33 Alternating-Current Circuits CHAPTER OUTLINE 33.1 AC Sources 33.2 Resistors in an AC Circuit 33.3 Inductors in an AC Circuit 33.4 Capacitors in an AC Circuit 33.5 The RLC Series Circuit 33.6 Power in an AC Circuit 33.7 Resonance in a Series RLC Circuit 33.8 The Transformer and Power Transmission 33.9 Rectifiers and Filters * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ33.1 Answer (c) When a power source, AC or DC, is first connected to a RL combination, the presence of the inductor impedes the buildup of a current in the circuit The value of the current starts at zero and increases as the back emf induced across the inductor decreases in magnitude OQ33.2 (i) Answer (e) Inductive reactance, X L = ω L , doubles when the frequency doubles, so the rms current is halved ( Irms = ΔVrms X L ) (ii) Answer (b) Capacitive reactance, X C = ω C , is cut in half when frequency doubles, so the rms current doubles ( Irms = ΔVrms XC ) (iii) Answer (d) The resistance remains unchanged ( I rms = ΔVrms R ) 518 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 519 OQ33.3 Answer (a) The voltage across the capacitor is proportional to the stored charge This charge, and hence the voltage ΔVC , is a maximum when the current has zero value and is in the process of reversing direction after having been flowing in one direction for a half cycle Thus, the voltage across the capacitor lags behind the current by 90° The capacitive reactance, X C = ω C , decreases as frequency increases, causing the impedance to decrease and the current to increase OQ33.4 (i) Answer (d) ΔVavg = ΔVmax (ii) Answer (c) The average of the squared voltage is ([ ΔV ] ) avg ΔVmax ) ( = Then its square root is ΔV rms = ΔVmax OQ33.5 Answer (d) If the voltage is a maximum when the current is zero, the voltage is either leading or lagging the current by 90° (or a quarter cycle) in phase Thus, the element could be either an inductor or a capacitor It could not be a resistor since the voltage across a resistor is always in phase with the current If the current and voltage were out of phase by 180°, one would be a maximum in one direction when the other was a maximum value in the opposite direction OQ33.6 (i) Answer (e) The voltage varies between +170 V and –170 V (ii) Answer (c) The average of a sine waveform is zero (iii) Answer (b) ΔVrms = ΔVmax OQ33.7 (i) = 170 V = 120 V Answer (a) We have: Z = R + ( X L − XC ) 2 ⎞ ⎛ = R + ⎜ωL − ⎟ ⎝ ωC ⎠ 2 ⎧⎪ ⎡ = ⎨( 20.0 Ω ) + ⎢ 2π ( 500 Hz )( 120 × 10−3 H ) ⎪⎩ ⎣ 1/2 Z = 51.5 Ω ⎤ ⎫⎪ − ⎥⎬ 2π ( 500 Hz )( 0.750 × 10−6 F ) ⎦⎥ ⎪⎭ and I rms = ΔVrms Z = ( 120 V ) ( 51.5 Ω ) = 2.33 A (ii) Answer (b) At the resonance frequency, XL = XC and the impedance is Z = R + ( X L − X C ) = R Thus, the rms current is I rms = ΔVrms Z = ( 120 V ) ( 20.0 Ω ) = 6.00 A © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 520 Alternating-Current Circuits OQ33.8 Answer (e) is false In an RLC circuit, the instantaneous voltages ΔvR , ΔvL , and ΔvC (across the resistance, inductance, and capacitance respectively) are not in phase with each other The instantaneous voltage ΔvR is in phase with the current, ∆vL leads the current by 90°, while ΔvC lags behind the current by 90° The instantaneous values of these three voltages add algebraically to give the instantaneous voltages across the RLC combination, but the rms voltages across these components not add algebraically The rms voltages across the three components must be added as vectors (or phasors) to obtain the correct rms voltage across the RLC combination OQ33.9 Answer (c) At resonance the inductive reactance and capacitive reactance cancel out OQ33.10 Answer (c) At resonance the inductive reactance and capacitive reactance add to zero: φ = tan–1[(XL – XC)/R] = OQ33.11 The ranking is (a) > (d) > (b) > (c) > (e) At the resonance frequency f0 = 000 Hz both XL and XC are equal: call their mutual value X A high-Q value means the resonance has a small width, so XL and XC are also much larger than R at f0 Inductive reactance XL is proportional to frequency, and capacitive reactance XC is inversely proportional to frequency In terms of X, the choices have the values: (a) f = f0 /2, so XC = 2X (b) f = 3f0 /2, so XC = 2X/3 (c) f = f0 /2, so XL = X/2 (d) f = 3f0 /2, so XL = 3X/2 (e) R is independent of frequency, and R is less than X Thus, we have (a) 2X > (d) 3X/2 > (b) 2X/3 > (c) X/2 > (e) less than X OQ33.12 Answer (e) The battery produces a constant current in the primary coil, which generates a constant flux through the secondary coil With no change in flux through the secondary coil, there is no induced voltage across the secondary coil OQ33.13 Answer (c) AC ammeters and voltmeters read rms values With an oscilloscope you can read a maximum voltage, or test whether the average is zero ANSWERS TO CONCEPTUAL QUESTIONS CQ33.1 (a) The Q factor determines the selectivity of the radio receiver For example, a receiver with a very low Q factor will respond to a wide range of frequencies and might pick up several adjacent radio stations at the same time To discriminate between 102.5 MHz and 102.7 MHz requires a high-Q circuit © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 CQ33.2 521 (b) Typically, lowering the resistance in the circuit is the way to get a higher quality resonance (a) The second letter in each word stands for the circuit element (b) CIVIL – in a capacitor C the current (I) leads voltage (represented by V), voltage leads current in an inductor L For an inductor L, the emf ε leads the current I—thus ELI For a capacitor C, the current leads the voltage across the device In a circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE CQ33.3 The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase CQ33.4 (a) In an RLC series circuit, the phase angle depends on the source frequency At very low frequency, the capacitor dominates the impedance and the phase angle is near –90° At very high frequencies, the inductor dominates the impedance and the phase angle is near –90° (b) When the inductive reactance equals the capacitive reactance, the frequency is the resonance frequency; the phase angle is zero CQ33.5 In 1881, an assassin shot President James Garfield The bullet was lost in his body Alexander Graham Bell invented the metal detector in an effort to save the President’s life The coil is preserved in the Smithsonian Institution The detector was thrown off by metal springs in Garfield’s mattress, a new invention itself Surgeons went hunting for the bullet in the wrong place Garfield died CQ33.6 (a) The person is doing work at a rate of P = Fv cos θ (b) Compare the previous equation to P = ΔVrms I rms cos φ One can consider the emf as the “force” that moves the charges through the circuit, and the current as the “speed” of the moving charges The cos θ factor measures the effectiveness of the cause in producing the effect Theta is an angle in real