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

            ( c)     At what time is the energy stored

             (i) completely electrical (i.e., stored in the capacitor)?

Answers (1)

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at any instant, the charge on the capacitor is:

Q=Q_0cos(w_{natural}t)=Q_0cos(2\pi f_{natural}t)=Q_0cos\left ( \frac{2\pi t}{T} \right )

Where time period :

 T=\frac{1}{f_{natural}}=\frac{1}{159}=6.28ms

Now, when the total energy is purely electrical, we can say that 

Q=Q_0 

Q_0=Q_0cos(\frac{2\pi}{T})

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

this is possible when

 t=0,\frac{T}{2},T,\frac{3T}{2}....

Hence Total energy will be purely electrical(stored in a capacitor) at 

t=0,\frac{T}{2},T,\frac{3T}{2}.....

Posted by

Pankaj Sanodiya

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