Series and Parallel Ac Circuits

boy30444_ch15.qxd 3/24/06 2:26 PM Page 637 SeriesSeries andand ParallelParallel acac CircuitsCircuits 1515 • Become familiar with the characteristics of series Objectives and parallel ac networks and be able to find current, voltage, and power levels for each element. • Be able to find the total impedance of any series or parallel ac network and sketch the impedance and admittance diagram of each. • Develop confidence in applying Kirchhoff’s current and voltage law to any series or parallel configuration. • Be able to apply the voltage divider rule or current divider rule to any ac network. • Become adept at finding the frequency response of a series or parallel combination of elements. 15.1 INTRODUCTION In this chapter, phasor algebra is used to develop a quick, direct method for solving both series and parallel ac circuits. The close relationship that exists between this method for solving for un- known quantities and the approach used for dc circuits will become apparent after a few simple examples are considered. Once this association is established, many of the rules (current divider rule, voltage divider rule, and so on) for dc circuits can be readily applied to ac circuits. SERIES ac CIRCUITS 15.2 IMPEDANCE AND THE PHASOR DIAGRAM Resistive Elements In Chapter 14, we found, for the purely resistive circuit in Fig. 15.1, that y and i were in phase, and the magnitude V I ϭ m or V ϭ I R m R m m i = Im sin qt + R v = Vm sin qt – a c FIG. 15.1 Resistive ac circuit. boy30444_ch15.qxd 3/24/06 2:27 PM Page 638 a c 638 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS In phasor form, ϭ ⇒ ϭ Є y Vm sin vt V V 0° ϭ where V 0.707Vm . Applying Ohm’s law and using phasor algebra, we have V Є0° V ϭ ϭ l Ϫ u I Є 0° R R uR R Since i and y are in phase, the angle associated with i also must be 0°. To ϭ satisfy this condition, uR must equal 0°. Substituting uR 0°, we find i V Є0° V V I ϭ ϭ l0° Ϫ 0° ϭ Є0° + R Є0° R R 5 ⍀ v = 100 sin qt – so that in the time domain, V i ϭ 12 sin vt a R b FIG. 15.2 ϭ We use the fact that uR 0° in the following polar format to ensure the Example 15.1. proper phase relationship between the voltage and current of a resistor: ϭ Є ZR R 0° (15.1) 100 V v The boldface roman quantity ZR , having both magnitude and an asso- i 3␲ ciated angle, is referred to as the impedance of a resistive element. It is 20 A ␲ ␲ 2 2 measured in ohms and is a measure of how much the element will “im- 0 ␲ ␻t 2 pede” the flow of charge through the network. The above format will prove to be a useful “tool” when the networks become more complex and phase relationships become less obvious. It is important to realize, how- Є ever, that ZR is not a phasor, even though the format R 0° is very simi- FIG. 15.3 lar to the phasor notation for sinusoidal currents and voltages. The term Waveforms for Example 15.1. phasor is reserved for quantities that vary with time, and R and its associated angle of 0° are fixed, nonvarying quantities. i = 4 sin(qt + 30°) EXAMPLE 15.1 Using complex algebra, find the current i for the cir- + cuit in Fig. 15.2. Sketch the waveforms of y and i. 2 ⍀ v – Solution: Note Fig. 15.3: y ϭ 100 sin vt 1 phasor form V ϭ 70.71 V Є0° Єu Є ϭ V ϭ V ϭ 70.71 V 0° ϭ Є FIG. 15.4 I Є ⍀ Є 14.14 A 0° Example 15.2. ZR R 0° 5 0° and i ϭ 12 14.14 sin vt ϭ 20 sin qt 1 2 8 V v EXAMPLE 15.2 Using complex algebra, find the voltage y for the cir- i cuit in Fig. 15.4. Sketch the waveforms of y and i. 4 A 3␲ 2 Solution: Note Fig. 15.5: 0 ␲ ␲ 2␲ ␻t 30° 2 i ϭ 4 sin(vt ϩ 30°) ⇒ phasor form I ϭ 2.828 A Є30° ϭ ϭ Є Є ϭ Є ⍀ Є V IZR (I u)(R 0°) (2.828 A 30°)(2 0°) 5.656 V Є30° and y ϭ 12 5.656 sin vt ϩ 30° ϭ 8.0 sin qt ؉ 30° FIG. 15.5 1 2 1 2 1 2 Waveforms for Example 15.2. boy30444_ch15.qxd 3/24/06 2:27 PM Page 639 a c IMPEDANCE AND THE PHASOR DIAGRAM ⏐⏐⏐ 