Respuesta :
Answer:
Explanation:
In the L-C oscillation , energy is transferred between capacitor and inductor with a certain periodicity.
Initial energy in the capacitor = 1/2X Q² / C
[tex]\frac{(6.3\times10^{-6})^2}{2\times2.9\times10^{-6}}[/tex]
= 6.84 x 10⁻⁶ J
Initial energy of inductor is zero.
Total energy = 6.84 x 10⁻⁶ J
When all the energy is stored in the inductor , it has maximum current . Let this current be I
Energy of inductor
= 1/2 L I²
Here I is maximum current in the inductor.
Conserving energy
1\2 L I² = 6.84 X 10⁻⁶
.5 X 30 X 10⁻³ I² = 6.84 X 10⁻⁶
I = 2.13 X 10⁻²
= 21.3 mA.
Time period of oscillation
T = [tex]2\pi\sqrt{LC}[/tex]
=[tex]2 X 3.14 \sqrt{30\times10^{-3}\times2.9\times10^{-6}}[[/tex]
188.4 10^{-5}.s
Current will be maximum after 1/ 4 of time period
= .25 x 188.1 x 10⁻⁵ s
47 X 10⁻⁵ s
Answer:
0.02654618617 seconds
Explanation:
[tex]I=I_{max} sin(ωt-Φ)[/tex]
In this case you want to find the maximum current. So [tex]I=I_{max}[/tex]
[tex]I/I_{max}=sin(ωt-Φ)[/tex] phi is 0
your resonance is going to be the capacitance and the inductance and then solve for t.[tex]sin^{-1}(1)=ωt[/tex]
