M30 YOUN7066 13 ISM c30 tủ tài liệu training

30 INDUCTANCE 30.1 IDENTIFY and SET UP: Apply Eq (30.4) di EXECUTE: (a) ε = M = (3.25 × 10−4 H)(830 A/s) = 0.270 V; yes, it is constant dt di2 ; M is a property of the pair of coils so is the same as in part (a) Thus ε1 = 0.270 V dt EVALUATE: The induced emf is the same in either case A constant di/dt produces a constant emf (b) 30.2 ε1 =M IDENTIFY: ε1 = M Δi2 and Δt ε2 = M N Φ Δi1 M = B , where ΦB is the flux through one turn of the Δt i1 second coil SET UP: M is the same whether we consider an emf induced in coil or in coil ε = 1.65 × 10−3 V = 6.82 × 10−3 H = 6.82 mH EXECUTE: (a) M = Δi1/Δt 0.242 A/s (b) ΦB = Δi2 = (6.82 × 10−3 H)(0.360 A/s) = 2.46 × 10−3 V = 2.46 mV Δt EVALUATE: We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil IDENTIFY: A coil is wound around a solenoid, so magnetic flux from the solenoid passes through the coil SET UP: Example 30.1 shows that the mutual inductance for this configuration of coils is μ NN A M = , where l is the length of coil l EXECUTE: Using the formula for M gives (4π × 10−7 Wb/m ⋅ A)(800)(50)π (0.200 × 10−2 m) = 6.32 × 10−6 H = 6.32 μ H M= 0.100 m EVALUATE: This result is a physically reasonable mutual inductance IDENTIFY: Changing flux from one object induces an emf in another object (a) SET UP: The magnetic field due to a solenoid is B = μ0 nI (c) 30.3 30.4 Mi1 (6.82 × 10−3 H)(1.20 A) = = 3.27 × 10−4 Wb N2 25 ε1 = M EXECUTE: The above formula gives (4π × 10−7 T ⋅ m/A)(300)(0.120 A) = 1.81 × 10−4 T 0.250 m The average flux through each turn of the inner solenoid is therefore B1 = ΦB = B1 A = (1.81 × 10−4 T)π (0.0100 m) = 5.68 × 10−8 Wb © 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 30-1 30-2 Chapter 30 (b) SET UP: The flux is the same through each turn of both solenoids due to the geometry, so M= N 2ΦB ,2 i1 = N 2ΦB,1 i1 −8 (25)(5.68 × 10 Wb) = 1.18 × 10−5 H 0.120 A di (c) SET UP: The induced emf is ε = − M dt EXECUTE: M = EXECUTE: 30.5 ε = −(1.18 × 10−5 H)(1750 A/s) = −0.0207 V EVALUATE: A mutual inductance around 10−5 H is not unreasonable IDENTIFY and SET UP: Apply Eq (30.5) N Φ 400(0.0320 Wb) EXECUTE: (a) M = B = = 1.96 H i1 6.52 A N1ΦB1 Mi (1.96 H)(2.54 A) so ΦB1 = = = 7.11 × 10−3 Wb i2 N1 700 EVALUATE: M relates the current in one coil to the flux through the other coil Eq (30.5) shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux IDENTIFY: One toroidal solenoid is wound around another, so the flux of one of them passes through the other N Φ μ Ni SET UP: B1 = 1 for a toroidal solenoid, M = B i1 2π r (b) M = 30.6 μ NiA μ0 N1i1 For each turn in the second solenoid the flux is ΦB = B1 A = 1 2π r 2π r N Φ μ NN A Therefore M = B = i1 2π r EXECUTE: (a) B1 = (b) M = 30.7 μ0 N1N A (500)(300)(0.800 × 10−4 m ) = (2 × 10−7 T ⋅ m/A) = 2.40 × 10−5 H = 24.0 μ H 2π r 0.100 m EVALUATE: This result is a physically reasonable mutual inductance IDENTIFY: We can relate the known self-inductance of the toroidal solenoid to its geometry to calculate the number of coils it has Knowing the induced emf, we can find the rate of change of the current μ N2A SET UP: Example 30.3 shows that the self-inductance of a toroidal solenoid is L = The voltage 2π r di across the coil is related to the rate at which the current in it is changing by ε = L dt EXECUTE: (a) Solving L = N= μ0 N A for N gives 2π r 2π rL 2π (0.0600 m)(2.50 × 10−3 H) = = 1940 turns μ0 A (4π × 10−7 T ⋅ m/A)(2.00 × 10−4 m ) di ε 2.00 V = = = 800 A/s dt L 2.50 × 10−3 H EVALUATE: The inductance is determined solely by how the coil is constructed The induced emf depends on the rate at which the current through the coil is changing IDENTIFY: A changing current in an inductor induces an emf in it μ N2A (a) SET UP: The self-inductance of a toroidal solenoid is L = 2π r (b) 30.8 EXECUTE: L = (4π × 10−7 T ⋅ m/A)(500) (6.25 × 10−4 m ) = 7.81 ì 104 H (0.0400 m) â 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 Inductance (b) SET UP: The magnitude of the induced emf is 30.9 30-3 ε = L di dt 00 A − 00 A ⎛ ⎞ EXECUTE: ε = (7.81 × 10−4 H) ⎜ ⎟ = 0.781 V ⎝ 3.00 × 10−3 s ⎠ (c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to b, making b at a higher potential than a EVALUATE: This is a reasonable value for self-inductance, in the range of a mH N ΦB Δi IDENTIFY: ε = L and L = i Δt SET UP: Δi = 0.0640 A/s Δt EXECUTE: (a) L = ε Δi/Δt = 0.0160 V = 0.250 H 0.0640 A/s Li (0.250 H)(0.720 A) = = 4.50 × 10−4 Wb 400 N EVALUATE: The self-induced emf depends on the rate of change of flux and therefore on the rate of change of the current, not on the value of the current IDENTIFY: Combine the two expressions for L: L = N ΦB /i and L = ε / di/dt (b) The average flux through each turn is ΦB = 30.10 SET UP: Φ B is the average flux through one turn of the solenoid (12.6 × 10−3 V)(1.40 A) = 238 turns (0.00285 Wb)(0.0260 A/s) EVALUATE: The induced emf depends on the time rate of change of the total flux through the solenoid IDENTIFY and SET UP: Apply ε = L di/dt Apply Lenz’s law to determine the direction of the induced EXECUTE: Solving for N we have N = ε i/ΦB di/dt = 30.11 emf in the coil EXECUTE: (a) 30.12 30.13 ε = L di/dt = (0.260 H)(0.0180 A/s) = 4.68 × 10−3 V (b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at a EVALUATE: The induced emf is directed so as to oppose the decrease in the current di IDENTIFY: Apply ε = − L dt SET UP: The induced emf points from low potential to high potential across the inductor EXECUTE: (a) The induced emf points from b to a, in the direction of the current Therefore, the current is decreasing and the induced emf is directed to oppose this decrease (b) ε = L di /dt , so di /dt = Vab /L = (1.04 V)/(0.260 H) = 4.00 A/s In 2.00 s the decrease in i is 8.00 A and the current at 2.00 s is 12.0 A −8.0 A = 4.0 A EVALUATE: When the current is decreasing the end of the inductor where the current enters is at the lower potential This agrees with our result and with Figure 30.6d in the textbook IDENTIFY: The inductance depends only on the geometry of the object, and the resistance of the wire depends on its length μ N2A SET UP: L = 2π r EXECUTE: (a) N = 2π rL (0.120 m)(0.100 × 10−3 H) = = 1.00 × 103 turns μ0 A (2 × 10−7 T ⋅ m/A)(0.600 × 10−4 m ) (b) A = π d /4 and c = π d , so c = 4π A = 4π (0.600 × 10−4 m ) = 0.02746 m The total length of the