EECE 210C Lecture, Fall 2006

6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200tA v = Ldig dt = −9.6e−300t + 38.4e−1200tV, t > 0+ v(0+) = −9.6 + 38.4 = 28.8 V [b] v = 0 when 38.4e−1200t = 9.6e−300t or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W [d] dp dt = 0 when e1800t − 12.5e900t + 16 = 0 Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0 x = 1.45, t = ln 1.45 900 = 411.05 µs x = 11.05, t = ln 11.05 900 = 2.67 ms p is maximum at t = 411.05 µs [e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W [f] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78 A wmax = (1/2)(4 × 10−3)(3.78)2 = 28.6 mJ [g] W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms. 6–1 6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance AP 6.2 [a] i = C dv dt = 24 × 10−6 d dt[e−15,000t sin 30,000t] = [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, i(0+) = 0.72 A [b] i � π 80 ms � = −31.66 mA, v � π 80 ms � = 20.505 V, p = vi = −649.23 mW [c] w = �1 2 � Cv2 = 126.13 µJ AP 6.3 [a] v = � 1 C � � t 0− i dx + v(0−) = 1 0.6 × 10−6 � t 0− 3 cos 50,000x dx = 100 sin 50,000t V [b] p(t) = vi = [300 cos 50,000t] sin 50,000t = 150 sin 100,000t W, p(max) = 150 W [c] w(max) = �1 2 � Cv2 max = 0.30(100)2 = 3000 µJ = 3 mJ AP 6.4 [a] Leq = 60(240) 300 = 48 mH [b] i(0+) = 3 + −5 = −2 A [c] i = 125 6 � t 0+(−0.03e−5x) dx − 2 = 0.125e−5t − 2.125 A [d] i1 = 50 3 � t 0+(−0.03e−5x) dx + 3 = 0.1e−5t + 2.9 A i2 = 25 6 � t 0+(−0.03e−5x) dx − 5 = 0.025e−5t − 5.025 A i1 + i2 = i AP 6.5 v1 = 0.5 × 106 � t 0+ 240 × 10−6e−10x dx − 10 = −12e−10t + 2 V v2 = 0.125 × 106 � t 0+ 240 × 10−6e−10x dx − 5 = −3e−10t − 2 V v1(∞) = 2 V, v2(∞) = −2 V W = �1 2(2)(4) + 1 2(8)(4) � × 10−6 = 20 µJ Problems 6–3 AP 6.6 [a] Summing the voltages around mesh 1 yields 4di1 dt + 8d(i2 + ig) dt + 20(i1 − i2) + 5(i1 + ig) = 0 or 4di1 dt + 25i1 + 8di2 dt − 20i2 = − � 5ig + 8dig dt � Summing the voltages around mesh 2 yields 16d(i2 + ig) dt + 8di1 dt + 20(i2 − i1) + 780i2 = 0 or 8di1 dt − 20i1 + 16di2 dt + 800i2 = −16dig dt [b] From the solutions given in part (b) i1(0) = −0.4 − 11.6 + 12 = 0; i2(0) = −0.01 − 0.99 + 1 = 0 These values agree with zero initial energy in the circuit. At infinity, i1(∞) = −0.4A; i2(∞) = −0.01A When t = ∞ the circuit reduces to ·. . i1(∞) = − �7.8 20 + 7.8 780 � = −0.4A; i2(∞) = − 7.8 780 = −0.01A From the solutions for i1 and i2 we have di1 dt = 46.40e−4t − 60e−5t di2 dt = 3.96e−4t − 5e−5t Also, dig dt = 7.84e−4t Thus 4di1 dt = 185.60e−4t − 240e−5t 6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 25i1 = −10 − 290e−4t + 300e−5t 8di2 dt = 31.68e−4t − 40e−5t 20i2 = −0.20 − 19.80e−4t + 20e−5t 5ig = 9.8 − 9.8e−4t 8dig dt = 62.72e−4t Test: 185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t +0.20 + 19.80e−4t − 20e−5t ?