Physics-
General
Easy

Question

Two capacitors C subscript 1 end subscript equals 2 mu F and C subscript 2 end subscript equals 6 mu F in series, are connected in parallel to a third capacitor C subscript 3 end subscript equals 4 mu F. This arrangement is then connected to a battery of e.m.f. = 2V, as shown in the figure. How much energy is lost by the battery in charging the capacitors

  1. 22 cross times 1 0 to the power of negative 6 end exponent J    
  2. 11 cross times 1 0 to the power of negative 6 end exponent J    
  3. open parentheses fraction numerator 32 over denominator 3 end fraction close parentheses cross times 1 0 to the power of negative 6 end exponent J    
  4. open parentheses fraction numerator 16 over denominator 3 end fraction close parentheses cross times 1 0 to the power of negative 6 end exponent J    

The correct answer is: 11 cross times 1 0 to the power of negative 6 end exponent J


    C subscript e q end subscript equals fraction numerator C subscript 1 end subscript C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction plus C subscript 3 end subscript equals fraction numerator 2 cross times 6 over denominator 2 plus 6 end fraction plus 4 equals 5.5 mu F
    Energy supplied left parenthesis E right parenthesis equals Q V equals C V to the power of 2 end exponent equals 22 cross times 1 0 to the power of negative 6 end exponent J
    P.E. storedleft parenthesis U right parenthesis equals fraction numerator 1 over denominator 2 end fraction C subscript e q end subscript V to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction cross times 5.5 cross times left parenthesis 2 right parenthesis to the power of 2 end exponent equals 11 cross times 1 0 to the power of negative 6 end exponent J
    Þ Energy lostequals E minus U equals 11 cross times 1 0 to the power of negative 6 end exponent J

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