Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T= [tex]2\pi\frac{q_0}{I_0}[/tex]
now putting the values we get
= [tex]2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}[/tex]
= 1.71×10^{-3} sec
b) note that the time [tex]t_{fc}[/tex] it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
[tex]t_{fc}= \frac{T}{4}[/tex]
[tex]t_{fc}= \frac{ 1.71\times10^{-3} }{4}[/tex]
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec