M31 YOUN7066 13 ISM c31 kho tài liệu bách khoa

ALTERNATING CURRENT 31.1 31 IDENTIFY: The maximum current is the current amplitude, and it must not ever exceed 1.50 A SET UP: I rms = I/ I is the current amplitude, the maximum value of the current 1.50 A = 1.06 A EVALUATE: The current amplitude is larger than the root-mean-square current IDENTIFY and SET UP: Apply Eqs (31.3) and (31.4) EXECUTE: (a) I = I rms = 2(2.10 A) = 2.97 A EXECUTE: I = 1.50A gives I rms = 31.2 (b) I rav = 31.3 31.4 π I= π (2.97 A) = 1.89 A EVALUATE: (c) The root-mean-square current 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 IDENTIFY and SET UP: Apply Eq (31.5) V 45.0 V EXECUTE: (a) Vrms = = = 31.8 V 2 (b) Since the voltage is sinusoidal, the average is zero EVALUATE: The voltage amplitude is larger than Vrms IDENTIFY: We want the phase angle for the source voltage relative to the current, and we want the capacitance if we know the current amplitude V SET UP: X C = and X C = 2π f C I EXECUTE: (a) φ = −90° The source voltage lags the current by 90° V 60.0 V = 11.3 Ω Solving X C = (b) X C = = for C gives I 5.30 A 2π fC C= 31.5 1 = = 1.76 × 10−4 F 2π fX C 2π (80.0 Hz)(11.3 Ω) EVALUATE: This is a 176-μ F capacitor, which is not unreasonable IDENTIFY: We want the phase angle for the source voltage relative to the current, and we want the inductance if we know the current amplitude V SET UP: X L = and X L = 2π fL I EXECUTE: (a) φ = +90° The source voltage leads the current by 90° V 45.0 V X 11.54 Ω (b) X L = = = 11.54 Ω Solving X L = 2π fL for f gives f = L = = 193 Hz I 3.90 A 2π L 2π (9.50 × 10−3 H) EVALUATE: The angular frequency is about 1200 rad/s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-1 31-2 31.6 Chapter 31 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 C = 1/ωC and X L = ω L 1 ⇒ω = ωC LC 1 = = 7560 rad/s (5.00 mH)(3.50 μ F) LC EXECUTE: (a) Equating the reactances gives ω L = (b) Using the numerical values we get ω = X C = X L = ω L = (7560 rad/s)(5.00 mH) = 37.8 Ω 31.7 31.8 EVALUATE: At other angular frequencies, the two reactances could be very different IDENTIFY and SET UP: Apply Eqs (31.18) and (31.19) V 170 V EXECUTE: V = IX C so X C = = = 200 Ω I 0.850 A 1 gives C = XC = = = 1.33 × 10−5 F = 13.3 μ F 2π fX C 2π (60.0 Hz)(200 Ω) ωC EVALUATE: The reactance relates the voltage amplitude to the current amplitude and is similar to Ohm’s law IDENTIFY: The reactance of an inductor is X L = ω L = 2π fL The reactance of a capacitor is 2π fC SET UP: The frequency f is in Hz EXECUTE: (a) At 60.0 Hz, X L = 2π (60.0 Hz)(0.450 H) = 170 Ω X L is proportional to f so at 600 Hz, XC = ωC = X L = 1700 Ω (b) At 60.0 Hz, X C = 2π (60.0 Hz)(2.50 × 10−6 F) = 1.06 × 103 Ω X C is proportional to 1/f , so at 600 Hz, X C = 106 Ω (c) X L = X C says 2π fL = EVALUATE: 31.9 X L increases when f increases X C decreases when f increases IDENTIFY and SET UP: Use Eqs (31.12) and (31.18) EXECUTE: (a) X L = ω L = 2π fL = 2π (80.0 Hz)(3.00 H) = 1510 Ω (b) X L = 2π fL gives L = (c) X C = (d) X C = ωC = XL 120 Ω = = 0.239 H 2π f 2π (80.0 Hz) 1 = = 497 Ω 2π fC 2π (80.0 Hz)(4.00 × 10−6 F) 1 = = 1.66 × 10−5 F gives C = 2π fC 2π fX C 2π (80.0 Hz)(120 Ω) EVALUATE: 31.10 1 and f = = = 150 Hz 2π fC 2π LC 2π (0.450 H)(2.50 × 10−6 F) X L increases when L increases; X C decreases when C increases IDENTIFY: VL = I ω L SET UP: ω is the angular frequency, in rad/s f = ω is the frequency in Hz 2π V (12.0 V) EXECUTE: VL = I ω L so f = L = = 1.63 × 106 Hz 2ω IL 2π (2.60 × 10−3 A)(4.50 × 10−4 H) EVALUATE: When f is increased, I decreases © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31.11 31-3 IDENTIFY: In an L-R ac circuit, we want to find out how the voltage across a resistor varies with time if we know how the voltage varies across the inductor SET UP: vL = − I ω L sin ωt and vR = VR cos(ωt ) EXECUTE: (a) vL = − I ω L sin ωt ω = 480 rad/s I ω L = 12.0 V I= 12.0 V 12.0 V = = 0.1389 A VR = IR = (0.1389 A)(90.0 Ω) = 12.5 V ωL (480 rad/s)(0.180 H) vR = VR cos(ωt ) = (12.5 V)cos[(480 rad/s)t ] (b) vR = (12.5 V)cos[(480 rad/s)(2.00 × 10−3 s)] = 7.17 V 31.12 EVALUATE: The instantaneous voltage (7.17 V) is less than the voltage amplitude (12.5 V) I IDENTIFY: Compare vC that is given in the problem to the general form vC = sin ωt and determine ω ωC SET UP: X C = vR = iR and i = I cos ωt ωC 1 EXECUTE: (a) X C = = = 1736 Ω ωC (120 rad/s)(4.80 × 10−6 F) (b) I = VC 7.60 V = = 4.378 × 10−3 A and i = I cos ωt = (4.378 × 10−3 A)cos[(120 rad/s)t ] Then X C 1736 Ω vR = iR = (4.38 × 10−3 A)(250 Ω)cos((120 rad/s)t ) = (1.10 V)cos((120 rad/s)t ) 31.13 EVALUATE: The voltage across the resistor has a different phase than the voltage across the capacitor IDENTIFY and SET UP: The voltage and current for a resistor are related by vR = iR Deduce the frequency of the voltage and use this in Eq (31.12) to calculate the inductive reactance Eq (31.10) gives the voltage across the inductor EXECUTE: (a) vR = (3.80 V)cos[(720 rad/s)t ] vR = iR, so i = vR ⎛ 3.80 V ⎞ =⎜ ⎟ cos[(720 rad/s)t ] = (0.0253 A)cos[(720 rad/s)t ] R ⎝ 150 Ω ⎠ (b) X L = ω L ω = 720 rad/s, L = 0.250 H, so X L = ω L = (720 rad/s)(0.250 H) = 180 Ω (c) If i = I cos ωt then vL = VL cos(ωt + 90°) (from Eq 31.10) VL = I ω L = IX L = (0.02533 A)(180 Ω) = 4.56 V vL = (4.56 V)cos[(720 rad/s)t + 90°] 31.14 But cos(a + 90°) = − sin a (Appendix B), so vL = −(4.56 V)sin[(720 rad/s)t ] EVALUATE: The current is the same in the resistor and inductor and the voltages are 90° out of phase, with the voltage across the inductor leading IDENTIFY: Calculate the reactance of the inductor and of the capacitor Calculate the impedance and use that result to calculate the current amplitude X V SET UP: With no capacitor, Z = R + X L2 and tan φ = L X L = ω L I = VL = IX L and VR = IR R Z For an inductor, the voltage leads the current EXECUTE: (a) X L = ω L = (250 rad/s)(0.400 H) = 100 Ω Z = (200 Ω) + (100 Ω)2 = 224 Ω (b) I = V 30.0 V = = 0.134 A Z 224 Ω (c) VR = IR = (0.134 A)(200 Ω) = 26.8 V VL = IX L = (0.134 A)(100 Ω) = 13.4 V X L 100 Ω = and φ = +26.6° Since φ is positive, the source voltage