Response of First Order RL and RC-Circuit And Network Analysis-Solution Manual, Exercises of Electrical Circuit Analysis

Download Response of First Order RL and RC-Circuit And Network Analysis-Solution Manual and more Exercises Electrical Circuit Analysis in PDF only on Docsity! 7 Response of First-Order RL and RC Circuits Assessment Problems AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 Ω resistor from the circuit. First combine the 30 Ω and 6 Ω resistors in parallel: 30‖6 = 5 Ω Use voltage division to find the voltage drop across the parallel resistors: v = 5 5 + 3 (120) = 75 V Now find the current using Ohm’s law: i(0−) = −v 6 = −75 6 = −12.5 A [b] w(0) = 1 2 Li2(0) = 1 2 (8 × 10−3)(12.5)2 = 625 mJ [c] To find the time constant, we need to find the equivalent resistance seen by the inductor for t > 0. When the switch opens, only the 2 Ω resistor remains connected to the inductor. Thus, τ = L R = 8 × 10−3 2 = 4 ms [d] i(t) = i(0−)et/τ = −12.5e−t/0.004 = −12.5e−250t A, t ≥ 0 [e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A 7–1 docsity.com 7–2 CHAPTER 7. Response of First-Order RL and RC Circuits So w (5 ms) = 12Li 2(5 ms) = 12(8) × 10−3(3.58)2 = 51.3 mJ w (dis) = 625 − 51.3 = 573.7 mJ % dissipated = (573.7 625 ) 100 = 91.8% AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor: Using current division, i(0−) = 10 10 + 6 (6.4) = 4 A Now use the circuit for t > 0 to find the equivalent resistance seen by the inductor, and use this value to find the time constant: Req = 4‖(6 + 10) = 3.2 Ω, ·. . τ = L Req = 0.32 3.2 = 0.1 s Use the initial inductor current and the time constant to find the current in the inductor: i(t) = i(0−)e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0 Use current division to find the current in the 10 Ω resistor: io(t) = 4 4 + 10 + 6 (−i) = 4 20 (−4e−10t) = −0.8e−10t A, t ≥ 0+ Finally, use Ohm’s law to find the voltage drop across the 10 Ω resistor: vo(t) = 10io = 10(−0.8e−10t) = −8e−10t V, t ≥ 0+ [b] The initial energy stored in the inductor is w(0) = 1 2 Li2(0−) = 1 2 (0.32)(4)2 = 2.56 J Find the energy dissipated in the 4 Ω resistor by integrating the power over all time: v4Ω(t) = L di dt = 0.32(−10)(4e−10t) = −12.8e−10t V, t ≥ 0+ docsity.com Problems 7–5 i(0−) = 24/2 = 12 A = i(0+) Note that i(0−) = i(0+) because the current in an inductor is continuous. [b] Use the circuit at t = 0+, shown below, to calculate the voltage drop across the inductor at 0+. Note that this is the same as the voltage drop across the 10 Ω resistor, which has current from two sources — 8 A from the current source and 12 A from the initial current through the inductor. v(0+) = −10(8 + 12) = −200 V [c] To calculate the time constant we need the equivalent resistance seen by the inductor for t > 0. Only the 10 Ω resistor is connected to the inductor for t > 0. Thus, τ = L/R = (200 × 10−3/10) = 20 ms [d] To find i(t), we need to find the final value of the current in the inductor. When the switch has been in position a for a long time, the circuit reduces to the one below: Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit. Thus, if = −8 A Now, i(t) = if + [i(0+) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02 = −8 + 20e−50t A, t ≥ 0 [e] To find v(t), use the relationship between voltage and current for an inductor: v(t) = L di(t) dt = (200 × 10−3)(−50)(20e−50t) = −200e−50t V, t ≥ 0+ docsity.com 7–6 CHAPTER 7. Response of First-Order RL and RC Circuits AP 7.6 [a] From Example 7.6, vo(t) = −60 + 90e−100t V Write a KVL equation at the top node and use it to find the relationship between vo and vA: vA − vo 8000 + vA 160,000 + vA + 75 40,000 = 0 20vA − 20vo + vA + 4vA + 300 = 0 25vA = 20vo − 300 vA = 0.8vo − 12 Use the above equation for vA in terms of vo to find the expression for vA: vA(t) = 0.8(−60 + 90e−100t) − 12 = −60 + 72e−100t V, t ≥ 0+ [b] t ≥ 0+, since there is no requirement that the voltage be continuous in a resistor. AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop across the capacitor: i = ( 40 × 103 125 × 103 ) (10 × 10−3) = 3.2 mA vc(0−) = (3.2 × 10−3)(25 × 103) = 80 V so vc(0+) = 80 V Now use the next circuit, valid for 0 ≤ t ≤ 10 ms, to calculate vc(t) for that interval: docsity.com Problems 7–7 For 0 ≤ t ≤ 100 ms: τ = RC = (25 × 103)(1 × 10−6) = 25 ms vc(t) = vc(0−)et/τ = 80e−40t V, 0 ≤ t ≤ 10 ms [b] Calculate the starting capacitor voltage in the interval t ≥ 10 ms, using the capacitor voltage from the previous interval: vc(0.01) = 80e−40(0.01) = 53.63 V Now use the next circuit, valid for t ≥ 10 ms, to calculate vc(t) for that interval: For t ≥ 10 ms : Req = 25 kΩ‖100 kΩ = 20 kΩ τ = ReqC = (20 × 103)(1 × 10−6) = 0.02 s Therefore vc(t) = vc(0.01+)e−(t−0.01)/τ = 53.63e−50(t−0.01) V, t ≥ 0.01 s [c] To calculate the energy dissipated in the 25 kΩ resistor, integrate the power absorbed by the resistor over all time. Use the expression p = v2/R to calculate the power absorbed by the resistor. w25k = ∫ 0.01 0 [80e−40t]2 25,000 dt + ∫ ∞ 0.01 [53.63e−50(t−0.01)]2 25,000 dt = 2.91 mJ [d] Repeat the process in part (c), but recognize that the voltage across this resistor is non-zero only for the second interval: w100kΩ = ∫ ∞ 0.01 [53.63e−50(t−0.01)]2 100,000 dt = 0.29 mJ We can check our answers by calculating the initial energy stored in the capacitor. All of this energy must eventually be dissipated by the 25 kΩ resistor and the 100 kΩ resistor. Check: wstored = (1/2)(1 × 10−6)(80)2 = 3.2 mJ wdiss = 2.91 + 0.29 = 3.2 mJ AP 7.8 [a] Note – the 30 Ω resistor should be a 3 Ω resistor; the resistor in parallel with the 8 A current source should be 9 Ω. Prior to switch a closing at t = 0, there are no sources connected to the inductor; thus, i(0−) = 0. At the instant A is closed, i(0+) = 0. docsity.com 7–10 CHAPTER 7. Response of First-Order RL and RC Circuits AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ τ = (160 × 103)(10 × 10−9) = 10−3; 1/τ = 625 vp = −2 + 2e−625t V; vn = vp Write a KVL equation at the inverting input, and use it to determine vo: vn 10,000 + vn − vo 40,000 = 0 ·. . vo = 5vn = 5vp = −10 + 10e−625t V The output will saturate at the negative power supply value: −10 + 10e−625t = −5; e−625t = 1/2; t = ln 2/625 = 1.11 ms [b] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V The analysis for vo is the same as in part (a): vo = 5vp = −10 + 15e−625t V The output will saturate at the negative power supply value: −10 + 15e−625t = −5; e−625t = 1/3; t = ln 3/625 = 1.76 ms docsity.com Problems 7–11 Problems P 7.1 [a] t < 0 2 kΩ‖6 kΩ = 1.5kΩ Find the current from the voltage source by combining the resistors in series and parallel and using Ohm’s law: ig(0−) = 40 (1500 + 500) = 20 mA Find the branch currents using current division: i1(0−) = 2000 8000 (0.02) = 5 mA i2(0−) = 6000 8000 (0.02) = 15 mA [b] The current in an inductor is continuous. Therefore, i1(0+) = i1(0−) = 5 mA i2(0+) = −i1(0+) = −5 mA (when switch is open) [c] τ = L R = 0.4 × 10−3 8 × 103 = 5 × 10 −5 s; 1 τ = 20,000 i1(t) = i1(0+)e−t/τ = 5e−20,000t mA, t ≥ 0 [d] i2(t) = −i1(t) when t ≥ 0+ ·. . i2(t) = −5e−20,000t mA, t ≥ 0+ [e] The current in a resistor can change instantaneously. The switching operation forces i2(0−) to equal 15 mA and i2(0+) = −5 mA. P 7.2 [a] i(0) = 60 V/(10 Ω + 5 Ω) = 4 A [b] τ = L R = 4 45 + 5 = 80 ms docsity.com 7–12 CHAPTER 7. Response of First-Order RL and RC Circuits [c] i = 4e−t/0.08 = 4e−12.5t A, t ≥ 0 v1 = −45i = −180e−12.5t V t ≥ 0+ v2 = L di dt = (4)(−12.5)(4e−12.5t) = −200e−12.5t V t ≥ 0+ [d] pdiss = i2(45) = 720e−25t W wdiss = ∫ t 0 720e−25x dx = 720 e−25x −25 ∣∣∣∣t 0 = 28.8 − 28.8e−25t J wdiss(40 ms) = 28.8 − 28.8e−1 = 18.205 J w(0) = 1 2 (4)(4)2 = 32 J % dissipated = 18.205 32 (100) = 56.89% P 7.3 [a] io(0−) = 0 since the switch is open for t < 0. [b] For t = 0− the circuit is: 120 Ω‖60 Ω = 40 Ω ·. . ig = 1210 + 40 = 0.24 A = 240 mA iL(0−) = (120 180 ) ig = 160 mA [c] For t = 0+ the circuit is: 120 Ω‖40 Ω = 30 Ω docsity.com Problems 7–15 P 7.7 [a] w(0) = 1 2 LI2g wdiss = ∫ to 0 I2gRe −2t/τ dt = I2gR e−2t/τ (−2/τ) ∣∣∣∣to 0 = 1 2 I2gRτ(1 − e−2to/τ ) = 1 2 I2gL(1 − e−2to/τ ) wdiss = σw(0) ·. . 1 2 LI2g (1 − e−2to/τ ) = τ (1 2 LI2g ) 1 − e−2to/τ = σ; e2to/τ = 1 (1 − σ) 2to τ = ln [ 1 (1 − σ) ] ; R(2to) L = ln[1/(1 − σ)] R = L ln[1/(1 − σ)] 2to [b] R = (30 × 10−3) ln[1/0.8] 30 × 10−6 R = 223.14 Ω P 7.8 [a] t < 0 iL(0−) = 150 180 (12) = 10 A t ≥ 0 τ = 1.6 × 10−3 8 = 200 × 10−6; 1/τ = 5000 io = −10e−5000t A t ≥ 0 docsity.com 7–16 CHAPTER 7. Response of First-Order RL and RC Circuits [b] wdel = 1 2 (1.6 × 10−3)(10)2 = 80 mJ [c] 0.95wdel = 76 mJ ·. . 76 × 10−3 = ∫ to 0 8(100e−10,000t) dt ·. . 76 × 10−3 = −80 × 10−3e−10,000t ∣∣∣∣to 0 = 80 × 10−3(1 − e−10,000to) ·. . e−10,000to = 4 × 10−3 so to = 552.1 µs ·. . to τ = 552.1 × 10−6 200 × 10−6 = 2.76 so to ≈ 2.76τ P 7.9 For t < 0+ ig = −48 6 + (18‖1.5) = −6.5 A iL(0−) = 18 18 + 1.5 (−6.5) = −6 A = iL(0+) For t > 0 iL(t) = iL(0+)e−t/τ A, t ≥ 0 τ = L R = 0.5 10 + 12.45 + (54‖26) = 0.0125 s; 1 τ = 80 iL(t) = −6e−80t A, t ≥ 0 io(t) = 54 80 (−iL(t)) = 5480(6e −80t) = 4.05e−80t V, t ≥ 0+ docsity.com Problems 7–17 P 7.10 From the solution to Problem 7.9, i54Ω = 26 80 (−iL) = −1.95e−80t A P54Ω = 54(i54Ω)2 = 205.335e−160tW wdiss = ∫ 0.0125 0 205.335e−160t dt = 205.335 −160 e −160t ∣∣∣∣0.0125 0 = 1.28(1 − e−2) = 1.11 J wstored = 1 2 (0.5)(−6)2 = 9 mJ. % diss = 1.11 9 × 100 = 12.3% P 7.11 [a] t < 0 : iL(0−) = iL(0+) = 70 70 + 4 (11.84) = 11.2 A i∆ = 70 160 iT = 0.4375iT vT = 30i∆ + iT (90)(70) 160 = 30(0.4375)iT + (90)(70) 160 iT = 52.5iT vT iT = RTh = 52.5 Ω docsity.com 7–20 CHAPTER 7. Response of First-Order RL and RC Circuits t > 0: Re = (10)(40) 50 + 10 = 18 Ω τ = L Re = 0.072 18 = 4 ms; 1 τ = 250 ·. . iL = 8e−250t A ·. . vo = −10iL − 0.072diL dt = −80e−250t + 144e−250t = 64e−250t A t ≥ 0+ P 7.15 w(0) = 1 2 (72 × 10−3)(8)2 = 2304 mJ p40Ω = v2o 40 = 642 40 e−500t = 102.4e−500t W w40Ω = ∫ ∞ 0 102.4e−500t dt = 204.8 mJ %diss = 204.8 2304 (100) = 8.89% P 7.16 [a] vo(t) = vo(0+)e−t/τ ·. . vo(0+)e−1×10−3/τ = 0.5vo(0+) ·. . e1×10−3/τ = 2 ·. . τ = L R = 1 × 10−3 ln 2 ·. . L = 10 × 10 −3 ln 2 = 14.43 mH docsity.com Problems 7–21 [b] vo(0+) = −10iL(0+) = −10(1/10)30 × 10−3 = −30 mV ·. . vo = −0.03e−t/τ V, t ≥ 0+ p10Ω = v2o 10 = 9 × 10−5e−2t/τ w10Ω(1 ms) = ∫ 10−3 0+ 9 × 10−5e−2t/τ dt = 4.5τ × 10−5(1 − e−2(0.001)/τ ) τ = 1 1000 ln 2 ·. . w10Ω(1 ms) = 48.69 nJ wL(0) = 1 2 Li2L(0) = 1 2 (14.43 × 10−3)(3 × 10−3)2 = 64.92 nJ %dissipated in 1 ms = 48.69 64.92 (100) = 75% P 7.17 [a] t < 0 : t = 0+: 33 = iab + 9 + 15, iab = 9 A, t = 0+ [b] At t = ∞: iab = 165/5 = 33 A, t = ∞ docsity.com 7–22 CHAPTER 7. Response of First-Order RL and RC Circuits [c] i1(0) = 9, τ1 = 12.5 × 10−3 5 = 2.5 ms i2(0) = 15, τ2 = 3.75 × 10−3 3 1.25 ms i1(t) = 9e−400t A, t ≥ 0 i2(t) = 15e−800t A, t ≥ 0 iab = 33 − 9e−400t − 15e−800t A, t ≥ 0+ 33 − 9e−400t − 15e−800t = 19 14 = 9e−400t + 15e−800t Let x = e−400t ·. . x2 = e−800t Substituting, 15x2 + 9x − 14 = 0 so x = 0.7116 = e−400t ·. . t = [ln(1/0.7116)] 400 = 850.6 µs P 7.18 [a] t < 0 1 kΩ‖4 kΩ = 0.8 kΩ 20 kΩ‖80 kΩ = 16 kΩ (105 × 10−3)(0.8 × 103) = 84 V docsity.com Problems 7–25 v1 = 1.25[(−1.5)(−2e−1.5t)] = 3.75e−1.5t V, vo = −v1 − vR = 11.25e−1.5t V t ≥ 0+ [b] io = 1 6 ∫ t 0 11.25e−1.5x dx + 0 = 1.25 − 1.25e−1.5t A t ≥ 0 P 7.20 [a] From the solution to Problem 7.19, iR = −2e−1.5t A pR = (−2e−1.5t)2(7.5) = 30e−3t W wdiss = ∫ ∞ 0 30e−3t dt = 30 e−3t −3 ∣∣∣∣∞ 0 = 10 J [b] wtrapped = 1 2 (10)(−1.25)2 + 1 2 (6)(1.25)2 = 12.5 J CHECK: w(0) = 12(1.25)(2) 2 + 12(10)(2) 2 = 22.5 J ·. . w(0) = wdiss + wtrapped P 7.21 [a] v1(0−) = v1(0+) = 40 V v2(0+) = 0 Ceq = (1)(4)/5 = 0.8 µF τ = (25 × 103)(0.8 × 10−6) = 20ms; 1 τ = 50 i = 40 25,000 e−50t = 1.6e−50t mA, t ≥ 0+ v1 = −1 10−6 ∫ t 0 1.6 × 10−3e−50x dx + 40 = 32e−50t + 8 V, t ≥ 0 v2 = 1 4 × 10−6 ∫ t 0 1.6 × 10−3e−50x dx + 0 = −8e−50t + 8 V, t ≥ 0 docsity.com 7–26 CHAPTER 7. Response of First-Order RL and RC Circuits [b] w(0) = 1 2 (10−6)(40)2 = 800 µJ [c] wtrapped = 1 2 (10−6)(8)2 + 1 2 (4 × 10−6)(8)2 = 160 µJ. The energy dissipated by the 25 kΩ resistor is equal to the energy dissipated by the two capacitors; it is easier to calculate the energy dissipated by the capacitors (final voltage on the equivalent capacitor is zero): wdiss = 1 2 (0.8 × 10−6)(40)2 = 640 µJ. Check: wtrapped + wdiss = 160 + 640 = 800µJ; w(0) = 800µJ. P 7.22 [a] Calculate the initial voltage drop across the capacitor: v(0) = (2.7 k‖3.3 k)(40 mA) = (1485)(40 × 10−3) = 59.4 V The equivalent resistance seen by the capacitor is Re = 3 k‖(2.4 k + 3.6 k) = 3 k‖6 k = 2 kΩ τ = ReC = (2000)(0.5) × 10−6 = 1000 µs; 1 τ = 1000 v = v(0)e−t/τ = 59.4e−1000t V t ≥ 0 io = v 2.4 k + 3.6 k = 9.9e−1000t mA, t ≥ 0+ [b] w(0) = 1 2 (0.5 × 10−6)(59.4)2 = 882.09 µJ i3k = 59.4e−1000t 3000 = 19.8e−1000t mA p3k = [(19.8 × 10−3)e−1000t]2(3000) = 1.176e−2000t w3k(500 µs) = 1.176 e−2000x −2000 ∣∣∣∣500×10 −6 0 = 1.176 −2000(e −1 − 1) = 371.72 µJ % = 371.72 882.09 × 100 = 42.14% P 7.23 [a] R = v i = 4 kΩ [b] 1 τ = 1 RC = 25; C = 1 (25)(4 × 103) = 10 µF [c] τ = 1 25 = 40 ms [d] w(0) = 1 2 (10 × 10−6)(48)2 = 11.52 mJ docsity.com Problems 7–27 [e] wdiss(60 ms) = ∫ 0.06 0 v2 R dt = ∫ 0.06 0 (48e−25t)2 (4 × 103) dt = 0.576 e−50t −50 ∣∣∣∣0.06 0 = −5.74 × 10−4 + 0.01152 = 10.95 mJ P 7.24 [a] t < 0: i1(0−) = i2(0−) = 3 V 30 Ω = 100 mA [b] t > 0: i1(0+) = 0.2 2 = 100 mA i2(0+) = −0.2 8 = −25 mA [c] Capacitor voltage cannot change instantaneously, therefore, i1(0−) = i1(0+) = 100 mA [d] Switching can cause an instantaneous change in the current in a resistive branch. In this circuit i2(0−) = 100 mA and i2(0+) = −25 mA [e] vc = 0.2e−t/τ V, t ≥ 0 Re = 2||(5 + 3) = 1.6 Ω τ = 1.6(2 × 10−6) = 3.2 × 10−6 s vc = 0.2e−312,500t V, t ≥ 0 i1 = vc 2 = 0.1e−312,500t A, t ≥ 0 [f] i2 = −vc 8 = −25e−312,500t mA, t ≥ 0+ docsity.com 7–30 CHAPTER 7. Response of First-Order RL and RC Circuits 4v∆ + v∆ − vo + 5v∆ = 0 ·. . v∆ = vo10 = −1.8e −25t V, t ≥ 0+ P 7.27 [a] pds = (16.2e−25t)(−450 × 10−6e−25t) = −7290 × 10−6e−50t W wds = ∫ ∞ 0 pds dt = −145.8 µJ. ·. . dependent source is delivering 145.8 µJ [b] w5k = ∫ ∞ 0 (5000)(0.36 × 10−3e−25t)2 dt = 648 × 10−6 ∫ ∞ 0 e−50t dt = 12.96 µJ w20k = ∫ ∞ 0 (16.2e−25t)2 20,000 dt = 13,122 × 10−6 ∫ ∞ 0 e−50t dt = 262.44 µJ wc(0) = 1 2 (0.8 × 10−6)(18)2 = 129.6 µJ ∑ wdiss = 12.96 + 262.44 = 275.4 µJ∑ wdev = 145.8 + 129.6 = 275.4 µJ. P 7.28 t < 0 docsity.com Problems 7–31 t > 0 vT = −5io − 15io = −20io = 20iT ·. . RTh = vT iT = 20 Ω τ = RC = 40 µs; 1 τ = 25,000 vo = 15e−25,000t V, t ≥ 0 io = − vo20 = −0.75e −25,000t A, t ≥ 0+ P 7.29 [a] The equivalent circuit for t > 0: τ = 2 ms; 1/τ = 500 vo = 10e−500t V, t ≥ 0 io = e−500t mA, t ≥ 0+ i24kΩ = e−500t (16 40 ) = 0.4e−500t mA, t ≥ 0+ p24kΩ = (0.16 × 10−6e−1000t)(24,000) = 3.84e−1000t mW w24kΩ = ∫ ∞ 0 3.84 × 10−3e−1000t dt = −3.84 × 10−6(0 − 1) = 3.84 µJ docsity.com 7–32 CHAPTER 7. Response of First-Order RL and RC Circuits w(0) = 1 2 (0.25 × 10−6)(40)2 + 1 2 (1 × 10−6)(50)2 = 1.45 mJ % diss (24 kΩ) = 3.84 × 10−6 1.45 × 10−3 × 100 = 0.26% [b] p400Ω = 400(1 × 10−3e−500t)2 = 0.4 × 10−3e−1000t w400Ω = ∫ ∞ 0 p400 dt = 0.40 µJ % diss (400Ω) = 0.4 × 10−6 1.45 × 10−3 × 100 = 0.03% i16kΩ = e−500t (24 40 ) = 0.6e−500t mA, t ≥ 0+ p16kΩ = (0.6 × 10−3e−500t)2(16,000) = 5.76 × 10−3e−1000t W w16kΩ = ∫ ∞ 0 5.76 × 10−3e−1000t dt = 5.76 µJ % diss (16kΩ) = 0.4% [c] ∑ wdiss = 3.84 + 5.76 + 0.4 = 10 µJ wtrapped = w(0) − ∑ wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ % trapped = 1.44 1.45 × 100 = 99.31% Check: 0.26 + 0.03 + 0.4 + 99.31 = 100% P 7.30 [a] Ce = (2 + 1)6 2 + 1 + 6 = 2 µF vo(0) = −5 + 30 = 25 V τ = (2 × 10−6)(250 × 103) = 0.5 s; 1 τ = 2 vo = 25e−2t V, t > 0+ docsity.com Problems 7–35 vo = 1 0.6 × 10−6 ∫ t 0 24 × 10−3e−5000x dx + 72 = (40,000) e−5000x −5000 ∣∣∣∣t 0 +72 = −8e−5000t + 8 + 72 vo = [−8e−5000t + 80] V, t ≥ 0 [c] wtrapped = (1/2)(0.3 × 10−6)(80)2 + (1/2)(0.6 × 10−6)(80)2 wtrapped = 2880 µJ. Check: wdiss = 1 2 (0.2 × 10−6)(24)2 = 57.6 µJ w(0) = 1 2 (0.3 × 10−6)(96)2 + 1 2 (0.6 × 10−6)(72)2 = 2937.6 µJ. wtrapped + wdiss = w(0) 2880 + 57.6 = 2937.6 OK. P 7.33 [a] t < 0 iL(0−) = −5 A t > 0 iL(∞) = 40 − 804 + 16 = −2 A τ = L R = 4 × 10−3 4 + 16 = 200 µs; 1 τ = 5000 docsity.com 7–36 CHAPTER 7. Response of First-Order RL and RC Circuits iL = iL(∞) + [iL(0+) − iL(∞)]e−t/τ = −2 + (−5 + 2)e−5000t = −2 − 3e−5000t A, t ≥ 0 vo = 16iL + 80 = 16(−2 − 3e−5000t) + 80 = 48 − 48e−5000t V, t ≥ 0+ [b] vL = L diL dt = 4 × 10−3(−5000)[−3e−5000t] = 60e−5000t V, t ≥ 0+ vL(0+) = 60 V From part (a) vo(0+) = 0 V Check: at t = 0+ the circuit is: vL(0+) = 40 + (5 A)(4 Ω) = 60 V, vo(0+) = 80 − (16 Ω)(5 A) = 0 V P 7.34 [a] t < 0 KVL equation at the top node: 50 = vo 8 + vo 40 + vo 10 Multiply by 40 and solve: 2000 = (5 + 1 + 4)vo; vo = 200 V ·. . io(0−) = vo10 = 200/10 = 20 A t > 0 docsity.com Problems 7–37 Use voltage division to find the Thévenin voltage: VTh = vo = 40 40 + 120 (800) = 200 V Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance: RTh = 10 + 120‖40 = 10 + 30 = 40 Ω The simplified circuit is: τ = L R = 40 × 10−3 40 = 1 ms; 1 τ = 1000 io(∞) = 20040 = 5 A ·. . io = io(∞) + [io(0+) − io(∞)]e−t/τ = 5 + (20 − 5)e−1000t = 5 + 15e−1000t A, t ≥ 0 [b] vo = 10io + L dio dt = 10(5 + 15e−1000t) + 0.04(−1000)(15e−1000t) = 50 + 150e−1000t − 600e−1000t vo = 50 − 450e−1000t V, t ≥ 0+ P 7.35 After making a Thévenin equivalent we have For t < 0, the 15 Ω resistor is bypassed: io(0−) = io(0+) = 50/5 = 10 A docsity.com 7–40 CHAPTER 7. Response of First-Order RL and RC Circuits ·. . 24vx = 1440 so vx = 60 V io(0−) = vx15 = 4 A t > 0 Find Thévenin equivalent with respect to a, b VTh − 320 5 − 0.8 ( VTh − 320 5 ) = 0 VTh = 320 V vT = (iT + 0.8vφ)(5) = ( iT + 0.8 vT 5 ) (5) docsity.com Problems 7–41 vT = 5iT + 0.8vT ·. . 0.2vT = 5iT vT iT = RTh = 25 Ω io(∞) = 320/40 = 8 A τ = 80 × 10−3 40 = 2 ms; 1/τ = 500 io = 8 + (4 − 8)e−500t = 8 − 4e−500t A, t ≥ 0 P 7.40 t > 0; calculate vo(0+) va 15 + va − vo(0+) 5 = 20 × 10−3 ·. . va = 0.75vo(0+) + 75 × 10−3 15 × 10−3 + vo(0 +) − va 5 + vo(0+) 8 − 9i∆ + 50 × 10−3 = 0 13vo(0+) − 8va − 360i∆ = −2600 × 10−3 i∆ = vo(0+) 8 − 9i∆ + 50 × 10−3 ·. . i∆ = vo(0 +) 80 + 5 × 10−3 ·. . 360i∆ = 4.5vo(0+) + 1800 × 10−3 docsity.com 7–42 CHAPTER 7. Response of First-Order RL and RC Circuits 8va = 6vo(0+) + 600 × 10−3 ·. . 13vo(0+) − 6vo(0+) − 600 × 10−3 − 4.5vo(0+)− 1800 × 10−3 = −2600 × 10−3 2.5vo(0+) = −200 × 10−3; vo(0+) = −80 mV vo(∞) = 0 Find the Thévenin resistance seen by the 4 mH inductor: iT = vT 20 + vT 8 − 9i∆ i∆ = vT 8 − 9i∆ ·. . 10i∆ = vT8 ; i∆ = vT 80 iT = vT 20 + 10vT 80 − 9vT 80 iT vT = 1 20 + 1 80 = 5 80 = 1 16 S ·. . RTh = 16Ω τ = 4 × 10−3 16 = 0.25 ms; 1/τ = 4000 ·. . vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV, t ≥ 0+ docsity.com Problems 7–45 [b] i1 = 1 60 × 10−3 ∫ t 0 −9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA, t ≥ 0 [c] i2 = 1 40 × 10−3 ∫ t 0 −9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA, t ≥ 0 P 7.45 [a] Let v be the voltage drop across the parallel branches, positive at the top node, then −Ig + v Rg + 1 L1 ∫ t 0 v dx + 1 L2 ∫ t 0 v dx = 0 v Rg + ( 1 L1 + 1 L2 ) ∫ t 0 v dx = Ig v Rg + 1 Le ∫ t 0 v dx = Ig 1 Rg dv dt + v Le = 0 dv dt + Rg Le v = 0 Therefore v = IgRge−t/τ ; τ = Le/Rg Thus i1 = 1 L1 ∫ t 0 IgRge −x/τ dx = IgRg L1 e−x/τ (−1/τ) ∣∣∣∣t 0 = IgLe L1 (1 − e−t/τ ) i1 = IgL2 L1 + L2 (1 − e−t/τ ) and i2 = IgL1 L1 + L2 (1 − e−t/τ ) [b] i1(∞) = L2 L1 + L2 Ig; i2(∞) = L1 L1 + L2 Ig P 7.46 For t < 0, i80mH(0) = 50 V/10 Ω = 5 A For t > 0, after making a Thévenin equivalent we have i = Vs R + ( Io − Vs R ) e−t/τ docsity.com 7–46 CHAPTER 7. Response of First-Order RL and RC Circuits 1 τ = R L = 8 100 × 10−3 = 80 Io = 5 A; If = Vs R = −80 8 = −10 A i = −10 + (5 + 10)e−80t = −10 + 15e−80t A, t ≥ 0 vo = 0.08 di dt = 0.08(−1200e−80t) = −96e−80t V, t > 0+ P 7.47 For t < 0 Simplify the circuit: 80/10,000 = 8 mA, 10 kΩ‖40 kΩ‖24 kΩ = 6 kΩ 8 mA − 3 mA = 5 mA 5 mA × 6 kΩ = 30 V