1 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 2 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 3 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 4 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 5 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 6 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 7 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 8 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: C =12µF Time (µs) 𝑖= 𝐶 𝑑𝑣 𝑑𝑡 𝑑𝑣 𝑑𝑡 (V/µs) i(t) (A) 0 ≤t≤6 2 24 6 ≤t ≤10 −5 −60 10 ≤t ≤16 1.33 16 t >16 0 0 Chapter 6: Capacitance and Inductance 9 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 10 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 11 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 12 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 𝑖= 𝐶 C =12µF 𝑑𝑣 Time(ms) 0 ≤t ≤2 2 ≤t ≤4 4 ≤t ≤7 𝑑𝑡 𝑑𝑣 𝑑𝑡 (V/µs) 2 −1 0 7 ≤t ≤9 9 ≤t ≤10 t >0 0 i(t) (A) 6 −3 0 3 − −4.5 2 1 3 0 Chapter 6: Capacitance and Inductance 13 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 6: Capacitance and Inductance 14 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 15 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 16 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 17 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 18 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 19 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 20 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: From the graph we know that 𝑣(𝑡) 𝑖(𝑡) 𝑝(𝑡) 𝑝(𝑡) = −2𝑡 + 24 0 ≤ 𝑡 ≤ 6 = 24 = 𝑣(𝑡) × 𝑖(𝑡) = −48𝑡 + 576 𝑡 𝐸(𝑡) = ∫ 𝑝(𝑥)𝑑𝑥 0 𝐸(𝑡) = −24𝑡 2 + 576𝑡 𝑚𝐽 = 3.455 𝑚𝐽 𝑎𝑡 𝑡 = 6𝑚𝑠 Chapter 6: Capacitance and Inductance 21 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 22 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 23 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 24 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 25 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 26 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 1 ∫ 𝑖𝑑𝑡 + 𝑉(𝑡0) 𝐶 1 ∫ 4𝑡 × 10−3 𝑑𝑡 + 10 = 50 × 10−3 2𝑡 2 = + 10 50 𝑣= Chapter 6: Capacitance and Inductance 27 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 6: Capacitance and Inductance 28 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 29 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 30 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 31 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 𝑡 𝑤 = 𝐿 ∫ 𝐿𝑖 2 𝑑𝑡 = 0 1 2 1 𝐿𝑡 (𝑡) − 𝐿𝑖 2 (−∞) 2 2 1 = × 80 × 10−3 × (60 × 10−3 )2 − 0 2 = 144µ𝐽 Chapter 6: Capacitance and Inductance 32 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 33 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: IR = 𝑉 = 6 𝑅 12 𝑑𝑉 IC= 𝐶 𝑑𝑡 𝑒 −2000𝑡 = 0.5𝑒 −2000𝑡 = 100 × 10−3 × 6(−2000)𝑒 −2000𝑡 = −120010−2000𝑡 I=IR+IC=-1199.5𝑒 −2000𝑡 𝑝 = 𝑣𝑖 = −7197𝑒 −4000𝑡 𝑊 Chapter 6: Capacitance and Inductance 34 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 1 𝑡 ∫ 6(1 − 𝑒 −𝑡 )𝑑𝑡 + 0 1 0 2 = 12(𝑡 + 𝑒 −𝑡 ) − 12 𝑣(2) = 12(2 + 𝑒 −𝑡 ) − 12 = 13.624𝑉 𝑝 = 𝑖𝑣 = [12(𝑡 + 𝑒 −𝑡 ) − 12]6(1 − 𝑒 −𝑡 ) 𝑝(2) = [12(2 + 𝑒 −2 ) − 12]6(1 − 𝑒 −2 ) = 70.66 𝑊 𝑣(𝑡) = Chapter 6: Capacitance and Inductance 35 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 36 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 37 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 38 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 39 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 40 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 41 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 42 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 43 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 44 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (See Next Page) Chapter 6: Capacitance and Inductance 45 