Chapter 8 Solutions - Soln0817 110224

Transcript

Chapter 8, Solution 17. i (0)  I 0  0, v(0)  V0  4 x5  20 di (0) 1   ( RI 0  V0 )  4(0  20)  80 dt L 1 1 o    10 LC 1 1 4 25 R 10    20, which is  o . 2L 2 1 4 s     2  o2  20  300  20  10 3  2.679,  37.32 i (t )  A1e  2.679t  A2 e 37.32t di (0)  2.679 A1  37.32 A2  80 dt This leads to A1  2.309   A2 i (0)  0  A1  A2 ,  i (t )  2.309 e 37.32t  e  2.679t Since, v(t )   1 t i (t )dt  20, we get C 0 v(t) = [21.55e-2.679t – 1.55e-37.32t] V