Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Step-by-step 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=
![2\pi(q_0)/(I_0)](https://img.qammunity.org/2020/formulas/physics/college/c8dv67lb05cm4qs5ajvuq4g49idwg0j6rp.png)
now putting the values we get
=
![2\pi(1.5*10^(-6))/(5.5*10^(-3))](https://img.qammunity.org/2020/formulas/physics/college/7n73cbp260g2jooo1765bp3mky5uag74pg.png)
= 1.71×10^{-3} sec
b) note that the time
it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
![t_(fc)= (T)/(4)](https://img.qammunity.org/2020/formulas/physics/college/n17jg0zc18qewkeydktulcwb76jmtqwfhq.png)
![t_(fc)= ( 1.71*10^(-3) )/(4)](https://img.qammunity.org/2020/formulas/physics/college/7oiadgwho1gg5504269ddofmkvxlr0jn4z.png)
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec