Respuesta :
Answer:
Part a)
[tex]Q = 320 \mu C[/tex]
Part b)
[tex]t = 8.52 \times 10^{-5} s[/tex]
Part c)
Rate of energy = 301.5 J/s
Explanation:
Part a)
Since energy is always conserved in LC oscillating system
So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor
[tex]\frac{1}{2}Li^2 = \frac{Q^2}{2C}[/tex]
now we have
[tex]Q = \sqrt{LC} i[/tex]
[tex]Q = \sqrt{(3.76 \times 10^{-3})(3.13 \times 10^{-6})} (2.95)[/tex]
[tex]Q = 320 \mu C[/tex]
Part b)
Energy stored in the capacitor is given as
[tex]U = \frac{q^2}{2C}[/tex]
now rate of energy stored is given as
[tex]\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}[/tex]
so here we also know that
[tex]q = Q sin(\omega t)[/tex]
[tex]\frac{dq}{dt} = Q\omega cos(\omega t)[/tex]
now from above equation
[tex]\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)[/tex]
so maximum rate of energy will be given when
[tex]sin\omega t = cos\omega t[/tex]
[tex]\omega t = \frac{\pi}{4}[/tex]
[tex]t = \frac{\pi}{4}\sqrt{LC}[/tex]
[tex]t = 8.52 \times 10^{-5} s[/tex]
Part c)
Greatest rate of energy is given as
[tex]\frac{dU}{dt} = \frac{Q^2\omega}{C}[/tex]
[tex]\frac{dU}{dt} = \frac{(320 \mu C)^2 \sqrt{\frac{1}{(3.76 mH)(3.13 \mu C)}}}{3.13 \mu C}[/tex]
[tex]\frac{dU}{dt} = 301.5 J/s[/tex]
