ALTERNATING CURRENT. with X C = 0.34 A. SET UP: The specified value is the root-mean-square current; I. EXECUTE: (a) V = (0.34 A) = 0.12 A.

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1 ATENATING UENT 3 3 IDENTIFY: i Icosωt and I I/ SET UP: The specified value is the root-mean-square current; I 34 A EXEUTE: (a) I 34 A (b) I I (34 A) 48 A (c) Since the current is positive half of the time and negative half of the time, its average value is zero (d) Since I is the square root of the average of i, the average square of the current is I (34 A) A EAUATE: The current amplitude is larger than its value 3 IDENTIFY and SET UP: Apply Eqs(33) and (34) EXEUTE: (a) I I (A) 97A (b) I rav I (97 A) 89 A π π EAUATE: (c) The root-mean-square voltage is always greater than the rectified average, because squaring the current before averaging, and then taking the square root to get the root-mean-square value will always give a larger value than just averaging 33 IDENTIFY and SET UP: Apply Eq(35) 45 EXEUTE: (a) 38 (b) Since the voltage is sinusoidal, the average is zero EAUATE: The voltage amplitude is larger than 34 IDENTIFY: IX with X ω SET UP: ω is the angular frequency, in rad/s I EXEUTE: (a) IX so I ω (6 )( rad s)( F) 3 A ω (b) I ω (6 )( rad s)( F) 3 A (c) I ω (6 )(, rad s)( F) 3 A (d) The plot of log I versus logω is given in Figure 34 EAUATE: I ω so log I log( ) + log ω A graph of log I versus logω should be a straight line with slope +, and that is what Figure 34 shows Figure 34 3-

2 3- hapter 3 35 IDENTIFY: IX with X ω SET UP: ω is the angular frequency, in rad/s EXEUTE: (a) 6 IX Iω and I A ω ( rad s)(5 H) 6 (b) I A ω ( rad s)(5 H) 6 (c) I A ω (, rad s)(5 H) (d) The plot of log I versus logω is given in Figure 35 EAUATE: I so log I log( / ) logω A graph of log I versus logω should be a straight line with ω slope, and that is what Figure 35 shows Figure IDENTIFY: The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance SET UP: The reactances are X / ω and X ω EXEUTE: (a) Equating the reactances gives ω ω ω (b) Using the numerical values we get ω 756 rad/s (5 mh)(35 µ F) X X ω (756 rad/s)(5 mh) 378 Ω EAUATE: At other angular frequencies, the two reactances could be very different 37 IDENTIFY and SET UP: For a resistor v i For an inductor, v cos( ωt+ 9 ) For a capacitor, v cos( ωt 9 ) EXEUTE: The graphs are sketched in Figures 37a-c The phasor diagrams are given in Figure 37d EAUATE: For a resistor only in the circuit, the current and voltage in phase For an inductor only, the voltage leads the current by 9 For a capacitor only, the voltage lags the current by 9 Figure 37a and b

3 Alternating urrent 3-3 Figure 37c Figure 37d 38 IDENTIFY: The reactance of an inductor is X ω π f The reactance of a capacitor is X ω π f SET UP: The frequency f is in Hz EXEUTE: (a) At 6 Hz, X π (6 Hz)(45 H) 7 Ω X is proportional to f so at 6 Hz, X 7 Ω 3 (b) At 6 Hz, X 6 Ω X is proportional to / f, so at 6 Hz, X 6 Ω π (6 Hz)(5 F) (c) X X says π f π f and EAUATE: X increases when f increases X 39 IDENTIFY and SET UP: Use Eqs(3) and (38) f 5 Hz π π (45 H)(5 F) increases when f increases EXEUTE: (a) X ω π f π(8 Hz)(3 H) 5 Ω X Ω (b) X π f gives 39 H π f π(8 Hz) (c) X 497 Ω ω π f π(8 Hz)(4 F) (d) X EAUATE: 3 IDENTIFY: gives π f π f X π(8 Hz)( Ω) X increases when increases; Iω 5 66 F X decreases when increases ω SET UP: ω is the angular frequency, in rad/s f is the frequency in Hz π ( ) 6 EXEUTE: Iω so f 63 Hz 3 4 ωi π(6 A)(45 H) EAUATE: When f is increased, I decreases 3 IDENTIFY and SET UP: Apply Eqs(38) and (39) 7 EXEUTE: IX so X Ω I 85 A 5 X gives 33 F 33 μf ω π f X π(6 Hz)( Ω) EAUATE: The reactance relates the voltage amplitude to the current amplitude and is similar to Ohm s law

4 3-4 hapter 3 I 3 IDENTIFY: ompare v that is given in the problem to the general form v sinωtand determine ω ω SET UP: X v i and i Icos ω ω EXEUTE: (a) X 736 Ω ω ( rad s)(48 F) 76 3 (b) 4378 I A 3 and i Icos ωt (4378 A)cos[( rad/s) t] Then X 736 Ω 3 (438 A)(5 )cos(( rad s) ) ( )cos(( rad s) ) v i Ω t t EAUATE: The voltage across the resistor has a different phase than the voltage across the capacitor 33 IDENTIFY and SET UP: The voltage and current for a resistor are related by v i Deduce the frequency of the voltage and use this in Eq(3) to calculate the inductive reactance Eq(3) gives the voltage across the inductor EXEUTE: (a) v (38 )cos[(7 rad/s) t] v 38 v i, so i cos[(7 rad/s) t] (53 A)cos[(7 rad/s) t] 5 Ω (b) X ω ω 7 rad/s, 5 H, so X ω (7 rad/s)(5 H) 8 Ω (c) If i Icosωt then v cos( ωt+ 9 ) (from Eq3) Iω IX (533 A)(8 Ω ) 456 v (456 ) cos[(7 rad/s) t+ 9 ] But cos( a+ 9 ) sin a (Appendix B), so v (456 )sin[(7 rad/s) t] EAUATE: The current is the same in the resistor and inductor and the voltages are 9 out of phase, with the voltage across the inductor leading 34 IDENTIFY: alculate the reactance of the inductor and of the capacitor alculate the impedance and use that result to calculate the current amplitude X SET UP: With no capacitor, Z + X and tan φ X ω I IXand I For an Z inductor, the voltage leads the current EXEUTE: (a) X ω (5 rad/s)(4 H) Ω Z ( Ω ) + ( Ω ) 4 Ω 3 (b) I 34 A Z 4 Ω (c) I (34 A)( Ω ) 68 IX (34 A)( Ω ) 34 X (d) tan Ω φ and 66 Ω φ + Since φ is positive, the source voltage leads the current (e) The phasor diagram is sketched in Figure 34 EAUATE: Note that + is greater than The loop rule is satisfied at each instance of time but the voltages across and reach their maxima at different times Figure IDENTIFY: v() t is given by Eq(38) v( t ) is given by Eq(3) SET UP: From Exercise 34, 3, 68, 34 and φ 66 EXEUTE: (a) The graph is given in Figure 35 (b) The different voltages are v (3 )cos(5t+ 66 ), v (68 )cos(5 t), (34 ) cos(5t+ 9 ) At t ms: v 5, v 76, v 85 Note that v + v v v