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 46 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 47 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 48 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 49 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 50 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 51 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 52 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 53 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 54 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 55 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 56 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 57 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 58 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 59 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 60 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 61 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (a) 3μF is in series with 6μF 3x6/(9) = 2μF V4µF= 1/2 x 120 = 60V V2µF = 60V V6µF= 3 6+3 (60) = 20𝑉 V3µF = 60 − 20 = 40 𝑉 1 (b) Hence 𝑤 = 𝐶𝑣 2 1 2 W4µF = × 4 × 10−6 × 3600 = 7.2𝑚𝐽 2 1 W2µF= × 2 × 10−6 × 3600 = 3.6𝑚𝐽 2 1 W6µF= × 2 × 10−6 × 400 = 1.2𝑚𝐽 2 1 W3µ= × 2 × 10−6 × 1600 = 2.4𝑚𝐽 2 Chapter 6: Capacitance and Inductance 62 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 63 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 64 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 65 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 66 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 67 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: The equivalent capacitor can be evaluated as following 8 × 6µ𝐹 8+6 Chapter 6: Capacitance and Inductance 68 Irwin, Engineering Circuit Analysis, 11e ISV 2µ𝐹 + 24µ𝐹 7 So the Cequivalent= 2.85µF Chapter 6: Capacitance and Inductance 69 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 70 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 71 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: C in series with C = C/(2) C/2 in parallel with C = 3C/2 3 in series with C = 2 3𝐶 2 3𝐶 2 𝐶 5 2 𝐶× In parallel with C = C+ 3𝐶 5 = 1.6 𝐶 Chapter 6: Capacitance and Inductance 72 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 73 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: We combine 10, 20, and 30μF capacitors in parallel to have 60μF. Then 60μF capacitor in series with another 60μF capacitor gives 30μF. 30 + 50 = 80μF, 80 + 40 = 120μF the circuit is reduced to that shown below. 120-μF capacitor in series with 80μF gives (80x120)/200 = 48μF 48 + 12 = 60μF 60μF capacitor in series with 12μF gives (60x12)/72 = 10μF Chapter 6: Capacitance and Inductance 74 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 75 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 76 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 77 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 78 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 79 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 80 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 81 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 82 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 83 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 84 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 85 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 86 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 87 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: The has been modified as following b From the figure in P6.53, 4C1=2C2 From figure above 3(C1+C2) = 3C2 Solving for C1 and C2.yields C1= 2µF C2 = 4µF Chapter 6: Capacitance and Inductance 88 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 89 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 90 