Answer:
[tex]1.68085\times 10^{-6}\ J[/tex]
0.00739 A
Explanation:
Q = Maximum charge on the capacitor = [tex]3.56\ \mu C[/tex]
C = Inductor capacitance = [tex]3.77\ \mu F[/tex]
L = Inductance = 61.5 mH
I = Current
Maximum energy in capacitor is given by
[tex]E=\frac{Q^2}{2C}\\\Rightarrow E=\frac{(3.56\times 10^{-6})^2}{2\times 3.77\times 10^{-6}}\\\Rightarrow E=1.68085\times 10^{-6}\ J[/tex]
Total energy in the circuit is [tex]1.68085\times 10^{-6}\ J[/tex]
Energy is also given by
[tex]E=\frac{LI^2}{2}\\\Rightarrow I=\sqrt{\frac{2E}{L}}\\\Rightarrow I=\sqrt{\frac{2\times 1.68085\times 10^{-6}}{61.5\times 10^{-3}}}\\\Rightarrow I=0.00739\ A[/tex]
The maximum current in the circuit is 0.00739 A
