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  • Solve it, - Electromagnetic Induction and Alternating currents - JEE Main-3

In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open.(q is charge on the capacitor and \tau = RC is Capacitive time constant). Which of the following statement is correct ?

  • Option 1)

    At\; t=\frac{\tau }{2},q=CV(1-e^{-1})

  • Option 2)

    Work done by the battery is half of the energy dissipated in the resistor.

     

  • Option 3)

    At\; t=\tau ,q=CV/2

  • Option 4)

    At\; t=2\tau ,q=CV(1-e^{-2})

 

Answers (1)

best_answer

Charge on capacitor at t = t is 

 q = cv (1- e^{-t/\tau })\\ at \: \: t = 2 \tau \\ q = cv (1-e^{-2})


Option 1)

At\; t=\frac{\tau }{2},q=CV(1-e^{-1})

Option 2)

Work done by the battery is half of the energy dissipated in the resistor.

 

Option 3)

At\; t=\tau ,q=CV/2

Option 4)

At\; t=2\tau ,q=CV(1-e^{-2})

Posted by

Avinash

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