wire is (1000)(0.02746 m) = 27.46 m Therefore R = (0.0760 Ω/m)(27.46 m) = 2.09 Ω EVALUATE: A resistance of Ω is large enough to be significant in a circuit © 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 30-4 Chapter 30 30.14 IDENTIFY: The changing current induces an emf in the solenoid N ΦB SET UP: By definition of self-inductance, L = The magnitude of the induced emf is i EXECUTE: L = 30.15 =L di dt N ΦB (800)(3.25 × 10−3 Wb) = = 0.8966 H i 2.90 A di ε 7.50 × 10−3 V = = = 8.37 × 10−3 A/s = 8.37 mA/s dt L 0.8966 H EVALUATE: An inductance of nearly a henry is rather large For ordinary laboratory inductors, which are around a few millihenries, the current would have to be changing much faster to induce 7.5 mV IDENTIFY: Use the definition of inductance and the geometry of a solenoid to derive its self-inductace N ΦB N SET UP: The magnetic field inside a solenoid is B = μ0 i, and the definition of self-inductance is L = i l N ΦB μ NAi N EXECUTE: (a) B = μ0 i, L = , and ΦB = Combining these expressions gives l i l L= N ΦB μ0 N A = i l (b) L = μ0 N A l A = π r = π (0.0750 × 10−2 m)2 = 1.767 × 10−6 m (4π × 10−7 T ⋅ m/A)(50) (1.767 × 10−6 m ) = 1.11 × 10−7 H = 0.111 μ H 5.00 × 10−2 m EVALUATE: This is a physically reasonable value for self-inductance IDENTIFY and SET UP: The stored energy is U = 12 LI The rate at which thermal energy is developed is L= 30.16 ε P = I R EXECUTE: (a) U = 12 LI = 12 (12.0 H)(0.300 A) = 0.540 J (b) P = I R = (0.300 A) (180 Ω) = 16.2 W = 16.2 J/s 30.17 EVALUATE: (c) No If I is constant then the stored energy U is constant The energy being consumed by the resistance of the inductor comes from the emf source that maintains the current; it does not come from the energy stored in the inductor IDENTIFY and SET UP: Use Eq (30.9) to relate the energy stored to the inductance Example 30.3 gives μ N2A , so once we know L we can solve for N the inductance of a toroidal solenoid to be L = 2π r 2U 2(0.390 J) EXECUTE: U = 12 LI so L = = = 5.417 × 10−3 H I (12.0 A) N= 30.18 2π rL 2π (0.150 m)(5.417 × 10−3 H) = = 2850 μ0 A (4π × 10−7 T ⋅ m/A)(5.00 × 10−4 m ) EVALUATE: L and hence U increase according to the square of N IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy μ NI (a) SET UP: The magnetic field inside a toroidal solenoid is B = 2π r μ0 (300)(5.00 A) −3 = 2.50 × 10 T = 2.50 mT EXECUTE: B = 2π (0.120 m) (b) SET UP: The self-inductance of a toroidal solenoid is L = EXECUTE: L = μ0 N A 2π r (4π × 10−7 T ⋅ m/A)(300)2 (4.00 × 10−4 m ) = 6.00 × 10−5 H 2π (0.120 m) © 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 Inductance 30-5 (c) SET UP: The energy stored in an inductor is U L = 12 LI EXECUTE: U L = 12 (6.00 × 10−5 H)(5.00 A) = 7.50 × 10−4 J (d) SET UP: The energy density in a magnetic field is u = EXECUTE: u = (2.50 × 10−3 T) 2(4π × 10−7 T ⋅ m/A) B2 μ0 = 2.49 J/m3 energy energy 7.50 × 10−4 J = = = 2.49 J/m3 volume 2π rA 2π (0.120 m)(4.00 × 10−4 m ) EVALUATE: An inductor stores its energy in the magnetic field inside of it IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy (a) SET UP: The magnetic field inside a solenoid is B = μ0nI (e) u = 30.19 EXECUTE: B = (4π × 10−7 T ⋅ m/A)(400)(80.0 A) = 0.161 T 0.250 m (b) SET UP: The energy density in a magnetic field is u = B2 μ0 (0.161T) = 1.03 × 104 J/m3 2(4π × 10−7 T ⋅ m/A) (c) SET UP: The total stored energy is U = uV EXECUTE: u = EXECUTE: U = uV = u (lA) = (1.03 × 104 J/m3 )(0.250 m)(0.500 × 10−4 m ) = 0.129 J (d) SET UP: The energy stored in an inductor is U = 12 LI 30.20 EXECUTE: Solving for L and putting in the numbers gives 2U 2(0.129 J) = 4.02 × 10−5 H L= = (80.0 A) I EVALUATE: An inductor stores its energy in the magnetic field inside of it IDENTIFY: Energy = Pt U = 12 LI SET UP: P = 200 W = 200 J/s EXECUTE: (a) Energy = (200 W)(24 h)(3600 s/h) = 1.73 × 107 J 30.21 2U 2(1.73 × 107 J) = 5.41 × 103 H (80.0 A) EVALUATE: A large value of L and a large current would be required, just for one light bulb Also, the resistance of the inductor would have to be very small, to avoid a large P = I R rate of electrical energy loss IDENTIFY: The energy density depends on the strength of the magnetic field, and the energy depends on the volume in which the magnetic field exists B2 SET UP: The energy density is u = μ0 (b) L = I = EXECUTE: First find the energy density: u = B2 (4.80 T) = = 9.167 × 106 J/m3 The energy μ0 2(4π × 10−7 T ⋅ m/A) U in a volume V is U = uV = (9.167 × 106 J/m3 )(10.0 × 10−6 m3 ) = 91.7 J 30.22 EVALUATE: A field of 4.8 T is very strong, so this is a high energy density for a magnetic field IDENTIFY and SET UP: The energy density (energy per unit volume) in a magnetic field (in vacuum) is U B2 (Eq 30.10) given by u = = V μ0 © 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 30-6 Chapter 30 EXECUTE: (a) V = (b) u = B= 30.23 μ0U B2 = 2(4π × 10−7 T ⋅ m/A)(3.60 × 106 J) (0.600 T) = 25.1 m3 U B2 = V μ0 2μ0U 2(4π × 10−7 T ⋅ m/A)(3.60 × 106 J) = = 11.9 T V (0.400 m)3 EVALUATE: Large-scale energy storage in a magnetic field is not practical The volume in part (a) is quite large and the field in part (b) would be very difficult to achieve IDENTIFY: Apply Kirchhoff’s loop rule to the circuit i(t) is given by Eq (30.14) SET UP: The circuit is sketched in Figure 30.23 di is positive as the current dt increases from its initial value of zero Figure 30.23 EXECUTE: ε − v R − vL = ε − iR − L di = so i = ε (1 − e−( R/L)t ) dt R di (a) Initially (t = 0), i = so ε − L = dt di ε 6.00 V = = = 2.40 A/s dt L 2.50 H di (b) ε − iR − L = (Use this equation rather than Eq (30.15) since i rather than t is given.) dt di ε − iR 6.00 V − (0.500 A)(8.00 Ω) Thus = = = 0.800 A/s dt L 2.50 H ε ⎛ 6.00 V ⎞ − (8.00 Ω /2.50 H)(0.250 s) (c) i = (1 − e− ( R /L )t ) = ⎜ ) = 0.750 A(1 − e−0.800 ) = 0.413 A ⎟ (1 − e Ω R 00 ⎝ ⎠ (d) Final steady state means t → ∞ and 6.00 V = 0.750 A 8.00 Ω EVALUATE: Our results agree with Figure 30.12 in the textbook The current is initially zero and increases to its final value of ε /R The slope of the current in the figure, which is di/dt, decreases with t IDENTIFY: With S1 closed and S open, the current builds up to a steady value Then with S1 open and i= 30.24 ε di → 0, so ε − iR = dt R = S closed, the current decreases exponentially SET UP: The decreasing current is i = I 0e −( R/L )t EXECUTE: (a) i = I 