Question

A series circuit contains an 80-?F capacitor, a 0.020-H inductor, and a switch. The resistance of the circuit is negligible. Initially, the switch is open, the capacitor voltage is 50 V, and the current in the inductor is zero. At time t = 0 s, the switch is closed. What is the maximum current in the circuit?

a.	2.8 A
b.	2.2 A
c.	2.4 A
d.	3.0 A
e.	3.2 A

Answer 1

answer is e . 3.2 A

given that,

inductance of the inductor L = 20 mH

                                           = (20 mH)(10^-3 H / 1 mH)

                                           = 20*10^-3 H

the capacitance of the capacitor C = 80 ?F

                                                    = (80 ?F)(10^-6 F / 1 ?F)

                                                    = 80*10^-6 F

the capacitor voltage V = 50 V

the initial charge on capacitor q = CV

                                                = (80*10^-6 F)(50 V)

                                                = 4*10^-3 C

the magnetic energy of the inductor is

                         U = (q²/2C)sin²?t

at maximum value , sin?t = 1

its means, ?t = 90^o (or) ?/2

therefore, the time t = ?/2?      ...... (1)

here, angular frequency ? = 1/?LC

hence, time t = [??LC]/2   

                    = [??(20*10^-3 H)(80*10^-6 F)]/2  

                    = 1.986*10^-3 s

                    = (1.986*10^-3 s)(1 ms/10^-3 s)

                    = 1.986 ms   

the maximum current in the circuit is

                        I = q/?LC

                          = (4*10^-3 C) / ?(20*10^-3 H)(80*10^-6 F)

                          = 3.16 A = 3.2 A