ECE2040D_HW2_Soln

ECE2040D_HW2_Soln - 6 6 12 12 4 V 1 6 3 5 4 18 3 5 12 2 V ;...

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Unformatted text preview: 6 6 12 12 4 V 1 6 3 5 4 18 3 5 12 2 V ; 12 V 2 3 18 18 3 4 8 12 V 4 18 3 v v v v g32 g32 g32 g14 g14 g14 g32 g32 g32 g32 g32 g32 10 P3.3-1 P3.3-4 1 3 3 1 Voltage division 16 12 8 V 16 8 4 12 4 V 4 8 KVL: 4 V v v v v v v g32 g32 g14 g32 g32 g14 g16 g16 g32 g32 g16 Page 1 / 7 ECE 2040D Solutions to HW#2 - Spring 2009 p i g32 g32 120 480 W a g11 g12 The power supplied by the dependent source is given by i g32 g32 0.2 20 4 A a g11 g12 Then g169 g185 g168 g184 g167 g183 10 g14 50 v g32 g32 120 20 V 10 a g11 g12 Use voltage division to get P3.3-8 1 1 1 6 4 4 1 1 1 1 1 1 2 3 6 3 6 3 2 1 1 2 3 4 A; 2 1 1 1 1 3 6 3 2 1 1 2 4 1 A 3 1 1 1 1 6 3 2 1 1 4 2 A 4 1 1 1 1 6 3 2 i i i i g32 g32 g14 g14 g14 g14 g14 g14 g32 g32 g14 g14 g14 g32 g32 g14 g14 g14 g32 g32 g14 g14 g14 A g32 P3.4-1 g11 g12 g11 g12 Current division: 8 6 2 A 1 16 8 8 6 3 A 2 8 8 1 A 1 2 i i i i i g32 g16 g32 g16 g14 g32 g16 g32 g16 g14 g32 g16 g32 g14 P3.4-4 Page 2 / 7 ECE 2040D Solutions to HW#2 - Spring 2009 1 2 2 1 2 2 1 1 1 1 1...
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ECE2040D_HW2_Soln - 6 6 12 12 4 V 1 6 3 5 4 18 3 5 12 2 V ;...

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