Answer:
The magnitude of the current in the circuit is 0.01159 A.
Explanation:
Given that,
Capacitor [tex]C = 0.2\mu F[/tex]
Inductance [tex]L=3 mH[/tex]
Voltage V = 2 V
We need to calculate the energy when the capacitor is charged
Using formula of energy
[tex]E=\dfrac{1}{2}\dfrac{q^2}{C}[/tex]
[tex]E=\dfrac{1}{2}\times\dfrac{C^2V^2}{C}[/tex]
[tex]E=\dfrac{1}{2}\timesCV^2[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2}\times0.2\times10^{-6}\times4[/tex]
[tex]E=4\times10^{-7}[/tex]
When the capacitor is discharged
Using formula of energy
[tex]E=\dfrac{1}{2}LI^2[/tex]
[tex]4\times10^{-7}=\dfrac{1}{2}\times3\times10^{-3}\times I^2[/tex]
[tex]I^2=\dfrac{2\times4\times10^{-7}}{3\times10^{-3}}[/tex]
[tex]I=\sqrt{0.00027}[/tex]
[tex]I=0.0164\ A[/tex]
rms value of current,
[tex]I_{rms}=\dfrac{I_{0}}{\sqrt{2}}[/tex]
[tex]I_{rms}=\dfrac{0.0164}{\sqrt{2}}[/tex]
[tex]I_{rms}=0.01159\ A[/tex]
Hence, The magnitude of the current in the circuit is 0.01159 A.
