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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    In a neon discharge tube\[2.9\times {{10}^{18}}N{{e}^{+}}\] ions move to the right each second while \[1.2\times {{10}^{18}}\] electrons move to the left per second, electrons charge is\[1.6\times {{10}^{19}}C\]. The current in the discharge tube is

    A)  1 A, towards right          

    B)  0.66 A, towards right

    C)  0.66 A, towards left

    D)  zero

    Correct Answer: B

    Solution :

    Direction of current is opposite to the direction of flow of electron and in the same direction of positive ion. Total current\[I={{I}_{Ne}}+{{I}_{electron}}\]                 \[=(2.9\times {{10}^{18}}+1.2\times {{10}^{18}})e\]                 \[=4.1\times {{10}^{18}}\times 1.6\times {{10}^{-19}}\]                 \[=0.66\,\,A\] Direction of current will be the directi6n of flow of\[N{{e}^{+}}\]ion i. e. towards right.


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