JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    An infinite number of identical capacitors each of capacitance \[1\mu F\] are connected as in adjoining figure. Then the equivalent capacitance between \[A\] and \[B\] is                                                                                           [EAMCET 1990]

    A)                    \[1\mu F\]

    B)                    \[2\mu F\]

    C)                    \[\frac{1}{2}\mu F\]

    D)                    \[\infty \]

    Correct Answer: B

    Solution :

               This combination forms a G.P.  \[S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}......\] Sum of infinite G.P. \[S=\frac{a}{1-r}\] Here a = first term = 1 and r = common ratio \[=\frac{1}{2}\] Þ \[S=\frac{1}{1-\frac{1}{2}}=2\]Þ \[{{C}_{eq}}=2\mu F\]


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