leads the current R 200 Ω (e) The phasor diagram is sketched in Figure 31.14 (d) tan φ = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-4 Chapter 31 EVALUATE: Note that VR + VL is greater than V The loop rule is satisfied at each instance of time but the voltages across R and L reach their maxima at different times Figure 31.14 31.15 IDENTIFY: Apply the equations in Section 31.3 SET UP: ω = 250 rad/s, R = 200 Ω, L = 0.400 H, C = 6.00 μ F and V = 30.0 V EXECUTE: (a) Z = R + (ω L − 1/ωC ) Z = (200 Ω)2 + ((250 rad/s)(0.400 H) − 1/((250 rad/s)(6.00 × 10−6 F))) = 601 Ω (b) I = V 30 V = = 0.0499 A Z 601 Ω ⎛ 100 Ω − 667 Ω ⎞ ⎛ ω L − 1/ωC ⎞ (c) φ = arctan ⎜ ⎟ = −70.6°, and the voltage lags the current ⎟ = arctan ⎜ R 200 Ω ⎝ ⎠ ⎝ ⎠ (d) VR = IR = (0.0499 A)(200 Ω) = 9.98 V; VL = I ω L = (0.0499 A)(250 rad/s)(0.400 H) = 4.99 V; VC = I (0.0499 A) = = 33.3 V ωC (250 rad/s)(6.00 × 10−6 F) EVALUATE: (e) At any instant, v = vR + vC + vL But vC and vL are 180° out of phase, so vC can be 31.16 larger than v at a value of t, if vL + vR is negative at that t IDENTIFY: For an L-R-C series ac circuit, we want to find the voltages and voltage amplitudes across all the circuit elements V X − XC SET UP: X C = and tan φ = L The , X L = ω L, Z = R + ( X L − X C ) , I = R ωC Z instantaneous voltages are vR = VR cos(ωt ) = IR cos(ωt ), vL = −VL sin(ωt ) = − IX L sin(ωt ), vC = VC sin(ωt ) = IX C sin(ωt ) and v = V cos(ωt + φ ) EXECUTE: XC = ωC = (250 rad/s)(6.00 × 10−6 F) = 666.7 Ω X L = ω L = (250 rad/s)(0.900 H) = 225 Ω Z = R + ( X L − X C ) = (200 Ω) + (225 Ω − 666.7 Ω) = 484.9 Ω V 30.0 V = = 0.06187 A = 61.87 mA Z 484.9 Ω X − X C 225 Ω − 666.7 Ω tan φ = L = = −2.2085 and φ = −1.146 rad R 200 Ω I= (a) vR = VR cos(ωt ) = IR cos(ωt ) = (0.06187 A)(200 Ω)cos[(250 rad/s)(20.0 × 10−3 s)] = 3.51 V vL = −VL sin(ωt ) = − IX L sin(ωt ) = −(0.06187 A)(225 Ω)sin[(250 rad/s)(20.0 × 10−3 s)] = 13.35 V © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-5 vC = VC sin(ωt ) = IX C sin(ωt ) = (0.06187 A)(666.7 Ω)sin[(250 rad/s)(20.0 × 10−3 s)] = −39.55 V v = V cos(ωt + φ ) = (30.0 V)cos[(250 rad/s)(20.0 × 10−3 s) − 1.146 rad] = −22.70 V vR + vL + vC = 3.51 V + 13.35 V + (−39.55 V) = −22.7 V vR + vL + vC is equal to v (b) VR = IR = 12.4 V VC = 41.2 V VL = 13.9 V VR + VC + VL = 12.4 V + 41.2 V + 13.9 V = 67.5 V VR + VC + VL is not equal to V 31.17 EVALUATE: The instantaneous voltages add up to v because they all occur at the same time, so they must add to v by Kirchhoff’s loop rule The amplitudes not add to V because the maxima not occur at the same time due to phase differences between the inductor, capacitor and resistor IDENTIFY and SET UP: Use the equation that preceeds Eq (31.20): V = VR2 + (VL − VC ) EXECUTE: V = (30.0 V) + (50.0 V − 90.0 V)2 = 50.0 V 31.18 EVALUATE: The equation follows directly from the phasor diagrams of Fig 31.13 (b or c) in the textbook Note that the voltage amplitudes not simply add to give 170.0 V for the source voltage IDENTIFY: For an L-R ac circuit, we want to use the resistance, voltage amplitude of the source and power in the resistor to find the impedance, the voltage amplitude across the inductor and the power factor X V SET UP: Pav = I R, Z = , VR = IR, and tan φ = L I R EXECUTE: (a) Pav = 2 Pav 2(216 W) V 500 V = = 1.20 A Z = = I R I = = 417 Ω I 1.20 A R 300 Ω (b) VR = IR = (1.20 A)(300 Ω) = 360 V VL = V − VR2 = (500 V)2 − (360 V)2 = 347 V X L VL 347 V = = gives φ = 43.95° The power factor is cos φ = 0.720 R VR 360 V EVALUATE: The voltage amplitude across the resistor cannot exceed the voltage amplitude (500 V) of the ac source IDENTIFY: For a pure resistance, Pav = Vrms I rms = I rms R (c) tan φ = 31.19 SET UP: 20.0 W is the average power Pav EXECUTE: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W 20.0 W P (b) I rms = av = = 0.167 A Vrms 120 V (c) R = Pav I rms = 20.0 W (0.167 A) = 720 Ω 31.20 VR2,rms (120 V) Vrms = = 20.0 W 720 Ω R R IDENTIFY: The average power supplied by the source is Pav = Vrms I rms cos φ The power consumed in the EVALUATE: We can also calculate the average power as Pav = = resistance is Pav = I rms R SET UP: ω = 2π f = 2π (1.25 × 103 Hz) = 7.854 × 103 rad/s X L = ω L = 157 Ω X C = ωC = 909 Ω EXECUTE: (a) First, let us find the phase angle between the voltage and the current: X − X C 157 Ω − 909 Ω tan φ = L = and φ = −65.04° The impedance of the circuit is R 350 Ω Z = R + ( X L − X C ) = (350 Ω) + (−752 Ω) = 830 Ω The average power provided by the generator is then Pav = Vrms I rms cos(φ ) = Vrms (120 V) cos(φ ) = cos(−65.04°) = 7.32 W Z 830 Ω © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-6 Chapter 31 31.21 ⎛ 120 V ⎞ (b) The average power dissipated by the resistor is PR = I rms R=⎜ ⎟ (350 Ω) = 7.32 W ⎝ 830 Ω ⎠ EVALUATE: Conservation of energy requires that the answers to parts (a) and (b) are equal IDENTIFY: Relate the power factor to R and Z for an L-R-C series ac circuit Then use this result to find the voltage amplitude across a resistor IR R SET UP and EXECUTE: (a) From Figure 31.13(a) or (b), cos φ = = IZ Z R V V cos φ (b) Using the result from (a) gives Z = I= = cos φ Z R VR = IR = V cos φ = (90.0 V)cos( −31.5°) = 76.7 V 31.22 EVALUATE: The voltage amplitude for the resistor is less than the voltage amplitude of the ac source IDENTIFY: We want to relate the average power delivered by the source in an L-R-C circuit to the rms current and resistance R SET UP: From Exercise 31.21 we know that the power factor is cos φ = We also know that Z Pav = Vrms I rms cos φ EXECUTE: (a) Pav = Vrms I rms cos φ cos φ = R R V so Pav = Vrms I rms But rms = I rms so Pav = I rms R Z Z Z ⎛ 36.0 V ⎞ ⎜ ⎟ Vrms ⎠ =⎝ = 6.75 W (b) I rms = and Vrms = V/ 2, so Pav R R 96.0 Ω EVALUATE: The instantaneous power can be greater than 6.75 W at times, but it can also be less than that at other times, giving an average of 6.75 W IDENTIFY and SET UP: Use the equations of Section 31.3 to calculate φ , Z and Vrms The average power V2 = rms 31.23 R delivered by the source is given by Eq (31.31) and the average power dissipated in the resistor is I rms