Physics-
General
Easy

Question

A network of six identical capacitors, each of valueC, is made as shown in the figure.
The equivalent capacitance between the points A space a n d blank B is

  1. fraction numerator 4 C over denominator 11 end fraction    
  2. fraction numerator 3 C over denominator 4 end fraction    
  3. fraction numerator 3 C over denominator 2 end fraction    
  4. 3 C    

The correct answer is: fraction numerator 4 C over denominator 11 end fraction


    In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

    fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction
    rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction
    This is connected in parallel with (4).
    therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction
    The three capacitor’s (5),fraction numerator 4 C over denominator 3 end fraction comma (6) are now connected in series.
    therefore Equivalent capacitance is
    fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction

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