space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf and the current in the circuit CQ33.7 The circuit can be considered an RLC series circuit (a) Yes The circuit is in resonance because the inductive reactance and capcitive reactance are equal, so the total impedance Z = R © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 522 Alternating-Current Circuits (b) Total power output by the emf Pemf = I2Rtotal, where Rtotal = 10 Ω (source resistance) + 10 Ω (load resistance) = 20 Ω Power delivered to the load Pload = I2RL , where RL = 10 Ω Fraction of average power delivered to the load to average power delivered by the source of emf: Pload I Rload R RL 10 Ω = = L = = = 0.5 Pemf I Rtotal Rtotal Rsource + RL 20 Ω (c) The resistance of the load could be increased to make a greater fraction of the emf’s power go to the load Then the emf would put out a lot less power and less power would reach the load CQ33.8 No A voltage is only induced in the secondary coil if the flux through the core changes in time No changing current, no changing flux, no induced voltage CQ33.9 (a) The capacitive reactance is proportional to the inverse of the frequency At higher and higher frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire (b) As the frequency goes to zero, the capacitive reactance approaches infinity—the resistance of an open circuit CQ33.10 The ratio of turns indicates the ratio of voltages: N 1/N = ΔV1/ΔV2 , where ΔV2 = 120 V; therefore, ΔV1 = 12 kV In its intended use, the transformer takes in energy by electric transmission at 12 kV and puts out nearly the same energy by electric transmission at 120 V With the small generator putting energy into the secondary side of the transformer at 120 V, the primary side has 12 kV induced across it It is deadly dangerous for the repairman SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 33.1 AC Sources Section 33.2 Resistors in an AC Circuit P33.1 (a) The rms voltage across the resistor is given by ΔVR, rms = I rms R = ( 8.00 A ) ( 12.0 Ω ) = 96.0 V (b) From Equation 33.5, ( ) ΔVR, max = ΔVR, rms = ( 96.0 V ) = 136 V © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 (c) 523 From Equation 33.4, I max = 2I rms = ( 8.00 A ) = 11.3 A (d) P33.2 Pavg = I rms R = ( 8.00 A ) ( 12.0 Ω ) = 768 W The rms voltage is ΔVrms = P33.3 170 V = 120 V (a) ΔVrms ) ( (120 V )2 = P= → R = (b) R= 75.0 W R (120 V )2 100 W 193 Ω = 144 Ω Each meter reads the rms value ( ΔVmax ΔVrms = R R ) = (100 V ) = 2.95 A (a) I rms = (b) The voltage across the resistor is the same as that across the power supply: ΔVrms = 24.0 Ω ΔVmax 100 V = = 70.7 V 2 ANS FIG P33.3 P33.4 (a) We compute the peak voltage from the rms voltage: ( ) ΔVR, max = ΔVR, rms = ( 120 V ) = 170 V (b) From the definition of power, Pavg = I rms R= ΔVrms R Solving for the resistance, R= ΔVrms ( 120 V ) = = 2.40 × 102 Ω Pavg 60.0 W © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 524 Alternating-Current Circuits (c) Because Pavg 2 ΔVrms ) ΔVrms ) ( ( = → R = , a 100-W bulb has less R Pavg resistance than a 60.0-W bulb P33.5 Δv ⎛ ΔVmax ⎞ = sin ω t Given the R ⎜⎝ R ⎟⎠ value of t, we want to identify a point with The current as a function of time is i = 0.600 ΔVmax ΔVmax = sin(ω t) R R ω t = sin–1 0.600 or To find the lowest frequency we choose the smallest angle satisfying this relation: ( 0.007 00 s )ω = sin −1 ( 0.600) = 0.644 rad Thus, P33.6 (a) ω = 91.9 rad s = 2π f so f = 14.6 Hz From Equation 