5 Alternating urrent 3-5 (c) At t 4 ms: v 5, v 49, v 79 Note that v + v v EAUATE: It is important to be careful with radians versus degrees in above expressions! Figure IDENTIFY: alculate the reactance of the inductor and of the capacitor alculate the impedance and use that result to calculate the current amplitude X X SET UP: With no resistor, Z ( X X) X X tan φ X X ω For an zero ω inductor, the voltage leads the current For a capacitor, the voltage lags the current EXEUTE: (a) X ω (5 rad/s)(4 H) Ω X 667 Ω ω (5 rad/s)(6 F) Z X X Ω67 Ω 567 Ω 3 (b) I 59 A Z 567 Ω (c) IX (59 A)(667 Ω ) 353 IX (59 A)( Ω ) 59 X X Ω67 Ω (d) tanφ and φ 9 The phase angle is negative and the source voltage lags zero zero the current (e) The phasor diagram is sketched in Figure 36 EAUATE: When X > X the phase angle is negative and the source voltage lags the current Figure IDENTIFY and SET UP: alculate the impendance of the circuit and use Eq(3) to find the current amplitude The voltage amplitudes across each circuit element are given by Eqs(37), (33), and (39) The phase angle is calculated using Eq(34) The circuit is shown in Figure 37a No inductor means X Ω, 6 F, 3, ω 5 rad/s Figure 37a

6 3-6 hapter Ω ω (5 rad/s)(6 F) EXEUTE: (a) X Z + ( X X ) ( Ω ) + (6667 Ω ) 696 Ω 3 (b) I 43 A 43 ma Z 696 Ω (c) oltage amplitude across the resistor: I (43 A)( Ω ) 86 oltage amplitude across the capacitor: IX (43 A)(6667 Ω ) 87 X X 667 Ω (d) tanφ 3333 so φ 733 Ω The phase angle is negative, so the source voltage lags behind the current (e) The phasor diagram is sketched qualitatively in Figure 37b Figure 37b EAUATE: The voltage across the resistor is in phase with the current and the capacitor voltage lags the current by 9 The presence of the capacitor causes the source voltage to lag behind the current Note that + > The instantaneous voltages in the circuit obey the loop rule at all times but because of the phase differences the voltage amplitudes do not 38 IDENTIFY: v() t is given by Eq(38) v ( t) is given by Eq(36) SET UP: From Exercise 37, 3, 86, 87 and φ 733 EXEUTE: (a) The graph is given in Figure 38 (b) The different voltage are: v (3 ) cos(5t 733 ), v (86 ) cos(5 t), v (87 ) cos(5t 9 ) At t ms: v 5, v 45, v 75 Note that v + v v (c) At t 4 ms: v 9, v 73, v 56 Note that v + v v EAUATE: It is important to be careful with radians vs degrees! Figure IDENTIFY: Apply the equations in Section 33 SET UP: ω 5 rad/s, Ω, 4 H, 6 μf and 3 EXEUTE: (a) Z + ( ω / ω) Z 3 (b) I 499 A Z 6Ω ( Ω ) + ((5 rad/s)(4 H) /((5 rad/s)(6 F))) 6Ω ω / ω Ω67 Ω (c) φ arctan arctan 76, and the voltage lags the current Ω

7 (d) I (499 A)( Ω ) 998 ; Alternating urrent 3-7 I (499 A) Iω (499 A)(5 rad s)(4 H) 499 ; 333 ω (5 rad/s)(6 F) EAUATE: (e) At any instant, v v + v + v But v and v are 8 out of phase, so v can be larger than v at a value of t, if v + v is negative at that t 3 IDENTIFY: v() t is given by Eq(38) v ( t ) is given by Eq(36) v( t ) is given by Eq(3) SET UP: From Exercise 39, 3, 499, 998, 333 and φ 76 EXEUTE: (a) The graph is sketched in Figure 3 The different voltages plotted in the graph are: v (3 )cos(5t 76 ), v (998 )cos(5 t), v (499 ) cos(5t+ 9 ) and v (333 )cos(5t 9 ) (b) At t ms: v 43, v 83, v 479, v 39 (c) At t 4 ms: v 38, v 837, v 7, v 8 EAUATE: In both parts (b) and (c), note that the source voltage equals the sum of the other voltages at the given instant Be careful with degrees versus radians! Figure 3 3 IDENTIFY and SET UP: The current is largest at the resonance frequency At resonance, X X and Z For part (b), calculate Z and use I / Z EXEUTE: (a) f 3 Hz I / 5 ma π (b) X / ω 5 Ω X ω 6 Ω Z + ( X X) ( Ω ) + (6 Ω 5 Ω ) 3945 Ω I / Z 76mA X > X so the source voltage lags the current EAUATE: ω π f 7 rad/s ω 4 rad/s and is less than ω When ω < ω, X > X Note that I in part (b) is less than I in part (a) 3 IDENTIFY: The impedance and individual reactances depend on the angular frequency at which the circuit is driven SET UP: The impedance is Z + ω, the current amplitude is I /Z, and the instantaneous ω values of the potential and current are v cos(ωt + φ), where tan φ (X X )/, and i I cos ωt EXEUTE: (a) Z is a minimum when ω, which gives ω 36 rad/s ω (8 mh)(5 µ F) 36 rad/s and Z 75 Ω (b) I /Z (5 )/(75 Ω) 43 A (c) i I cos ωt I/, so cosωt, which gives ωt 6 π/3 rad v cos(ωt + φ), where tan φ (X X )/ / So, v (5 ) cosωt (5 )(/) 5 v i (75 Ω)(/)(43 A) 5 43 A v cos(ωt 9 ) IX cos(ωt 9 ) cos(6 9 ) +33 (36 rad/s)(5 µ F)