EXECUTE: (a) X L = ω L = 2π fL = 2π (400 Hz)(0.120 H) = 301.6 Ω XC = ωC tan φ = = 1 = = 54.51 Ω 2π fC 2π (400 Hz)(7.3 × 10−6 F) X L − X C 301.6 Ω − 54.41 Ω = , so φ = +45.8° The power factor is cosφ = +0.697 R 240 Ω (b) Z = R + ( X L − X C ) = (240 Ω) + (301.6 Ω − 54.51 Ω) = 344 Ω (c) Vrms = I rms Z = (0.450 A)(344 Ω) = 155 V (d) Pav = I rmsVrms cos φ = (0.450 A)(155 V)(0.697) = 48.6 W (e) Pav = I rms R = (0.450 A) (240 Ω) = 48.6 W 31.24 EVALUATE: 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 V R IDENTIFY and SET UP: Pav = Vrms I rms cos φ I rms = rms cos φ = Z Z 80.0 V 75.0 Ω EXECUTE: I rms = = 0.762 A cos φ = = 0.714 Pav = (80.0 V)(0.762 A)(0.714) = 43.5 W 105 Ω 105 Ω EVALUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also R = (0.762 A) (75.0 Ω) = 43.5 W calculate the average power as Pav = I rms © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31.25 31-7 IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance X C of the capacitor The angular frequency and the inductance can be used to calculate the reactance X L of the inductor Calculate the phase angle φ and then the power factor is cos φ Calculate the impedance of the circuit and then the rms current in the circuit The average power is Pav = Vrms I rms cos φ On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor V 45 V SET UP: The source has rms voltage Vrms = = = 31.8 V 2 EXECUTE: (a) X L = ω L = (360 rad/s)(15 × 10−3 H) = 5.4 Ω XC = = ωC (360 rad/s)(3.5 × 10−6 F) The power factor is cos φ = 0.302 = 794 Ω tan φ = X L − X C 5.4 Ω − 794 Ω = and φ = −72.4° R 250 Ω (b) Z = R + ( X L − X C ) = (250 Ω) + (5.4 Ω − 794 Ω) = 827 Ω I rms = Pav = Vrms I rms cos φ = (31.8 V)(0.0385 A)(0.302) = 0.370 W Vrms 31.8 V = = 0.0385 A Z 827 Ω (c) The average power delivered to the resistor is Pav = I rms R = (0.0385 A) (250 Ω) = 0.370 W The 31.26 average power delivered to the capacitor and to the inductor is zero EVALUATE: 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 IDENTIFY: At resonance in an L-R-C ac circuit, we know the reactance of the capacitor and the voltage amplitude across it From this information, we want to find the voltage amplitude of the source SET UP: At resonance, Z = R VC = IX C EXECUTE: I = V 600 V = = 3.00 A Z = R = 300 Ω V = IZ = (3.00 A)(300 Ω) = 900 V X 200 Ω EVALUATE: At resonance, Z = R, but X C is not zero 31.27 IDENTIFY and SET UP: The current is largest at the resonance frequency At resonance, X L = X C and Z = R For part (b), calculate Z and use I = V/Z EXECUTE: (a) f = = 113 Hz I = V/R = 15.0 mA 2π LC (b) X C = 1/ωC = 500 Ω X L = ω L = 160 Ω Z = R + ( X L − X C ) = (200 Ω) + (160 Ω − 500 Ω) = 394.5 Ω I = V/Z = 7.61 mA X C > X L so the source voltage lags the current EVALUATE: ω0 = 2π f = 710 rad/s ω = 400 rad/s and is less than ω0 When ω < ω0 , X C > X L Note 31.28 that I in part (b) is less than I in part (a) IDENTIFY: The impedance and individual reactances depend on the angular frequency at which the circuit is driven ⎞ ⎛ SET UP: The impedance is Z = R + ⎜ ω L − ⎟ , the current amplitude is I = V/Z and the instantaneous C⎠ ω ⎝ values of the potential and current are v = V cos(ωt + φ ), where tan φ = ( X L − X C )/R, and i = I cos ωt EXECUTE: (a) Z is a minimum when ω L = ω= ωC , which gives 1 = = 3162 rad/s, which rounds to 3160 rad/s Z = R = 175 Ω LC (8.00 mH)(12.5 μ F) (b) I = V/Z = (25.0 V)/(175 Ω) = 0.143 A (c) i = I cos ωt = I/2, so cos ωt = 12 , which gives ωt = 60° = π /3 rad v = V cos(ωt + φ ), where tan φ = ( X L − X C )/R = 0/R = So, v = (25.0 V)cos ωt = (25.0 V)(1/2) = 12.5 V © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-8 Chapter 31 vR = Ri = (175 Ω)(1/2)(0.143 A) = 12.5 V vC = VC cos(ωt − 90°) = IX C cos(ωt − 90°) = 0.143 A cos(60° − 90°) = +3.13 V (3162 rad/s)(12.5 μ F) vL = VL cos(ωt + 90°) = IX L cos(ωt + 90°) = (0.143 A)(3162 rad/s)(8.00 mH)cos(60° + 90°) vL = −3.13 V 31.29 (d) vR + vL + vC = 12.5 V + (−3.13 V) + 3.13 V = 12.5 V = vsource EVALUATE: 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 180° out of phase, so they add to zero, leaving all the potential difference across the resistor IDENTIFY and SET UP: At the resonance frequency, Z = R Use that V = IZ , VR = IR, VL = IX L and VC = IX C Pav is given by Eq (31.31) (a) EXECUTE: V = IZ = IR = (0.500 A)(300 Ω) = 150 V (b) VR = IR = 150 V X L = ω L = L(1/ LC ) = L/C = 2582 Ω; VL = IX L = 1290 V X C = 1/(ωC ) = L/C = 2582 Ω; VC = IX C = 1290 V (c) Pav = 12 VI cos φ = 12 I R, since V = IR and cos φ = at resonance Pav = 12 (0.500 A) (300 Ω) = 37.5 W 31.30 EVALUATE: At resonance VL = VC Note that VL + VC > V However, at any instant vL + vC = IDENTIFY: The current is maximum at the resonance frequency, so choose C such that ω = 50.0 rad/s is the resonance frequency At the resonance frequency Z = R SET UP: VL = I ω L V EXECUTE: (a) The amplitude of the current is given by I = Thus, the current will ⎞ ⎛ R + ⎜ωL − ωC ⎟⎠ ⎝ 1 have a maximum amplitude when ω L = Therefore, C = = = 44.4 μ F ωC ω L (50.0 rad/s)2 (9.00 H) (b) With the capacitance calculated above we find that Z = R, and the amplitude of the current is V 120 V I= = = 0.300 A Thus, the amplitude of the voltage across the inductor is R 400 Ω VL = I (ω L) = (0.300 A)(50.0 rad/s)(9.00 H) = 135 V EVALUATE: Note that VL is greater than the source voltage amplitude 31.31 IDENTIFY and SET UP: At resonance X L = X C , φ = and Z = R R = 150 Ω, L = 0.750 H, C = 0.0180 μ F, V = 150 V EXECUTE: (a) At the resonance frequency X L = X C and from tan φ = and the power factor is cos φ = 1.00 X L − XC we have that φ = 0° R (b) Pav = 12 VI cos φ (Eq 31.31) At the resonance frequency Z = R, so I = V V = Z R V (150 V) ⎛V ⎞ Pav = 12 V ⎜ ⎟ cos φ = 12 = 12 = 75.0 W R R 150 Ω ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-9 (c) EVALUATE: When C and f are changed but the circuit is kept on resonance, nothing changes in Pav = V /(2 R ), so the average power is unchanged: Pav = 75.0 W The resonance frequency changes but 31.32 31.33 since Z = R at resonance the current doesn’t change VC = IX C V = IZ IDENTIFY: ω0 = LC SET UP: At resonance, Z = R 1 