33.3, ΔvR = ΔVmax sin ω t To find the angular frequency, we write ΔvR = 0.250 ( ΔVmax ) so sin ω t = 0.250 or ω t = sin −1 ( 0.250 ) The smallest angle for which this is true is ω t = 0.253 rad Thus, if t = 0.010 s, ω= (b) 0.253 rad = 25.3 rad s 0.010 s The second time when ΔvR = 0.250 ( ΔVmax ) , ω t = sin −1 ( 0.250 ) again For this occurrence, ω t = π − 0.253 rad = 2.89 rad (to understand why this is true, recall the identity sin (π − θ ) = sin θ from trigonometry) Thus, t= 2.89 rad = 0.114 s 25.3 rad s © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 P33.7 525 We are given ΔVmax = 15.0 V and Rtotal = 8.20 Ω + 10.4 Ω = 18.6 Ω The maximum current in the circuit is I max = ΔVmax 15.0 V = = 0.806 A = 2I rms Rtotal 18.6 Ω And the power delivered to the speakers is ⎛ 0.806 A ⎞ Pspeaker = I rms Rspeaker = ⎜ ⎟ ( 10.4 Ω ) = 3.38 W ⎝ ⎠ P33.8 All lamps are connected in parallel with the voltage source, so ΔVrms = 120 V for each lamp Also, the current is I rms = Pavg ΔVrms and the resistance is R = ΔVrms I rms (a) For the 150-W bulbs, I rms = 150 W = 1.25 A 120 V For the 100-W bulb, I rms = 100 W = 0.833 A 120 V The rms current in each 150-W bulb is 1.25 A The rms current in the 100-W bulb is 0.833 A (b) The resistance in bulbs and is R1 = R2 = 120 V = 96.0 Ω 1.25 A and the resistance in bulb is R3 = (c) 120 V = 144 Ω 0.833 A The bulbs are in parallel, so 1 1 1 = + + = + + Req R1 R2 R3 96.0 Ω 96.0 Ω 144 Ω Req = 36.0 Ω © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 526 Alternating-Current Circuits Section 33.3 P33.9 Inductors in an AC Circuit Inductive reactance is proportional to frequency At 50.0 Hz, X′ L f ′ = XL f X′ L = X L f ′ 50.0 Hz = ( 54.0 Ω ) = 45.0 Ω f 60.0 Hz The maximum current is I max = P33.10 (a) XL = ω L = L= (b) ΔVmax = XL ( ΔVrms ) = XL ( 100 V ) = 3.14 A 45.0 Ω ΔVmax I max ΔVmax 100 V = = 0.0424 H ω I max 2π ( 50.0 Hz ) ( 7.50 A ) ΔVmax ΔVmax = , we see that is current inversely XL ωL proportional to angular frequency: From I max = I max ω ′ = I′ max ω I 7.50 A ω ′ = ω max = [ 2π ( 50.0 Hz )] = 942 rad/s I′ max 2.50 A P33.11 The inductive reactance is XL = ω L = (65.0 π s–1)(70.0 × 10–3 V · s/A) = 14.3 Ω The amplitude of the current is I max = ΔVmax 80.0 V = = 5.60 A XL 14.3 Ω Equation 33.7 lets us evaluate the current: i = −I max cos ω t = − ( 5.60 A ) cos ⎡⎣( 65.0π s −1 )( 0.015 s )⎤⎦ = − ( 5.60 A ) cos ( 3.17 rad ) = +5.60 A © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 P33.12 527 The relationship between current, inductance, and maximum voltage is ΔVL, rms I rms = XL = ( ΔV L, max ωL )= ΔVL, max ( 2π ) fL In order to restrict the current to I rms < 2.00 × 10−3 A, we require ΔVL,max < 2.00 × 10−3 A ( 2π ) fL Solving for the inductance then gives ΔVL,max ( 2π ) f ( 2.00 × 10−3 A ) L> = L > 0.750 H or P33.13 P33.14 4.00 V ( 2π )( 300.0 Hz )( 2.00 × 10−3 A ) (a) X L = 2π f L = 2π ( 80.0 Hz ) ( 25.0 × 10−3 H ) = 12.6 Ω (b) I rms = (c) I max = I rms = ( 6.21 A ) = 8.78 A ΔVL, rms XL = 78.0 V = 6.21 A XL In the inductor, because U B = LiL = when t = 0, iL = I max sin (ω t ) Then, I rms = and ΔVrms ΔVrms 120 V = = = 15.9 A XL ωL ⎡⎣ 2π ( 60.0 ) s −1 ⎤⎦ ( 0.020 H ) I max = 2I rms = ( 15.9 A ) = 22.5 A the current in the inductor is ⎡ ⎛ ⎞⎤ iL = I max sin ω t = ( 22.5 A ) sin ⎢ 2π ( 60.0 ) s −1 ⋅ ⎜ s ⎝ 180 ⎟⎠ ⎥⎦ ⎣ = ( 22.5 A ) sin 120° = 19.5 A and the energy stored is UB = 2 LiL = ( 0.020 H )( 19.5 A ) = 3.80 J 2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 (b) 557 One possibility is T = 10.9 N and d = 0.200 m Any values of T and d related according to this expression will work We did not need to use the value of the current and magnetic field P33.70 (a) See the graph in ANS FIG P33.70(a) ANS FIG P33.70(a) (b) ⎛ ω L − 1/ ω C ⎞ φ = tan −1 ⎜ ⎟⎠ changes from –90° for ω = to at the ⎝ R resonance frequency to +90° as ω goes to infinity The slope of the graph is dφ/dω : dφ = dω = (c) 1⎛ 1 ⎞ ⎜⎝ L − (−1) ⎟⎠ C ω ⎛ ω L − 1/ ω C ⎞ R 1+ ⎜ ⎟ ⎝ ⎠ R R ⎞ ⎛ ⎟ ⎜L+ ω 2C ⎠ R + (ω L − 1/ ω C ) ⎝ At resonance we have ω 02 = LC ; substituting, we find the slope at the resonance point is dφ dω = ω0 ⎛ LC ⎞ 2L 2Q L+ = ⎟= ⎜ R+0 ⎝ C ⎠ R ω0 where Q = ω L R P33.71 (a) When ω L is very large, Z is large for the bottom branch, so it carries negligible current Also, will be negligible compared ωC to R for the top branch, so I= V V 45.0 V ≈ = = 0.225 A Z R 200 Ω flows in the power supply and the top branch © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 558 Alternating-Current Circuits (b) is very large in the top branch and ω L is very small ωC compared to R in the bottom branch; the generator and bottom branch carry Now I= P33.72 (a) V V 45.0 V ≈ = = 0.450 A Z R 100 Ω With both switches closed, the current goes only through the generator and resistor i= ΔVmax cos ω t R (b) ( ΔVmax ) P= R (c) i= ⎡ −1 ⎛ ω L ⎞ ⎤ cos ω t + tan ⎜⎝ ⎟ ⎢ R ⎠ ⎥⎦ ⎣ R + ω L2 ΔVmax ⎛ ω L − ( ω 0C ) ⎞ (d) For = φ = tan −1 ⎜ ⎟, R ⎝ ⎠ We require ω L = (e) 1 , so C = ω0L ω 0C The frequency is the resonance frequency: Z = R For parts (f) and (g), the circuit is at resonance, so Z = R and XC = X L = ω 0L (f) To find the maximum energy stored in the capacitor, we start with U= 1 2 C ( ΔVC ) = C ( IX C ) 2 When I = Imax, U max (g) U max ⎛ ΔV ⎞ = CI max X C2 = C ⎜ max ⎟ (ω L ) = 2 ⎝ R ⎠ ( ΔVmax ) L = LI max = R2 ( ΔVmax )2 L 2R 2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 (h) Now ω = 2ω = 559 , so LC ⎛ ⎞ ⎛ ω L − (1 ω C ) ⎞ −1 L C − ( ) L C φ = tan −1 ⎜ = tan ⎜ ⎟ ⎟⎠ R R ⎝ ⎝ ⎠ ⎛ L⎞ = tan −1 ⎜ ⎝ 2R C ⎟⎠ P33.73 1 , so ω = 2ωC ω = 2LC (i) Now ω L = (a) The inductive reactance of the circuit is X L = 2π f L = 2π ( 50.0 Hz )( 0.250 H ) = 78.5 Ω (b) The capacitive reactance of the circuit is XC = 1 = 2π f C 2π ( 50.0 Hz )( 2.00 × 10−6 F ) = 1.59 × 103 Ω = 1.59 kΩ (c) The impedance of the circuit is Z = R + ( X L − X C ) = ( 150 Ω ) + ( 78.5 Ω − 1590 Ω ) 2 = 1.52 × 103 Ω = 1.52 kΩ (d) The maximum current is I max = ΔVmax 2.10 × 102 V = = 0.138 A = 138 mA Z 1.52 × 103 Ω (e) ⎛ X − XC ⎞ ⎛ 78.5 Ω − 1590 Ω ⎞ φ = tan −1 ⎜ L = tan −1 ⎜ ⎟ ⎟⎠ = −84.3° ⎝ ⎝ R ⎠ 150 Ω (f) ⎡ ⎛ 78.5 Ω − 590 Ω ⎞ ⎤ cosφ = cos ⎢ tan −1 ⎜ ⎟⎠ ⎥ = 0.098 ⎝ 150 Ω ⎣ ⎦ (g) The power input into the circuit is P = I rms ΔVrms ΔVrms ) ( ΔVmax ( cosφ = cosφ = ΔVmax ) ( = cosφ Z Z ) cosφ 2Z ( 210 V )2 P= ( 0.098 ) = 1.43 W ( 1.52 × 103 Ω ) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 560 P33.74 Alternating-Current Circuits (a) We are given XL = XC = 884 Ω when f = 000 Hz The impedance is then L= 884 Ω XL = = 0.150 H 2π f 000π rad s and the capacitance is C= 1 = ( 2π f ) XC ( 000π rad s )(1 884 Ω) = 42.2 nF therefore, X L = 2π f ( 0.150 H ) XC = ( 2π f )( 4.22 × 10−8 F ) Z= ( 40.0 Ω )2 + ( X L − XC ) and TABLE P33.74 lists the inductive reactance, the capacitive reactance, and the impedance for the frequencies listed in the problem statement f (Hz) 300 600 800 000 500 000 000 000 000 10 000 X L (Ω) X C (Ω) Z (Ω) 283 12 600 12 300 565 