Physics-
General
Easy

Question

The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch S subscript 1 end subscript while keeping switch S subscript 2 end subscript open. The capacitor can be connected in series with an inductor ‘L’ by closing switch S subscript 2 end subscript and opening S subscript 1 end subscript .

Initially, the capacitor was uncharged. Now, switch S subscript 1 end subscript is closed and S subscript 2 end subscript is kept open. If time constant of this circuit is tau, then

  1. after time interval tau, charge on the capacitor is CV/2    
  2. after time interval 2tau, charge on the capacitor is CV(1 – e to the power of negative 2 end exponent)    
  3. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged    
  4. after time interval 2tau, charge on the capacitor is CV(1 – e to the power of negative 1 end exponent)    

The correct answer is: after time interval 2tau, charge on the capacitor is CV(1 – e to the power of negative 2 end exponent)


    table row cell Q equals 0 subscript 0 end subscript open parentheses 1 minus e to the power of negative t divided by 3 end exponent close parentheses end cell row cell Q equals C V open parentheses 1 minus e to the power of negative 4 r end exponent close parentheses text end text text a end text text f end text text t end text text e end text text r end text text end text text t end text text i end text text m end text text e end text text end text text i end text text n end text text t end text text e end text text r end text text v end text text a end text text l end text text end text 2 t text end text text. end text text end text end cell end table

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