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Bài giải phần giải mạch P10

Chapter 10, Solution 1. ω=1 10 cos( t − 45°) → 10∠ - 45° 5 sin( t + 30°) → 5∠ - 60° 1H → 1F → jωL = j 1 = -j jωC The circuit becomes as shown below. 3Ω Vo jΩ 10∠-45° V + − 2 Io + − 5∠-60° V Applying nodal analysis, (10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo + = 3 j -j j10∠ - 45° + 15∠ - 60° = j Vo Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9° Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V Chapter 10, Solution. | Chapter 10 Solution 1. 1 10cos t - 45 --- 10Z -45 5sin t 30 ---- 5 Z -60 1 H ----- j L j 1 1F ------ -j j C The circuit becomes as shown below. Applying nodal analysis 10Z -45 - Vo 5Z -60 - Vo V 3 j -j j10Z-45 15Z-60 j Vo Vo 10Z-45 15Z-150 15.73Z247.9 Therefore vo t 15.73 cos t 247.9 V Chapter 10 Solution 2. 10 4cos 10t -n 4 ---- 4Z -45 20sin 10t n 3 --- 20Z -150 1H ------ j L j10 11 0.02 F ----- -j5 j C j0.2 The circuit becomes that shown below. 20Z-150 V 10 n AMr V_ TIO g jio n Tf -j5 n 4Z-45 A Applying nodal analysis 20Z-150 - Vo 4Z-45 V. V. 10 j10 -j5 20Z-150 4Z-45 0.1 1 j Vo I o V 2Z-150 4Z-45 - ----------------- 2.816Z150.98 j10 j 1 j Therefore io t 2.816 cos 10t 150.98 A Chapter 10 Solution 3. 4 2cos 4t -- 2Z0 16sin 4t - 16Z -90 -j16 2 H ----- j L j8 11 1 j C j 4 112 -j Applying nodal analysis -j16 - V V V ----- 2 - 4 - j3 1 6 j8 -j16 I 1 1 I 2 11 -------Iv 4 - j3 4 - j3 6 j8 o Vo Therefore 3.92 - j2.56 4.682Z-33.15 ----------- -------------- 3.835 Z - 35.02 1.22 j0.04 1.2207Z1.88 vo t 3.835 cos 4t - 35.02 V Chapter 10 Solution 4. 16sin 4t -10 ---- 16Z -10 o 4 1 H ------ joL j4 16Z -10 - V1 1 V1 j4 2I x 1-j But i 16Z -10 - V1 x j4 So j8 3 16Z -10 - V1 _V_ .