JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Grouping of Resistances

  • question_answer
    The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 ohm is  [MP PMT 1984; EAMCET 1994]

    A) \[1/250\,\Omega \]                      

    B) \[1/200\,\Omega \]

    C) \[1/100\,\Omega \]                      

    D) \[1/10\,\Omega \]

    Correct Answer: C

    Solution :

      Lowest resistance will be in the case when all the resistors are connected in parallel. \[\frac{1}{R}=\frac{1}{0.1}+\frac{1}{0.1}.......\] 10 times \[\frac{1}{R}=10+10.......10\]times  \[\frac{1}{R}=100\] i.e. \[R=\frac{1}{100}\Omega \]


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