= −[9.8 − 9.8e−4t + 62.72e−4t] −9.8 + (300 − 240 − 40 − 20)e−5t +(185.60 − 290 + 31.68 + 19.80)e−4t ?= −(9.8 + 52.92e−4t) −9.8 + 0e−5t + (237.08 − 290)e−4t ?= −9.8 − 52.92e−4t −9.8 − 52.92e−4t = −9.8 − 52.92e−4t (OK) Also, 8di1 dt = 371.20e−4t − 480e−5t 20i1 = −8 − 232e−4t + 240e−5t 16di2 dt = 63.36e−4t − 80e−5t 800i2 = −8 − 792e−4t + 800e−5t 16dig dt = 125.44e−4t Test: 371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t −8 − 792e−4t + 800e−5t ?= −125.44e−4t (8 − 8) + (800 − 480 − 240 − 80)e−5t +(371.20 + 232 + 63.36 − 792)e−4t ?= −125.44e−4t (800 − 800)e−5t + (666.56 − 792)e−4t ?= −125.44e−4t −125.44e−4t = −125.44e−4t (OK) Problems 6–5 Problems P 6.1 [a] i = 0 t < 0 i = 50t A 0 ≤ t ≤ 5 ms i = 0.5 − 50t A 5 ≤ t ≤ 10 ms i = 0 10 ms < t [b] v = Ldi dt = 20 × 10−3(50) = 1 V 0 ≤ t ≤ 5 ms v = 20 × 10−3(−50) = −1 V 5 ≤ t ≤ 10 ms v = 0 t < 0 v = 1 V 0 < t < 5 ms v = −1 V 5 < t < 10 ms v = 0 10 ms < t p = vi p = 0 t < 0 p = (50t)(1) = 50t W 0 < t < 5 ms p = (0.5 − 50t)(−1) = 50t − 0.5 W 5 < t < 10 ms p = 0 10 ms < t w = 0 t < 0 w = � t 0 (50x) dx = 50x2 2 ���� t 0= 25t2 J 0 < t < 5 ms w = � t 0.005(50x − 0.5) dx + 0.625 × 10−3 = 25x2 − 0.5x ���� t 0.005 +0.625 × 10−3 = 25t2 − 0.5t + 2.5 × 10−3 J 5 < t < 10 ms w = 0 10 ms < t 6–6 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance P 6.14 iC = C(dv/dt) 0 < t < 0.5 : vc = 30t2 V iC = 20 × 10−6(60)t = 1.2t mA 0.5 < t < 1 : vc = 30(t − 1)2 V iC = 20 × 10−6(60)(t − 1) = 1.2(t − 1) mA P 6.15 [a] 0 ≤ t ≤ 5 µs C = 5 µF 1 C = 2 × 105 v = 2 × 105 � t 0 4 dx + 12 v = 8 × 105t + 12 V 0 ≤ t ≤ 5 µs v(5 µs) = 4 + 12 = 16 V [b] 5 µs ≤ t ≤ 20 µs v = 2 × 105 � t 5×10−6 −2 dx + 16 = −4 × 105t + 2 + 16 v = −4 × 105t + 18V 5 ≤ t ≤ 20 µs v(20 µs) = −4 × 105(20 × 10−6) + 18 = 10 V Problems 6–7 [c] 20 µs ≤ t ≤ 25 µs v = 2 × 105 � t 20×10−6 6 dx + 10 = 12 × 105t − 24 + 10 v = 12 × 105t − 14 V, 20 µs ≤ t ≤ 25 µs v(25 µs) = 12 × 105(25 × 10−6) − 14 = 16 V [d] 25 µs ≤ t ≤ 35 µs v = 2 × 105 � t 25×10−6 4 dx + 16 = 8 × 105t − 20 + 16 v = 8 × 105t − 4 V, 25 µs ≤ t ≤ 35 µs v(35 µs) = 8 × 105(35 × 10−6) − 4 = 24 V [e] 35 µs ≤ t < ∞ v = 2 × 105 � t 35×10−6 0 dx + 24 = 24 v = 24 V, 35 µs ≤ t < ∞ [f] P 6.21 30∥20 = 12 H 80∥(8 + 12) = 16 H 60∥(14 + 16) = 20 H 15∥(20 + 10) = 20 H Lab = 5 + 10 = 15 H 6–8 