Question

[24] In an oscillating LC circuit, L 10-3 H and C-4 x 10-5 F. The maximum charge on the capacitor is qmaz-2x 10-6 C. Find the maximum current. 2 x 10-6A 5 x 10-3 A max ?max- Insufficient information

Answer 1

Given that Rightarrow The oscillating LC circuit L = 10^- 3 H, C = 4 times 10^- 5 F, Rightarrow q_max = 2 times 10^- 2 C, & rightarrow Now the energy in the system is (1/2) L . 1^2 where l is the current at the time capacitor charge is zero will equal the maximum energy stored on the capacitor Rightarrow (1/2) L . 1^2 = (1/2) cv max^2 Rightarrow a_max = cv max = c squareroot (L/C) 1^2 Rightarrow squareroot r L . C Vertical-bar ^2) rightarrow so I_max = squareroot (Q^2/LC) = squareroot (4 times 10^- 12/4 times 10^- 8) Rightarrow I_max = 10^- 2 A Answers I_max = 0.01 Amp