EXECUTE: (a) ω0 = = = 1.54 × 104 rad/s LC (0.350 H)(0.0120 × 10−6 F) ⎛V ⎞ ⎛V ⎞ 1 (b) V = IZ = ⎜ C ⎟ Z = ⎜ C ⎟ R X C = = = 5.41 × 103 Ω −6 C ω X X × × (1 54 10 rad/s)(0 0120 10 F) ⎝ C⎠ ⎝ C⎠ 550 V ⎛ ⎞ V =⎜ ⎟ (400 Ω) = 40.7 V ⎝ 5.41 × 103 Ω ⎠ EVALUATE: The voltage amplitude for the capacitor is more than a factor of 10 times greater than the voltage amplitude of the source 1 X L = ω L X C = IDENTIFY and SET UP: The resonance angular frequency is ω0 = and ω C LC Z = R + ( X L − X C )2 At the resonance frequency X L = X C and Z = R EXECUTE: (a) Z = R = 115 Ω (b) ω0 = (4.50 × 10 −3 = 1.33 × 104 rad/s ω = 2ω0 = 2.66 × 104 rad/s H)(1.26 × 10−6 F) X L = ω L = (2.66 × 104 rad/s)(4.50 × 10−3 H) = 120 Ω X C = 1 = = 30 Ω ωC (2.66 × 10 rad/s)(1.25 × 10−6 F) Z = (115 Ω) + (120 Ω − 30 Ω) = 146 Ω (c) ω = ω0 /2 = 6.65 × 103 rad/s X L = 30 Ω X C = = 120 Ω ωC Z = (115 Ω) + (30 Ω − 120 Ω) = 146 Ω, the same value as in part (b) EVALUATE: For ω = 2ω0 , X L > X C For ω = ω0 /2, X L < X C But ( X L − X C ) has the same value at 31.34 these two frequencies, so Z is the same IDENTIFY: At resonance Z = R and X L = X C V = IZ VR = IR, VL = IX L and VC = VL LC 1 EXECUTE: (a) ω0 = = = 945 rad/s LC (0.280 H)(4.00 × 10−6 F) SET UP: ω0 = (b) I = 1.70 A at resonance, so R = Z = V 120 V = = 70.6 Ω I 1.70 A (c) At resonance, VR = 120 V, VL = VC = I ω L = (1.70 A)(945 rad/s)(0.280 H) = 450 V EVALUATE: At resonance, VR = V and VL − VC = 31.35 IDENTIFY and SET UP: Eq (31.35) relates the primary and secondary voltages to the number of turns in 2 each I = V/R and the power consumed in the resistive load is I rms = Vrms /R Let I1, V1 and I , V2 be rms values for the primary and secondary V N N V 120 V = 10 EXECUTE: (a) = so = = V1 N1 N V2 12.0 V © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-10 Chapter 31 (b) I = V2 12.0 V = = 2.40 A R 5.00 Ω (c) Pav = I 22 R = (2.40 A) (5.00 Ω) = 28.8 W (d) The power drawn from the line by the transformer is the 28.8 W that is delivered by the load Pav = V12 V (120 V)2 = 500 Ω so R = = R Pav 28.8 W ⎛N ⎞ And ⎜ ⎟ (5.00 Ω) = (10) (5.00 Ω) = 500 Ω, as was to be shown ⎝ N2 ⎠ EVALUATE: The resistance is “transformed.” A load of resistance R connected to the secondary draws the same power as a resistance ( N1/N ) R connected directly to the supply line, without using the transformer 31.36 IDENTIFY: Pav = Vrms I rms and Pav,1 = Pav,2 N1 V1 = Let I1, V1 and I , V2 be rms values for the N V2 primary and secondary SET UP: V1 = 120 V V2 = 13,000 V EXECUTE: (a) N V2 13,000 V = = = 108 N1 V1 120 V (b) Pav = V2 I = (13,000 V)(8.50 × 10−3 A) = 110 W Pav 110 W = = 0.917 A V1 120 V EVALUATE: Since the power supplied to the primary must equal the power delivered by the secondary, in a step-up transformer the current in the primary is greater than the current in the secondary IDENTIFY: Let I1, V1 and I , V2 be rms values for the primary and secondary A transformer transforms (c) I1 = 31.37 voltages according to Reff = R ( N /N1 ) V2 N = The effective resistance of a secondary circuit of resistance R is V1 N1 Resistance R is related to Pav and Vrms by Pav = Vrms Conservation of energy requires R Pav,1 = Pav,2 so V1I1 = V2 I SET UP: Let V1 = 240 V and V2 = 120 V, so P2,av = 1600 W These voltages are rms EXECUTE: (a) V1 = 240 V and we want V2 = 120 V, so use a step-down transformer with N /N1 = 12 (b) Pav = V1I1, so I1 = Pav 1600 W = = 6.67 A V1 240 V (c) The resistance R of the blower is R = Reff = 9.00 Ω (1/2) V12 (120 V)2 = = 9.00 Ω The effective resistance of the blower is Pav 1600 W = 36.0 Ω EVALUATE: I 2V2 = (13.3 A)(120 V) = 1600 W Energy is provided to the primary at the same rate that it 31.38 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 IDENTIFY: Z = R + ( X L − X C ) 2, with X L = ω L and X C = ωC SET UP: The woofer has a R and L in series and the tweeter has a R and C in series EXECUTE: (a) Z tweeter = R + (1/ωC ) (b) Z woofer = R + (ω L)2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-12 Chapter 31 EXECUTE: C = L= μ0 AN l ω0 = 31.42 = ε0 A d = (8.85 × 10−12 C2 /N ⋅ m )(4.50 × 10−2 m) 8.00 × 10−3 m (4π × 10−7 T ⋅ m/A)π (0.250 × 10−2 m)2 (1125)2 9.00 × 10−2 m (3.47 × 10 −4 H)(2.24 × 10−12 F) = 2.24 × 10−12 F = 3.47 × 10−4 H = 3.59 × 107 rad/s EVALUATE: The result is a rather high angular frequency IDENTIFY: Use geometry to calculate the self-inductance of the toroidal solenoid Then find its reactance and use this to find the impedance, and finally the current amplitude, of the circuit μ N2A SET UP: L = , X L = 2π fL, Z = R + X L2 , and I = V/Z 2π r EXECUTE: L = μ0 N A (2900) (0.450 × 10−4 m ) = (2 × 10−7 T ⋅ m/A) = 8.41 × 10−4 H 2π r 9.00 × 10−2 m X L = 2π fL = (2π )(365 Hz)(8.41× 10−4 H) = 1.929 Ω Z = R + X L2 = 3.40 Ω I = 31.43 EVALUATE: The inductance is physically reasonable IDENTIFY: An L-R-C ac circuit operates at resonance We know L, C, and V and want to find R 1 SET UP: At resonance, Z = R and ω = ω0 = XC = , I = V/ Z ωC LC 1 EXECUTE: ω = = 633.0 rad/s X C = = = 329.1 Ω ωC (633 rad/s)(4.80 × 10−6 F) LC I= 31.44 V 24.0 V = = 7.06 A Z 3.40 Ω V 56.0 V V VC 80.0 V = 230 Ω = = 0.2431 A At resonance Z = R, so I = R = = R I 0.2431 A X C 329.1 Ω EVALUATE: At resonance, the impedance is a minimum IDENTIFY: X L = ω L Pav = Vrms I rms cos φ SET UP: f = 120 Hz; ω = 2π f EXECUTE: (a) X L = ω L ⇒ L = XL ω = 250 Ω = 0.332 H 2π (120 Hz) (b) Z = R + X L2 = (400 Ω)2 + (250 Ω) = 472 Ω cos φ = Vrms = Z V2 R R V and I rms = rms Pav = rms , so Z Z Z Z Pav 800 W = (472 Ω) = 668 V R 400 Ω EVALUATE: I rms = Vrms 668 V = = 1.415 A We can calculate Pav as 472 Ω Z I rms R = (1.415 A) (400 Ω) = 800 W, which checks 31.45 (a) IDENTIFY and SET UP: Source voltage lags current so it must be that X C > X L and we must add an inductor in series with the circuit When X C = X L the power factor has its maximum value of unity, so calculate the additional L needed to raise X L to equal X C (b) EXECUTE: Power factor cos φ equals so φ = and X C = X L Calculate the present value of X C − X L to see how much more X L is needed: R = Z cosφ = (60.0 Ω)(0.720) = 43.2 Ω X L − XC so X L − X C = R tan