639 It is often helpful in the analysis of networks to have a phasor dia- j gram, which shows at a glance the magnitudes and phase relations among the various quantities within the network. For example, the phasor diagrams of the circuits considered in the two preceding examples would be as shown in Fig. 15.6. In both cases, it is immediately obvious that y and i are in phase since they both have the same phase angle. IV Inductive Reactance + We learned in Chapter 13 that for the pure inductor in Fig. 15.7, the volt- 14.14 A 70.7 V age leads the current by 90° and that the reactance of the coil XL is determined by vL. (a) ϭ ⇒ ϭ Є y Vm sin vt phasor form V V 0° j i + X = qL v = V sin qt 5.565 V L m V – 2.828 A I 30° + FIG. 15.7 Inductive ac circuit. By Ohm’s law, (b) V Є0° V FIG. 15.6 ϭ ϭ l0° Ϫ u I Є L Phasor diagrams for Examples 15.1 and 15.2. XL uL XL Since y leads i by 90°, i must have an angle of Ϫ90° associated with it. ϩ ϭ To satisfy this condition, uL must equal 90°. Substituting uL 90°, we obtain Є ϭ V 0° ϭ V l Ϫ ϭ V ЄϪ I Є 0° 90° 90° XL 90° XL XL so that in the time domain, V i ϭ 12 sin vt Ϫ 90° a XL b 1 2 ϭ We use the fact that uL 90° in the following polar format for inductive reactance to ensure the proper phase relationship between the voltage and current of an inductor: ϭ Є ZL XL 90° (15.2) The boldface roman quantity ZL , having both magnitude and an associated angle, is referred to as the impedance of an inductive element. It is measured in ohms and is a measure of how much the inductive element “controls or impedes” the level of current through the network (always keep in mind that inductive elements are storage devices and do not dis- sipate like resistors). The above format, like that defined for the resistive element, will prove to be a useful tool in the analysis of ac networks. Again, be aware that ZL is not a phasor quantity, for the same reasons in- dicated for a resistive element. boy30444_ch15.qxd 3/24/06 2:27 PM Page 640 a c 640 ⏐⏐⏐ SERIES AND PARALLEL ac CIRCUITS i EXAMPLE 15.3 Using complex algebra, find the current i for the cir- + cuit in Fig. 15.8. Sketch the y and i curves. ⍀ XL = 3 v = 24 sin qt Solution: Note Fig. 15.9: – y ϭ 24 sin vt 1 phasor form V ϭ 16.968 V Є0° V V Єu 16.968 VЄ0° I ϭ ϭ ϭ ϭ 5.656 A ЄϪ90° FIG. 15.8 Є ⍀ Є ZL XL 90° 3 90° Example 15.3. (and i ϭ 12 5.656 sin vt Ϫ 90° ϭ 8.0 sin(qt ؊ 90° 1 2 1 2 24 V v EXAMPLE 15.4 Using complex algebra, find the voltage y for the cir- 8 A i cuit in Fig. 15.10. Sketch the y and i curves. 0 ␲ ␲ 3␲ 2␲ 5␲ ␻ t Solution: Note Fig. 15.11: 2 2 2 ° 90 i ϭ 5 sin(vt ϩ 30°) ⇒ phasor form I ϭ 3.535 A Є30° ϭ ϭ Є Є ϭ Є ⍀ Єϩ V IZL (I u)(XL 90°) (3.535 A 30°)(4 90°) FIG. 15.9 14.140 V Є120° Waveforms for Example 15.3. (and y ϭ 12 14.140 sin vt ϩ 120° ϭ 20 sin(qt ؉ 120° 1 2 1 2 The phasor diagrams for the two circuits of the two preceding exam- i = 5 sin(qt + 30°) ples are shown in Fig. 15.12. Both indicate quite clearly that the voltage leads the current by 90°. + j X = 4 ⍀ v L j – V Leading FIG. 15.10 A I 16.968 A 14.140 V Example 15.4. 3.535 30° V + + 5.656 A I Leading 20 V v FIG. 15.12 5 A i Phasor diagrams for Examples 15.3 and 15.4. ␲ – 0 ␲ ␲ 3␲ 2␲ ␻ t 2 2 2 Capacitive Reactance 90° 30° We learned in Chapter 13 that for the pure capacitor in Fig. 15.13, the current leads the voltage by 90° and that the reactance of the capacitor XC is determined by 1/vC. FIG. 15.11 Waveforms for Example 15.4. i + XC = 1/qC v = Vm sin qt – FIG. 15.13 Capacitive ac circuit. boy30444_ch15.qxd 3/24/06 2:27 PM Page 641 a c IMPEDANCE AND THE PHASOR DIAGRAM ⏐⏐⏐ 641 ϭ ⇒ ϭ Є y Vm sin vt phasor form V V 0° Applying Ohm’s law and using phasor algebra, we find V Є0° V ϭ ϭ l Ϫ u I Є 0° C XC uC XC Since i leads y by 90°, i must have an angle of ϩ90° associated with it.

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