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Q7.12 (d)    An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

At what times is the total energy shared equally between the inductor and the capacitor?

 

Answers (1)

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The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store.i.e.

\frac{Q^2}{2C}=\frac{1}{2}\frac{Q_0^2}{2C}

From Here, we got

Q=\frac{Q_0}{\sqrt{2}}

So Now, we know the charge on the capacitor, we can calculate time for which

\frac{Q_0}{\sqrt{2}}=Q_0cos\left ( \frac{2\pi t}{T} \right )

\frac{1}{\sqrt{2}}=cos\left ( \frac{2\pi t}{T} \right )

From here,

 t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}..

Hence for these times, the total energy will be shared equally between capacitor and inductor. 

Posted by

Pankaj Sanodiya

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