Thus, for t < 0 ·. . vo(0−) = vo(0+) = 30 V t > 0 docsity.com Problems 7–47 Simplify the circuit: 8 mA + 2 mA = 10 mA 10 k‖40 k‖24 k = 6 kΩ (10 mA)(6 kΩ) = 60 V Thus, for t > 0 vo(∞) = −60 V τ = RC = (10 k)(0.05 µ) = 0.5 ms; 1 τ = 2000 vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = −60 + [30 − (−60)]e−2000t = −60 + 90e−2000t V t ≥ 0 P 7.48 [a] Simplify the circuit for t > 0 using source transformation: Since there is no source connected to the capacitor for t < 0 vo(0−) = vo(0+) = 0 V From the simplified circuit, vo(∞) = 60 V τ = RC = (20 × 103)(0.5 × 10−6) = 10 ms 1/τ = 100 vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = (60 − 60e−100t) V, t ≥ 0 docsity.com 7–50 CHAPTER 7. Response of First-Order RL and RC Circuits vo = −19.8 + 19.8e−250t V, t ≥ 0 w(t) = 1 2 (0.25 × 10−6)v2o = w(∞)(1 − e−250t)2 J (1 − e−250t)2 = 0.36w(∞) w(∞) = 0.36 1 − e−250t = 0.6 e−250t = 0.4 ·. . t = 3.67 ms P 7.53 [a] io(0+) = −36 5000 = −7.2 mA [b] io(∞) = 0 [c] τ = RC = (5000)(0.8 × 10−6) = 4 ms [d] io = 0 + (−7.2)e−250t = −7.2e−250t mA, t ≥ 0+ [e] vo = −[36 + 1800(−7.2 × 10−3e−250t)] = −36 + 12.96e−250t V, t ≥ 0+ P 7.54 [a] vo(0−) = vo(0+) = 120 V vo(∞) = −150 V; τ = 2 ms; 1 τ = 500 vo = −150 + (120 − (−150))e−500t vo = −150 + 270e−500t V, t ≥ 0 [b] io = −0.04 × 10−6(−500)[270e−500t] = 5.4e−500t mA, t ≥ 0+ docsity.com Problems 7–51 [c] vg = vo − 12.5 × 103io = −150 + 202.5e−500t V [d] vg(0+) = −150 + 202.5 = 52.5 V Checks: vg(0+) = io(0+)[37.5 × 103] − 150 = 202.5 − 150 = 52.5 V i50k = vg 50k = −3 + 4.05e−500t mA i150k = vg 150k = −1 + 1.35e−500t mA -io + i50k + i150k + 4 = 0 (ok) P 7.55 For t < 0, vo(0) = (−3 m)(15 k) = −45 V t > 0: VTh = −20 × 103i∆ + 1050(75) = −20 × 10 3 ( −75 50 × 103 ) + 15 = 45 V vT = −20 × 103i∆ + 8 × 103iT = −20 × 103(0.2)iT + 8 × 103iT = 4 × 103iT RTh = vT iT = 4 kΩ t > 0 vo = 45 + (−45 − 45)e−t/τ docsity.com 7–52 CHAPTER 7. Response of First-Order RL and RC Circuits τ = RC = (4000) ( 1 16 × 10−6 ) = 250 µs; 1 τ = 4000 vo = 45 − 90e−4000t V, t ≥ 0 P 7.56 vo(0) = 45 V; vo(∞) = −45 V RTh = 20 kΩ τ = (20 × 103) ( 1 16 × 10−6 ) = 1.25 × 10−3; 1 τ = 800 v = −45 + (45 + 45)e−800t = −45 + 90e−800t V, t ≥ 0 P 7.57 t < 0; io(0−) = 20 100 (10 × 10−3) = 2 mA; vo(0−) = (2 × 10−3)(50,000) = 100 V t = ∞: io(∞) = −5 × 10−3 ( 20 100 ) = −1 mA; vo(∞) = io(∞)(50,000) = −50 V RTh = 50 kΩ‖50 kΩ = 25 kΩ; C = 16 nF τ = (25,000)(16 × 10−9) = 0.4 ms; 1 τ = 2500 ·. . vo(t) = −50 + 150e−2500t V, t ≥ 0 ic = C dvo dt = −6e−2500t mA, t ≥ 0+ docsity.com Problems 7–55 [c] v1 = −1 0.2 × 10−6 ∫ t 0 −6.4 × 10−3e−1000x dx + 32 = 64 − 32e−1000t V, t ≥ 0 [d] v2 = −1 0.8 × 10−6 ∫ t 0 −6.4 × 10−3e−1000x dx + 8 = 16 − 8e−1000t V, t ≥ 0 [e] wtrapped = 1 2 (0.2 × 10−6)(64)2 + 1 2 (0.8 × 10−6)(16)2 = 512 µJ. P 7.61 [a] vc(0+) = 50 V [b] Use voltage division to find the final value of voltage: vc(∞) = 2020 + 5(−30) = −24 V [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: VTh = −24 V, RTh = 20‖5 = 4 Ω, Therefore τ = ReqC = 4(25 × 10−9) = 0.1 µs The simplified circuit for t > 0 is: [d] i(0+) = −24 − 50 4 = −18.5 A [e] vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ = −24 + [50 − (−24)]e−t/τ = −24 + 74e−107t V, t ≥ 0 [f] i = C dvc dt = (25 × 10−9)(−107)(74e−107t) = −18.5e−107t A, t ≥ 0+ P 7.62 [a] Use voltage division to find the initial value of the voltage: vc(0+) = v9k = 9 k 9 k + 3 k (120) = 90 V [b] Use Ohm’s law to find the final value of voltage: vc(∞) = v40k = −(1.5 × 10−3)(40 × 103) = −60 V docsity.com 7–56 CHAPTER 7. Response of First-Order RL and RC Circuits [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: VTh = −60 V, RTh = 10 k + 40 k = 50 kΩ τ = RThC = 1 ms = 1000 µs [d] vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ = −60 + (90 + 60)e−1000t = −60 + 150e−1000t V, t ≥ 0 We want vc = −60 + 150e−1000t = 0: Therefore t = ln(150/60) 1000 = 916.3 µs P 7.63 [a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and right of the 400-mH inductor. We get i(0−) = −60 − 200 15 k + 5 k = −13 mA i(0−) = i(0+) = −13 mA [b] For t > 0, the circuit reduces to Therefore i(∞) = −60/5,000 = −12 mA [c] τ = L R = 400 × 10−3 5000 = 80 µs [d] i(t) = i(∞) + [i(0+) − i(∞)]e−t/τ = −12 + [−13 + 12]e−12,500t = −12 − e−12,500t mA, t ≥ 0 P 7.64 [a] From Example 7.10, Leq = L1L2 − M2 L1 + L2 − 2M = 36 − 16 20 − 8 = 5 3 H τ = Leq R = (5/3) (50/3) = 1 10 docsity.com Problems 7–57 io = 100 (50/3) − 100 (50/3) e−10t = 6 − 6e−10t A t ≥ 0 [b] vo = 100 − 503 io = 100 − 50 3 (6 − 6e−10t) = 100e−10t V, t ≥ 0+ [c] vo = 2 di1 dt + 4 di2 dt io = i1 + i2 dio dt = di1 dt + di2 dt di2 dt = dio dt − di1 dt = 60e−10t − di1 dt ·. . 