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CMC Medical CMC-Medical VELLORE Solved Paper-2010

  • question_answer
    A moving oil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliamp ere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V the resistance in ohm needed to be connected in series with the coil will be

    A)  99995                  

    B)  9995

    C)  \[{{10}^{3}}\]                                   

    D)  \[{{10}^{5}}\]

    E)  None of these

    Correct Answer: B

    Solution :

                    Voltage sensitivity \[=\frac{\text{Current sensitivity}}{\text{Resistance}\,\text{of}\,\text{galvanometer}\,\,\text{G}}\] \[\Rightarrow \]               \[G=\frac{10}{2}=5\,\Omega \] Here, \[{{I}_{g}}=\]Full scale deflection current  \[=\frac{150}{10}=15\,mA\] V = voltage to be measured\[=150\times 1=150\,V\] Hence,                  \[R=\frac{V}{{{I}_{g}}}-G\]                                 \[=\frac{150}{15\times {{10}^{-3}}}-5\]                                 \[=9995\,\Omega \]


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