8 3-8 hapter 3 v cos(ωt + 9 ) IX cos(ωt + 9 ) (43 A)(36 rad/s)(8 mh) cos(6 + 9 ) v 33 (d) v + v + v 5 + ( 33 ) v source EAUATE: The instantaneous potential differences across all the circuit elements always add up to the value of the source voltage at that instant In this case (resonance), the potentials across the inductor and capacitor have the same magnitude but are 8 out of phase, so they add to zero, leaving all the potential difference across the resistor 33 IDENTIFY and SET UP: Use the equation that preceeds Eq(3): + ( ) EXEUTE: (3 ) + (5 9 ) 5 EAUATE: The equation follows directly from the phasor diagrams of Fig33 (b or c) Note that the voltage amplitudes do not simply add to give 7 for the source voltage 34 IDENTIFY and SET UP: X ωand X ω EXEUTE: (a) If ω ω, then X ω and X ω (b) When ω > ω, X > (c) When ω > ω, X < (d) The graph of X versus ω is given in Figure 34 EAUATE: Z + X and tan φ X / Figure IDENTIFY: For a pure resistance, Pav I I SET UP: W is the average power P av EXEUTE: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 4 W Pav W (b) I 67 A Pav W (c) 7 Ω I (67 A), ( ) EAUATE: We can also calculate the average power as P av W 75 Ω 36 IDENTIFY: The average power supplied by the source is P I cosφ The power consumed in the resistance is P I 3 3 SET UP: ω π f π(5 Hz) 7854 rad/s X ω 57 Ω X 99 Ω ω EXEUTE: (a) First, let us find the phase angle between the voltage and the current: X X 57 Ω 99 Ω tanφ and φ 654 The impedance of the circuit is 35 Ω Z + ( X X) (35 Ω ) + ( 75 Ω ) 83 Ω The average power provided by the generator is then ( ) P I cos( φ) cos( φ) cos( 654 ) 73 W Z 83 Ω

9 Alternating urrent 3-9 (b) The average power dissipated by the resistor is P I (35 Ω ) 73 W 83 Ω EAUATE: onservation of energy requires that the answers to parts (a) and (b) are equal 37 IDENTIFY: The power factor is cos φ, where φ is the phase angle in Fig33 The average power is given by Eq(33) Use the result of part (a) to rewrite this expression (a) SET UP: The phasor diagram is sketched in Figure 37 EXEUTE: From the diagram I cos φ, IZ Z as was to be shown Figure 37 (b) Pav I cos φ I I But I, so Pav I Z Z Z EAUATE: In an -- circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements The power delivered by the source equals the power dissipated in the resistor 38 IDENTIFY and SET UP: P I cos φ av I Z cos φ Z 8 75 Ω EXEUTE: I 76 A cosφ 74 P av (8 )(76 A)(74) 435 W 5 Ω 5 Ω EAUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also calculate the average power as Pav I (76 A) (75 Ω ) 435 W 39 IDENTIFY and SET UP: Use the equations of Section 33 to calculate φ, Z and The average power delivered by the source is given by Eq(33) and the average power dissipated in the resistor is I EXEUTE: (a) X ω π f π(4 Hz)( H) 36 Ω X 545 Ω ω π f π(4 Hz)(73 Hz) X X 36 Ω 544 Ω tan φ, so φ The power factor is cosφ Ω (b) Z + ( X X) (4 Ω ) + (36 Ω 545 Ω ) 344 Ω (c) I Z (45 A)(344 Ω ) 55 (d) P I av cos φ (45 A)(55 )(697) 486 W (e) Pav I (45 A) (4 Ω ) 486 W EAUATE: The average electrical power delivered by the source equals the average electrical power consumed in the resistor (f) All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another part of the cycle There is no net dissipation of energy in the capacitor (g) The answer is the same as for the capacitor Energy is repeatedly being stored and released in the inductor, but no net energy is dissipated there 33 IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance X of the capacitor The angular frequency and the inductance can be used to calculate the reactance X of the inductor alculate the phase angle φ and then the power factor is cos φ alculate the impedance of the circuit and then the current in the circuit The average power is P av I cos φ On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor 45 SET UP: The source has voltage 38

