Answer:
0.054 A
Explanation:
We have given inductance [tex]L=1.44mH=1.44\times 106{-3}H[/tex]
Capacitance [tex]C=4.52\mu F=4.52\times 10^{-6}F[/tex]
Charge [tex]Q=4.40\mu C=4.40\times 10^{-6}C[/tex]
The energy stored in the LC circuit is given by [tex]E=\frac{1}{2}Li^2=\frac{Q^2}{2C}[/tex]
So energy [tex]E=\frac{Q^2}{2C}=\frac{(4.40\times 10^{-6})^2}{2\times 4.52\times 10^{-6}}=2.1415\times 10^{-6}J[/tex]
This energy is also equal to [tex]\frac{1}{2}Li^2[/tex]
So [tex]\frac{1}{2}Li^2=2.1415\times 10^{-6}[/tex]
[tex]\frac{1}{2}\times 1.44\times 10^{-3}i^2=2.1415\times 10^{-6}[/tex]
i=0.054 A
So maximum current will be 0.054 A