280 720 754 710 960 942 770 830 410 510 100 880 880 40 830 260 570 770 942 830 650 628 020 420 377 040 TABLE P33.74 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 (b) 561 ANS FIG P33.74(b) shows a graph of XL , XC , and Z as a function of the frequency f ANS FIG P33.74(b) P33.75 The resonance frequency is f0 = 2π LC = 2π (0.050 0 H)(5.00 × 10 –6 F) = 318 Hz The operating frequency is f = f0/2 = 159 Hz We can calculate the impedance at this frequency The inductive reactance is XL = π f L = π (159 Hz)(0.050 H) = 50.0 Ω The capacitive reactance is XC = 1 = = 200 Ω 2π fC 2π (159 Hz)(5.00 × 10 –6 F) The impedance is Z = R + ( X L – XC ) = 8.002 + (50.0 – 200)2 Ω = 150 Ω So the current is I rms = ΔVrms = Z 400 V = 2.66 A 150 Ω The power delivered by the source is the power delivered to the resistor: P = ( 2.66 A ) ( 8.00Ω ) = 56.7 W P33.76 (a) At resonance, ω = LC = (1.00 × 10 –3 )( H 1.00 × 10 F –9 ) = 1.00 × 106 rad/s © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 562 Alternating-Current Circuits At that point, Z = R = 1.00 Ω and I = 1.00 V = 1.00 A 1.00 Ω The power is P = I2R = (1.00 A)2(1.00 Ω ) = 1.00 W We compute the power at some other angular frequencies, listed in TABLE P33.76 ω ω L (Ω) (Ω ) Z (Ω ) P = Irms R (W ) ω0 ωC 0.9990 999.0 1001.0 2.24 0.19984 0.9991 999.1 1000.9 2.06 0.23569 0.9993 999.3 1000.7 1.72 0.33768 0.9995 999.5 1000.5 1.41 0.49987 0.9997 999.7 1000.3 1.17 0.73524 0.9999 999.9 1000.1 1.02 0.96153 1.0000 1000 1000.0 1.00 1.00000 1.0001 1000.1 999.9 1.02 0.96154 1.0003 1000.3 999.7 1.17 0.73535 1.0005 1000.5 999.5 1.41 0.50012 1.0007 1000.7 999.3 1.72 0.33799 1.0009 1000.9 999.1 2.06 0.23601 1.0010 1001 999.0 2.24 0.20016 TABLE P33.76 ANS FIG P33.76 shows a graph of the results tabulated above ANS FIG P33.76 (b) The angular frequencies giving half the maximum power are 0.999 × 106 rad/s and 1.000 ì 106 rad/s â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 563 so the full width at half the maximum is Δω = (1.000 – 0.999 5) × 106 rad/s Δω = 1.00 × 10 rad/s Since Δω = 2π Δf , Δf = 159 Hz and for comparison, R 1.00 Ω = = 159 Hz 2π L 2π ( 1.00 × 10 –3 H ) The two quantities agree Challenge Problems P33.77 We start with Δvout IR R R = = = Δvin IZ Z R + ( X L − XC ) 8.00 Ω = ( 8.00 Ω )2 + [ 2π fL − 2π f C ] Then, at 200 Hz, 2 ⎛ Δvout ⎞ 8.00 Ω ) ( ⎛ R⎞ ⎜⎝ Δv ⎟⎠ = ⎜⎝ Z ⎟⎠ = = ( 8.00 Ω )2 + [ 400π L − 400π C ] in and at 000 Hz, 2 ⎛ Δvout ⎞ 8.00 Ω ) ( ⎛ R⎞ ⎜⎝ Δv ⎟⎠ = ⎜⎝ Z ⎟⎠ = = ( 8.00 Ω )2 + [ 8000π L − 8000π C ] in At the low frequency, XL – XC < This reduces to 400π L − = −13.9 Ω 400π C [1] For the high frequency half-voltage point, 000π L − = +13.9 Ω 000π C [2] Solving equations [1] and [2] simultaneously gives (a) L = 580 àH â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 564 Alternating-Current Circuits ANS FIG P33.77(a) (b) C = 54.6 µF (c) When XL = XC, Δvout ⎛ Δvout ⎞ = = 1.00 Δvin ⎜⎝ Δvin ⎟⎠ max (d) XL = XC requires f0 = (e) 2π LC = 2π ( 5.80 × 10 −4 H ) ( 5.46 × 10−5 F ) = 894 Hz Δvout R = = and XC > XL, so the phasor diagram is Δvin Z as shown in ANS FIG P33.77(e) At 200 Hz , ⎛ R⎞ ⎛ 1⎞ φ = − cos −1 ⎜ ⎟ = − cos −1 ⎜ ⎟ = −60.0° ⎝ Z⎠ ⎝ 2⎠ so ( Δvout leads Δvin ) At f0 , XL = XC so φ = 0 (Δvout is in phase with Δvin ) At 000 Hz , Δvout R = = and XL – XC > Δvin Z ANS FIG P33.77(e) ⎛ 1⎞ Thus, φ = cos −1 ⎜ ⎟ = +60.0° ⎝ 2⎠ or Δvout lags Δvin © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 (f) 565 At 200 Hz and at kHz, ( Δv P= ) ( Δv P= ) = ( Δv