10 3- hapter 3 EXEUTE: X 794 Ω ω (36 rad/s)(35 F) 3 ω (36 rad/s)(5 H) 54 Ω X X X 54 Ω 794 Ω tanφ and φ 74 The power factor is cosφ 3 5 Ω 38 (b) Z + ( X X) (5 Ω ) + (54 Ω 794 Ω ) 87 Ω I 385 A Z 87 Ω P I av cos φ (38 )(385 A)(3) 37 W (c) The average power delivered to the resistor is Pav I (385 A) (5 Ω ) 37 W The average power delivered to the capacitor and to the inductor is zero EAUATE: On average the power delivered to the circuit equals the power consumed in the resistor The capacitor and inductor store electrical energy during part of the current oscillation but each return the energy to the circuit during another part of the current cycle 33 IDENTIFY and SET UP: At the resonance frequency, Z Use that IZ, I, IX and IX Pav is given by Eq(33) (a) EXEUTE: IZ I (5 A)(3 Ω ) 5 (b) I 5 X ω (/ ) / 58 Ω ; IX 9 X /( ω) / 58 Ω ; IX 9 (c) P Icos φ I, since I and cosφ at resonance av P av (5 A) (3 Ω ) 375 W EAUATE: At resonance Note that + > However, at any instant v + v 33 IDENTIFY: The current is maximum at the resonance frequency, so choose such that ω 5 rad/s is the resonance frequency At the resonance frequency Z SET UP: Iω EXEUTE: (a) The amplitude of the current is given by I Thus, the current will have a + ( ω ) ω maximum amplitude when ω Therefore, 444 F ω μ ω (5 rad/s) (9 H) (b) With the capacitance calculated above we find that Z, and the amplitude of the current is I 3 A Thus, the amplitude of the voltage across the inductor is 4 Ω I( ω) (3 A)(5 rad/s)(9 H) 35 EAUATE: Note that is greater than the source voltage amplitude 333 IDENTIFY and SET UP: At resonance X X, φ and Z 5 Ω, 75 H, 8 μf, 5 X X EXEUTE: (a) At the resonance frequency X X and from tanφ we have that φ and the power factor is cosφ (b) Pav Icosφ (Eq33) At the resonance frequency Z, so I Z (5 ) Pav cosφ 75 W 5 Ω (c) EAUATE: When and f are changed but the circuit is kept on resonance, nothing changes in Pav /( ), so the average power is unchanged: P av 75 W The resonance frequency changes but since Z at resonance the current doesn t change 334 IDENTIFY: ω IX IZ SET UP: At resonance, Z

11 Alternating urrent 3-4 EXEUTE: (a) ω 54 rad/s (35 H)( F) 3 (b) IZ Z X 54 Ω 4 X X ω (54 rad/s)( F) 55 (4 ) 47 3 Ω 54 Ω EAUATE: The voltage amplitude for the capacitor is more than a factor of times greater than the voltage amplitude of the source 335 IDENTIFY and SET UP: The resonance angular frequency is ω X ω X and ω Z + ( X X) At the resonance frequency X X and Z EXEUTE: (a) Z 5 Ω 4 (b) ω 33 rad/s 3 (45 H)(6 F) 4 ω ω 66 rad/s 4 3 X ω (66 rad/s)(45 H) Ω X 3 Ω 4 ω (66 rad/s)(5 F) Z (5 Ω ) + ( Ω 3 Ω ) 46 Ω 3 (c) ω ω / 665 rad/s 3 X Ω X Ω Z (5 Ω ) + (3 Ω Ω ) 46 Ω, the ω same value as in part (b) EAUATE: For ω ω, X > X For ω ω /, X < X But ( X X) has the same value at these two frequencies, so Z is the same 336 IDENTIFY: At resonance Z and X X SET UP: ω IZ I, IX and EXEUTE: (a) ω 945 rad s ( 8 H)( 4 F) (b) I A at resonance, so Z 76 Ω I 7 A (c) At resonance, Iω ( )( )( ), 7 A 945 rad s 8 H 45 EAUATE: At resonance, and 337 IDENTIFY and SET UP: Eq(335) relates the primary and secondary voltages to the number of turns in each I / and the power consumed in the resistive load is I / N N EXEUTE: (a) so N N (b) I 4 A 5 Ω (c) Pav I (4 A) (5 Ω ) 88 W (d) The power drawn from the line by the transformer is the 88 W that is delivered by the load ( ) P av so 5 P 88 W Ω av N And (5 Ω ) () (5 Ω ) 5 Ω, as was to be shown N EAUATE: The resistance is transformed A load of resistance connected to the secondary draws the same power as a resistance ( N / N ) connected directly to the supply line, without using the transformer

12 3- hapter IDENTIFY: Pav I and P P av, av, N N SET UP: 3, N 3, EXEUTE: (a) 8 N 3 (b) Pav I (3, )(85 A) W Pav W (c) I 97 A EAUATE: Since the power supplied to the primary must equal the power delivered by the secondary, in a stepup transformer the current in the primary is greater than the current in the secondary N 339 IDENTIFY: A transformer transfo voltages according to The effective resistance of a secondary N circuit of resistance is eff ( N/ N) esistance is related to P and by P av av onservation of energy requires Pav, Pav, so I I SET UP: et 4 and, so P,av 6 W These voltages are EXEUTE: (a) 4 and we want, so use a step-down transformer with N/ N Pav 6 W (b) Pav I, so I 667 A 4 ( ) (c) The resistance of the blower is 9 Ω The effective resistance of the blower is P 6 W 9 Ω eff 36 Ω (/ ) EAUATE: I (33 A)( ) 6 W Energy is provided to the primary at the same rate that it is consumed in the secondary Step-down transformers step up resistance and the current in the primary is less than the current in the secondary 34 IDENTIFY: Z + ( X X), with X ω and X ω SET UP: The woofer has a and in series and the tweeter has a and in series EXEUTE: (a) Ztweeter + ( ω) (b) Z ( ) woofer + ω (c) If Z, tweeter Z woofer then the current splits evenly through each branch (d) At the crossover point, where currents are equal, ( ) ( ) + ω + ω ω and ω f π π EAUATE: The crossover frequency corresponds to the resonance frequency of a -- circuit, since the crossover frequency is where X X 34 IDENTIFY and SET UP: Use Eq(34) to relate and to φ The voltage across the coil leads the current in it by 53, so φ + 53 X X X EXEUTE: tan φ But there is no capacitance in the circuit so X Thus tan φ and X X 6 Ω tan φ (48 Ω ) tan 53 6 Ω X ω π f so 4 H π f π(8 Hz) φ > when ( X X ) >, which is the case here EAUATE: IDENTIFY: SET UP: Z + ( X X) I I, I X, I X Z 3