0e− ( R/L )t = L=− ε e−( R/L)t R e−( R/L )t = iR ε = Rt (0.320 A)(15.0 Ω) = − ln(0.7619) = 0.7619 L 6.30 V Rt (15.0 Ω)(2.00 × 10−3 s) =− = 0.110 H ln(0.7619) ln(0.7619) © 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 Inductance (b) 30-7 Rt i = − ln(0.0100) = e −( R/L )t e−( R/L )t = 0.0100 L I0 ln(0.0100) L ln(0.0100)(0.110 H) =− = 0.0338 s = 33.8 ms R 15.0 Ω EVALUATE: Typical LR circuits change rapidly compared to human time scales, so 33.8 ms is not unusual IDENTIFY: i = ε /R(1 − e−t/τ ), with τ = L/R The energy stored in the inductor is U = 12 Li t=− 30.25 SET UP: The maximum current occurs after a long time and is equal to EXECUTE: (a) imax = ε /R so i = imax /2 when (1 − e−t/τ ) = t= ε /R and e−t/τ = 12 −t/τ = ln ( 12 ) L ln (ln 2)(1.25 × 10−3 H) = = 17.3 μs R 50.0 Ω (b) U = 12 U max when i = imax / − e−t/τ = 1/ 2, so e−t/τ = − 1/ = 0.2929 t = − L ln(0.2929)/R = 30.7 μs 30.26 EVALUATE: τ = L/R = 2.50 × 10−5 s = 25.0 μ s The time in part (a) is 0.692τ and the time in part (b) is 1.23τ IDENTIFY: With S1 closed and S open, i (t ) is given by Eq (30.14) With S1 open and S closed, i (t ) is given by Eq (30.18) SET UP: U = 12 Li After S1 has been closed a long time, i has reached its final value of I = ε /R EXECUTE: (a) U = 12 LI and I = 2U 2(0.260 J) = = 2.13 A 0.115 H L (b) i = Ie−( R/L )t and U = 12 Li = 12 LI 2e−2( R/L )t = 12 U = 12 t=− 30.27 L ln 2R ( LI 2 ) ε = IR = (2.13 A)(120 Ω) = 256 V e−2( R/L )t = 12 , so 0.115 H ln ( 12 ) = 3.32 × 10−4 s ( 12 ) = − 2(120 Ω) EVALUATE: τ = L/R = 9.58 × 10−4 s The time in part (b) is τ ln (2)/2 = 0.347τ IDENTIFY: Apply the concepts of current decay in an R-L circuit Apply the loop rule to the circuit i (t ) is given by Eq (30.18) The voltage across the resistor depends on i and the voltage across the inductor depends on di/dt SET UP: The circuit with S1 closed and S open is sketched in Figure 30.27a ε − iR − L di = dt Figure 30.27a Constant current established means ε di = dt 60.0 V = = 0.250 A R 240 Ω (a) SET UP: The circuit with S closed and S1 open is shown in Figure 30.27b EXECUTE: i = i = I 0e−( R/L )t At t = 0, i = I = 0.250 A Figure 30.27b © 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 30-8 Chapter 30 The inductor prevents an instantaneous change in the current; the current in the inductor just after S is closed and S1 is opened equals the current in the inductor just before this is done (b) EXECUTE: i = I 0e− ( R/L )t = (0.250 A)e− (240 Ω/0.160 H)(4.00 × 10 −4 s) = (0.250 A)e−0.600 = 0.137 A (c) SET UP: See Figure 30.27c Figure 30.27c EXECUTE: If we trace around the loop in the direction of the current the potential falls as we travel through the resistor so it must rise as we pass through the inductor: vab > and vbc < So point c is at a higher potential than point b vab + vbc = and vbc = −vab Or, vcb = vab = iR = (0.137 A)(240 Ω) = 32.9 V (d) i = I 0e−( R/L )t i = 12 I says = I 0e−( R/L )t and 1I 2 = e−( R/L )t Taking natural logs of both sides of this equation gives ln ( 12 ) = − Rt/L ⎛ 0.160 H ⎞ −4 t =⎜ ⎟ ln = 4.62 × 10 s Ω 240 ⎝ ⎠ EVALUATE: The current decays, as shown in Figure 30.13 in the textbook The time constant is τ = L/R = 6.67 × 10−4 s The values of t in the problem are less than one time constant At any instant the 30.28 potential drop across the resistor (in the direction of the current) equals the potential rise across the inductor IDENTIFY: Apply Eq (30.14) di SET UP: vab = iR vbc = L The current is increasing, so di/dt is positive dt EXECUTE: (a) At t = 0, i = vab = and vbc = 60 V (b) As t → ∞, i → ε /R and di/dt → vab → 60 V and vbc → (c) When i = 0.150 A, vab = iR = 36.0 V and vbc = 60.0 V − 36.0 V = 24.0 V 30.29 EVALUATE: At all times, ε = vab + vbc , as required by the loop rule IDENTIFY: i (t ) is given by Eq (30.14) SET UP: The power input from the battery is ε i The rate of dissipation of energy in the resistance is i R The voltage across the inductor has magnitude Ldi/dt , so the rate at which energy is being stored in the inductor is iLdi/dt EXECUTE: (a) P = εi = ε I (1 − e − ( R/L )t ) = P = (4.50 W)(1 − e−(3.20 s (b) PR = i R = (c) PL = iL ε R −1 )t ε2 R (1 − e − ( R/L )t ) = (6.00 V)2 (1 − e− (8.00 Ω/2.50 H)t ) 8.00 Ω ) (1 − e − ( R/L )t ) = −1 (6.00 V) (1 − e− (8.00 Ω/2.50 H)t ) = (4.50 W)(1 − e− (3.20 s )t ) 8.00 Ω di ε ⎛ε ⎞ ε = (1 − e −( R /L )t ) L ⎜ e −( R /L )t ⎟ = (e − ( R /L )t − e−2( R /L )t ) dt R ⎝L ⎠ R PL = (4.50 W)(e −(3.20 s −1 )t − e −(6.40 s −1 )t ) © 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 Inductance 30.30 EVALUATE: (d) Note that if we expand the square in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor Conservation of energy requires that this be so IDENTIFY: With S1 closed and S open, the current builds up to a steady value SET UP: Applying Kirchhoff’s loop rule gives 30.31 ε − iR − L di = dt di EXECUTE: vR = ε − L = 18.0 V − (0.380 H)(7.20 A/s) = 15.3 V dt EVALUATE: The rest of the 18.0 V of the emf is across the inductor d 2q IDENTIFY: Evaluate and insert into Eq (30.20) dt SET UP: Eq (30.20) is d 2q dt + q = LC EXECUTE: q = Q cos(ωt + φ ) ⇒ d 2q dq d 2q = −ωQ sin(ωt + φ ) ⇒ = −ω 2Q cos(ωt + φ ) dt dt Q 1 q = −ω 2Q cos(ωt + φ ) + cos(ωt + φ ) = ⇒ ω = ⇒ω = LC LC LC LC dt EVALUATE: The value of φ depends on the initial conditions, the value of q at t = IDENTIFY: An L -C circuit oscillates, with the energy going back and forth between the inductor and capacitor 1 ω (a) SET UP: The frequency is f = and ω = , giving f = 2π LC 2π LC EXECUTE: f = = 2.13 × 103 Hz = 2.13 kHz −3 −6 2π (0.280 × 10 H)(20.0 × 10 F) 30.32 30-9 + (b) SET UP: The energy stored in a capacitor is U = 12 CV EXECUTE: U = 12 (20.0 × 10−6 F)(150.0 V)2 = 0.225 J (c) SET UP: The current in the circuit is i = −ωQ sin ωt , and the energy stored in the inductor is U = 12 Li EXECUTE: First find ω and Q ω = 2π f = 1.336 × 10 rad/s Q = CV = (20.0 × 10−6 F)(150.0 V) = 3.00 × 10−3 C Now calculate the current: i = −(1.336 × 10 rad/s)(3.00 × 10−3 C)sin[(1.336 × 104 rad/s)(1.30 × 10−3 s)] Notice that the argument of the sine is in radians, so convert it to degrees if necessary The result is i = 39.92 A Now find the energy in the inductor: U = 12 Li = 12 (0.280 × 10−3 H)(39.92 A)2 = 0.223 J 30.33 EVALUATE: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the capacitor IDENTIFY: The energy moves back and forth between