100e−10t = 2di1 dt + 4 ( 60e−10t − di1 dt ) ·. . di1 dt = 70e−10t di1 = 70e−10t dt∫ i1 0 dx = 70 ∫ t 0 e−10y dy ·. . i1 = 70e −10y −10 ∣∣∣∣t 0 = 7 − 7e−10t A, t ≥ 0 [d] i2 = io − i1 = 6 − 6e−10t − 7 + 7e−10t = −1 + e−10t A, t ≥ 0 [e] vo = L2 di2 dt + M di1 dt = 18(−10e−10t) + 4(70e−10t) = 100e−10t V, t ≥ 0+ (checks) Also, vo = L1 di1 dt + M di2 dt = 2(70e−10t) + 4(−10e−10t) = 100e−10t V, t ≥ 0+ CHECKS docsity.com 7–60 CHAPTER 7. Response of First-Order RL and RC Circuits vo = 10 di2 dt − 5di1 dt = 80e−20t V, t ≥ 0+ (checks) vo(0+) = 80 V, which agrees with io(0+) = 0 A io(∞) = 4 A; io(∞)Leq = (4)(1) = 4 Wb-turns i1(∞)L1 + i2(∞)M = (2.4)(5) + (1.6)(−5) = 4 Wb-turns (ok) i2(∞)L2 + i1(∞)M = (1.6)(10) + (2.4)(−5) = 4 Wb-turns (ok) Therefore, the final values of io, i1, and i2 are consistent with conservation of flux linkage. Hence, the answers make sense in terms of known circuit behavior. P 7.67 [a] Leq = 5 + 10 − 2.5(2) = 10 H τ = L R = 10 40 = 1 4 ; 1 τ = 4 i = 2 − 2e−4t A, t ≥ 0 [b] v1(t) = 5 di1 dt − 2.5di dt = 2.5 di dt = 2.5(8e−4t) = 20e−4t V, t ≥ 0+ [c] v2(t) = 10 di1 dt − 2.5di dt = 7.5 di dt = 7.5(8e−4t) = 60e−4t V, t ≥ 0+ [d] i(0) = 2 − 2 = 0, which agrees with initial conditions. 80 = 40i1 + v1 + v2 = 40(2 − 2e−4t) + 20e−4t + 60e−4t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. P 7.68 [a] Leq = 5 + 10 + 2.5(2) = 20 H τ = L R = 20 40 = 1 2 ; 1 τ = 2 i = 2 − 2e−2t A, t ≥ 0 [b] v1(t) = 5 di1 dt + 2.5 di dt = 7.5 di dt = 7.5(4e−2t) = 30e−2t V, t ≥ 0+ [c] v2(t) = 10 di1 dt + 2.5 di dt = 12.5 di dt = 12.5(4e−2t) = 50e−2t V, t ≥ 0+ [d] i(0) = 0, which agrees with initial conditions. 80 = 40i1 + v1 + v2 = 40(2 − 2e−2t) + 30e−2t + 50e−2t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. docsity.com Problems 7–61 P 7.69 Use voltage division to find the initial voltage: vo(0) = 60 40 + 60 (50) = 30 V Use Ohm’s law to find the final value of voltage: vo(∞) = (−5 mA)(20 kΩ) = −100 V τ = RC = (20 × 103)(250 × 10−9) = 5 ms; 1 τ = 200 vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = −100 + (30 + 100)e−200t = −100 + 130e−200t V, t ≥ 0 P 7.70 [a] t < 0: Using Ohm’s law, ig = 800 40 + 60‖40 = 12.5 A Using current division, i(0−) = 60 60 + 40 (12.5) = 7.5 A = i(0+) [b] 0 ≤ t ≤ 1 ms: i = i(0+)e−t/τ = 7.5e−t/τ 1 τ = R L = 40 + 120‖60 80 × 10−3 = 1000 i = 7.5e−1000t i(200µs) = 7.5e−10 3(200×10−6) = 7.5e−0.2 = 6.14 A docsity.com 7–62 CHAPTER 7. Response of First-Order RL and RC Circuits [c] i(1ms) = 7.5e−1 = 2.7591 A 1 ms ≤ t < ∞ 1 τ = R L = 40 80 × 10−3 = 500 i = i(1 ms)e−(t−1 ms)/τ = 2.7591e−500(t−0.001) A i(6ms) = 2.7591e−500(0.005) = 2.7591e−2.5 = 226.48 mA [d] 0 ≤ t ≤ 1 ms: i = 7.5e−1000t v = L di dt = (80 × 10−3)(−1000)(7.5e−1000t) = −600e−1000t V v(1−ms) = −600e−1 = −220.73 V [e] 1 ms ≤ t ≤ ∞: i = 2.7591e−500(t−0.001) v = L di dt = (80 × 10−3)(−500)(2.591e−500(t−0.001)) = −110.4e−500(t−0.001) V v(1+ms) = −110.4 V P 7.71 Note that for t > 0, vo = (4/6)vc, where vc is the voltage across the 0.5 µF capacitor. Thus we will find vc first. t < 0 vc(0) = 3 15 (−75) = −15 V docsity.com Problems 7–65 P 7.73 For t < 0: i(0) = 10 15 (15) = 10 A 0 ≤ t ≤ 10 ms: i = 10e−100t A i(10ms) = 10e−1 = 3.68 A 10 ms ≤ t ≤ 20 ms: Req = (5)(20) 25 = 4 Ω 1 τ = R L = 4 50 × 10−3 = 80 i = 3.68e−80(t−0.01) A 20 ms ≤ t ≤ ∞: i(20ms) = 3.68e−80(0.02−0.01) = 1.65 A docsity.com 7–66 CHAPTER 7. Response of First-Order RL and RC Circuits i = 1.65e−100(t−0.02) A vo = L di dt ; L = 50 mH di dt = 1.65(−100)e−100(t−0.02) = −165e−100(t−0.02) vo = (50 × 10−3)(−165)e−100(t−0.02) = −8.26e−100(t−0.02) V, t > 20+ ms vo(25ms) = −8.26e−100(0.025−0.02) = −5 V P 7.74 From the solution to Problem 7.73, the initial energy is w(0) = 1 2 (50 mH)(10 A)2 = 2.5 J 0.04w(0) = 0.1 J ·. . 1 2 (50 × 10−3)i2L = 0.1 so iL = 2 A Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms since i(10 ms) = 3.68 A and i(20 ms) = 1.65 A For 10 ms ≤ t ≤ 20 ms: i = 3.68e−80(t−0.01) = 2 e80(t−0.01) = 3.68 2 so t − 0.01 = 0.0076 ·. . t = 17.6 ms P 7.75 0 ≤ t ≤ 10 µs: τ = RC = (4 × 103)(20 × 10−9) = 80 µs; 1/τ = 12,500 docsity.com Problems 7–67 vo(0) = 0 V; vo(∞) = −20 V vo = −20 + 20e−12,500t V 0 ≤ t ≤ 10 µs 10 µs ≤ t ≤ ∞: t = ∞: i = −50 V 20 kΩ = −2.5 mA vo(∞) = (−2.5 × 10−3)(16,000) + 30 = −10 V vo(10 µs) = −20 + 20−0.125 = −2.35 V vo = −10 + (−2.35 + 10)e−(t − 10×10−6)/τ RTh = 4 kΩ‖16 kΩ = 3.2 kΩ τ = (3200)(20 × 10−9) = 64 µs; 1/τ = 15,625 vo = −10 + 7.65e−15,625(t − 10×10−6) 10 µs ≤ t ≤ ∞ P 7.76 0 ≤ t ≤ 200 µs; Re = 150‖100 = 60 kΩ; τ = (10 3 × 10−9 ) (60,000) = 200µs docsity.com 7–70 CHAPTER 7. Response of First-Order RL and RC Circuits 0 ≤ t ≤ 5: τ = 5/0 = ∞ iL(t) = 2e−t/∞ = 2e−0 = 2 iL(t) = 2 A, 0 ≤ t ≤ 5 s 5 ≤ t ≤ ∞: τ = 5 1 = 5 s; 1/τ = 0.2 iL(t) = 2e−0.2(t −5) A, t ≥ 5 s P 7.79 [a] 0 ≤ t ≤ 2.5 ms vo(0+) = 80 V; vo(∞) = 0 τ = L R = 2 ms; 1/τ = 500 vo(t) = 80e−500t V, 0+ ≤ t ≤ 2.5 ms vo(2.5− ms) = 80e−1.25 = 22.92 V io(2.5− ms) = (80 − 22.92) 20 = 2.85 A vo(2.5+ ms) = −20(2.85) = −57.08 V vo(∞) = 0; τ = 2 ms; 1/τ = 500 vo = −57.08e−500(t − 0.0025) V 2.5+ ms ≤ t ≤ ∞ docsity.com Problems 7–71 [b] [c] vo(5 ms) = −16.35 V io = +16.35 20 = 817.68 mA P 7.80 [a] io(0) = 0; io(∞) = 25 mA 1 τ = R L = 2000 250 × 103 = 8000 io = (25 − 25e−8000t) mA, 0 ≤ t ≤ 75 µs vo = 0.25 dio dt = 50e−8000t V, 0+ ≤ t ≤ 75− µs 75+ µs ≤ t ≤ ∞: io(75µs) = 25 − 25e−0.6 = 11.28 mA; io(∞) = 0 io = 11.28e−8000(t−75×10 −6) mA