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Manipal Medical Manipal Medical Solved Paper-2002

  • question_answer
    In the given network capacitance\[{{C}_{2}}=10\mu F,{{C}_{1}}=5\mu F\]and\[{{C}_{3}}=4\mu F\]. The resultant capacitance between P and Q will be:

    A)  \[4.7\mu F\]           

    B)  \[1.2\mu F\]

    C)  \[3.2\mu F\]           

    D)  \[2.2\mu F\]

    Correct Answer: C

    Solution :

     Seeing in the given circuit\[{{l}_{1}}\]and\[{{l}_{2}}\]are connected in parallel. Hence, their equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=5+10=15\mu F\] As\[{{C}_{eq}}\]and\[{{C}_{3}}\]are connected in series, Hence, resultant capacitance between P and Q is given by \[\frac{1}{{{C}_{PQ}}}=\frac{1}{{{C}_{eq}}}+\frac{1}{{{C}_{3}}}=\frac{1}{15}+\frac{1}{4}=\frac{19}{60}\] \[{{C}_{PQ}}=\frac{60}{19}=3.2\,\mu F\]


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