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.

21-4: X L and R in Parallel Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. = = 24Ω 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

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.

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.

21-5: Q of a Coil  Figure Fig 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 is determined as follows: Q = X L /r i Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 21-7:

21-6: AF and RF Chokes Fig  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.

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.