In the LC circuit in the figure below, the inductance is L-17.6 mH and the capac
ID: 1787559 • Letter: I
Question
In the LC circuit in the figure below, the inductance is L-17.6 mH and the capacitance is C = 17.8 mF. At some moment, UB = UE = 80.0 m Capacitor initially fully charged; switch open (a) What is the maximum charge stored by the capacitor? 7.55e-2 C (b) What is the maximum current in the circuit? Your response differs from the correct answer by more than 10%. Double check your calculations. A (c) At t0, the capacitor is fully charged. Write an expression for the charge stored by the capacitor as a function of time. (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume Q is in coulombs and t is in seconds. Do not include units in your answer.) Q(t) (d) Write an expression for the current as a function of time. (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume I is in amperes and t is in seconds. Do not include units in your answer. Assume that the positive current is in clockwise direction.) I(t)Explanation / Answer
L = 17.6 mH
C = 17.8 mF
Ub = Ue = 80 mJ
a. maximum energy stored in the capacitor = Ub + Ue = 160 mJ
hence maximum charge = q
0.5q^2/C = 160*10^-3
q = 75.471*10^-3 C
b. maximum current in the circuit = Io
160*10^-3 = 0.5*L*io^2
Io = 4.264 A
c. at t = 0, capacitor is fully charged Qo = 75.471*10^-3
so Q(t) = Qocos(wt)
where w = 1/sqroot(LC) = 56.498 per s
hence
Q(t) = 75.471*10^-3*cos(56.498t)
d. I(t) = -d(Q(t))/dt = 4.263sin(56.498t)