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BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    Two resistors\[400\,\Omega \]and\[800\,\Omega \]are connected in series with a 6V battery. The potential difference measured by voltmeter of\[10\,\Omega \]across\[400\,\Omega \]resistor is

    A)  2V                         

    B)  1.95V

    C)  3.8V                     

    D)  4V

    Correct Answer: B

    Solution :

                     Here, the resistances of\[400\,\Omega \]and\[10000\,\Omega \]are in parallel, their effective resistance\[{{R}_{P}}\]will be \[{{R}_{P}}=\frac{400\times 10000}{400+10000}\]                 \[=\frac{5000}{13}\Omega \] Total resistance of the circuit,                 \[=\frac{5000}{13}+800=\frac{15400}{13}\Omega \] Current in the circuit                                 \[I=\frac{6}{15400/13}=\frac{39}{7700}A\] Potential difference across voltmeter \[=I{{R}_{p}}=\frac{39}{7700}\times \frac{5000}{13}=1.95\,V\]


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