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BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    Two coils are wound on the same iron rod so that the flux generated by one passes through the other. The primary coil has\[{{N}_{p}}\]turns in it and when a current 2 A flows through it the flux in it is\[2.5\times {{10}^{-4}}Wb\]. If the secondary coil has 12 turns the mutual inductance of the coils is (assume the secondary coil is in open circuit)

    A)  \[10\times {{10}^{-4}}H\]           

    B)  \[15\times {{10}^{-4}}H\]

    C)  \[20\times {{10}^{-4}}H\]           

    D)  \[25\times {{10}^{-4}}H\]

    Correct Answer: B

    Solution :

                     \[M=\frac{{{N}_{s}}\phi }{i}\] \[=\frac{12\times 2.5\times {{10}^{-4}}}{2}=15\times {{10}^{-4}}H\]


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