Physics-
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Question

Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

  1. open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus open parentheses fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close parentheses close square brackets    
  2. open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus open parentheses fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close parentheses close square brackets    
  3. open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 plus open parentheses fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close parentheses close square brackets    
  4. open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 plus open parentheses fraction numerator C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close parentheses close square brackets    

The correct answer is: open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus open parentheses fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close parentheses close square brackets


    Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
    equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
    C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    Since, V is the voltage of the battery, so charge on this system
    q equals C V
    q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
    C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
    therefore New charge stored
    q to the power of ´ end exponent equals C ´ V
    q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
    Change in total charge
    increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets

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