the inductor and capacitor 1 2π (a) SET UP: The period is T = = = = 2π LC f ω /2π ω EXECUTE: Solving for L gives T2 (8.60 × 10−5 s) = 2.50 × 10−2 H = 25.0 mH 4π C 4π (7.50 × 10−9 C) (b) SET UP: The charge on a capacitor is Q = CV L= = EXECUTE: Q = CV = (7.50 × 10−9 F)(12.0 V) = 9.00 × 10 –8 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 30-10 Chapter 30 (c) SET UP: The stored energy is U = Q /2C (9.00 × 10−8 C)2 = 5.40 × 10−7 J 2(7.50 × 10−9 F) (d) SET UP: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial energy U L + U C = U Total 12 LI + = U Total EXECUTE: U = EXECUTE: Solve for the current: 2U Total 2(5.40 × 10−7 J) = = 6.58 × 10−3 A = 6.58 mA L 2.50 × 10−2 H EVALUATE: The energy oscillates back and forth forever However, if there is any resistance in the circuit, no matter how small, all this energy will eventually be dissipated as heat in the resistor IDENTIFY: The circuit is described in Figure 30.14 of the textbook SET UP: The energy stored in the inductor is U L = 12 Li and the energy stored in the capacitor is I= 30.34 U C = q /2C Initially, U C = 12 CV , with V = 22.5 V The period of oscillation is T = 2π LC = 2π (12.0 × 10−3 H)(18.0 × 10−6 F) = 2.92 ms EXECUTE: (a) Energy conservation says U L (max) = U C (max), and imax = V C/L = (22.5 V) 18 × 10−6 F 12 × 10−3 H Li 2 max = 12 CV = 0.871 A The charge on the capacitor is zero because all the energy is in the inductor (b) From Figure 30.14 in the textbook, q = at t = T/4 = 0.730 ms and at t = 3T/4 = 2.19 ms (c) q0 = CV = (18 μ F)(22.5 V) = 405 μ C is the maximum charge on the plates The graphs are sketched in Figure 30.34 q refers to the charge on one plate and the sign of i indicates the direction of the current EVALUATE: If the capacitor is fully charged at t = it is fully charged again at t = T/2, but with the opposite polarity Figure 30.34 30.35 IDENTIFY and SET UP: The angular frequency is given by Eq (30.22) q (t ) and i (t ) are given by Eqs (30.21) and (30.23) The energy stored in the capacitor is U C = 12 CV = q /2C The energy stored in the inductor is U L = 12 Li EXECUTE: (a) ω = 1 = = 105.4 rad/s, which rounds to 105 rad/s The LC (1.50 H)(6.00 × 10−5 F) 2π = 0.0596 s 105.4 rad/s (b) The circuit containing the battery and capacitor is sketched in Figure 30.35 period is given by T = 2π ω = ε −Q =0 C Q = ε C = (12.0 V)(6.00 × 10−5 F) = 7.20 × 10−4 C Figure 30.35 © 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 30-18 30.54 Chapter 30 IDENTIFY: This is a decaying R -L circuit with I = ε /R i (t ) = I 0e− ( R/L )t SET UP: ε = 60.0 V, R = 240 Ω and L = 0.160 H The rate at which energy stored in the inductor is decreasing is iLdi/dt ⎛ 60 V ⎞ ⎛ε ⎞ −3 LI = L ⎜ ⎟ = (0.160 H) ⎜ ⎟ = 5.00 × 10 J 2 ⎝R⎠ ⎝ 240 Ω ⎠ EXECUTE: (a) U = (b) i = ε e−( R/L)t ⇒ di = − R i ⇒ dU L R dt L dt = iL di ε e−2( R/L)t = − Ri = dt R dU L (60 V) −2(240/0.160)(4.00×10−4 ) =− = −4.52 W e dt 240 Ω (c) In the resistor, PR = (d) PR (t ) = i R = dU R ε −2( R/L)t (60 V)2 −2(240/0.160)(4.00 × 10−4 ) e e = i2R = = = 4.52 W dt R 240 Ω ε e−2( R/L )t R UR = ε2 ∞ −2( R/L )t e R ∫0 = ε2 L (60 V) (0.160 H) = = 5.00 × 10−3 J, which is R 2R 2(240 Ω) the same as part (a) EVALUATE: During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor 30.55 IDENTIFY and SET UP: Follow the procedure specified in the problem Li 2 is the energy stored in the inductor and q /2C is the energy stored in the capacitor The equation is −iR − L di q − = dt C di qi + = dt C d di ⎛ di ⎞ = 12 L (i ) = 12 L ⎜ 2i ⎟ = Li , the second term dt dt ⎝ dt ⎠ EXECUTE: Multiplying by –i gives i R + Li d d UL = dt dt ( Li 2 ) d d ⎛ q2 ⎞ d dq qi = , the third term i R = PR , the rate at which electrical UC = ⎜ (q ) = (2q ) ⎟⎟ = ⎜ dt dt ⎝ 2C ⎠ 2C dt 2C dt C d energy is dissipated in the resistance U L = PL , the rate at which the amount of energy stored in the dt d inductor is changing U C = PC , the rate at which the amount of energy stored in the capacitor is dt changing EVALUATE: The equation says that PR + PL + PC = 0; the net rate of change of energy in the circuit is zero Note that at any given time one of PC or PL is negative If the current and U L are increasing the charge on the capacitor and U C are decreasing, and vice versa 30.56 IDENTIFY: The energy stored in a capacitor is U C = 12 Cv The energy stored in an inductor is U L = 12 Li Energy conservation requires that the total stored energy be constant SET UP: The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero EXECUTE: (a) Initially v = 16.0 V and i = U L = and U C = 12 Cv = 12 (7.00 × 10−6 F)(16.0 V)2 = 8.96 × 10−4 J The total energy stored is 0.896 mJ (b) The current is maximum when q = and U C = U C + U L = 8.96 × 10−4 J so U L = 8.96 × 10−4 J Li 2 max = 8.96 × 10−4 J and imax = 2(8.96 × 10−4 J) = 0.691 A 3.75 × 10−3 H EVALUATE: The maximum charge on the capacitor is Q = CV = 112 μ 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 Inductance 30.57 30-19 IDENTIFY and SET UP: Use U C = 12 CVC2 (energy stored in a capacitor) to solve for C Then use Eq (30.22) and ω = 2π f to solve for the L that gives the desired current oscillation frequency EXECUTE: VC = 12.0 V; U C = 12 CVC2 so C = 2U C /VC2 = 2(0.0160 J)/(12.0 V)2 = 222 μ F 1 so L = 2π LC (2π f ) C f = 3500 Hz gives L = 9.31 μ H EVALUATE: f is in Hz and ω is in rad/s; we must be careful not to confuse the two IDENTIFY: Apply energy conservation to the circuit SET UP: For a capacitor V = q/C and U = q /2C For an inductor U = 12 Li f = 30.58 EXECUTE: (a) Vmax = (b) Q 6.00 × 10−6 C = = 0.0240 V C 2.50 × 10−4 F Q 6.00 × 10−6 C Q2 = = 1.55 × 10−3 A Limax = , so imax = −4 2C LC (0.0600 H)(2.50 × 10 F) (c) U max = Limax = (0.0600 H)(1.55 × 10−3 A) = 7.21 × 10−8 J 2 1 (d) If i = imax then U L = U max = 1.80 × 10−8 J and U C = U max = 4 ( (3/4)Q 2C ) = q2 This gives 2C q= Q = 5.20 × 10−6 C q2 for all times Li + 2C IDENTIFY: The initial energy stored in the capacitor is shared between the inductor and the capacitor q di SET UP: The potential across the capacitor and inductor is always the same, so =L The capacitor C dt EVALUATE: U max = 30.59 energy is U C = q2 and the inductor energy is U L = Li 2C q 2 Qmax + Li = Qmax = (84.0 × 10−9 F)(12.0 V) = 1.008 × 10−6 C 2C 2C 1 − q2 ) = Li = (Qmax ((1.008 × 10−6 C) − (0.650 × 10−6 C) ) = 3.533 × 10−6 J 2C 2(84.0 × 10−9 F) EXECUTE: i= 30.60 