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NEET NEET SOLVED PAPER 2018

  • question_answer
    A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is               [NEET - 2018]

    A)  20                              

    B)  11

    C)  10                              

    D)  9

    Correct Answer: C

    Solution :

    \[\text{I=}\frac{\text{E}}{\text{nR+R}}\]                                              ?(i) \[\text{10}\,\,\text{I=}\frac{\text{E}}{\frac{\text{R}}{\text{n}}\text{+R}}\]                                          ?(ii) Dividing (ii) by (i), \[\text{10=}\frac{\text{(n+1)R}}{\left( \frac{\text{1}}{\text{n}}\text{+1} \right)\text{R}}\] After solving the equation, \[\text{n=10}\]


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