out, rms ( ⎡( ) Δvin, rms ⎤⎦ (1 4) Δvin, max =⎣ = R R 2 ) R 2 )( 10.0 V ) ( = = 1.56 W ( 8.00 Ω ) At f0, out, rms R ) = ( Δv in, rms R in, max ) = (1 )(10.0 V ) 2 ( 8.00 Ω ) R = 6.25 W (g) We take −4 ω L 2π f0 L 2π ( 894 Hz ) ( 5.80 × 10 H ) Q= = = = 0.408 R R 8.00 Ω P33.78 ΔVrms 100 V = = 1.25 A R 80.0 Ω (a) I R, rms = (b) The total current will lag the applied voltage as seen in the phasor diagram shown in ANS FIG P33.78 I L, rms = = ANS FIG P33.78 ΔVrms XL 100 V = 1.33 A 2π ( 60.0 s −1 )( 0.200 H ) Thus, the phase angle is: ⎛I ⎞ ⎛ 1.33 A ⎞ φ = tan −1 ⎜ L, rms ⎟ = tan −1 ⎜ = 46.7° ⎝ 1.25 A ⎟⎠ ⎝ I R, rms ⎠ P33.79 We are given L = 2.00 H, C = 10.0 × 10–6 F, R = 10.0 Ω, and Δv = ( 100sin ω t ) (a) The resonance frequency ω produces the maximum current and thus the maximum power delivery to the resistor ω0 = = LC ( 2.00 H ) (10.0 × 10 −6 F) −1 = 224 s © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 566 Alternating-Current Circuits (b) At the resonance frequency ω , the impedance Z = R, and P = I rms R ( 2 ΔVmax ΔVrms ) ΔVrms ) ( ( ⎛ ΔVrms ⎞ =⎜ R= R = R = ⎟ ⎝ Z ⎠ Z2 R2 R 100 V ) ( = = ( 10.0 Ω ) (c) ) 500 W Now, ( ΔVrms ) R = ( ΔVrms ) P = Pmax → Z2 R 2 So, Z2 = 2R2, or ⎛ ⎞ R + ⎜ω L − = 2R ⎟ ω C⎠ ⎝ which is a fourth order equation in ω But this can be simplified to two equations: ωL− = ±R → ω LC ω CR − = ωC The angular frequency ω must be positive, so we solve for the positive roots (In the following, we suppress all units.) For ω LC − ω CR − = 0, − ( − CR ) + ( −CR ) − 4LC ( −1) ω = 2LC = R + R + 4L C 2L 10.0 + ( 10.0 ) + ( 2.00 ) ( 10.0 × 10−6 ) = ( 2.00 ) = 226 s −1 For ω LC + ω CR − = 0, − ( CR ) + ( −CR ) − 4LC ( −1) 2LC ω = −R + R + 4L C = 2L −10.0 + ( 10.0 ) + ( 2.00 ) ( 10.0 × 10−6 ) = ( 2.00 ) = 221 s −1 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 P33.80 567 The currents in the three components of the circuit are IR = (a) ΔVrms ΔVrms ΔVrms ΔVrms ΔVrms , IL = = , and IC = = R XL ωL XC (ω C )−1 Then, 12 I rms = ⎡⎣ I + ( IC − I L ) ⎤⎦ R (b) tan φ = ⎡⎛ ⎞ ⎛ ⎞ ⎤ ⎥ = ΔVrms ⎢⎜ ⎟ + ⎜ ω C − ω L ⎟⎠ ⎥ ⎢⎣⎝ R ⎠ ⎝ ⎦ 12 ⎞ ⎡ IC − I L ⎤⎛ = ΔVrms ⎢ − ⎥⎜ ⎟ IR ⎣ X C X L ⎦ ⎝ ΔVrms R ⎠ ⎛ 1 ⎞ tan φ = R ⎜ − ⎝ X C X L ⎟⎠ ANS FIG P33.80 ( ΔVrms ) R , and ⎛ ΔVrms ⎞ R=⎜ R= ⎟ ⎝ Z ⎠ Z2 P33.81 We have P = I rms ⎛ ⎞ Z = R + ⎜ω L − ω C ⎟⎠ ⎝ 2 Therefore, Z 2 ΔVrms ) R ( = P ⎛ ( ΔVrms ) R ⎞ R + ⎜ω L − = ⎟ ω C⎠ P ⎝ → 2 ⎛ ( ΔVrms ) R − R ⎞ ⎜⎝ ω L − ω C ⎟⎠ = P 2 which is a fourth order equation in ω But this can be simplified to two equations: ωL− = ±A ωC → ω LC ω CA − = © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 568 Alternating-Current Circuits where A= ( ΔVrms )2 R − R P We will solve for ω when ΔVrms = 100 V and P = 250 W From Figure P33.24, we have R = 40.0 Ω, L = 185 mH = 0.185 H, and C = 65.0 àF = 65.0 ì 10–6 F The quantity A is A= ( ΔVrms )2 R − R = (120 V )2 ( 40.0 Ω) − ( 40.0 Ω)2 250 W P = 704 Ω The angular frequency ω must be positive, so we solve for the positive roots (In the following, we suppress all units.) For ω LC − ω CA − = 0, ω= = = − ( −CA ) + ( −CA )2 − 4LC ( −1) 2LC A + A2 + L C 2L 704 + 704 + ( 0.185 ) ( 65.0 × 10−6 ) ( 0.185 ) = 226 s −1 = 2π f → f = 58.7 Hz For ω LC + ω CA − = 0, ω= − ( CA ) + ( −CA )2 − 4LC ( −1) 2LC −A + A + L C = = 2L − 704 + 704 + ( 0.185 ) ( 65.0 × 10−6 ) ( 0.185 ) = 225 s −1 = 2π f → f = 35.9 Hz There are two answers because the circuit can be either above or below resonance © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 569 ANSWERS