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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    Four resistance of 10\[\Omega \], 60 \[\Omega \] 100 \[\Omega \] and 200 \[\Omega \] respectively taken in order are used to form a. Wheatstones bridge. A 15 V battery is connected to the ends of a 200\[\Omega \], resistance the current through it will be

    A)  \[7.5\times {{10}^{-5}}A\]      

    B)  \[7.5\times {{10}^{-4}}A\]

    C)  \[7.5\times {{10}^{-3}}A\]      

    D)  \[7.5\times {{10}^{-2}}A\]

    Correct Answer: D

    Solution :

     Here, the resistances\[10\text{ }\Omega ,\text{ }60\text{ }\Omega \]and\[\text{100 }\Omega \]are in series and they together are in parallel to\[\text{200 }\Omega \], resistance. When a potential difference of 15 V is applied across\[\text{200 }\Omega ,\]then current through it is \[I=\frac{15}{200}=7.5\times {{10}^{-2}}A\]


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