Chapter 8 Solutions - Soln0822 050416

Transcript

Chapter 8, Solution 22. Compare the characteristic equation with eq. (8.8), i.e. R 1 s2  s  0 L LC we obtain R R 2000  100   L   20H L 100 100 1  106 LC  C 1 10 6   50 nF 20 106 L