Chapter 09 Part 1-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 09 Part 1-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! February 5, 2006 CHAPTER 9 P.P.9.1 amplitude = 5 phase = -60° angular frequency (ω) = 4π = 12.57 rad/s period (T) = ω π2 = 0.5 s frequency (f) = T 1 = 2 Hz P.P.9.2 )9025tcos(4)25tsin(4-i1 °+°+ω=°+ω= )115tcos(4i1 °+ω= , 377=ω rad/s Compare this with )40tcos(5i2 °−ω= indicates that the phase angle between and is 1i 2i 115° + 40° = 155° Thus, i1 leads i2 by 155° P.P.9.3 (a) (5 + j2)(-1 + j4) = -5 + j20 – j2 – 8 = -13 + j18 5∠60° = 2.5 + j4.33 (5 + j2)(-1 + j4) – 5∠60° = -15.5 + j13.67 [ (5 + j2)(-1 + j4) – 5∠60 ]* = -15.5 – j13.67 = 20.67∠221.41° (b) 3∠40° = 2.298 + j1.928 10 + j5 + 3∠40° = 12.298 + j6.928 = 14.115∠29.39° -3 + j4 = 5∠126.87° °∠= °∠ °∠ = + °∠++ 48.97-823.2 87.1265 39.29115.14 4j3- 4035j10 2.823∠-97.48° = -0.3675 – j2.8 10∠30° = 8.66 + j5 =°∠+ + °∠++ 3010 4j3- 4035j10 8.293 + j2.2 P.P.9.4 (a) -cos(A) = cos(A – 180°) = cos(A + 180°) Hence, v = -7 cos(2t + 40°) = 7 cos(2t + 40° + 180°) v = 7 cos(2t + 220°) The phasor form is V = 7∠220° V (b) Since sin(A) = cos(A – 90°), docsity.com i = 4 sin(10t + 10°) = 4 cos(10t+10° – 90°) i = 4 cos(10t – 80°) The phasor form is I = 4∠-80° A P.P.9.5 (a) Since -1 = 1∠-180° = 1∠180° V = -10∠30° = 10∠(30°+180°) = 10∠210° The sinusoid is v(t) = 10 cos(ωt + 210°) V (b) I = j (5 – j12) = 12 + j5 = 13∠22.62° The sinusoid is i(t) = 13 cos(ωt + 22.62°) A P.P.9.6 Let )45tcos(20)30tsin(10-V °−ω+°+ω= Then, )45tcos(20)9030tcos(10V °−ω+°+°+ω= Taking the phasor of each term V = 10∠120° + 20∠-45° V = -5 + j8.66 + 14.14 – j14.14 V = 9.14 – j5.48 = 10.66∠-30.95° Converting V to the time domain v(t) = 10.66 cos(ωt – 30.95°)V P.P.9.7 Given that )30t5cos(20dtv10v5 dt dv 2 °−=++ ∫ we take the phasor of each term to get 2jω V +5 V + ωj 10 V = 20∠-30°, ω = 5 V [j10 + 5 – j(10/5)] = V (5 + j8) = 20∠-30° V = °∠ °∠ = + °∠ 58434.9 30-20 8j5 30-20 V = 2.12∠-88° Converting V to the time domain v(t) = 2.12 cos(5t – 88°)V P.P.9.8 For the capacitor, V = I / (jωC),where V = 6∠-30°, ω = 100 I = jωC V = (j100)(50x10-6)(6∠-30°) I = 30∠60° mA i(t) = 30 cos(100t + 60°) mA P.P.9.9 Vs = 5∠0°, ω = 10 docsity.com P.P.9.13 To show that the circuit in Fig. (a) meets the requirement, consider the equivalent circuit in Fig. (b). Z = -j10 || (10 – j10) = j2010 j10)(10j10- − − = j21 j10)(10j- − − = 2 – j6 Ω + Vo − 10 Ω -j10 Ω + Vi − (a) 10 Ω -j10 Ω V1 + − 10 Ω Z = 2−j6 ΩVi = 10 V + V1 − (b) docsity.com V1 = )j1(3 10 )10( 6j210 6j2 −= −+ − Vo = 10j10 10j- − V1 = )j1(3 10 j1 j- −⎟ ⎠ ⎞ ⎜ ⎝ ⎛⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 3 10 j- Vo = °∠ 90-3 10 This implies that the RC circuit provides a 90° lagging phase shift. The output voltage is 3 10 = 3.333 V P.P.9.14 the 1-mH inductor is jωL = = j31.42 )101)(105)(2(j 3-3 ××π the 2-mH inductor is jωL = = j62.83 )102)(105)(2(j 3-3 ××π Consider the circuit shown below. Z = 10 || (50 + j62.83) = 83.62j60 )83.62j50)(10( + + Z = 9.205 + j0.833 = 9.243∠5.17° V1 = Z / (Z + j31.42) Vi = )1(253.32j205.9 17.5243.9 + °∠ = 0.276∠-68.9° Vo = 83.62j50 50 + V1 = °∠ °∠ 49.51297.80 )9.68-276.0(50 = 0.172∠-120.4° Therefore, magnitude = 0.172 phase = 120.4° phase shift is lagging + Vo − j31.42 Ω 10 Ω + Vi − j62.83 Ω 50 Ω V1 docsity.com P.P.9.15 Zx = (Z3 / Z1) Z2 Z3 = 12 kΩ Z1 = 4.8 kΩ Z2 = 10 + jωL = = 10 + j9.425 )1025.0()106)(2(j10 6-6 ××π+ Zx = k8.4 k12 (10 + j9.425) = 25+ j23.5625 Ω Rx = 25, Xx = 23.5625 = ωLx H625.0 )106(2 5625.23 f2 X L 6 x x μ=×π = π = i.e. a 25-Ω resistor in series with a 0.625-μH inductor. docsity.com