vo = (0.25) dio dt = −22.56e−8000(t−75µs) ·. . t < 0 : vo = 0 0+ ≤ t ≤ 75− µs : vo = 50e−8000t V 75+ µs ≤ t ≤ ∞ : vo = −22.56e−8000(t−75µs) [b] vo(75−µs) = 50e−0.6 = 27.44 V vo(75+µs) = −22.56 V [c] io(75−µs) = io(75+µs) = 11.28 mA docsity.com 7–72 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.81 [a] 0 ≤ t < 1 ms: vc(0+) = 0; vc(∞) = 50 V; RC = 400 × 103(0.01 × 10−6) = 4 ms; 1/RC = 250 vc = 50 − 50e−250t vo = 50 − 50 + 50e−250t = 50e−250t V, 0 ≤ t ≤ 1 ms 1 ms < t ≤ ∞: vc(1 ms) = 50 − 50e−0.25 = 11.06 V vc(∞) = 0 V τ = 4 ms; 1/τ = 250 vc = 11.06e−250(t − 0.001) V vo = −vc = −11.06e−250(t − 0.001) V, 1 ms < t ≤ ∞ [b] P 7.82 [a] t < 0; vo = 0 0 ≤ t ≤ 4 ms: τ = (200 × 103)(0.025 × 10−6) = 5 ms; 1/τ = 200 vo = 100 − 100e−200t V, 0 ≤ t ≤ 4 ms vo(4 ms) = 100(1 − e−0.8) = 55.07 V 4 ms ≤ t ≤ 8 ms: vo = −100 + 155.07e−200(t−0.004) V vo(8 ms) = −100 + 155.07e−0.8 = −30.32 V 8 ms ≤ t ≤ ∞: vo = −30.32e−200(t−0.008) V docsity.com Problems 7–75 [c] vo(0−) = 0 vo(0+) = 64 V vo(2− ms) = 320 − 256e−0.5 = 164.73 V vo(2+ ms) = 100.73 [d] [e] P 7.84 [a] Using Ohm’s law, vT = 5000iσ Using current division, iσ = 20,000 20,000 + 5000 (iT + βiσ) = 0.8iT + 0.8βiσ docsity.com 7–76 CHAPTER 7. Response of First-Order RL and RC Circuits Solve for iσ: iσ(1 − 0.8β) = 0.8iT iσ = 0.8iT 1 − 0.8β ; vT = 5000iσ = 4000iT (1 − 0.8β) Find β such that RTh = −5 kΩ: RTh = vT iT = 4000 1 − 0.8β = −5000 1 − 0.8β = −0.8 ·. . β = 2.25 [b] Find VTh; Write a KCL equation at the top node: VTh − 40 5000 + VTh 20,000 − 2.25iσ = 0 The constraint equation is: iσ = (VTh − 40) 5000 = 0 Solving, VTh = 50 V Write a KVL equation around the loop: 50 = −5000i + 0.2di dt Rearranging: di dt = 250 + 25,000i = 25,000(i + 0.01) docsity.com Problems 7–77 Separate the variables and integrate to find i; di i + 0.01 = 25,000 dt ∫ i 0 dx x + 0.01 = ∫ t 0 25,000 dx ·. . i = −10 + 10e25,000t mA di dt = (10 × 10−3)(25,000)e25,000t = 250e25,000t Solve for the arc time: v = 0.2 di dt = 50e25,000t = 45,000; e25,000t = 900 ·. . t = ln 900 25,000 = 272.1 µs P 7.85 Find the Thévenin equivalent with respect to the terminals of the capacitor. RTh calculation: iT = vT 2000 + vT 5000 − 4 vT 5000 ·. . iT vT = 5 + 2 − 8 10,000 = − 1 10,000 vT iT = −10,000 1 = −10 kΩ Open circuit voltage calculation: docsity.com 7–80 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.88 [a] τ = (25)(2) × 10−3 = 50 ms; 1/τ = 20 vc(0+) = 80 V; vc(∞) = 0 vc = 80e−20t V ·. . 80e−20t = 5; e20t = 16; t = ln 16 20 = 138.63 ms [b] 0+ < t < 138.63 ms: i = (2 × 10−6)(−1600e−20t) = −3.2e−20t mA 138.63+ ms < t ≤ ∞: τ = (2)(4) × 10−3 = 8 ms; 1/τ = 125 vc(138.63+ ms) = 5 V; vc(∞) = 80 V vc = 80 − 75e−125(t−0.13863) V, 138.63+ ms ≤ t ≤ ∞ i = 2 × 10−6(9375)e−125(t−0.13863) = 18.75e−125(t−0.13863) mA, 138.63+ ms ≤ t ≤ ∞ [c] 80 − 75e−125∆t = 0.85(80) = 68 80 − 68 = 75e−125∆t = 12 e125∆t = 6.25; ∆t = ln 6.25 12.5 ∼= 14.66 ms P 7.89 Use voltage division to find the voltage at the non-inverting terminal: vp = 80 100 (−45) = −36 V = vn docsity.com Problems 7–81 Write a KCL equation at the inverting terminal: −36 − 14 80,000 + 2.5 × 10−6 d dt (−36 − vo) = 0 ·. . 2.5 × 10−6 dvo dt = −50 80,000 Separate the variables and integrate: dvo dt = −250 ·. . dvo = −250dt ∫ vo(t) vo(0) dx = −250 ∫ t 0 dy ·. . vo(t) − vo(0) = −250t vo(0) = −36 + 56 = 20 V vo(t) = −250t + 20 Find the time when the voltage reaches 0: 0 = −250t + 20 ·. . t = 20 250 = 80 ms P 7.90 The equation for an integrating amplifier: vo = 1 RC ∫ t 0 (vb − va) dy + vo(0) Find the values and substitute them into the equation: RC = (100 × 103)(0.05 × 10−6) = 5 ms 1 RC = 200; vb − va = −15 − (−7) = −8 V vo(0) = −4 + 12 = 8 V vo = 200 ∫ t 0 −8 dx + 8 = (−1600t + 8) V, 0 ≤ t ≤ tsat RC circuit analysis for v2: v2(0+) = −4 V; v2(∞) = −15 V; τ = RC = (100 k)(0.05 µ) = 5 ms docsity.com 7–82 CHAPTER 7. Response of First-Order RL and RC Circuits v2 = v2(∞) + [v2(0+) − v2(∞)]e−t/τ = −15 + (−4 + 15)e−200t = −15 + 11e−200t V, 0 ≤ t ≤ tsat vf + v2 = vo ·. . vf = vo − v2 = 23 − 1600t − 11e−200t V, 0 ≤ t ≤ tsat Note that −1600tsat + 8 = −20 ·. . tsat = −28−1600 = 17.5 ms so the op amp operates in its linear region until it saturates at 17.5 ms. P 7.91 vo = − 1 R(0.5 × 10−6) ∫ t 0 4 dx + 0 = −4t R(0.5 × 10−6) −4(15 × 10−3) R(0.5 × 10−6) = −10 ·. . R = −4(15 × 10 −3) −10(0.5 × 10−6) = 12 kΩ P 7.92 vo = −4t R(0.5 × 10−6) + 6 = −4(40 × 10−3) R(0.5 × 10−6) + 6 = −10 ·. . R = −4(40 × 10 −3) −16(0.5 × 10−6) = 20 kΩ P 7.93 [a] Cdvp dt + vp − vb R = 0; therefore dvp dt + 1 RC vp = vb RC vn − va R + C d(vn − vo) dt = 0; therefore dvo dt = dvn dt + vn RC − va RC But vn = vp Therefore dvn dt + vn RC = dvp dt + vp RC = vb RC Therefore dvo dt = 1 RC (vb − va); vo = 1 RC ∫ t 0 (vb − va) dy [b] The output is the integral of the difference between vb and va and then scaled by a factor of 1/RC. docsity.com Problems 7–85 vo(1 µs) = −125 × 1010(1 × 10−6)2 = −1.25 V 1 µs ≤ t ≤ 3 µs: vg = 4 − 2 × 106t vo = −125 × 104 ∫ t 1×10−6 (4 − 2 × 106x) dx − 1.25 = −125 × 