13 Alternating urrent 3-3 EXEUTE: (a) ω rad/s, so X ω ( rad/s)(4 H) 8 Ω and X 833 Ω Z ( Ω ) + (8 Ω 833 Ω ) 779 Ω ω ( rad/s)(6 F) I 7 A reads, I (7 A)( Ω ) 544 reads Z 779 Ω, I X (7 A)(8 Ω ) 8 3 reads, I X (7 A)(833 Ω ) 7 4 reads,, reads 833 Ω (b) ω rad/s so X ω (5)(8 Ω ) 4 Ω and X 67 Ω ω 5 Z ( Ω ) + (4 Ω7 Ω ) 37 Ω I 69 A reads, 38 reads Z 37 Ω, 76 3 reads, 5 4 reads,, reads EAUATE: The resonance frequency for this circuit is ω 645 rad/s rad/s is less than the resonance frequency and X > X rad/s is greater than the resonance frequency and X > X 343 IDENTIFY and SET UP: The rectified current equals the absolute value of the current i Evaluate the integral as specified in the problem EXEUTE: (a) From Fig33b, the rectified current is zero at the same values of t for which the sinusoidal current is zero At these t, cosωt and ωt ± π /, ± 3 π /, The two smallest positive times are t π / ω, t 3 π / ω (b) t t I A idt Icosωtdt I sin ωt (sinωt sin ωt) t t ω ω sinωt sin[ ω( π / ω)] sin( π /) sinωt sin[ ω(3 π / ω)] sin(3 π / ) t t I I A ( ( )) ω ω (c ) Irav( t t) I/ ω I I I Irav, which is Eq(33) ω( t t) ω(3 π / ω π / ω) π EAUATE: We have shown that Eq(33) is correct The average rectified current is less than the current amplitude I, since the rectified current varies between and I The average of the current is zero, since it has both positive and negative values 344 IDENTIFY: X ω P I cosφ av SET UP: EXEUTE: (a) f Hz; ω π f X 5 Ω X ω 33 Ω ω π Hz ( ) (b) Z + X ( 4 Ω ) + ( 5 Ω ) 47 Ω cos φ and I Pav, Z Z Z Z so Pav 8 W Z ( 47 Ω ) Ω 668 EAUATE: 45 A I We can calculate Pav as I (45 A) (4 Ω ) 8 W, which Z 47 Ω checks 345 (a) IDENTIFY and SET UP: Source voltage lags current so it must be that X > X and we must add an inductor in series with the circuit When X X the power factor has its maximum value of unity, so calculate the additional needed to raise X to equal X

14 3-4 hapter 3 (b) EXEUTE: power factor cosφ equals so φ and X X alculate the present value of X X to see how much more X is needed: Zcos φ (6 Ω )(7) 43 Ω X X tan φ so X X tanφ cosφ 7 gives φ 4395 (φ is negative since the voltage lags the current) Then X X tan φ (43 Ω) tan( 4395 ) 464 Ω Therefore need to add 464 Ω of X X 464 Ω X ω π f and 33 H, amount of inductance to add π f π(5 Hz) EAUATE: From the information given we can t calculate the original value of in the circuit, just how much to add When this is added the current in the circuit will increase 346 IDENTIFY: Use IZ to calculate Z and then find Pav I SET UP: X 5 Ω 4 EXEUTE: Z 8 Ω + X ( 5 ) + Ω Thus, I 3 A ( 8 ) ( 5 ) 64 by the resistor, which is P I ( ) ( ) Ω Ω Ω The average power supplied to this circuit is equal to the power dissipated 3 A 64 Ω 56 W X X 5 Ω tanφ and φ Ω P I cos (4 )(3 A)cos( 387 ) 56 W, which checks EAUATE: av 347 IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values SET UP: The voltage amplitudes are I, X I, and X I, where I /Z and Z + ω ω EXEUTE: (a) ω πf π(5 Hz) 7854 rad/s arrying extra figures in the calculator gives X ω (7854 rad/s)(35 mh) 75 Ω; X /ω /[(7854 rad/s)( µf)] 7 Ω; Z + ( X X) (5 Ω ) + (75 Ω 7 Ω ) 5 Ω; I /Z (6 )/(5 Ω) 5 A; I (5 Ω)(5 A) 575 ; X I (75 Ω)(5 A) 36 ; X I (7 Ω)(5 A) 47 The voltage amplitudes can add to more than 6 because these voltages do not all occur at the same instant of time At any instant, the instantaneous voltages all add to 6 (b) All of them will change because they all depend on ω X ω will double to 55 Ω, and X /ω will decrease by half to 635 Ω Therefore Z (5 Ω ) + (55 Ω35 Ω ) 698 Ω; I /Z (6 )/(698 Ω) 86 A; I (86 A)(5 Ω) 43 ; IX (86 A)(55 Ω) 473 ; IX (86 A)(635 Ω) 547 EAUATE: The new amplitudes in part (b) are not simple multiples of the values in part (a) because the impedance and reactances are not all the same simple multiple of the angular frequency 348 IDENTIFY and SET UP: X X ω ω EXEUTE: (a) ω ω and ω At angular frequency ω, X ( ) 4 X ω ω ω / ω ω X > X (b) At angular frequency ω, X ω 3 ω3 X > X X 3 ω 9 EAUATE: When ω increases, X increases and X decreases When ω decreases, X decreases and X increases (c) The resonance angular frequency ω is the value of ω for which X X, so ω ω

15 Alternating urrent IDENTIFY and SET UP: Express Z and I in te of ω,, and The voltages across the resistor and the inductor are 9 out of phase, so out + EXEUTE: The circuit is sketched in Figure 349 X ω, X ω Z + ω ω s s I Z + ω ω Figure ω out I + X I + ω s + ω ω out + ω s + ω ω ω small As ω gets small, + ω, + ω ω ω out Therefore ω as ω becomes small s (/ ω ) ω large As ω gets large, + ω + ω ω, + ω ω ω out ω Therefore, as ω becomes large ω s EAUATE: out s / as ω becomes small, so there is out only when the frequency ω of s is large If the source voltage contains a number of frequency components, only the high frequency ones are passed by this filter 35 IDENTIFY: IX I / Z SET UP: X ω, X ω EXEUTE: I out out ω s + ω If ω is large: out s ( ) ω ω ω ω ω ( ) ( ) ( ) + ω out ω If ω is small: s ω ( ω) ω EAUATE: When ω is large, X is small and X is large so Z is large and the current is small Both factors in IX are small When ω is small, X is large and the voltage amplitude across the capacitor is much larger than the voltage amplitudes across the resistor and the inductor 35 IDENTIFY: I / Z and P I av SET UP: Z + ( ω / ω) EXEUTE: (a) I Z + ω ω ( )

