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NEET AIPMT SOLVED PAPER SCREENING 2011

  • question_answer
    A current of 2 A flows through a \[2\,\Omega \] resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a \[9\,\Omega \] resistor. The internal resistance of the battery is

    A)  \[1/3\,\,\Omega \]     

    B)        \[1/4\,\,\Omega \]        

    C)         \[1\,\Omega \]              

    D)         \[0.5\,\Omega \]

    Correct Answer: A

    Solution :

    Current \[i=\frac{E}{R+r}\] \[2=\frac{E}{2+r}\]                          ?(i) \[0.5=\frac{E}{9+r}\]                       ?(ii) From Eqs. (i) and (ii), We have                 \[=\frac{2}{0.5}=\frac{9+r}{2+r}\]                 \[4=\frac{9+r}{2+r}\]                 \[3r=1\]                 \[r=\frac{1}{3}\Omega \]


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