2(3.533 × 10−6 J) = 0.0130 A = 13.0 mA 0.0420 H di q 0.650 × 10−6 C = = = 184 A/s dt LC (0.0420 H)(84.0 × 10−9 F) EVALUATE: The current is only 13 mA but is changing at a rate of 184 A/s However, it only changes at that rate for a tiny fraction of a second IDENTIFY: The total energy is shared between the inductor and the capacitor q di SET UP: The potential across the capacitor and inductor is always the same, so =L The capacitor C dt energy is U C = q2 and the inductor energy is U L = Li 2C © 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 30-20 Chapter 30 EXECUTE: The total energy is q = LC q 2 Qmax + Li = = CVmax 2C 2C di = (0.330 H)(5.90 × 10−4 F)(89.0 A/s) = 1.733 × 10−2 C dt (1.733 × 10−2 C) 2 CVmax = + (0.330 H)(2.50 A)2 = 1.286 J 2(5.90 × 10−4 F) Vmax = 30.61 2(1.286 J) 5.90 × 10−4 F) = 66.0 V EVALUATE: By energy conservation, the maximum potential across the inductor will also be 66.0 V, but that will occur only at the instants when the capacitor is uncharged IDENTIFY: The current through an inductor doesn’t change abruptly After a long time the current isn’t changing and the voltage across each inductor is zero SET UP: First combine the inductors EXECUTE: (a) Just after the switch is closed there is no current in the inductors There is no current in the resistors so there is no voltage drop across either resistor A reads zero and V reads 20.0 V (b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits The circuit becomes equivalent to the circuit shown in Figure 30.61a I = (20.0 V)/(75.0 Ω) = 0.267 A The voltage between points a and b is zero, so the voltmeter reads zero (c) Combine the inductor network into its equivalent, as shown in Figure 30.61b R = 75.0 Ω is the equivalent resistance Eq (30.14) says i = (ε /R)(1 − e −t/τ ) with τ = L/R = (10.8 mH)/(75.0 Ω) = 0.144 ms ε = 20.0 V, R = 75.0 Ω, t = 0.115 ms so i = 0.147 A VR = iR = (0.147 A)(75.0 Ω) = 11.0 V 20.0 V − VR − VL = and VL = 20.0 V − VR = 9.0 V The ammeter reads 0.147 A and the voltmeter reads 9.0 V EVALUATE: The current through the battery increases from zero to a final value of 0.267 A The voltage across the inductor network drops from 20.0 V to zero Figure 30.61 30.62 IDENTIFY: i (t ) is given by Eq (30.14) SET UP: The graph shows V = at t = and V approaches the constant value of 25 V at large times EXECUTE: (a) The voltage behaves the same as the current Since VR is proportional to i, the scope must be across the 150-Ω resistor (b) From the graph, as t → ∞, VR → 25 V, so there is no voltage drop across the inductor, so its internal ⎛ 1⎞ resistance must be zero VR = Vmax (1 − e −t/r ) When t = τ , VR = Vmax ⎜1 − ⎟ ≈ 0.63Vmax From the graph, ⎝ e⎠ V = 0.63Vmax = 16 V at t ≈ 0.5 ms Therefore τ = 0.5 ms L/R = 0.5 ms gives L = (0.5 ms)(150 Ω) = 0.075 H (c) The graph if the scope is across the inductor is sketched in Figure 30.62 © 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 Inductance 30-21 EVALUATE: At all times VR + VL = 25.0 V At t = all the battery voltage appears across the inductor since i = At t → ∞ all the battery voltage is across the resistance, since di/dt = Figure 30.62 30.63 IDENTIFY and SET UP: The current grows in the circuit as given by Eq (30.14) In an R-L circuit the full emf initially is across the inductance and after a long time is totally across the resistance A solenoid in a circuit is represented as a resistance in series with an inductance Apply the loop rule to the circuit; the voltage across a resistance is given by Ohm’s law EXECUTE: (a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero In the graph, the voltage drops, so the oscilloscope is across the solenoid (b) At t → ∞ the current in the circuit approaches its final, constant value The voltage doesn’t go to zero because the solenoid has some resistance RL The final voltage across the solenoid is IRL , where I is the final current in the circuit (c) The emf of the battery is the initial voltage across the inductor, 50 V Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit (d) As t → ∞, ε − IR − IRL = ε = 50 V and from the graph IRL = 15 V (the final voltage across the inductor), so IRL = 35 V and I = (35 V)/R = 3.5 A (e) IRL = 15 V, so RL = (15 V)/(3.5 A) = 4.3 Ω ε − VL − iR = 0, ε ⎡ ⎤ R (1 − e−t/τ ), so VL = ε ⎢1 − (1 − e−t/τ ) ⎥ R tot ⎣ ⎦ = 50 V, R = 10 Ω, Rtot = 14.3 Ω, so when t = τ , VL = 27.9 V From the graph, VL has this value when VL = ε − iR, i = ε where VL includes the voltage across the resistance of the solenoid Rtot t = 3.0 ms (read approximately from the graph), so τ = L/Rtot = 3.0 ms Then L = (3.0 ms)(14.3 Ω) = 43 mH EVALUATE: At t = there is no current and the 50 V measured by the oscilloscope is the induced emf due to the inductance of the solenoid As the current grows, there are voltage drops across the two resistances in the circuit We derived an equation for VL , the voltage across the solenoid At t = it gives 30.64 VL = ε and at t → ∞ it gives VL = ε R/Rtot = iR IDENTIFY: At t = 0, i = through each inductor At t → ∞, the voltage is zero across each inductor SET UP: In each case redraw the circuit At t = replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire EXECUTE: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is shown ε 50 V = 0.333 A V = (100 Ω)(0.333 A) = 33.3 V in Figure 30.64a i = = R 150 Ω V4 = (50 Ω)(0.333 A) = 16.7 V V3 = since no current flows through it V2 = V4 = 16.7 V, since the inductor is in parallel with the 50-Ω resistor A1 = A3 = 0.333 A, A2 = (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it The 50 V = 0.385 A simplified circuit is sketched in Figure 30.64b i = ε /R = 130 Ω V1 = (100 Ω)(0.385 A) = 38.5 V; V2 = 0; V3 = V4 = 50 V − 38.5 V = 11.5 V i1 = 0.385 A; i2 = 11.5 V 11.5 V = 0.153 A; i3 = = 0.230 A 75 Ω 50 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 30-22 Chapter 30 EVALUATE: Just after the switch is closed the current through the battery is 0.333 A After a long time the current through the battery is 0.385 A After a long time there is an additional current path, the equivalent resistance of the circuit is decreased and the current has increased Figure 30.64 30.65 IDENTIFY and SET UP: Just after the switch is closed, the current in each branch containing an inductor is zero and the voltage across any capacitor is zero The inductors can be treated as