16 3-6 hapter 3 (b) P I Z + ω ω ( ) av (c) The average power and the current amplitude are both greatest when the denominator is smallest, which occurs for ω, soω ω ( ) ( Ω) 5ω (d) Pav ( Ω ) + ( ω( H) [ ω(5 F)] ) 4, ω + ( ω,, ) The graph of Pav versus ω is sketched in Figure 35 EAUATE: Note that as the angular frequency goes to zero, the power and current are zero, just as they are when the angular frequency goes to infinity This graph exhibits the same strongly peaked nature as the light purple curve in Figure 39 in the textbook Figure 35 I 35 IDENTIFY: Iωand ω SET UP: Problem 35 shows that I + ( ω /[ ω]) ω ω EXEUTE: (a) Iω Z + ( ω /[ ω] ) I I (b) ω ω Z ω + ( ω [ ω ] ) (c) The graphs are given in Figure 35 EAUATE: (d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of (equal to the total source voltage) At very high frequencies, the capacitor voltage goes to zero, while the inductor s voltage goes to At resonance, ω rad s, the two voltages are equal, and are a maximum, Figure 35

17 353 IDENTIFY: UB i UE v SET UP: et x denote the average value of the quantity x that I + ( ω /[ ω]) Problem 35 shows that 4 I EXEUTE: (a) UB i UB i I I E E, 4 U v U v (b) Using Problem 35a UB I ( ) Alternating urrent 3-7 i I and v Problem 35 shows ω + ( ω /[ ω]) 4 4 ( ω ω ) 4 + ( ω ω + ) Using Problem (347b): UE 4 4 ω ( + ( ω ω) ) 4ω ( + ( ω ω) ) (c) The graphs of the magnetic and electric energies are given in Figure 353 EAUATE: (d) When the angular frequency is zero, the magnetic energy stored in the inductor is zero, while the electric energy in the capacitor is U E 4 As the frequency goes to infinity, the energy noted in both inductor and capacitor go to zero The energies equal each other at the resonant frequency where ω and U B UE 4 Figure IDENTIFY: At any instant of time the same rules apply to the parallel ac circuit as to parallel dc circuit: the voltages are the same and the currents add SET UP: For a resistor the current and voltage in phase For an inductor the voltage leads the current by 9 and for a capacitor the voltage lags the current by 9 EXEUTE: (a) The parallel -- circuit must have equal potential drops over the capacitor, inductor and resistor, so v v v v Also, the sum of currents entering any junction must equal the current leaving the junction Therefore, the sum of the currents in the branches must equal the current through the source: i i + i + i (b) i v is always in phase with the voltage i v lags the voltage by 9, and i ω vω leads the voltage by 9 The phase diagram is sketched in Figure 354 (c) From the diagram, ( ) (d) From part (c): I I + I I + ω ω I + ω But I, so + ω ω Z Z ω

18 3-8 hapter 3 EAUATE: For large ω, Z The current in the capacitor branch is much larger than the current in the ω other branches For small ω, Z ω The current in the inductive branch is much larger than the current in the other branches Figure IDENTIFY: Apply the expression for I from problem 354 when ω / SET UP: From Problem 354, I + ω ω EXEUTE: (a) At resonance, I I ω ω ω I I and I is a minimum (b) Pav cos φ at resonance where < Z so power is a maximum Z (c) At ω ω, I and are in phase, so the phase angle is zero, which is the same as a series resonance EAUATE: (d) The parallel circuit is sketched in Figure 355 At resonance, i i and at any instant of time these two currents are in opposite directions Therefore, the net current between a and b is always zero (e) If the inductor and capacitor each have some resistance, and these resistances aren t the same, then it is no longer true that i + i and the statement in part (d) isn t valid Figure IDENTIFY: efer to the results and the phasor diagram in Problem 354 The source voltage is applied across each parallel branch SET UP: 3 EXEUTE: (a) I A 4 Ω 3 36 rad s 6 F 67 A (b) I ω ( )( )( ) I 67 A (c) φ arctan arctan 48 I 778 A (d) I I I ( ) ( ) A + 67 A 3 A (e) eads since φ > EAUATE: The phasor diagram shows that the current in the capacitor always leads the source voltage 357 IDENTIFY and SET UP: efer to the results and the phasor diagram in Problem 354 The source voltage is applied across each parallel branch EXEUTE: (a) I ; I ω; I ω (b) The graph of each current versus ω is given in Figure 357a (c) ω : I ; I ω : I ; I At low frequencies, the current is not changing much so the inductor s back-emf doesn t resist This allows the current to pass fairly freely However, the current in the capacitor goes to zero because it tends to fill up over the slow period, making it less effective at passing charge At high frequency, the induced emf in the inductor resists the violent changes and passes little current The capacitor never gets a chance to fill up so passes charge freely

19 Alternating urrent 3-9 (d) ω rad sec and f 59 Hz The phasor diagram is sketched in ( H)(5 F) Figure 357b (e) I + ω ω I + ( )( s )(5 F) 5 A Ω ( s )( H) (f) At resonance I I ω ( )( s )(5 F) 5 A and I 5 A Ω EAUATE: At resonance i i at all times and the current through the source equals the current through the resistor Figure IDENTIFY: The average power depends on the phase angle φ SET UP: The average power is P av I cos φ, and the impedance is Z + ω ω EXEUTE: (a) P av I cos φ ( I ), which gives cos φ, so φ π/3 6 tan φ (X X )/, which gives tan 6 (ω /ω)/ Using 75 Ω, 5 mh, and 5 µf and solving for ω we get ω 876 rad/s 8,8 rad/s (b) Z + ( X X ), where X ω (8,76 rad/s)(5 mh) 44 Ω and X /ω /[(8,76 rad/s)(5 µf)] 39 Ω, giving Z (75 Ω ) + (44 Ω 39 Ω ) 5 Ω; I /Z (5 )/(5 Ω) A and P av I cos φ (5 )( A)(/) 375 W EAUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero 359 IDENTIFY: We know, X and φ so Eq(34) tells us X Use Pav I from Exercise 37 to calculate I Then calculate Z and use Eq(36) to calculate for the source SET UP: Source voltage lags current so φ 54 X 35 Ω, 8 Ω, Pav 4 W X X EXEUTE: (a) tanφ X tan φ + X (8 Ω)tan( 54 ) + 35 Ω 48 Ω+ 35 Ω Ω Pav 4 W (b) Pav I cosφ I (Exercise 37) I 88 A 8 Ω (c) Z + ( X X ) (8 Ω ) + ( Ω 35 Ω ) 36 Ω I Z (88 A)(36 Ω ) 7 EAUATE: We could also use Eq(33): P I cosφ av Pav 4 W 7, which agrees The source voltage lags the current when I cos φ (88 A)cos( 54 ) X > X and this agrees with what we found,