φ R cos φ = 0.720 gives φ = −43.95° (φ is negative since the voltage lags the current) tan φ = Then X L − X C = R tan φ = (43.2 Ω) tan(−43.95°) = −41.64 Ω Therefore need to add 41.64 Ω of X L © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-13 XL 41.64 Ω = = 0.133 H, amount of inductance to add 2π f 2π (50.0 Hz) EVALUATE: From the information given we can’t calculate the original value of L in the circuit, just how much to add When this L is added the current in the circuit will increase IDENTIFY: We know R, X L , X C , and VL for a series L-R-C ac circuit We want to find VR , VC , V and X L = ω L = 2π fL and L = 31.46 the power delivered by the source V SET UP: I = L , V = IX , Pav = I rms R XL EXECUTE: (a) I = VL 450 V = = 0.500 A VR = IR = (0.500 A)(300 Ω) = 150 V X L 900 Ω (b) VC = IX C = (0.500 A)(500 Ω) = 250 V (c) V = VR2 + (VL − VC ) = (150 V) + (450 V − 250 V) = 250 V VR2 (150 V) I R= = = 37.5 W 2 R 300 Ω EVALUATE: The voltage amplitude of the source is not the sum of the voltage amplitudes of the other circuit elements since the voltages have their maxima at different times and are hence out of phase IDENTIFY: We know the impedances and the average power consumed From these we want to find the power factor and the rms voltage of the source R R cos φ = Z = R + ( X L − X C ) Vrms = I rms Z SET UP: P = I rms Z R= (d) Pav = I rms 31.47 P 60.0 W = = 0.447 A Z = (300 Ω) + (500 Ω − 300 Ω) = 361 Ω R 300 Ω EXECUTE: (a) I rms = R 300 Ω = = 0.831 Z 361 Ω (b) Vrms = I rms Z = (0.447 A)(361 Ω) = 161 V cos φ = EVALUATE: The voltage amplitude of the source is Vrms = 228 V 31.48 IDENTIFY: Use Vrms = I rms Z to calculate Z and then find R Pav = I rms R SET UP: X C = 50.0 Ω EXECUTE: Z = Vrms 240 V = = 80.0 Ω = R + X C2 = R + (50.0 Ω) Thus, I rms 3.00 A R = (80.0 Ω) − (50.0 Ω) = 62.4 Ω The average power supplied to this circuit is equal to the power dissipated by the resistor, which is P = I rms R = (3.00 A) (62.4 Ω) = 562 W X L − X C −50.0 Ω = and φ = −38.7° R 62.4 Ω Pav = Vrms I rms cos φ = (240 V)(3.00 A)cos(−38.7°) = 562 W, which checks EVALUATE: tan φ = 31.49 IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values SET UP: The voltage amplitudes are VR = RI , VL = X L I , and VC = X C I , where I = V/Z and ⎞ ⎛ Z = R2 + ⎜ ωL − ωC ⎟⎠ ⎝ EXECUTE: (a) ω = 2π f = 2π (1250 Hz) = 7854 rad/s Carrying extra figures in the calculator gives X L = ω L = (7854 rad/s)(3.50 mH) = 27.5 Ω; XC = 1/ωC = 1/[(7854 rad/s)(10.0 µF)] = 12.7 Ω; Z = R + ( X L − X C ) = (50.0 Ω)2 + (27.5 Ω − 12.7 Ω) = 57.5 Ω; I = V/Z = (60.0 V)/(52.1 Ω) = 1.15 A; VR = RI = (50.0 Ω)(1.15 A) = 57.5 V; © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-14 Chapter 31 VL = X L I = (27.5 Ω)(1.15 A) = 31.6 V; VC = X C I = (12.7 Ω)(1.15 A) = 14.6 V The voltage amplitudes can add to more than 60.0 V because the voltage maxima not all occur at the same instant of time At any instant, the instantaneous voltages across the resistor, inductor and capacitor all add to equal the instantaneous source voltage (b) All of them will change because they all depend on ω X L = ω L will double to 55.0 Ω, and X C = 1/ωC will decrease by half to 6.35 Ω Therefore Z = (50.0 Ω)2 + (55.0 Ω − 6.35 Ω)2 = 69.8 Ω; I = V/Z = (60.0V)/(69.8 Ω) = 0.860 A; VR = IR = (0.860 A)(50.0 Ω) = 43.0 V; VL = IX L = (0.860 A)(55.0 Ω) = 47.3 V; VC = IX C = (0.860 A)(6.35 Ω) = 5.46 V 31.50 EVALUATE: 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 IDENTIFY and SET UP: X C = X L = ω L ωC 1 EXECUTE: (a) = ω1L and LC = At angular frequency ω2 , ω1C ω1 XL ω L = = ω22 LC = (2ω1 ) 2 = X L > X C X C 1/ω2C ω1 (b) At angular frequency ω3 , EVALUATE: When ω increases, X L ⎛ ⎞ ⎜⎜ ⎟⎟ = X C > X L ⎝ ω1 ⎠ increases and X C decreases When ω decreases, X L decreases XL ⎛ω ⎞ = ω32 LC = ⎜ ⎟ XC ⎝ ⎠ and X C increases 31.51 (c) The resonance angular frequency ω0 is the value of ω for which X C = X L , so ω0 = ω1 IDENTIFY and SET UP: Express Z and I in terms of ω , L, C and R The voltages across the resistor and the inductor are 90° out of phase, so Vout = VR2 + VL2 EXECUTE: The circuit is sketched in Figure 31.51 X L = ω L, X C = ωC ⎞ ⎛ Z = R2 + ⎜ ωL − ωC ⎟⎠ ⎝ I= Vs = Z Vs ⎞ ⎛ R2 + ⎜ ω L − ωC ⎟⎠ ⎝ Figure 31.51 Vout = I R + X L2 = I R + ω L2 = Vs Vout = Vs R + ω L2 ⎞ ⎛ R2 + ⎜ ω L − ωC ⎟⎠ ⎝ R + ω L2 ⎞ ⎛ R2 + ⎜ ω L − ωC ⎟⎠ ⎝ ω small ⎞ ⎛ 2 2 As ω gets small, R + ⎜ ω L − ⎟ → 2 , R +ω L → R ωC ⎠ ω C ⎝ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-15 Vout R2 → = ω RC as ω becomes small Vs (1/ω 2C ) Therefore ω large ⎞ ⎛ 2 2 2 2 2 As ω gets large, R + ⎜ ω L − ⎟ → R +ω L →ω L , R +ω L →ω L C ω ⎝ ⎠ Therefore, 31.52 ω L2 Vout → = as ω becomes large Vs ω L2 EVALUATE: Vout /Vs → as ω becomes small, so there is Vout only when the frequency ω of Vs is large If the source voltage contains a number of frequency components, only the high frequency ones are passed by this filter IDENTIFY: V = VC = IX C I = V/Z SET UP: X L = ω L, X C = ωC I EXECUTE: Vout = VC = ωC ⇒ Vout = Vs ωC R + (ω L − 1/ωC ) If ω is large: Vout 1 = ≈ = 2 Vs LC ( )ω ωC R + (ω L − 1/ωC ) ωC (ω L) If ω is small: Vout ωC ≈ = = Vs C ω ωC (1/ωC ) EVALUATE: When ω is large, X C is small and X L is large so Z is large and the current is small Both 31.53 factors in VC = IX C are small When ω is small, X C is large and the voltage amplitude across the capacitor is much larger than the voltage amplitudes across the resistor and the inductor IDENTIFY: I = V/Z and Pav = 12 I R SET UP: Z = R + (ω L − 1/ωC )2 EXECUTE: (a) I = V = Z V R + (ω L − 1/ωC ) (b) Pav = V R/2 ⎛V ⎞ I R= ⎜ ⎟ R= 2⎝Z⎠ R + (ω L − 1/ωC ) (c) The average power and the current amplitude are both greatest when the denominator is smallest, which 1 occurs for ω0 L = , so ω0 = ω0C LC (d) Pav = (100 V) (200 Ω)/2 (200 Ω) + (ω (2.00 H) − 1/[ω (0.500 × 10−6 F)]) = 25ω 40,000ω + (2ω − 2,000,000 s −2 ) W The graph of Pav versus ω is sketched in Figure 31.53 EVALUATE: 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 31.19 in the textbook © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-16 Chapter 31 Figure 31.53 31.54 IDENTIFY: VL = I ω L and VC = I ωC V