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 91 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 92 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 93 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 94 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: We can simplify the above figure by as following L1=L3=L6= 1 mH L2= 12 mH, L4 = 6mH L5=4 mH, L7= L8= L9=2 mH Leq1 = L6 + L7 = 3 Mh Leq2 = L4Lwq1/(L4 + Leq1) = 2Mh Leq3 = L8 + L9 + 4 Mh Leq4 = Leq3L5/(Leq3+Leq5)= 2Mh Leq5 = 𝐿2(𝐿𝑒𝑞2+𝐿𝑒𝑞4) 𝐿2+𝐿𝑒𝑞2+𝐿𝑒𝑞4 = 5𝑚𝐻 Chapter 6: Capacitance and Inductance 95 Irwin, Engineering Circuit Analysis, 11e ISV LAB = 5 mH Chapter 6: Capacitance and Inductance 96 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 97 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: The figure in the question can be drawn as follows 4𝐿 + 2) 𝐿 4+𝐿 =2 4𝐿 +2+𝐿 4+𝐿 6𝐿2 + 8𝐿 = 2𝐿2 + 20𝐿 + 16 (𝑙 + 4)(𝐿 + 1) = 0 𝐿 = 4𝑚𝐻 ( Chapter 6: Capacitance and Inductance 98 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 99 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 100 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 101 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: (a) VL= 𝐿 = −2𝑒 𝑑𝑖 𝑑𝑡 −400𝑡 = 100 × 10−3 (−400) × 50 × 10−3 𝑒 −400𝑡 𝑉) (b) R and L are parallel to each other so VR = VL 1 (C) 𝑤 = 𝐿𝑖 2 = 0.5 × 100 × 10−3 (0.05)2 2 = 125µ𝐽 Chapter 6: Capacitance and Inductance 102 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 103 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 104 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 105 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Rs = 10KΩ 𝑣 = −10 ∫ 𝑣𝑠 𝑑𝑡 Req = Rs + 70K = 80k Considering i1 = i2 (ideal op-amp) 𝑣𝑠 − 0 𝑑𝑣 = −𝐶 𝑅𝑒𝑞 𝑑𝑡 1 ∫ 𝑣𝑠 𝑑𝑡 𝑣𝑜 = − 𝑅𝑒𝑞 𝐶 1 𝑅𝑒𝑞 𝐶 = 10 𝐶 = 1.25µ𝐹 Chapter 6: Capacitance and Inductance 106 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 107 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 108 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 109 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 110 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 111 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: 1 1 2 2 (a) W5 = 𝐿𝑖 2 = × 5 × (4)2 = 40 𝐽 1 W20 = (20)(−2)2 = 40𝐽 2 (b) w= w5+w20 = 80J (c) i1 = I2 == 1 𝑡 ∫ −50𝑒 −200𝑡 𝐿2 0 𝑑𝑡 + 𝑖1(0) 1 𝑡 ∫ −50𝑒 −200𝑡 𝑑𝑡 𝐿2 0 = 5 × 10−5 (𝑒 −200𝑡 − 1) + 4 𝐴 + 𝑖2(0) = 1.25 × 10−5 (𝑒 −200𝑡 − 1) − 2 𝐴 (d) I = i1+i2 = 6.25 × 10−5 (𝑒 −200𝑡 − 1) + 2 𝐴 Chapter 6: Capacitance and Inductance 112 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: If we differentiate the wave form of I it will give the wave form similar to the voltage wave form So 𝑑𝑖 𝑑𝑡 −4𝑡, 0 < 𝑡 < 1 𝑚𝑠 𝑖={ 8 − 4𝑡, 1 < 𝑡 < 2 𝑚𝑠 4000𝐿, 0 < 𝑡 < 1 𝑚𝑠 𝑣={ −4000𝐿, 1 < 𝑡 < 2 𝑚𝑠 𝑣=𝐿 From the graph 4000L = 5 L = 1.25 mH Chapter 6: Capacitance and Inductance 113 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: Chapter 6: Capacitance and Inductance 114 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 115 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: i = iR+ iC 𝑣𝑖 − 0 0 − 𝑣0 𝑑(0 − 𝑣0 ) = +𝐶 𝑅 𝑅𝐹 𝑑𝑡 𝑅𝐹 = 106 × 10−6 = 1 𝑣𝑖 = −(𝑣0 + The graph of 𝑣0 is shown below Chapter 6: Capacitance and Inductance 𝑑𝑣0 ) 𝑑𝑡 116 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance 117 Irwin, Engineering Circuit Analysis, 11e ISV SOLUTION: The given equation can be modeled with a op amp summer, integrator and an inverter. Summer equation 𝑣0 = −( 𝑅𝐹 𝑅𝐹2 𝑣1 + 𝑣 ) 𝑅1 𝑅2 2 Integrator equation 𝑣0 = − 1 ∫ 𝑣𝑠 𝑑𝑡 𝑅𝐶 Inverter equation 𝑣0 = − 𝑅𝐹 𝑣 𝑅 𝑠 Comparing the above equation with the given equation we get 𝑅𝐹 = 10𝑅 1 𝐶= 2𝑅 and for inverter 𝑅𝐹 = 𝑅 So the circuit of the given equation is following Chapter 6: Capacitance and Inductance 118 Irwin, Engineering Circuit Analysis, 11e ISV Chapter 6: Capacitance and Inductance