TO EVEN-NUMBERED PROBLEMS P33.2 (a) 193 Ω; (b) 144 Ω P33.4 (a) 170 V; (b) 2.40 × 102 Ω; (c) Because Pavg 2 ΔVrms ) ΔVrms ) ( ( = → R = , a 100-W bulb has less R Pavg resistance than a 60.0-W bulb P33.6 (a) 25.3 rad/s; (b) 0.114 s P33.8 (a) The rms current in each 150-W bulb is 1.25 A The rms current in the 100-W bulb is 0.833 A; (b) R1 = 96.0 Ω, R2 = 96.0 Ω, and R3 = 144 Ω; (c) 36.0 Ω P33.10 (a) 0.0424 H; (b) 942 rad/s P33.12 0.750 H P33.14 3.80 J P33.16 (a) 15.0 Hz; (b) 84.9 V; (c) 47.1 Ω; (d) 1.80 A; (e) 2.55 A P33.18 (a) 221 Ω; (b) 0.163 A; (c) 0.230 A; (d) no P33.20 (a) 69.3 V; (b) 40.0 Hz; (c) 20.3 µF P33.22 2C ( ΔVrms ) P33.24 (a) 146 V ; (b) 212 V; (c) 179 V; (d) 33.4 V P33.26 (a) 109 Ω; (b) Imax = 0.367 A; (c) ω = 100 rad/s; (d) φ = −0.896 rad = −51.3° P33.28 2.79 kHz P33.30 See ANS FIG P33.30 P33.32 (a) 88.4 Ω; (b) 107 Ω; (c) 1.12 A; (d) the voltage lags behind the current by 55.8°; (e) Adding an inductor will change the impedence, and hence the current in the circuit The current could be larger or smaller, depending on the inductance added The largest current would result when the inductive reactance equals the capacitive reactance, the impedance has its minimum value, equal to 60.0 Ω, and the current in the circuit is I max P33.34 ΔVmax ΔVmax 1.20 × 102 V = = = = 2.00 A Z R 60.0 Ω In order for the power factor to be equal to 1.00, we would have to have XL = 0, which would require either L or f to be zero Because this © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 570 Alternating-Current Circuits is not the case, the situation is impossible P33.36 8.00 W P33.38 (a) 66.8 Ω; (b) 0.953 A; (c) 45.4 W P33.40 (a) 5.43 A; (b) 0.905; (c) 281 µF; (d) 109 V P33.42 (a) 3.56 kHz; (b) 5.00 A; (c) 22.4; (d) 2.24 kV P33.44 (a) 156 pH; (b) 8.84 Ω P33.46 4π RC LC ( ΔVrms ) 4R 2C + 9L P33.48 (a) 9.23 V; (b) 2.40 W P33.50 (a) 29.0 kW; (b) 5.80 × 10–3; (c) It is impossible to transmit so much power at such low voltage P33.52 1.88 V P33.54 (a) P33.56 (a) P33.58 The resonance frequency for this circuit is not in the North American AM frequency range P33.60 (a) 0.200 A, 36.8°; (b) 40.0 V at φ = 0° ; (c) 20.0 V at φ = −90.0° ; (d) 50.0 V at φ = +90.0° P33.62 R R2 + (1 ω C ) 1ωC R2 + (1 ω C ) ρCu P π f ( ΔVrms ) 2 ; (b) 0; (c) 2π RC ; (b) 1; (c) 0; (d) P33.64 See P33.64 for full explanation P33.66 (a) R = 99.6 Ω; (b) 24.9 µF; (c) 164 mH or 402 mH; (d) Only one value for R and only one value for C are possible Two values for L are possible P33.68 (a) Higher At the resonance frequency, XL = XC As the frequency increases, XL goes up and XC goes down; (b) It is possible We have three independent equations in the three unknowns L, C, and the certain f ; (c) L = 4.90 mH and C = 51.0 µF P33.70 (a) See ANS FIG P33.70(a); (b) R ⎞ ⎛ ⎟ ; (c) See ⎜L+ ⎝ ω 2C ⎠ R + (ω L − 1/ ω C ) P33.70(c) for full explanation © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 33 571 ΔVmax ( ΔVmax ) cos ω t; (b) P = (a) i = ; R R P33.72 (c) i = ⎡ −1 ⎛ ω L ⎞ ⎤ C = ; (e) R; (d) cos ω t + tan ; ⎜ ⎟ ⎢ ⎥ ⎝ R ⎠⎦ ω0L ⎣ R + ω L2 ΔVmax ΔVmax ) L ( (f) ; (g) 2R ⎛ L⎞ ( ΔVmax ) L ; (h) tan −1 ⎜ ; (i) 2 R ⎝ 2R C ⎟⎠ 2LC P33.74 (a) See Table P33.74; (b) See ANS FIG P33.74(b) P33.76 (a) See ANS FIG P33.76; (b) See P33.76 for full explanation P33.78 (a) 1.25 A; (b) lag, 46.7° P33.80 ⎡⎛ ⎞ ⎛ ⎞ ⎤ (a) ΔVrms ⎢⎜ ⎟ + ⎜ ω C − ⎥ ω L ⎟⎠ ⎥ ⎢⎣⎝ R ⎠ ⎝ ⎦ 12 ⎛ 1 ⎞ ; (b) tan φ = R ⎜ − ⎝ X C X L ⎟⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part