breaks in the circuit and the capacitors can be replaced by wires After a long time there is no voltage across each inductor and no current in any branch containing a capacitor The inductors can be replaced by wires and the capacitors by breaks in the circuit EXECUTE: (a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor, so V3 = V2 + V3 = V4 = V5 , and since V5 = and V3 = 0, V4 and V2 are also zero V4 = means V3 reads zero V1 then must equal 40.0 V, and this means the current read by A1 is (40.0 V)/(50.0 Ω) = 0.800 A A2 + A3 + A4 = A1, but A2 = A3 = so A4 = A1 = 0.800 A A1 = A4 = 0.800 A; all other ammeters read zero V1 = 40.0 V and all other voltmeters read zero (b) After a long time the capacitor is fully charged so A4 = 0, The current through the inductor isn’t changing, so V2 = The currents can be calculated from the equivalent circuit that replaces the inductor by a short circuit, as shown in Figure 30.65a Figure 30.65a I = (40.0 V)/(83.33 Ω) = 0.480 A; A1 reads 0.480 A V1 = I (50.0 Ω) = 24.0 V The voltage across each parallel branch is 40.0 V − 24.0 V = 16.0 V V2 = 0, V3 = V4 = V5 = 16.0 V V3 = 16.0 V means A2 reads 0.160 A V4 = 16.0 V means A3 reads 0.320 A A4 reads zero Note that A2 + A3 = A1 (c) V5 = 16.0 V so Q = CV = (12.0 μ F)(16.0 V) = 192 μ C (d) At t = and t → ∞, V2 = As the current in this branch increases from zero to 0.160 A the voltage V2 reflects the rate of change of the current The graph is sketched in Figure 30.65b © 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 Inductance 30-23 Figure 30.65b 30.66 EVALUATE: This reduction of the circuit to resistor networks only apply at t = and t → ∞ At intermediate times the analysis is complicated IDENTIFY: At all times v1 + v2 = 25.0 V The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing SET UP: Immediately after closing the switch the current through the inductor is zero After a long time the current is no longer changing EXECUTE: (a) i = so v1 = and v2 = 25.0 V The ammeter reading is A = (b) After a long time, v2 = and v1 = 25.0 V v1 = iR and i = 30.67 v1 25.0 V = = 1.67 A The ammeter reading R 15.0 Ω is A = 1.67A (c) None of the answers in (a) and (b) depend on L so none of them would change EVALUATE: The inductance L of the circuit affects the rate at which current reaches its final value But after a long time the inductor doesn’t affect the circuit and the final current does not depend on L IDENTIFY: At t = 0, i = through each inductor At t → ∞, the voltage is zero across each inductor SET UP: In each case redraw the circuit At t = replace each inductor by a break in the circuit and at t → ∞ replace each inductor by a wire EXECUTE: (a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit The current is the same through the 40.0-Ω and 15.0-Ω resistors and is equal to (25.0 V)/(40.0 Ω + 15.0 Ω) = 0.455 A A1 = A4 = 0.455 A; A2 = A3 = (b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit The circuit is equivalent to the circuit sketched in Figure 30.67 I = (25.0 V)/(42.73 Ω) = 0.585 A A1 reads 0.585 A The voltage across each parallel branch is 25.0 V − (0.585 A)(40.0 Ω) = 1.60 V A2 reads (1.60 V)/(5.0 Ω) = 0.320 A A3 reads (1.60 V)/(10.0 Ω) = 0.160 A A4 reads (1.60 V)/(15.0 Ω) = 0.107 A EVALUATE: Just after the switch is closed the current through the battery is 0.455 A After a long time the current through the battery is 0.585 A After a long time there are additional current paths, the equivalent resistance of the circuit is decreased and the current has increased Figure 30.67 30.68 IDENTIFY: Closing S and simultaneously opening S1 produces an L-C circuit with initial current through the inductor of 3.50 A When the current is a maximum the charge q on the capacitor is zero and © 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 30-24 Chapter 30 when the charge q is a maximum the current is zero Conservation of energy says that the maximum energy qmax stored in the capacitor Li stored in the inductor equals the maximum energy max C SET UP: imax = 3.50 A, the current in the inductor just after the switch is closed EXECUTE: (a) Li 2 max = 2 qmax C qmax = ( LC )imax = (2.0 × 10−3 H)(5.0 × 10−6 F)(3.50 A) = 3.50 × 10−4 C = 0.350 mC (b) When q is maximum, i = 30.69 EVALUATE: In the final circuit the current will oscillate IDENTIFY: Apply the loop rule to each parallel branch The voltage across a resistor is given by iR and the voltage across an inductor is given by L di/dt The rate of change of current through the inductor is limited SET UP: With S closed the circuit is sketched in Figure 30.69a The rate of change of the current through the inductor is limited by the induced emf Just after the switch is closed the current in the inductor has not had time to increase from zero, so i2 = Figure 30.69a EXECUTE : (a) ε − vab = 0, so vab = 60.0 V (b) The voltage drops across R, as we travel through the resistor in the direction of the current, so point a is at higher potential (c) i2 = so vR = i2 R2 = ε − vR 2 − vL = so vL = ε = 60.0 V (d) The voltage rises when we go from b to a through the emf, so it must drop when we go from a to b through the inductor Point c must be at higher potential than point d di (e) After the switch has been closed a long time, → so vL = Then ε − vR2 = and i2 R2 = ε dt ε 60.0 V = = 2.40 A so i2 = R2 25.0 Ω SET UP: The rate of change of the current through the inductor is limited by the induced emf Just after the switch is opened again the current through the inductor hasn’t had time to change and is still i2 = 2.40 A The circuit is sketched in Figure 30.69b EXECUTE: The current through R1 is i2 = 2.40 A in the direction b to a Thus vab = −i2 R1 = −(2.40 A)(40.0 Ω) vab = −96.0 V Figure 30.69b (f) Point where current enters resistor is at higher potential; point b is at higher potential © 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 Inductance 30-25 (g) vL − vR1 − vR2 = vL = vR1 + vR2 vR1 = −vab = 96.0 V; vR2 = i2 R2 = (2.40 A)(25.0 Ω) = 60.0 V Then vL = vR1 + vR2 = 96.0 V + 60.0 V = 156 V As you travel counterclockwise around the circuit in the direction of the current, the voltage drops across each resistor, so it must rise across the inductor and point d is at higher potential than point c The current is decreasing, so the induced emf in the inductor is directed in the direction of