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ECE2040D_HW8_Soln

ECE2040D_HW8_Soln - ECE 2040D Solutions to HW#8 Spring 2009...

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Unformatted text preview: ECE 2040D Solutions to HW#8 - Spring 2009 P10.5-2 Apply KCL at node a: v 4 cos 2 t d 0.25 v i 1 dt 0 Apply KVL to the right mesh: 4i 4 d iv dt 0 v 4i 4 d i dt After some algebra: d2 d i5 i5i 2 dt dt 4 cos 2t Now use i j 2 t T I m Re{e } and 4 cos 2 t 4 Re{e j 2 t } to write 4 Re{e j 2 t } d2 d I Re{e j 2 t T } 5 I m Re{e j 2 t T } 5 I m Re{e j 2 t T } 2 m dt dt d2 d Re 2 I m e j 2 t T 5 I m e j 2 t T 5 I m e j 2 t T Re{4 e j 2 t } dt dt Re 4 e j T I m e j 2 t 5 j 2 e jT I m e j 2 t 5 e j T I m e j 2 t 4 e j T I m 5 j 2 e j T I m 5 e j T I m ^ ` Re{4 e j 2t } 4 I m e jT 4 4 5 j 2 5 4 1 j 10 4 10.0584 0.398 84q i t 0.398 cos 2 t 85q A (checked 7/6/05) Page 1 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 P10.5-3 VS ZR 2 90 qV j = j 16000 : (500)(0.125u106 ) R; Z C j ZC j 16000 1600090 q 2 90 q V (Z ) = 290 q 2561239q 20000 j 16000 therefore v t 1.25cos 500t 141q V 1.25 141 q V Page 2 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 Page 3 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 Page 4 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 Page 5 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 Page 6 / 7 ECE 2040D Solutions to HW#8 - Spring 2009 Page 7 / 7 ...
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