Physics-
General
Easy

Question

A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

  1. 0 %    
  2. 20 %    
  3. 75%    
  4. 80%    

The correct answer is: 80%


    q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
    This charge will remain constant after switch is shifted from position 1 to position 2.
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
    thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.

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