20 3- hapter 3 36 IDENTIFY and SET UP: alculate Z and I / Z EXEUTE: (a) For ω 8 rad s: 7 Z + ( ω ω) (5 Ω ) + ((8 rad/s)( H) ((8 rad/s)(5 F))) Z 3 Ω 97 A I 97 A I (97 A)(5 Ω ) 486, 43 7 Z 3 Ω ω (8 rad s)(5 F) ω ( ω) and Iω (97 A)(8 rad s)( H) 55 φ arctan 69 The graph of each voltage versus time is given in Figure 36a (b) epeating exactly the same calculations as above for ω rad/s: Z 5 Ω ; φ ; I A; ; 4 The graph of each voltage versus time is given in Figure 36b (c) epeating exactly the same calculations as part (a) for ω 5 rad/s: Z 3 Ω ; φ + 69 ; I 97 A; 486 ; 55 ; 43 The graph of each voltage versus time is given in Figure 36c EAUATE: The resonance frequency is ω rad/s For ω < ω the phase ( H)(5 μf) angle is negative and for ω > ω the phase angle is positive Figure 36

21 Alternating urrent 3-36 IDENTIFY and SET UP: Eq(39) allows us to calculate I and then Eq(3) gives Z Solve Eq(3) for 36 EXEUTE: (a) IX so I 75 A X 48 Ω (b) IZ so Z 6 Ω I 75 A (c) Z + ( X X ) X X Z ±, so X X Z X 69 Ω or 34 Ω ± 48 Ω ± (6 Ω) (8 Ω ) 48 Ω ± 39 Ω (d) EAUATE: X and X ω At resonance, X X As the frequency is lowered below the ω resonance frequency X increases and X decreases Therefore, for ω < ω, X < X So for X 34 Ω the angular frequency is less than the resonance angular frequency ω is greater than ω when X 69 Ω But at these two values of X, the magnitude of X X is the same so Z and I are the same In one case ( X 69 Ω ) the source voltage leads the current and in the other ( X 34 Ω ) the source voltage lags the current 36 IDENTIFY and SET UP: The maximum possible current amplitude occurs at the resonance angular frequency because the impedance is then smallest EXEUTE: (a) At the resonance angular frequency ω /, the current is a maximum and Z, giving I max / At the required frequency, I I max /3 I /Z I max /3 (/)/3, which means that Z 3 Squaring gives + (ω /ω) 9 Solving for ω gives ω 39 5 rad/s and ω rad/s Imax 495 (b) (35 ) 495 I 3 A 3 3 3(5 Ω) 4 For ω 835 rad/s: 5 Ω and I 65 Ω ; X ω 5 Ω and 65 ; X 479 Ω and 63 ω 5 For ω 39 rad/s: 5 Ω and I 65 Ω ; X ω 479 Ω and 63 ; X 5 Ω and 65 ω EAUATE: For the lower frequency, X > X and > For the higher frequency, X > X and > 363 IDENTIFY and SET UP: onsider the cycle of the repeating current that lies between t τ / and t 3 τ / In I this interval i ( t τ ) τ I t t av i dt and I t i dt t t t t t 3 τ / t 3 τ /I I EXEUTE: Iav idt ( t τ) dt t τt t t t τ τ / τ τ I 9τ 3τ τ τ I Iav ( + I) 8 (9 + 4) (3 3) τ / t 4I I ( I ) τ av i dt ( t τ ) dt t t t τ τ / τ / 4 I τ 3 3 / 4 4 ( ) I τ 3 / 3 3 ( ) I τ τ I t τ dt t τ τ τ / 3 τ τ 3τ I I [+ ] 3 I 6 I I I 3 EAUATE: In each cycle the current has as much negative value as positive value and its average is zero i is always positive and its average is not zero The relation between I and the current amplitude for this current is different from that for a sinusoidal current (Eq34) τ /

22 3- hapter IDENTIFY: Apply IZ SET UP: ω and Z + ( X X) EXEUTE: (a) ω 786 rad s 7 (8 H)(9 F) 7 (b) Z + ( ω ω) Z (3 Ω ) + ((786 rad s)(8 H) ((786 rad s)(9 F))) 3 Ω 6 I- A Z 3 Ω 4 (c) We want I I- + ( ω ω) Z + ( ω ω) I- 4 4 ω + + and ( ω ) + ω + ω I- I- 6 Substituting in the values for this problem, the equation becomes ( ω ) (34) + ω ( 47 ) Solving this quadratic equation in ω we find ω 89 rad s or 48 rad /s and ω 943 rad s or 654 rad s (d) (i) 3 Ω, I- A, ω ω 89 rad s (ii) 3 Ω, I- A, ω ω 8 rad/s (iii) 3 Ω, I - A, ω ω 88 rad/s EAUATE: The width gets smaller as gets smaller; I- gets larger as gets smaller 365 IDENTIFY: The resonance frequency, the reactances, and the impedance all depend on the values of the circuit elements SET UP: The resonance frequency is ω /, the reactances are X ω and X /ω, and the impedance is Z + ( X X ) EXEUTE: (a) ω / becomes /, so ω decreases by (b) Since X ω, if is doubled, X increases by a factor of (c) Since X /ω, doubling decreases X by a factor of (d) Z + ( X X) Z ( ) + ( X X), so Z does not change by a simple factor of or EAUATE: The impedance does not change by a simple factor, even though the other quantities do N 366 IDENTIFY: A transformer transfo voltages according to The effective resistance of a secondary N circuit of resistance is eff ( N/ N) SET UP: N 75 and 5 EXEUTE: (a) ( N/ N) (5 )(834/ 75) Ω (b) eff 36 Ω ( N/ N) (834/ 75) EAUATE: The voltage across the secondary is greater than the voltage across the primary since N > N The effective load resistance of the secondary is less than the resistance connected across the secondary 367 IDENTIFY: The resonance angular frequency is ω and the resonance frequency is f π SET UP: ω is independent of EXEUTE: (a) ω (or f ) depends only on and so change these quantities (b) To double ω, decrease and by multiplying each of them by EAUATE: Increasing and decreases the resonance frequency; decreasing and increases the resonance frequency 368 IDENTIFY: At resonance, Z I / I, IX and IX UE and U I SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities

