Physics

5. Energy

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Energy on Inductor

Emag=12Li2E_{mag} = \dfrac{1}{2}Li^2

Energy on Capacitor

Eelec=12q2CE_{elec} = \dfrac{1}{2}\dfrac{q^2}{C}

Total Energy

12Li2+12q2C=12Q2C\dfrac{1}{2}Li^2 + \dfrac{1}{2}\dfrac{q^2}{C} = \dfrac{1}{2}\dfrac{Q^2}{C}

or

12Li2+12q2C=12LI2\dfrac{1}{2}Li^2 + \dfrac{1}{2}\dfrac{q^2}{C} = \dfrac{1}{2}LI^2

Current

i=±Q2q2LCi = \pm \sqrt{\dfrac{Q^2 - q^2}{LC}}

Charge

q=±Q2LCi2q = \pm \sqrt{Q^2 - \dfrac{LC}{i^2}}