SET UP: Problem 31.53 shows that I = EXECUTE: (a) VL = I ω L = (b) VC = I ωC = R + (ω L − 1/[ωC ]) VωL R + (ω L − 1/[ωC ]) V ωC R + (ω L − 1/[ωC ]) (c) The graphs are given in Figure 31.54 EVALUATE: (d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of 100 V (equal to the total source voltage) At very high frequencies, the capacitor voltage goes to zero, while the inductor’s voltage goes to 100 V At resonance, ω0 = = 1000 rad/s, the two voltages LC are equal, and are a maximum, 1000 V Figure 31.54 31.55 IDENTIFY: We know R, X C and φ so Eq (31.24) tells us X L Use Pav = I rms R to calculate I rms Then calculate Z and use Eq (31.26) to calculate Vrms for the source SET UP: Source voltage lags current so φ = −54.0° X C = 350 Ω, R = 180 Ω, Pav = 140 W EXECUTE: (a) tan φ = X L − XC R © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-17 X L = R tan φ + X C = (180 Ω) tan(−54.0°) + 350 Ω = −248 Ω + 350 Ω = 102 Ω Pav 140 W = = 0.882 A R 180 Ω (b) Pav = Vrms I rms cos φ = I rms R (Exercise 31.22) I rms = (c) Z = R + ( X L − X C ) = (180 Ω) + (102 Ω − 350 Ω) = 306 Ω Vrms = I rms Z = (0.882 A)(306 Ω) = 270 V EVALUATE: We could also use Eq (31.31): Pav = Vrms I rms cos φ Vrms = 31.56 Pav 140 W = = 270 V, which agrees The source voltage lags the current I rms cos φ (0.882 A)cos(−54.0°) when X C > X L , and this agrees with what we found IDENTIFY: At any instant of time the same rules apply to the parallel ac circuit as to the 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 90° and for a capacitor the voltage lags the current by 90° EXECUTE: (a) The parallel L-R-C circuit must have equal potential drops over the capacitor, inductor and resistor, so vR = vL = vC = 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 = iR + iL + iC v v is always in phase with the voltage iL = lags the voltage by 90°, and iC = vωC leads the R ωL voltage by 90° The phase diagram is sketched in Figure 31.56 (b) iR = 2 V ⎞ ⎛V ⎞ ⎛ (c) From the diagram, I = I R2 + ( I C − I L ) = ⎜ ⎟ + ⎜ V ωC − ω L ⎟⎠ ⎝R⎠ ⎝ (d) From part (c): I = V R2 ⎞ V ⎛ + ⎜ ωC − ⎟ But I = , so = Z Z ωL ⎠ ⎝ EVALUATE: For large ω , Z → ⎞ ⎛ + ωC − ⎜ ω L ⎟⎠ R ⎝ The current in the capacitor branch is much larger than the current ωC in the other branches For small ω , Z → ω L The current in the inductive branch is much larger than the current in the other branches Figure 31.56 31.57 IDENTIFY: Apply the expression for I from Problem 31.56 when ω0 = 1/ LC SET UP: From Problem 31.56, I = V EXECUTE: (a) At resonance, ω0 = R 2 ⎞ ⎛ + ⎜ ωC − ⎟ L⎠ ω ⎝ 1 V ⇒ ω0C = ⇒ I C = V ω0C = = I L so ω0 L ω0 L LC I = I R and I is a minimum © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-18 Chapter 31 Vrms V2 at resonance where R < Z so power is a maximum cos φ = Z R (c) At ω = ω0 , I and V are in phase, so the phase angle is zero, which is the same as a series resonance (b) Pav = EVALUATE: (d) The parallel circuit is sketched in Figure 31.57 At resonance, iC = iL 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 iC + iL = and the statement in part (d) isn’t valid Figure 31.57 31.58 IDENTIFY: Refer to the results and the phasor diagram in Problem 31.56 The source voltage is applied across each parallel branch SET UP: V = 2Vrms = 311 V EXECUTE: (a) I R = V 311 V = = 0.778 A R 400 Ω (b) I C = V ωC = (311 V)(360 rad/s)(6.00 × 10−6 F) = 0.672 A ⎛I ⎞ ⎛ 0.672 A ⎞ (c) φ = arctan ⎜ C ⎟ = arctan ⎜ ⎟ = 40.8° ⎝ 0.778 A ⎠ ⎝ IR ⎠ (d) I = I R2 + I C2 = (0.778 A) + (0.672 A) = 1.03 A 31.59 (e) Leads since φ > EVALUATE: The phasor diagram shows that the current in the capacitor always leads the source voltage IDENTIFY and SET UP: Refer to the results and the phasor diagram in Problem 31.56 The source voltage is applied across each parallel branch V V EXECUTE: (a) I R = ; I C = V ωC ; I L = R ωL (b) The graph of each current versus ω is given in Figure 31.59a (c) ω → : I C → 0; I L → ∞ ω → ∞: I C → ∞; I L → 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 1 (d) ω = = = 1000 rad/sec and f = 159 Hz The phasor diagram is sketched LC (2.0 H)(0.50 × 10−6 F) in Figure 31.59b 2 V ⎞ ⎛V ⎞ ⎛ (e) I = ⎜ ⎟ + ⎜ V ωC − ⎟ R L⎠ ω ⎝ ⎠ ⎝ 2 ⎞ ⎛ 100 V ⎞ ⎛ 100 V −1 −6 I= ⎜ ⎟ = 0.50 A ⎟ + ⎜⎜ (100 V)(1000 s )(0.50 × 10 F) − − (1000 s )(2.0 H) ⎟⎠ ⎝ 200 Ω ⎠ ⎝ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current (f) At resonance I L = I C = V ωC = (100 V)(1000 s −1 )(0.50 × 10−6 F) = 0.0500 A and I R = 31-19 V 100 V = = 0.50 A R 200 Ω EVALUATE: At resonance iC = iL = at all times and the current through the source equals the current through the resistor Figure 31.59 31.60 IDENTIFY: The circuit is in resonance, and we know R, L, C and V We want the resonance angular frequency, the current amplitude through the source and resistor and the maximum energy stored in the inductor and capacitor V SET UP: ω0 = and at resonance, Z = R I = VR = VC = VL = V VR = I R R, VC = I C X C , Z LC VL = I L X L The maximum energy stored in the inductor is U L = 12 LI L The maximum energy stored in the capacitor is U C = 12 CVC EXECUTE: (a) ω0 = (0.300 H)(0.100 × 10 −6 F) = 5.77 × 103 rad/s V V 240 V = = = 2.40 A Z R 100 Ω V (c) I R = = 2.40 A R (b) I = (d) X L = ω L = (5.77 × 103 rad/s)(0.300 H) = 1.73 × 103 Ω IL = V 240 V = = 0.139 A X L 1.73 × 103 Ω (e) X C = ωC = −6 (5.77 × 10 rad/s)(0.100 × 10 F) = 1.73 × 103 Ω I C = I L = 0.139 A (f) U L = 12 LI L = 12 (0.300 H)(0.139 A) = 2.90 × 10−3 J U C = 12 CVC = 12 (0.100 × 10−6 F)(240 V) = 2.90 × 10−3 31.61 = 2.90 mJ J = 2.90 mJ EVALUATE: The maximum energy stored in the inductor and capacitor is the same, but not at the same time 1 IDENTIFY: The resonance angular frequency is ω0 = and the resonance frequency is f = LC 2π LC SET UP: ω0 is independent of R EXECUTE: (a) ω0 (or f ) depends only on L and C so change these quantities (b) To double ω0 , decrease L and C by multiplying each of them by © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-20 31.62 Chapter 31 EVALUATE: Increasing L and C decreases the resonance frequency; decreasing L and C increases the resonance frequency IDENTIFY: The average power depends on the phase angle