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance P 6.22 [a] i(t) = −1 2 � t 012e−x dx + 6 = 6e−x ���� t 0 +6 = 6e−t − 6 + 6 i(t) = 6e−t A, t ≥ 0 [b] i1(t) = −1 3 � t 012e−x dx + 2 = 4e−x ���� t 0 +2 = 4(e−t − 1) + 2 i1(t) = 4e−t − 2 A, t ≥ 0 [c] i2(t) = −1 6 � t 012e−x dx + 4 = 2e−x ���� t 0 +4 = 2(e−t − 1) + 4 i2(t) = 2e−t + 2 A, t ≥ 0 [d] p = vi = (12e−t)(6e−t) = 72e−2t W w = � ∞ 0 p dt = � ∞ 0 72e−2t dt = 72e−2t −2 ���� ∞ 0 = 36 J [e] w = 1 2(3)(2)2 + 1 2(6)(4)2 = 54 J [f] wtrapped = 1 2(3)(−2)2 + 1 2(6)(2)2 = 18 J wtrapped = 54 − 36 = 18 J checks [g] Yes, they agree. Problems 6–9 P 6.26 Work from the right hand side of the circuit, simplifying step by step: 1. 48 µF in series with 16 µF : 1/C = 1/16 µ + 1/48 µ ·. . C = 12 µF The voltages add in series, so the 12 µF capacitor has a voltage of 20 V, negative at the top. 2. Previous 12 µF in parallel with 3 µF : C = 12 µ + 3 µ = 15 µF The voltage is 20 V, negative at the top. 3. Previous 15 µF in series with 30 µF : 1/C = 1/15 µ + 1/30 µ ·. . C = 10 µF The voltages add in series, so the 10 µF capacitor has a voltage of 10 V, positive at the right. 4. Previous 10 µF in parallel with 10 µF : C = 10 µ + 10 µ = 20 µF The voltage is 10 V, negative at the top. 5. Previous 20 µF in series with 5 µF and 4 µF : 1/C = 1/20 µ + 1/5 µ + 1/4 µ ·. . C = 2 µF The voltages in series add: 5V − 10V + 30V = 25V positive at the top. The equivalent capacitance is 2 µF with a voltage of 25 V, positive at the top. P 6.27 [a] vo = − 1 2 × 10−6 � t 020 × 10−6e−x dx + 10 = 10e−x ���� t 0 +10 = 10e−t V, t ≥ 0 [b] v1 = − 1 3 × 10−6(20 × 10−6)e−x ���� t 0 +4 = 6.67e−t − 2.67 V, t ≥ 0 [c] v2 = − 1 6 × 10−6(20 × 10−6)e−x ���� t 0 +6 = 3.33e−t + 2.67 V, t ≥ 0 [d] p = vi = (10e−t)(20 × 10−6)e−t = 200 × 10−6e−2t 6–10 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance w = � ∞ 0 200 × 10−6e−2t dt = 200 × 10−6e−2t −2 ���� ∞ 0 = −100 × 10−6(0 − 1) = 100 µJ [e] w = 1 2(3 × 10−6)(4)2 + 1 2(6 × 10−9)(6)2 = 132 µJ [f] wtrapped = 1 2(3 × 10−6)(8/3)2 + 1 2(6 × 10−6)(8/3)2 = 32 µJ CHECK: 100 + 32 = 132 µJ [g] Yes, they agree. P 6.34 [a] −2dig dt + 16di2 dt + 32i2 = 0 16di2 dt + 32i2 = 2dig dt [b] i2 = e−t − e−2t A di2 dt = −e−t + 2e−2t A/s ig = 8 − 8e−t A dig dt = 8e−t A/s ·. . −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t [c] v1 = 4dig dt − 2di2 dt = 4(8e−t) − 2(−e−t + 2e−2t) = 34e−t − 4e−2t V, t > 0 [d] v1(0) = 34 − 4 = 30 V; Also v1(0) = 4dig dt (0) − 2di2 dt (0) = 4(8) − 2(−1 + 2) = 32 − 2 = 30 V Yes, the initial value of v1 is consistent with known circuit behavior.