the current Thus, vcd = −156 V (h) Point d is at higher potential EVALUATE: The voltage across R1 is constant once the switch is closed In the branch containing R2 , just after S is closed the voltage drop is all across L and after a long time it is all across R2 Just after S is 30.70 opened the same current flows in the single loop as had been flowing through the inductor and the sum of the voltage across the resistors equals the voltage across the inductor This voltage dies away, as the energy stored in the inductor is dissipated in the resistors IDENTIFY: Apply the loop rule to the two loops The current through the inductor doesn’t change abruptly di SET UP: For the inductor ε = L and ε is directed to oppose the change in current dt EXECUTE: (a) Switch is closed, then at some later time di di = 50.0 A/s ⇒ vcd = L = (0.300 H)(50.0 A/s) = 15.0 V dt dt 60.0 V The top circuit loop: 60.0 V = i1R1 ⇒ i1 = = 1.50 A 40.0 Ω The bottom loop: 60.0 V − i2 R2 − 15.0 V = ⇒ i2 = (b) After a long time: i2 = 45.0 V = 1.80 A 25.0 Ω 60.0 V = 2.40 A, and immediately when the switch is opened, the inductor 25.0 Ω maintains this current, so i1 = i2 = 2.40 A EVALUATE: The current through R1 changes abruptly when the switch is closed 30.71 IDENTIFY and SET UP: The circuit is sketched in Figure 30.71a Apply the loop rule Just after S1 is closed, i = After a long time i has reached its final value and di/dt = The voltage across a resistor depends on i and the voltage across an inductor depends on di/dt Figure 30.71a EXECUTE: (a) At time t = 0, i0 = so vac = i0 R0 = By the loop rule ε − vac − vcb = so vcb = ε − vac = ε = 36.0 V (i0 R = so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the changing current.) di di (b) After a long time → so the potential − L across the inductor becomes zero The loop rule dt dt gives ε − i0 ( R0 + 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 30-26 Chapter 30 i0 = ε R0 + R = 36.0 V = 0.180 A 50.0 Ω + 150 Ω vac = i0 R0 = (0.180 A)(50.0 Ω) = 9.0 V di0 = (0.180 A)(150 Ω) + = 27.0 V (Note that vac + vcb = ε ) dt − vac − vcb = Thus vcb = i0 R + L (c) ε ε − iR0 − iR − L di = dt ⎛ L ⎞ di ε ⎜ ⎟ = −i + R R dt R R0 + + 0⎠ ⎝ di ⎛ R + R0 ⎞ =⎜ ⎟ dt −i + ε /(R + R0 ) ⎝ L ⎠ L di = ε − i ( R0 + R ) and dt Integrate from t = 0, when i = 0, to t, when i = i0 : i0 ∫0 i ⎡ di R + R0 t ε ⎤ = ⎛ R + R0 ⎞ t , so ln dt i = = − − + ⎢ ⎥ ⎜ ⎟ L ∫0 R + R0 ⎦ ⎝ L ⎠ −i + ε /(R + R0 ) ⎣ ⎛ ε ⎞ − ln ⎛ ε ⎞ = − ⎛ R + R0 ⎞ t ln ⎜ −i0 + ⎟ ⎜ ⎟ ⎜ ⎟ R + R0 ⎠ ⎝ L ⎠ ⎝ ⎝ R + R0 ⎠ ⎛ −i + ε /(R + R0 ) ⎞ ⎛ R + R0 ⎞ ln ⎜ ⎟ = −⎜ ⎟t ⎝ L ⎠ ⎝ ε /(R + R0 ) ⎠ Taking exponentials of both sides gives −i0 + ε /(R + R0 ) ε (1 − e−( R + R0 )t/L ) = e− ( R + R0 )t/L and i0 = ε /(R + R0 ) R + R0 36.0 V (1 − e − (200 Ω/4.00 H)t ) = (0.180 A)(1 − e −t/0.020 s ) 50 Ω + 150 Ω At t → 0, i0 = (0.180 A)(1 − 1) = (agrees with part (a)) At t → ∞, i0 = (0.180 A)(1 − 0) = 0.180 A (agrees Substituting in the numerical values gives i0 = with part (b)) vac = i0 R0 = ε R0 (1 − e−( R + R0 )t/L ) = 9.0 V(1 − e−t/0.020 s ) R + R0 vcb = ε − vac = 36.0 V − 9.0 V(1 − e −t/0.020 s ) = 9.0 V(3.00 + e−t/0.020 s ) At t → 0, vac = 0, vcb = 36.0 V (agrees with part (a)) At t → ∞, vac = 9.0 V, vcb = 27.0 V (agrees with part (b)) The graphs are given in Figure 30.71b Figure 30.71b EVALUATE: The expression for i (t ) we derived becomes Eq (30.14) if the two resistors R0 and R in series are replaced by a single equivalent resistance R0 + R 30.72 IDENTIFY: Apply the loop rule The current through the inductor doesn’t change abruptly SET UP: With S closed, vcb must be zero © 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 Inductance 30-27 EXECUTE: (a) Immediately after S is closed, the inductor maintains the current i = 0.180 A through R The loop rule around the outside of the circuit yields ε + ε L − iR − i0 R0 = 36.0 V + (0.18 A)(150 Ω) − (0.18 A)(150 Ω) − i0 (50 Ω) = vac = (0.72 A)(50 Ω) = 36.0 V and vcb = (b) After a long time, vac = 36.0 V, and vcb = Thus i0 = is = 0.720 A (c) i0 = 0.720 A, iR (t ) = ε Rtotal ε R0 = 36 V = 0.720 A 50 Ω 36.0 V = 0.720 A, iR = and 50 Ω e − ( R/L )t and iR (t ) = (0.180 A)e− (12.5 s −1 i0 = −1 )t −1 is (t ) = (0.720 A) − (0.180 A)e−(12.5 s )t = (0.180 A)(4 − e−(12.5 s )t ) The graphs of the currents are given in Figure 30.72 EVALUATE: R0 is in a loop that contains just ε and R0 , so the current through R0 is constant After a long time the current through the inductor isn’t changing and the voltage across the inductor is zero Since vcb is zero, the voltage across R must be zero and iR becomes zero Figure 30.72 30.73 IDENTIFY: Follow the steps specified in the problem SET UP: Find the flux through a ring of height h, radius r and thickness dr Example 28.10 shows that μ Ni inside the toroid B= 2π r b b ⎛ μ Ni ⎞ μ Nih b dr μ0 Nih EXECUTE: (a) ΦB = ∫ B (h dr ) = ∫ ⎜ ⎟ ( h dr ) = = ln(b /a ) a a ⎝ 2π r ⎠ 2π ∫a r 2π (b) L = N ΦB μ0 N h = ln(b/a ) i 2π © 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 30-28 Chapter 30 b − a (b − a )2 μ0 N h ⎛ b − a ⎞ + + " ⇒ ≈ L ⎜ ⎟ a 2π ⎝ a ⎠ 2a EVALUATE: h(b − a ) is the cross-sectional area A of the toroid and a is approximately the radius r, so this (c) ln(b /a ) = ln(1 − (b − a )/a ) ≈ 30.74 result is approximately the same as the result derived in Example 30.3 IDENTIFY: At steady state with the switch in position 1, no current flows to the capacitors and the inductors can be replaced by wires Apply conservation of energy to the circuit with the switch in position SET UP: Replace the series combinations of inductors and capacitors by their equivalents ε 75.0 V = 0.600 A EXECUTE: (a) At steady state i = = R 125 Ω (b) The equivalent circuit capacitance of the two capacitors is given by 1 = + and Cs 25 μF 35 μF Cs = 14.6 μ F Ls = 15.0 mH + 5.0 mH = 20.0 mH The equivalent circuit is sketched in Figure 30.74a q2 = Li0 q = i0 LC = (0.600 A) (20 × 10−3 H)(14.6 × 10−6 F) = 3.24 × 10−4 C 2C As shown in Figure 30.74b, the capacitors have their maximum charge at t = T/4 Energy conservation: π π LC = (20 × 10−3 H)(14.6 × 10−6 F) = 8.49 × 10−4 s 2 EVALUATE: With the switch closed the battery stores energy in the inductors This then is the energy in the L-C circuit when the switch is in position t = 14 T = 14 (2π LC ) = Figure 30.74 30.75 (a) IDENTIFY and SET UP: With switch S closed the circuit is shown in Figure 