φ ⎞ ⎛ SET UP: The average power is Pav = Vrms I rms cos φ , and the impedance is Z = R + ⎜ ω L − ⎟ ω C⎠ ⎝ EXECUTE: (a) Pav = Vrms I rms cos φ = 12 (Vrms I rms ), which gives cos φ = 12 , so φ = π /3 = 60° tan φ = ( X L − X C )/R, which gives tan 60° = (ω L − 1/ωC )/R Using R = 75.0 Ω, L = 5.00 mH and C = 2.50 µF and solving for ω we get ω = 28760 rad/s = 28,800 rad/s (b) Z = R + ( X L − X C ) , where X L = ω L = (28,760 rad/s)(5.00 mH) = 144 Ω and X C = 1/ωC = 1/[(28,760 rad/s)(2.50 µF)] = 13.9 Ω, giving Z = (75 Ω)2 + (144 Ω − 13.9 Ω)2 = 150 Ω; I = V/Z = (15.0 V)/(150 Ω) = 0.100 A and Pav = 12 VI cos φ = 12 (15.0 V)(0.100 A)(1/2) = 0.375 W 31.63 EVALUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero IDENTIFY and SET UP: Eq (31.19) allows us to calculate I and then Eq (31.22) gives Z Solve Eq (31.21) for X L EXECUTE: (a) VC = IX C so I = (b) V = IZ so Z = VC 360 V = = 0.750 A X C 480 Ω V 120 V = = 160 Ω I 0.750 A (c) Z = R + ( X L − X C ) X L − X C = ± Z − R , so X L = X C ± Z − R = 480 Ω ± (160 Ω) − (80.0 Ω) = 480 Ω ± 139 Ω X L = 619 Ω or 341 Ω (d) EVALUATE: XC = ωC and X L = ω L At resonance, X C = X L As the frequency is lowered below the resonance frequency X C increases and X L decreases Therefore, for ω < ω0 , X L < X C So for X L = 341 Ω the angular frequency is less than the resonance angular frequency ω is greater than ω0 when X L = 619 Ω But at these two values of X L , the magnitude of X L − X C is the same so Z and I are the same In one case ( X L = 691 Ω) the source voltage leads the current and in the other ( X L = 341 Ω) the 31.64 source voltage lags the current IDENTIFY and SET UP: Calculate Z and I = V/Z EXECUTE: (a) For ω = 800 rad/s: Z = R + (ω L − 1/ωC ) = (500 Ω)2 + ((800 rad/s)(2.0 H) − 1/((800 rad/s)(5.0 × 10−7 F))) Z = 1030 Ω I= V 100 V = = 0.0971 A VR = IR = (0.0971 A)(500 Ω) = 48.6 V, Z 1030 Ω VC = 0.0971 A = = 243 V and VL = I ω L = (0.0971 A)(800 rad/s)(2.00 H) = 155 V ωC (800 rad/s)(5.0 × 10−7 F) ⎛ ω L − 1/(ωC ) ⎞ ⎟ = −60.9° The graph of each voltage versus time is given in Figure 31.64a R ⎝ ⎠ (b) Repeating exactly the same calculations as above for ω = 1000 rad/s: φ = arctan ⎜ Z = R = 500 Ω; φ = 0; I = 0.200 A; VR = V = 100 V; VC = VL = 400 V The graph of each voltage versus time is given in Figure 31.64b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current 31-21 (c) Repeating exactly the same calculations as part (a) for ω = 1250 rad/s: Z = 1030 Ω; φ = +60.9°; I = 0.0971 A; VR = 48.6 V; VC = 155 V; VL = 243 V The graph of each voltage versus time is given in Figure 31.64c EVALUATE: The resonance frequency is ω0 = 1 = = 1000 rad/s For ω < ω0 the LC (2.00 H)(0.500 μ F) phase angle is negative and for ω > ω0 the phase angle is positive Figure 31.64 31.65 IDENTIFY and SET UP: Consider the cycle of the repeating current that lies between t2 t2 1 2I t1 = τ /2 and t2 = 3τ /2 In this interval i = (t − τ ) I av = i dt and I rms = i dt ∫ ∫ τ t2 − t1 t1 t2 − t1 t1 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-22 Chapter 31 EXECUTE: I av = 3τ /2 t2 1 3τ / 2 I 2I ⎡ ⎤ i dt = ∫ (t − τ ) dt = 20 ⎢ t − τ t ⎥ ∫ τ t /2 τ τ t2 − t1 τ ⎣2 ⎦ τ /2 3τ τ τ ⎞ I ⎛ I ⎞ ⎛ 9τ − − + ⎟ = (2 I ) 18 (9 − 12 − + 4) = (13 − 13) = I av = ⎜ 20 ⎟ ⎜ ⎜ ⎟⎠ ⎝ τ ⎠⎝ = ( I )av = I rms = I rms I 02 τ3 t2 1 3τ / I 02 i dt = ∫ (t − τ ) dt ∫ τ τ /2 τ t2 − t1 t1 3τ /2 ∫τ /2 (t − τ ) dt = 3τ /2 I 02 ⎡ I ⎡⎛ τ ⎞ ⎛ τ ⎞ = 03 ⎢⎜ ⎟ − ⎜ − ⎟ (t − τ ) ⎤ ⎣3 ⎦ τ /2 τ 3τ ⎢⎣⎝ ⎠ ⎝ ⎠ 3⎤ ⎥ ⎥⎦ I 02 [1 + 1] = 13 I 02 I I rms = I rms = EVALUATE: 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 rms and the current amplitude for I rms = 31.66 this current is different from that for a sinusoidal current (Eq 31.4) IDENTIFY: Apply Vrms = I rms Z and Z = R + ( X L − X C ) LC 1 EXECUTE: (a) ω0 = = = 786 rad/s LC (1.80 H)(9.00 × 10−7 F) SET UP: ω0 = (b) Z = R + (ω L − 1/ωC ) Z = (300 Ω) + ((786 rad/s)(1.80 H) − 1/((786 rad/s)(9.00 × 10−7 F))) = 300 Ω I rms-0 = Vrms 60 V = = 0.200 A Z 300 Ω (c) We want I = ω L2 + ω 2C − V I rms-0 = rms = Z Vrms R + (ω L − 1/ωC ) R + (ω L − 1/ωC ) = 4Vrms I rms-0 ⎞ ⎛ 2L 4V 2 L 4Vrms + R − rms = and (ω ) L2 + ω ⎜ R − − ⎟⎟ + = ⎜ C C I rms-0 ⎠ C I rms-0 ⎝ Substituting in the values for this problem, the equation becomes (ω ) (3.24) + ω ( −4.27 × 106 ) + 1.23 × 1012 = Solving this quadratic equation in ω we find ω = 8.90 × 105 rad /s or 4.28 × 105 rad /s and ω = 943 rad/s or 654 rad/s (d) (i) R = 300 Ω, I rms-0 = 0.200 A, ω1 − ω2 = 289 rad/s (ii) R = 30 Ω, I rms-0 = 2A, ω1 − ω2 = 28 rad/s (iii) R = Ω, I rms-0 = 20 A, ω1 − ω2 = 2.88 rad/s EVALUATE: The width gets smaller as R gets smaller; I rms − gets larger as R gets smaller 31.67 IDENTIFY: The resonance frequency, the reactances, and the impedance all depend on the values of the circuit elements SET UP: The resonance frequency is ω0 = 1/ LC , the reactances are X L = ω L and X C = 1/ωC , and the impedance is Z = R + ( X L − X C ) EXECUTE: (a) ω0 = 1/ LC becomes (b) Since X L = ω L, if L is doubled, X L → 1/2, so ω0 decreases by L 2C increases by a factor of 2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current (c) Since X C = 1/ωC , doubling C decreases X C by a factor of 31-23 (d) Z = R + ( X L − X C ) → Z = (2 R ) + (2 X L − 12 X C ) , so Z does not change by a simple factor of or 31.68 EVALUATE: The impedance does not change by a simple factor, even though the other quantities IDENTIFY: At resonance, Z = R I = V/R VR = IR, VC = IX C and VL = IX L U C = 12 CVC2 and U L = 12 LI SET UP: The amplitudes of each time-dependent quantity correspond to the maximum values of those quantities V V V EXECUTE: (a) I = = and I max = At resonance ω L = 2 C ω R Z R + (ω L − 1/ωC ) (b) VC = IX C = (c) VL = IX L = V Rω0C = V L R C V V L ω0 L = R R C 1 V2 L V2 (d) U C = CVC2 = C = L 2 R C R V2 LI = L 2 R EVALUATE: At resonance VC = VL and the maximum energy stored in the inductor equals the maximum (e) U L = 31.69 energy stored in the capacitor IDENTIFY: I = V/R VR = IR, VC = IX C and VL = IX L U C = 12 CVC2 and U L = 12 LI SET UP: The amplitudes of each time-dependent quantity correspond to the maximum values of