104 [ 4x ∣∣∣∣t 1×10−6 −2 × 106 x 2 2 ∣∣∣∣t 1×10−6 ] − 1.25 = −5 × 106t + 5 + 125 × 1010t2 − 1.25 − 1.25 = 125 × 1010t2 − 5 × 106t + 2.5 V, 1 µs ≤ t ≤ 3 µs vo(3 µs) = 125 × 1010(3 × 10−6)2 − 5 × 106(3 × 10−6) + 2.5 = −1.25 3 µs ≤ t ≤ 4 µs: vg = −8 + 2 × 106t vo = −125 × 104 ∫ t 3×10−6 (−8 + 2 × 106x) dx − 1.25 = −125 × 104 [ −8x ∣∣∣∣t 3×10−6 +2 × 106 x 2 2 ∣∣∣∣t 3×10−6 ] − 1.25 = 107t − 30 − 125 × 1010t2 + 11.25 − 1.25 = −125 × 1010t2 + 107t − 20 V, 3 µs ≤ t ≤ 4 µs vo(4 µs) = −125 × 1010(4 × 10−6)2 + 107(4 × 10−6) − 20 = 0 [b] [c] The output voltage will also repeat. This follows from the observation that at t = 4 µs the output voltage is zero, hence there is no energy stored in the capacitor. This means the circuit is in the same state at t = 4 µs as it was at t = 0, thus as vg repeats itself, so will vo. docsity.com 7–86 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.97 [a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal. At the instant T1 turns ON the capacitor C2 is connected across b2 − e2, thus vbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity of the voltage on C2 starts to reverse, that is, the right-hand terminal of C2 starts to charge toward +VCC . At the same time, C1 is charging toward VCC , positive on the right. At the instant the charge on C2 reaches zero, vbe2 is zero, T2 turns ON. This makes vbe1 = −VCC and T1 snaps OFF. Now the capacitors C1 and C2 start to charge with the polarities to turn T1 ON and T2 OFF. This switching action repeats itself over and over as long as the circuit is energized. At the instant T1 turns ON, the voltage controlling the state of T2 is governed by the following circuit: It follows that vbe2 = VCC − 2VCCe−t/R2C2 . [b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the voltage across C1, thus It follows that vce2 = VCC − VCCe−t/RLC1 . [c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will become positive and turn T2 ON. vbe2 = 0 when VCC − 2VCCe−t/R2C2 = 0, or when t = R2C2 ln 2. [d] When t = R2C2 ln 2, we have vce2 = VCC − VCCe−[(R2C2 ln 2)/(RLC1)] = VCC − VCCe−10 ln 2 ∼= VCC [e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have docsity.com Problems 7–87 ib1 = VCC R1 + VCC RL e−t/RLC1 [f] At the instant T2 turns back ON, t = R2C2 ln 2; therefore ib1 = VCC R1 + VCC RL e−10 ln 2 ∼= VCC R1 When T2 turns OFF, ib1 drops to zero instantaneously. [g] [h] P 7.98 [a] tOFF2 = R2C2 ln 2 = 14.43 × 103(1 × 10−9) ln 2 ∼= 10 µs [b] tON2 = R1C1 ln 2 ∼= 10 µs [c] tOFF1 = R1C1 ln 2 ∼= 10 µs [d] tON1 = R2C2 ln 2 ∼= 10 µs [e] ib1 = 10 1000 + 10 14,430 ∼= 10.69 mA [f] ib1 = 10 14,430 + 10 1000 e−10 ∼= 0.693 mA docsity.com 7–90 CHAPTER 7. Response of First-Order RL and RC Circuits So, (tc − to) = (19,640)(10 × 10−6) ln 4 − 0.1081 − 0.108 = 0.289 s The flash lasts for 0.289 s. P 7.104 [a] At t = 0 we have τ = (800)(25) × 10−3 = 20 sec; 1/τ = 0.05 vc(∞) = 40 V; vc(0) = 5 V vc = 40 − 35e−0.05t V, 0 ≤ t ≤ to 40 − 35e−0.05to = 15; ·. . e0.05to = 1.4 to = 20 ln 1.4 s = 6.73 s At t = to we have The Thévenin equivalent with respect to the capacitor is τ = (800 81 ) (25) × 10−3 = 20 81 s; 1 τ = 81 20 = 4.05 vc(to) = 15 V; vc(∞) = 4081 V vc(t) = 40 81 + ( 15 − 40 81 ) e−4.05(t−to) V = 40 81 + 1175 81 e−4.05(t−to) ·. . 40 81 + 1175 81 e−4.05(t−to) = 5 docsity.com Problems 7–91 1175 81 e−4.05(t−to) = 365 81 e4.05(t−to) = 1175 365 = 3.22 t − to = 14.05 ln 3.22 ∼= 0.29 s One cycle = 7.02 seconds. N = 60/7.02 = 8.55 flashes per minute [b] At t = 0 we have τ = 25R × 10−3; 1/τ = 40/R vc = 40 − 35e−(40/R)t 40 − 35e−(40/R)to = 15 ·. . to = R40 ln 1.4, R in kΩ At t = to: vTh = 10 R + 10 (40) = 400 R + 10 ; RTh = 10R R + 10 kΩ τ = (25)(10R) × 10−3 R + 10 = 0.25R R + 10 ; 1 τ = 4(R + 10) R vc = 400 R + 10 + ( 15 − 400 R + 10 ) e− 4(R+10) R (t−to) ·. . 400 R + 10 + [15R − 250 R + 10 ] e− 4(R+10) R (t−to) = 5 or (15R − 250 R + 10 ) e− 4(R+10) R (t−to) = 5R − 350 (R + 10) docsity.com 7–92 CHAPTER 7. Response of First-Order RL and RC Circuits ·. . e 4(R+10)R (t−to) = 3R − 50 R − 70 ·. . t − to = R4(R + 10) ln (3R − 50 R − 70 ) At 12 flashes per minute to + (t − to) = 5 s ·. . R 40 ln 1.4︸ ︷︷ ︸+ R 4(R + 10) ln (3R − 50 R − 70 ) = 5 dominant term Start the trial-and-error procedure by setting (R/40) ln 1.4 = 5, then R = 200/(ln 1.4) or 594.40 kΩ. If R = 594.40 kΩ then t − to ∼= 0.29 s. Second trial set (R/40) ln 1.4 = 4.7 s or R = 558.74 kΩ. With R = 558.74 kΩ, t − to ∼= 0.30 s The procedure converges to R = 559.3 kΩ P 7.105 [a] to = RC ln ( Vmin − Vs Vmax − Vs ) = (3700)(250 × 10−6) ln (−700 −100 ) = 1.80 s tc − to = RCRL R + RL ln ( Vmax − VTh Vmin − VTh ) RL R + RL = 1.3 1.3 + 3.7 = 0.26 RC = (3700)(250 × 10−6) = 0.925 s VTh = 1000(1.3) 1.3 + 3.7 = 260 V RTh = 3.7 k||1.3 k = 962 Ω ·. . tc − to = (0.925)(0.26) ln(640/40) = 0.67 s ·. . tc = 1.8 + 0.67 = 2.47 s flashes/min = 60 2.47 = 24.32 [b] 0 ≤ t ≤ to: vL = 1000 − 700e−t/τ1 τ1 = RC = 0.925 s to ≤ t ≤ tc: vL = 260 + 640e−(t−to)/τ2 τ2 = RThC = 962(250) × 10−6 = 0.2405 s docsity.com