30.75a Apply the loop rule to loops and EXECUTE: loop ε − i1R1 = i1 = ε R1 (independent of t) Figure 30.75a loop (2) ε − i2 R2 − L di2 dt =0 This is in the form of Eq (30.12), so the solution is analogous to Eq (30.14): i2 = ε R2 (1 − e− R2t/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 Inductance (b) EVALUATE: The expressions derived in part (a) give that as t → ∞, i1 = ε R1 and i2 = ε R2 30-29 Since di2 → at steady state, the inductance then has no effect on the circuit The current in R1 is constant; the dt current in R2 starts at zero and rises to ε /R2 (c) IDENTIFY and SET UP: The circuit now is as shown in Figure 30.75b Let t = now be when S is opened At t = 0, i = ε R2 Figure 30.75b Apply the loop rule to the single current loop di di EXECUTE: −i ( R1 + R2 ) − L = (Now is negative.) dt dt di di ⎛ R + R2 ⎞ L = −i ( R1 + R2 ) gives = −⎜ ⎟ dt dt i ⎝ L ⎠ Integrate from t = 0, when i = I = ε /R2 , to t ⎛ i ⎞ di ⎛ R + R2 ⎞ t ⎛ R1 + R2 ⎞ = −⎜ ⎟t ⎟ ∫0 dt and ln ⎜ ⎟ = − ⎜ i I ⎝ L ⎠ ⎝ L ⎠ ⎝ 0⎠ i ∫I Taking exponentials of both sides of this equation gives i = I 0e−( R1 + R2 )t/L = ε R2 e−( R1 + R2 )t/L (d) IDENTIFY and SET UP: Use the equation derived in part (c) and solve for R2 and ε EXECUTE: L = 22.0 H PR1 = V2 V (120 V) = 40.0 W gives R1 = = = 360 Ω R1 PR1 40.0 W We are asked to find R2 and ε Use the expression derived in part (c) I = 0.600 A so ε /R2 = 0.600 A i = 0.150 A when t = 0.080 s, so i = ε R2 e−( R1 + R2 )t/L gives 0.150 A = (0.600 A)e − ( R1 + R2 )t/L = e−( R1 + R2 )t/L so ln = ( R1 + R2 )t/L L ln (22.0 H)ln − R1 = − 360 Ω = 381.2 Ω − 360 Ω = 21.2 Ω t 0.080 s Then ε = (0.600 A) R2 = (0.600 A)(21.2 Ω) = 12.7 V R2 = (e) IDENTIFY and SET UP: Use the expressions derived in part (a) ε = 12.7 V = 0.0353 A R1 360 Ω EVALUATE: When the switch is opened the current through the light bulb jumps from 0.0353 A to 0.600 A Since the electrical power dissipated in the bulb (brightness) depend on i , the bulb suddenly becomes much brighter IDENTIFY: Follow the steps specified in the problem SET UP: The current in an inductor does not change abruptly EXECUTE: (a) Using Kirchhoff’s loop rule on the left and right branches: EXECUTE: The current through the light bulb before the switch is opened is i1 = 30.76 © 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 30-30 Chapter 30 Left: ε − (i1 + i2 ) R − L di1 = ⇒ R(i1 + i2 ) + L di1 = ε Right: ε dt dt q2 q2 − (i1 + i2 ) R − = ⇒ R (i1 + i2 ) + = ε C C (b) Initially, with the switch just closed, i1 = 0, i2 = ε and q2 = R (c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise but straightforward exercise We will show that the initial conditions are satisfied: At t = 0, q2 = i1 (t ) = ε R ε ωR e− β t sin(ωt ) = ε ωR sin(0) = (1 − e − βt [(2ω RC ) −1 sin(ωt ) + cos(ωt )] ⇒ i1(0) = (d) When does i2 first equal zero? ω = i2 (t ) = = ε R ε R (1 − [cos(0)]) = 1 − = 625 rad/s LC (2 RC )2 e − βt [− (2ω RC )−1 sin(ωt ) + cos(ωt )] ⇒ −(2ω RC ) −1 tan(ωt ) + = and tan(ωt ) = +2ω RC = +2(625 rad/s)(400 Ω)(2.00 × 10−6 F) = +1.00 ωt = arctan ( + 1.00) = +0.785 ⇒ t = 0.785 = 1.256 × 10−3 s 625 rad/s EVALUATE: As t → ∞, i1 → ε /R, q2 → and i2 → 30.77 IDENTIFY: Apply L = N ΦB to calculate L i μ Ni In the liquid, BL = W W μ Ni K μ0 Ni EXECUTE: (a) ΦB = BA = BL AL + BAir AAir = (( D − d )W ) + (dW ) = μ0 Ni [ ( D − d ) + Kd ] W W N ΦB d d ⎛ L − L0 ⎞ L= = μ0 N [( D − d ) + Kd ] = L0 − L0 + Lf = L0 + ⎜ f ⎟ d i D D ⎝ D ⎠ SET UP: In the air the magnetic field is BAir = μ0 Ni ⎛ L − L0 ⎞ 2 d =⎜ ⎟ D, where L0 = μ0 N D, and Lf = K μ0 N D L L − 0⎠ ⎝ f d⎞ ⎛ (b) and (c) Using K = χ m + we can find the inductance for any height L = L0 ⎜1 + χ m ⎟ D ⎝ ⎠ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury d = D/4 0.63024 H 0.63000 H d = D/2 0.63048 H 0.62999 H d = 3D/4 d=D 0.63072 H 0.63096 H 0.62999 H 0.62998 H The values χ m (O ) = 1.52 × 10−3 and χ m (Hg) = −2.9 × 10−5 have been used 30.78 EVALUATE: (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury IDENTIFY: The induced emf across the two coils is due to both the self-inductance of each and the mutual inductance of the pair of coils di SET UP: The equivalent inductance is defined by ε = Leq , where ε and i are the total emf and current dt across the combination © 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 Inductance 30-31 di1 di di di di + L2 + M 21 + M12 ≡ Leq dt dt dt dt dt di di di di di But i = i1 + i2 ⇒ = + and M12 = M 21 ≡ M , so ( L1 + L2 + M ) = Leq and dt dt dt dt dt Leq = L1 + L2 + 2M EXECUTE: Series: L1 di1 di di di di di di di di + M12 = Leq and L2 + M 21 = Leq , with + = and dt dt dt dt dt dt dt dt dt di di di M12 = M 21 ≡ M To simplify the algebra let A = , B = , and C = So dt dt dt L1 A + MB = LeqC , L2 B + MA = LeqC , A + B = C Now solve for A and B in terms of C Parallel: We have L1 ( L1 − M ) A + ( M − L2 ) B = using A = C − B ( L1 − M )(C − B) + ( M − L2 ) B = ( L1 − M )C − ( L1 − M ) B + ( M − L2 ) B = (2 M − L1 − L2 ) B = ( M − L1 )C and B = A=C−B=C− ( M − L1 ) C But (2M − L1 − L2 ) M − L2 ( M − L1 )C (2 M − L1 − L2 ) − M + L1 C Substitute A in B = C , or A = M − L1 − L2 (2 M − L1 − L2 ) (2 M − L1 − L2 ) back into original equation: M − L1L2 L1 ( M − L2 )C M ( M − L1 ) C = LeqC Finally, + C = LeqC and M − L1 − L2 M − L1 − L2 (2M − L1 − L2 ) L1L2 − M L1 + L2 − 2M EVALUATE: If the flux of one coil doesn’t pass through the other coil, so M = 0, then the results reduce to those of inductors in parallel IDENTIFY: Apply Kirchhoff’s loop rule to the top and bottom branches of the circuit SET UP: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is zero di ε EXECUTE: (a) ε − i1R1 − L = ⇒ i1 = (1 − e− ( R1/L )t ) dt R1 Leq = 30.79 ε − i2 R2 − q2 = ⇒ − di2 R2 − i2 C t ε R2 q2 = ∫ i2 dt ′ = − (b) i1 (0) ε R1 dt C R2Ce− (1/R2C )t ′ (1 − e0 ) = 0, i2 = (c) As t → ∞: i1 (∞) = ε ε R2 = ⇒ i2 = t ex = + x + ε R1 e0 = (1 − e−∞ ) = (1 − e−( R1/L)t ) = ε R2 R2 e−(1/R2C )t ) = εC (1 − e− (1/R2C )t ) 48.0 V = 9.60 × 10−3 A 5000 Ω ε R1 R1 time” is many time constants later (d) i1 = i2 ⇒ ε = 48.0 V ε e−∞ = A good definition of a “long = 1.92 A, i2 = 25.0 Ω R2 e−(1/R2C )t ⇒ (1 − e−( R1/L)t ) = R1 −(1/R2C )t Expanding the exponentials like e R2 ⎞ x x3 R 1⎛ R ⎞ R ⎛ t t2 + + ", we find: t − ⎜ ⎟ t + " = ⎜1 − + 2 − " ⎟ and ⎟ 3! L 2⎝ L ⎠ R2 ⎜⎝ RC R C ⎠ ⎛R R ⎞ R t ⎜ + 21 ⎟ + O(t ) + " = , if we have assumed that t