23 EXEUTE: (a) (b) I IX ω (c) IX ω Z + ( ω / ω ) At resonance ω and Imax ω Alternating urrent 3-3 (d) U (e) U I EAUATE: At resonance and the maximum energy stored in the inductor equals the maximum energy stored in the capacitor 369 IDENTIFY: I / I, IX and IX UE and U I SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities ω EXEUTE: ω (a) I Z ω 9 + / ω + 4 (b) IX ω ω (c) IX (d) U (e) U I EAUATE: For ω < ω, > and the maximum energy stored in the capacitor is greater than the maximum energy stored in the inductor 37 IDENTIFY: I / I, IX and IX UE and U I SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities EXEUTE: ω ω (a) I Z + (ω / ω) (b) IX ω (c) IX ω (d) U

24 3-4 hapter 3 (e) U I EAUATE: For, > and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor 37 IDENTIFY and SET UP: Assume the angular frequency ω of the source and the resistance of the resistor are known I EXEUTE: onnect the source, capacitor, resistor, and inductor in series Measure and ω I and can be calculated EAUATE: There are a number of other approaches The frequency could be varied until, and then this frequency is equal to / If is known, then can be calculated 37 IDENTIFY: P I cosφ av I alculate Z Zcos φ Z SET UP: f 5 Hz and ω π f The power factor is cos φ cos φ ( ) (56) EXEUTE: (a) Pav cos φ Z 367 Ω Z Pav ( W) Zcos φ (367 Ω )(56) 6 Ω (b) Z + X X Z (367 Ω) (6 Ω ) 34 Ω But φ is at resonance, so the inductive and capacitive reactances equal each other Therefore we need to add X 34 Ω X therefore gives ω 4 5 F ωx π f X π(5 Hz)(34 Ω) ( ) (c) At resonance, Pav 699 W 6 Ω EAUATE: Pav I and I is maximum at resonance, so the power drawn from the line is maximum at resonance di q p i p i p i dt SET UP: i Icosωt 373 IDENTIFY: EXEUTE: (a) p i I cos ( ωt ) Icos ( ωt) I(+ cos( ωt)) T I T I T Pav( ) p ( cos( )) [ ] dt t dt t I T ω T + T di T (b) p i ωi cos( ωt)sin( ωt) I sin( ωt) But sin( ) av( ) dt ω tdt P q T (c) p i vi Isin( ωt)cos( ωt) Isin( ωt) But sin( ) av( ) ω tdt P (d) p p + p + pc Icos ( ωt) Isin( ωt) + Isin( ωt) and p Icos( ωt)( cos( ωt) sin( ωt) + sin( ωt)) But cosφ and sin φ, so p I cos( ωt)(cosφcos( ωt) sinφsin( ωt)), at any instant of time EAUATE: At an instant of time the energy stored in the capacitor and inductor can be changing, but there is no net consumption of electrical energy in these components 374 IDENTIFY: d IX dω at the ω where d is a maximum IX dω at the ω where is a maximum SET UP: Problem 35 shows that I + ( ω / ω) EXEUTE: (a) maximum when ω ω

25 d (b) maximum when dω Therefore: d d ω dω dω + ( ω ω) ω ( ω )( + ω ) 4 + ( ω ω) ω ( ω ) 3 + ( ω ω) ( + ( ω ω) ) + ω ω ω and ω Alternating urrent 3-5 d (c) maximum when dω Therefore: d d dω dω ω + ( ω ω) ( ω )( + ω ) 4 + ( ω ω) ω ( ω ) 3 ω + ( ω ω) ( + ( ω ω) ) + ω ω and ω + ω ω EAUATE: is maximum at a frequency greater than the resonance frequency and X is a maximum at a frequency less than the resonance frequency These frequencies depend on, as well as on and on 375 IDENTIFY: Follow the steps specified in the problem SET UP: In part (a) use Eq(33) to calculate Z and then I / Z φ is given by Eq(34) In part (b) let Z + ix EXEUTE: (a) From the current phasors we know that Z + ( ω ω) (4 ) ( rad s)(5 H) 5 Z Ω + Ω ( rad s)(5 F) I 4 A Z 5 Ω ω ( ω) ( rad s)(5 H) ( rad s)(5 F) (b) φ arctan φ arctan Ω (c) Zcpx + i ω Zcpx 4 Ω i ( rad s)(5 H) 4 3 i Ω Ω ω ( rad s)(5 F) Z (4 Ω ) + ( 3 Ω ) 5 Ω 8 + 6i (d) Icpx A (3 A) + (4 A) i Zcpx (4 3 i) Ω 5 Im( Icpx ) 6 5 (e) tanφ 75 φ e( I ) 8 5 cpx 8+ 6i (f) cpx Icpx (4 Ω ) ( i) i cpx iicpxω i ( rad s)(5 H) ( + 6 i) 5 Icpx 8+ 6i cpx i i ( i) ω 5 ( rad s)(5 F) (g) cpx cpx + cpx + cpx ( i) + ( + 6 i) + (9 56 i) EAUATE: Both approaches yield the same value for I and for φ 8+ 6i 8i I 4 A 5 5

26

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