those quantities EXECUTE: ω = (a) I = V = Z ω0 V ⎛ω L ⎞ R + ⎜ − 2/ω0C ⎟ ⎝ ⎠ (b) VC = IX C = (c) VL = IX L = ω0C ω0 L V R2 + 9L 4C V 9L R2 + 4C = = = V R2 + L C L C 9L 4C 2V R2 + 9L 4C V/2 L R + 4C LV (d) U C = CVC2 = 9L R2 + 4C LV LI = 2 R2 + L 4C EVALUATE: For ω < ω0 , VC > VL and the maximum energy stored in the capacitor is greater than the (e) U L = maximum energy stored in the inductor © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-24 31.70 Chapter 31 IDENTIFY: I = V/R VR = IR, VC = IX C and VL = IX L U C = 12 CVC2 and U L = 12 LI SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities EXECUTE: ω = 2ω0 (a) I = V = Z V R + (2ω0 L − 1/2ω0C ) (b) VC = IX C = 2ω0C V 9L R2 + 4C V (c) VL = IX L = 2ω0 L (d) U C = CVC2 = (e) U L = R2 + 9L 4C 9L R + 4C L C = = V = L C V/2 L R + 4C 2V R2 + 9L 4C LV 9L R2 + 4C LV 2 LI = 9L R + 4C EVALUATE: For ω > ω0 , VL > VC and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor 31.71 IDENTIFY: A transformer transforms voltages according to secondary circuit of resistance R is Reff = R ( N /N1) V2 N = The effective resistance of a V1 N1 SET UP: N = 275 and V1 = 25.0 V EXECUTE: (a) V2 = V1 ( N /N1 ) = (25.0 V)(834/275) = 75.8 V (b) Reff = R 125 Ω = 13.6 Ω ( N /N1) (834/275) EVALUATE: The voltage across the secondary is greater than the voltage across the primary since N > N1 The effective load resistance of the secondary is less than the resistance R connected across the = secondary 31.72 Vrms Calculate Z R = Z cos φ Z f = 50.0 Hz and ω = 2π f The power factor is cosφ IDENTIFY: Pav = Vrms I rms cos φ and I rms = SET UP: EXECUTE: (a) Pav = V cos φ (120 V) (0.560) Vrms cos φ Z = rms = = 36.7 Ω Pav (220 W) Z R = Z cos φ = (36.7 Ω)(0.560) = 20.6 Ω (b) Z = R + X L2 ⋅ X L = Z − R = (36.7 Ω) − (20.6 Ω) = 30.4 Ω But φ = is at resonance, so the inductive and capacitive reactances equal each other Therefore we need to add X C = 30.4 Ω X C = therefore gives C = ω XC = ωC 1 = = 1.05 × 10−4 F 2π fX C 2π (50.0 Hz)(30.4 Ω) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Alternating Current (c) At resonance, Pav = 31-25 V (120 V) = = 699 W R 20.6 Ω EVALUATE: Pav = I rms R and I rms is maximum at resonance, so the power drawn from the line is maximum at resonance 31.73 pR = i R pL = iL IDENTIFY: SET UP: i = I cos ωt di q pC = i C dt EXECUTE: (a) pR = i R = I cos (ωt ) R = VR I cos (ωt ) = VR I (1 + cos(2ωt )) T VR I T VR I T Pav ( R ) = ∫ pR dt = (1 + cos(2ωt )) dt = [t ]0 = VR I T 2T ∫ 2T T di (b) pL = Li = −ω LI cos(ωt )sin(ωt ) = − 12 VL I sin(2ωt ) But ∫ sin(2ωt )dt = ⇒ Pav (L) = 0 dt T q (c) pC = i = vC i = VC I sin(ωt )cos(ωt ) = 12 VC I sin(2ωt ) But ∫ sin(2ωt )dt = ⇒ Pav (C ) = 0 C (d) p = pR + pL + pc = VR I cos (ωt ) − 12 VL I sin(2ωt ) + 12 VC I sin(2ωt ) and VR V −V and sin φ = L C , so V V p = VI cos(ωt )(cos φ cos(ωt ) − sin φ sin(ωt )), at any instant of time p = I cos(ωt )(VR cos(ωt ) − VL sin(ωt ) + VC sin(ωt )) But cos φ = 31.74 EVALUATE: 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 dVL dVC IDENTIFY: VL = IX L = at the ω where VL is a maximum VC = IX C = at the ω where dω dω VC is a maximum SET UP: Problem 31.53 shows that I = V R + (ω L − 1/ωC ) LC dVL dVL d ⎛⎜ VωL =0= (b) VL = maximum when = Therefore: dω dω dω ⎜ R + (ω L − 1/ωC ) ⎝ EXECUTE: (a) VR = maximum when VC = VL ⇒ ω = ω0 = 0= R2 + VL R + (ω L − 1/ωC ) 2 ω C − − ( R + (ω L − 1/ωC )2 )3/ R + (ω L − 1/ωC ) = ω ( L2 − 1/ω 4C ) 2L 1 R 2C = − 2 = LC − and ω = C ω C ω (c) VC = maximum when 0=− V ω L( L − 1/ω 2C )( L + 1/ω 2C ) ⎞ ⎟ ⎟ ⎠ LC − R 2C /2 dVC d ⎛⎜ V dVC = Therefore: =0= dω dω ⎜ ωC R + (ω L − 1/ωC ) dω ⎝ V ω 2C R + (ω L − 1/ωC ) − V ( L − 1/ω 2C )( L + 1/ω 2C ) C ( R + (ω L − 1/ωC ) )3/ R + ω L2 − 2L = −ω L2 and ω = C R + ω L2 − 2L = −ω L2 C ⎞ ⎟ ⎟ ⎠ R + (ω L − 1/ωC ) = −ω ( L2 − 1/ω 4C ) R2 − LC L © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 31-26 31.75 Chapter 31 EVALUATE: VL is maximum at a frequency greater than the resonance frequency and VC is a maximum at a frequency less than the resonance frequency These frequencies depend on R, as well as on L and on C IDENTIFY: Follow the steps specified in the problem SET UP: In part (a) use Eq (31.23) to calculate Z and then I = V/Z φ is given by Eq (31.24) In part (b) let Z = R + iX EXECUTE: (a) From the current phasors we know that Z = R + (ω L − 1/ωC ) ⎛ ⎞ Z = (400 Ω) + ⎜⎜ (1000 rad/s)(0.50 H) − ⎟ = 500 Ω −6 (1000 rad/s)(1.25 × 10 F) ⎟⎠ ⎝ I= V 200 V = = 0.400 A Z 500 Ω ⎛ (1000 rad/s)(0.500 H) − 1/(1000 rad/s)(1.25 × 10−6 F) ⎞ ⎛ ω L − 1/(ωC ) ⎞ (b) φ = arctan ⎜ ⎟⎟ = +36.9° ⎟ φ = arctan ⎜⎜ R 400 Ω ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎞ ⎛ (c) Z cpx = R + i ⎜ ω L − ⎟⎟ = ⎟ Z cpx = 400 Ω − i ⎜⎜ (1000 rad/s)(0.50 H) − −6 ω C (1000 rad/s)(1.25 10 F) × ⎝ ⎠ ⎝ ⎠ 400 Ω − 300 Ωi Z = (400 Ω) + (−300 Ω) = 500 Ω (d) I cpx = V 200 V ⎛ + 6i ⎞ ⎛ + 6i ⎞ ⎛ − 6i ⎞ = =⎜ ⎟ A = (0.320 A) + (0.240 A)i I = ⎜ ⎟⎜ ⎟ A = 0.400 A Z cpx (400 − 300i ) Ω ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ (e) tan φ = Im( I cpx ) Re( I cpx ) = 6/25 = 0.75 ⇒ φ = +36.9° 8/25 ⎛ + 6i ⎞ (f) VRcpx = I cpx R = ⎜ ⎟ (400 Ω) = (128 + 96i )V ⎝ 25 ⎠ ⎛ + 6i ⎞ VLcpx = iI cpxω L = i ⎜ ⎟ (1000 rad/s)(0.500 H) = (−120 + 160i )V ⎝ 25 ⎠ I cpx ⎛ + 6i ⎞ = i⎜ = (+192 − 256i )V VCcpx = i ⎟ ωC ⎝ 25 ⎠ (1000 rad/s)(1.25 × 10−6 F) (g) Vcpx = VRcpx + VLcpx + VCcpx = (128 + 96i) V + ( −120 + 160i)V + (192 − 256i) V = 200 V EVALUATE: Both approaches yield the same value for I and for φ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... Z = (200 Ω) + (100 Ω)2 = 224 Ω (b) I = V 30.0 V = = 0 .134 A Z 224 Ω (c) VR = IR = (0 .134 A)(200 Ω) = 26.8 V VL = IX L = (0 .134 A)(100 Ω) = 13. 4 V X L 100 Ω = and φ = +26.6° Since φ is positive,... 90°) = +3 .13 V (3162 rad/s)(12.5 μ F) vL = VL cos(ωt + 90°) = IX L cos(ωt + 90°) = (0.143 A)(3162 rad/s)(8.00 mH)cos(60° + 90°) vL = −3 .13 V 31.29 (d) vR + vL + vC = 12.5 V + (−3 .13 V) + 3 .13 V =... −22.70 V vR + vL + vC = 3.51 V + 13. 35 V + (−39.55 V) = −22.7 V vR + vL + vC is equal to v (b) VR = IR = 12.4 V VC = 41.2 V VL = 13. 9 V VR + VC + VL = 12.4 V + 41.2 V + 13. 9 V = 67.5 V VR + VC + VL