Answer:
a) 926 μJ
b) 3.802 mC
c) 8.61 A
d) 0.0721
e) 3.2137
Explanation:
The energy in the inductor is
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J[/tex]
The energy store in a capacitor is
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
The voltage in a capacitor is
V = Q/C
[tex]V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V[/tex]
Therefore:
[tex]Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J[/tex]
The total energy is Et = 925.6 + 1.1 = 926.7 μJ
At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
V = Q/C
[tex]Ec = \frac{1}{2}*C*(\frac{Q}{C})^2[/tex]
[tex]Ec = \frac{1}{2}*\frac{Q^2}{C}[/tex]
[tex]Q^2 = 2*Ec*C[/tex]
[tex]Q = \sqrt{2*Ec*C}[/tex]
[tex]Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC[/tex]
When the capacitor is completely empty all the energy will be in the inductor and current will be maximum
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]I^2 = 2*\frac{El}{L}[/tex]
[tex]I = \sqrt{2*\frac{El}{L}}[/tex]
[tex]I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A[/tex]
At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC
q = Q * cos(vt + f)
q(0) = Q * cos(v*0 + f)
3.8 = 3.81 * cos(f)
cos(f) = 3.8/3.81
f = arccos(3.8/3.81) = 0.0721
If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137
